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NOTES ON DYNAMICS.
FoR THE SENIOR CLASS or ARTILLERY OFFICERs,
woolwich,
4, —f
ºw-cº.
I. FUNDAMENTAL UNITS.
1. The Fundamental Units in Dynamics are the Units of Length, Time, and Weight ; and from
these units are derived the units of area, volume, density, velocity, acceleration, momentum, energy,
work, power; also the electrical units, the ohm, volt, ampère, watt, joule, coulomb, farad, &c. -
The three Indefinables in Nature are Space, Time, and Matter; but although they cannot be
defined, they are capable of measurement, in terms of the above-mentioned Fundamental-Units.
+
LENGTH.
2. The British Unit of Length, generally employed in Dynamics, is the foot, the third part of
the Standard Yard, defined by Act of Parliament, 18 and 19 Vict., 1855, as “the straight line
distance at 62°F. between the centres of the transverse lines in the two gold plugs in the bronze
bar deposited in the Office of the Exchequer”; but now, under the Standards Act, 1866, trans-
ferred to the custody of the Warden of the Standards. - - -
The British Unit of Area is then the square foot, and of Volume is the cubic foot.
The Metric Unit of Length is the metre, defined originally as the ten-millionth part of the
distance on a meridian of the Earth from the equator to the pole; but now defined practicaliy by
the length between the ends of a rod of platinum made by Borda, the rod being at the temperature
of melting ice. - - -
The Metric Unit of Area is then the square metre, and of Volume is the cubic metre.
* But for purposes of minute scientific (electrical) measurement, the centimetre, the one-
hundredth part of a metre, is usually taken as the Unit of Length, and the square centimetre and
cubic centimetre as the Units of Area and Volume. - -
TIME.
3. The Unit of Time in universal use in Dynamics is the second, the mean solar sexagesimal.
second. -
The Mean Solar Day measures day and night equably; it is divided into 24 hours, each hour
into 60 minutes, and each minute into 60 seconds; and in general it is requisite to convert any
measure of time into seconds before making use of it in a dynamical formula.
When the Metric System was brought out (1795) it was proposed, for completeness of decimal
and centesimal measurement, that the mean solar day should be divided into 40 centesimal hours,
each hour into 100 centesimal minutes, and each minute into 100 centesimal seconds. -'
At the same time, for purposes of Navigation, the equator was to be divided into 400 grades
(degrees) of longitude, and the quadrant of the meridian into 100 grades of latitude; and each
grade divided into 100 centesimal minutes, and each minute into 100 centesimal seconds; so that the
factor 10 (instead of 15 as at present) should convert centesimal time into longitude, and vice
Q967°SO,.
The centesimal minute of latitude is thus the kilometre, of 1,000 metres; and the kilometre
was intended to displace the present nautical mile or sexagesimal minute of latitude, about 6,080 ft.
long, of which 90 x 60 = 5,400 go to the quadrant of the Earth. º
But as the centesimal mode of reckoning time was never adopted, centesimal grades, minutes,
and seconds, and kilometres are useless in Navigation. -
To measure Time, a clock or chronometer is employed, regulated by a pendulum or spiral
spring on a balance wheel, to preserve as equable a motion as possible. -
The chief function of the Greenwich Observatory was formerly to keep mean Solar Time, for
the benefit of Navigation, with accuracy, checked by observation of the Sun and Stars.
WEIGHT.
4. The British Unit of Weight is the pound, defined by the Act of Parliament cited above as a
weight of platinum, marked “P.S., 1844, 1 lb.,” deposited in the Office of the Exchequer, or with
the Warden of the Standards. - - -
(Wt. f ſº-º: 6873)
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Notice that it is not necessary to specify the temperature, as in the case of the Standard of
Length; nor is it necessary to mention the height of the barometer, the height above sea-level,
the latitude, &c. (Rankine, Rules and Tables, Part II, Section VI); these details only require
specification for the purpose of allowing for the buoyancy of the air, when a copy of the platinum
pound is to be made in some other metal, brass, for instance; and to cover this refinement, the words
“in vacuo" are added in recent Acts of Parliament, as defining the standard condition of the weight.
The gallon, however, is defined in the Act as the volume of 10 lb. of distilled water, weighed
in air by brass weights when the barometer stands at 30 inches and the temperature is 62°F; so
that if this water was placed in a closed vessel, and the weighing was performed in an exhausted
receiver, the water would weigh more than 10 lb. by an amount equal to the difference of the
weights of air displaced by equal to the difference of the air displaced by the water and by the
brass weights.
According to the latest determination (H. J. Chaney, Phil. Trans., 1892) a cubic inch of water,
weighed in air against brass weights of specific gravity 8:143, is equilibrated by 252-286 grains,
when the barometer stands at 30 inches and the thermometer at 62° F.; and this makes the
volume of the gallon.
70000 –– 252-286 = 277.463 cubic inches.
Taking a cubic foot of this air as weighing 534.22 grains, or a cubic inch as weighing
534.22 –– 1728 = 0.309 grains, then the brass weights will displace 0:309 -- 8:143 = 0.038 grains;
and the difference, 0.271 grains, will be the extra weight required to restore equilibrium if the
weighing is performed in vacuo; so that 252.557 grains is the true weight of a cubic inch of water
under the above conditions, although 252-286 grains would be given in the Act as more useful for
commercial purposes.
The standard pound avoirdupois is divided into 16 ounces, or into 7,000 grains; and besides
these we have other measures of weight, the ton of 2,240 lbs, the cwt. of 112 lbs, &c., but the
Americans are beginning to use a short ton of 2,000 lbs, and a short cwt of 100 lbs, instead of
Our long ton and cwt.
The Metric Unit of Weight is the kilogramme of 1,000 grammes, defined originally as the
weight of a decimetre cube of water at its maximum density, when the temperature is 4°C; so
that a gramme is the weight of a centimetre cube of water.
But the actual determination of the weight of a decimetre cube of water is an operation which
requires great care as a reference to Mr. Chaney's paper, referred to above, will show ; and the
differences between the results obtained by the most skilful observers, though small, are a thousand
times greater than the results of a comparison of standard weights by weighing them ; so that
nowadays the practical Metric Standard of Weight is the Kilogramme des Archives, made of
platinum by Borda, preserved in the Conservatoire des Arts et Métiers, Paris, Intended to represent
the weight of a decimetre cube of distilled water at the temperature 4°C. -
A metric tonne, of 1,000 kilogrammes (2,203 lbs.), the weight of a metre cube of water, is also
employed, while for scientific purposes the gramme, the weight of a centimetre cube of water is the
Unit of Weight.
5. Convenient abbreviations in general use for the metric units are:—
Kilometre = km. Metre square == m”. Tonne = t.
Metre = Ill. Metro cube - n°. Rilogramme = kg.
Centimetre = CII). Centimetre square = cnn”. Gramme - §.
Millimetre = In Ill. Centimetre cube = Cnh”.
etc.
We may also write ft.*, ft“, in”, in”, for square feet. cubic feet, square inches, cubic inches.
There are only three Systems of Fundamental Units which need be considered in Dynamics:—
(i) The British F.P.S. (foot-pound-second) System.
(ii) The Metric M.K.S. (metre-kilogramme-second) System, in use in Continental commerce.
(iii) The C.G.S. (centimetre-gramme-second) System, in use for electrical and scientific
measurements.
All quantities in Nature can be measured by means of the three Fundamental Units of Length,
Time, and Weight; so that, as Maxwell points out (Theory of Heat, p. 75), the whole system of
modern civilized life may be fitly symbolized by a foot or metre rule, a clock or chronometer, and
a set of weights, British or Metric. *- 2
According to the most careful determinations, legalised by Act of Parliament, 1878,
1 ft = 0.304794 m, 1 m = 3-280869 ft = 39°370432 in.
1 lb = 0-45359 kg, 1 kg = 2.2046212 lb
1 lb = 453.59265 g, 1 g = 0.0022040 lb
With a quadrant of the Earth of 10 million metres, this makes the minute of latitude about
6075-7 feet.
According to Col. Clarke's geodetic measurements, the mean length of a minute of latitude is
6076'7 ft, and the length of a minute of longitude at the equator is 6087.1 ft. This makes the
quadrant of a meridian 6076.7 × 5400 ft; and if this is divided into ten million metres, it makes
the metre equal to 3:281148 ft.
But with the metre of 3:280869 ft, the quadrant of a meridian is 10001.7 km, and a quadrant
of the equator is 10019 km.
The mean value of the minute of latitude or longitude at the equator, in round numbers, 6080
ft, is taken as the length of the Admiralty nautical mile.
Sailors divide the sea-mile into 10 cable-lengths or cables, and the cable into 100 fathoms (so
that the fathom is 6 ft stretched about 1 in); the fathom is the stretch of the arms (the natural
way of measuring the length of a rope), French, brasse; Spanish, brazo; Greek, öpyvua; so that the
stadium is the cable, and the Parasang of 30 stadiums is the league (Xenophon, “Retreat of the
Ten Thousand.” Evtej6ev ééeXavvov Trévita Tapaordy'yas).
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DENSITY.
6. The density of a substance is the weight of the unit of volume.
Thus if V cubic feet of the substance weighs W lb, the density is W/V.
If the body is not homogeneous, this gives the average density of the body; to determine the
actual density at any point, we must find AW, the weight of a small volume AV enclosing the
8. then the density is the lt (AW/AW) (= dW/dV in the notation of the Differential
alculus).
With Metric Units, the density of a substance is the weight in g/cm", or the weight in thm”.
Denoting the density by w, then
w = W/W, or W = wV,
gives W the weight in g (or t) of W cm3 (or mº) of the substance; but
W = 1,000 wV,
gives the weight in kg of V mº.
Thus a litre of hydrogen weighing (in vacuo) 0:089 g, the density of hydrogen is
| 0000089, or 10-3 × 48.
SPECIFIC GRAVITY.
The specific gravity of a substance is the old-fashioned name for the density relative to
water.
In the metric system, the density and specific gravity of a substance are the same, when a
centimetre and gramme (or metre and tonne) are the units of length and weight.
For denoting the specific gravity by s, the equation W = 8 V gives the weight in g (or t) of
W cm3 (or mº).
But W = 1000 sV gives the weight in kg of V m” so that the density is 1000 s, when the
metre and kilogramme are the units of length and weight, as in the M.K.S. system.
Considering that s is in general given to 3 decimals only, 1000 s will be given as an integer; for
instance—
s = 7-207 for cast iron, so that 1000 s = 7207.
With British units, the equation—
W = 1000 SW,
gives the weight in oz of V ft” of a substance of S.G. (specific gravity) s, if we assume roughly
that a cubic foot of water weighs 1,000 Oz. ; or
W = 62.5 s V, more nearly 62-4 sy,
give the weight in lb of V ft”; thus with British units, the density is 62.4 times the specific
gravity, a great numerical inconvenience.
Numerical examples. To be worked with four figure logarithms.
1. Calculate in British and Metric Units the surface, volume, and weight of the Earth, taking
a geographical mile as 6,080 feet, and a mean density of 5:576.
2. Calculate, in kilometres, and in nautical miles or feet, the range of the Moon with a
parallax of 57, and of the Sun, with a parallax of 8”8. -
(Take cosec 57' = 60, cosec 8”.8 = 23440, cosec 1" = 206265).
Calculate their semi-diameters, supposing them each to subtend 16'.
Calculate the mean density of the Moon, supposing its weight 1 = 80 the weight of the
Earth.
3. Find the weight of the atmosphere, the mean height of the barometer being 75 cm, and
the density of mercury 13-6.
II. LINEAR DYNAMICS.
7. We begin Dynamics by considering motion in a straight line, as of a train on a straight piece
of railway, of a shot in the bore of a gun, of the piston of a steam engine, of water in a straight
uniform pipe, &c. &
We thus avoid all difficulties of geometry until the fundamental laws of Dynamics are clearly
established, and have become a reality to the reader's mind; afterwards the composition of velocities
and forces, and curvilinear motion may be considered, and the correspondingly more complicated
problems introduced.
An acknowledgment must be made here to the Rev. F. Twisden's Practical Mechanics, as the
source of many of the illustrative examples.
WELOCITY.
The velocity of a body is its rate of change of position, or the rate of growth of its
distance from a fixed point or terminus, measured in a straight or curved path.
The velocity of a body is measured, if constant, by the number of units of length (feet,
metres, or centimetres) described in the unit of time (the second).
Thus if s feet (or metres) is described in t seconds, the velocity v is given by
v = s/t f/s (feet per second).
or, v = s/t m/s (metres per second).
If s is given in centimetres, the velocity v is given in cm per second, that is, in kines, according
to the C.G.S. terminology.
An ordinary rate of walking is 6 km an hour, or a kilometre in 10 minutes, a hectometre in
1 minute.
If the velocity is variable, then s/t is the average velocity in f's with which s ft. is described in
t seconds, and the actual velocity at any point is the average velocity over a very small distance
enclosing the point.
In the notation of the Differential Calculus, if A s fº. additional described in At seconds, the
average velocity over A s is(A s/A t) f/s; and the actual velocity v at the point of time t is
* = lt (A 8/A t) = ds/dt, f/s.
So with other units: thus if the railway journey from London to Edinburgh is performed in
8 hours, a distance of 400 miles, the average speed is 400-80 = 50 miles an hour, the unit employed
for train speeds; but in consequence of the retardation due to stoppages and inclines, the actual
Velocity will fluctuate, and for most of the time will be more than 50 miles an hour.
At sea the speed is always measured in knots (French, nauds; Spanish, nudos; Italian, modi,
German, knoten; Dutch knoopen, &c.), the sailor's knot being the cosmopolitan unit of speed, of one
nautical mile or mean sexagesimal minute of latitude per hour.
(N.B.-Not knots an hour, as landsmen say, which would properly mean an acceleration.)
The length of the Admiralty sea mile being 6,080 ft., a knot is a speed of 6,080 –- 3,600 = 1.69
fls or roughly, a knot is a speed of 100 ft. a minute, more nearly 101 to 102, say 101.4 ft a minute.
In Artillery velocity is always measured in f/s or in m/s, and numerical exercises can be
found in the Records of the Bashforth Earperiments,
The name quad has been proposed by Professor Oliver Lodge for a velocity of one quadrant of
the Earth, 10" kines or cm/s, the quad being a suitable unit for very-high velocities; thus the
olºn is a velocity of one quad; while the velocity of light and electromagnetic propagation in air
of a vacuum is about 3 x 10" kines, or 30 quads, and in glass about 20 quads.
Since 1 quad = 10° kines = 90 x 60 × 3,600 knots,
therefore 1 kine = 0-01944 knots, 1 knot = 51.44 kines.
*** * 3. * *
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.”
EXAMPLES.
1. Determine, in knots, in fſs, and in kines, the velocity of a point on the Equator, due to the
Earth's rotation; the velocity of the Moon relative to the Earth, with a period of 27 days; the
velocity of the Earth relative to the Sun, with a period of 365 days.
2. A steamer takes a minutes on the measured mile with the tide, and b minutes against the
tide: prove that the speed through the water is 30 (a + b)/ab knots, and determine the speed of
the tide.
3. Determine the speed in knots to go round the world in 80 days, and of a postcard in
53 days. **-
Hºw long would it take to go round the world on latitude 50° at this speed 2
Determine the parallel of latitude in which a steamer going westward at V knots will keep
up with the Sun; and determine on any other latitude the daily gain or loss of the chronometer
on the mean Sun at noon; on the westward and the eastward passage.
What is the gain in apparent speed on the actual speed through the water, in a voyage of
3,000 miles to New York, 5 hours W longitude from Greenwich.
4. Prove that the maximum piston speed is about ; T times the average piston speed.
5. Prove that the product of the diameter of a wheel by the number of revolutions in
a minute divided by 28, is very nearly the velocity in miles an hour.
6. Express a velocity of 2,200 f/s in miles an hour.
Also 60, 45, 30, 15, 7% miles an hour in f/s.
7. Determine the velocity at the screens, 150 ft. apart, which were cut at times given by the
following chronograph records in decimals of a second (Rounds 479,482, Final Report on Bashforth
Eaperiments, 1878–80).

Screen. 1. 2 3 4 5 6 7 8 9 10 11 12
Round 479 || 0000 || 0666 *1343 *2031 •2729 '3439 *4159 *4891 ‘5633 6.387 715.1 | "7927
, 482 || 0000 || 0657 ‘lā24 *2004 || 2695 '3397 *4110 *4834 ‘5569 -6314 '7069 '7835
8. A train passes two men walking beside the railway in 3% secs. and 3; secs, respectively;
a second train passes the men in 43 secs. and 4% secs, respectively. Show that this train will
overtake the former, and if on a different line could pass it completely in 36 secs.
9. A ship is sailing at 63 knots directly towards a battery, where artillery practice is going
on. The man at the helm observes the flash of a gun and hears the report 15" afterwards; five
minutes after the first flash he observes a second flash and hears the report 12" afterwards,
Find the velocity of sound, assuming that a sea mile is equal to 6,080 feet.
10. Two ships, A B, of which B bears from A 35° east of north, lie eastwards of a coast
which presents seawards a long vertical precipice running north and south. A cannon is fired
from B and 5 seconds after the flash is seen from A the report is heard, and 7 seconds later still
its echo from the precipice. Taking the velocity of sound at 1100 f/s, find the distance of A from
the coast.
11. Determine the length of the wave seen in a column on the march, keeping step with a
drum marking 100 paces a minute, taking 1100 f/s as the velocity of sound.
(6873) C
ACCELERATION.
The rate of change or rate of growth of velocity is called the acceleration. .*
The acceleration is measured, if constant, by the growth of velocity (in f/s or m/s or kines)
per second. º {º sº
Thus, if the velocity grows at a constant rate from V to v in t seconds, the acceleration a is
given by a = (v- W )ſt
so that w = W-F at . . . . . . . . . . . . . . (1)
If the velocity is given in f's then a is given in f's added per second, which we may denote by
f/s” (or by celoes, to use the word invented by Rev. J. B. Lock, veloes being used by him to denote
velocity estimated in fſs, or feet per second). s
If the velocity v is given in kines, then the acceleration a is said to be so many spouds, in the
terminology of the C.G.S. System.
We use the letter a instead off to denote acceleration, as f suggests the old term accelerating
force, now obsolete.
If, in the small interval of time, At seconds, the velocity grows Av (f/s, veloes, or kines), then
the average acceleration is Av/At (f/s”, celoes, or spouds), and the actual acceleration at the
instant is
a = It (Av|At) = dv/dt
(f/s”, celoes, or spouds) in the notation of the Differential Calculus. . . .
9. With constant acceleration a, the velocity grows uniformly, so that the average velocity U
over any distance s is the A.M. (arithmetic mean) of the initial velocity W and the final velocity
w; or
U = } (V -- v).
Then if the distance s is described in t seconds, with average velocity U,
s|t = U = } (V--v), or s = } (V -- v) t.
But with constant acceleration
a = (v — V)/t, or v — W = at ;
therefore U = V + # at,
and 8 = Wi + 4 at” . . . . . . . . . . . . (2)
Again, since
v — W = aff,
# (v + V) = s/t,
therefore by multiplication—
# *- : W* = as. . . . . . . . . . . . . (3),
the equation connecting a, s, and v.; and (1), (2), (3) are the equations employed with Motion
under Constant Acceleration.
The acceleration of gravity is the constant acceleration with which we are most familiar :
denoting it as usual by g, then the velocity of a body falling freely will grow g units of velocity
per ºnd or, as Clifford expresses it, velocity is poured into the body at the rate of g units per
SeCOI) Oi.
With a foot as unit of length, g = 32-1912 at sea level in London; usually replaced by 32 in
practical problems.
With a metre as unit of length, g = 9.81 at Paris, very nearly 10; so that, in C.G.S. units
g = 981 spouds. y
One great reason for choosing such a small unit of time as the second appears to be that g is
then represented by a small number easily remembered.
f %.
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If we had taken the minute as unit of time, then we should have
g = 32 x 60° = 115200;
while, if the mile and hour are units,
g = 32 x 60' + 5280 = 78555#.
If the kilometre and hour are units, \
g = 9.81 × 60' + 1,000 = 117137.6.
In knots an hour, the acceleration of gravity would be
g = 32 x 60' + 6,080 = 68210.5.
"We are now prepared, with the aid of these formulas, to solve some problems on uniform
motion and uniformly accelerated motion of a practical nature.
EXAMPLES.
1. Consider first the following question, derived from a sentence taken out of a model essay
on the Alps: “We reflected, in looking over the fearful precipices that, make a false step, and
we should have fallen 10,000 feet in 10 seconds.”
(i) What must be the numerical value of g for a body to fall 10,000 ft. in 10 secs. ?
(ii) With g = 32 how many feet would a body fall in 10 secs. ?
(iii) How many seconds would it take to fall 10,000 ft.
(iv) With what velocity must it be projected so as to fall 10,000 feet in 10 secs. ?
1. A balloon, 6 miles high, ascending with velocity, 1,500 ft. per min., was found after
13 mins. to be descending with velocity 2,000 ft. per min. Find the greatest height ascended;
find the apparent acceleration of g, supposed constant; find the time it will take for the balloon to
reach the ground, and the velocity with which the balloon will strike the ground (Glaisher and
Coxwell, 1862).
2. An express train reduced speed from 60 to 20 miles an hour (m/h) in 800 yards, the distance
between the distant and home signals, as at Abbots Ripton. How much further out should the
distant signal be placed, or how much should the brake power be increased ?
With full brake power it is calculated that a train going 30 m/h can be pulled up in 225 ft.,
at 60 m/h in 900 ft., at 80 m/h in 1,600 ft., &c., Determine the number of seconds it takes to pull
up, and the distance overshot due to a delay of a second in the action of the brakes.
3. An express train, timed to run at full speed of 60 miles an hour, is checked by signal to
20 miles an hour, over a mile of road under repair. Prove that if the train takes one mile from
rest to get up full speed, and half a mile to pull up, the train will be 2 mins. 40 secs. late.
4. A body shot vertically upwards with velocity W fls will, in the absence of any resistance,
attain a height of V*/g ft. in W/g secs, reaching the ground again after 2W/g s, with
reversed velocity V fls.
Ex.: Suppose W = 1600, g = 32.
5. Prove that if a body, shot vertically upwards, takes t seconds to reach a height of h feet,
and tº seconds more to come down to the point of projection again,
h = }gtt.' (Col. Sladen's formula.)
If H denotes the greatest height attained in feet, and T = t + tº the whole time, in seconds,
the body is in the air, prove that
H = }gT2 = 4 Tº = (2 T)”,
and the velocity of projection is #gT = 16 T f/s.
Determine the height at half time, and the time at half height, also the time of falling equal
vertical distances. §
Graduate the bar of a gravity chronograph (i), for equal intervals of time (ii), for velocity.
Ex. A body was a minute in the air when projected vertically; determine the velocity of
projection, and the greatest height attained; find also the height after 20 and 40 secs, and the
time to attain a height of 960 ft.
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6. (i) A body falling from the top of a house was observed to take t secsin falling past a window
h ft. high. Prove that the distance of the top of the house from the sill of the window is
(h + ft.)” 29t' ft, and from the top of the window is (h I-39*) /2gt” ft. e
(ii) A stone let fall from the top of a house was observed tº take one-eighth of a second to
pass a window 74 ft high; determine the depth of the window sill below the top of the house.
(iii) A body falling from a mast down a hatchway, took t seconds to fall from the hatchway
to the bottom of the hold, a distance of h ft.
Prove that the body fell (h -- $gt)/2gt ft, and struck with velocity.
h/t + #gt, f/s.
7. Assuming that the resistance of the air is constant, and one nth of the weight of a body
#Vºlg fi
prove that a body shot vertically upwards with velocity V fis will reach a height
n — 1\ º
7, -H i) f's after
taking +}, seconds, and will reach the ground again with velocity V A/ (
another } vº-I) seconds.
Ex. Take W = 900, g = 32, and n = #: prove that the whole time in the air is 62-5 seconds,
&c.
8. A body under constant acceleration passes over two successive intervals of l ft in t, and t,
seconds; prove that the acceleration is
21(t, — tº)
→4–H– celoes;
tito (t + ta)
and that the velocities at the beginning, middle, and end of the interval, 21 feet, are
l( — tº + 2t2t + tº) !(t,” + tº) (* + 2*,t, - tº)
tiº, (t, + te) 5 tit, (t. + te)' tut, (t, + t.)
f.s.
Ex. A train, timed past quarter-mile posts, was found to take 2 minutes and 1 minute; deter-
mine the acceleration and the velocities.
Apply the same formula to the reduction of Bashforth’s screen records.
Determine also the expressions for the acceleration and velocities, when successive unequal
intervals of l, and l, ft are described in t, and to seconds.
9. Assuming that the velocity of rebound is e times the striking velocity, prove that an
elastic ball let fall from a height of h ft on a smooth horizontal pavement will, before coming to
rest on the pavement, describe
1 2 e 2/
H. h ft, in } + € A/ l, seconds,
Ex. h = 16, e = #.
10. A goods, train going uniformly at 15 miles an hour, and a passenger train going also at
an average velocity of 15 miles an hour, but stopping every mile at a station for two minutes, are
travelling on parallel rails (e.g., from New Cross to Croydon). Determine when and where the
trains will pass, the goods train having a start of, say one minute, and the passenger train taking
one quarter of a minute to pull up.
Give a graphical solution of this, and of the similar general problem.
cº
*.
“.
§
º
:
i
*.
*
ºr
*
\*...
$
.-čº
:§
***
WARIABLE ACCELERATION.
10A. When the acceleration a is not constant, we have recourse to the methods of the
Differential and Integral Calculus. *
We have already shown that if s ft is described in t seconds, the velocity v, in f/s, and the
acceleration a, in fſs”, or celoes, are given by
ds dv
* = i, q = i,
so that the definitions of velocity and acceleration may be given together, thus—
The (ºr } is the rate of change of {{...} or is the rate of growth of
distance distance
velocity velocity
if variable by the number of units of {º} which would be ſº in the next unit of
time, if the {.. }were to continue constant, and equal in value to the value it has at
}: measured, if constant by the growth of per unit of time (the second);
acceleration
the instant considered.
The expression for a may be variously transformed; thus—
dº dº dº — dº o dº
dt ds dt ds ds
If a is a given function of t,
w = W -- |ad.
0
and • = |* = v. 4 ||ade
0 0 0
supposing V is the initial velocity, when t = 0.
Thus for instance, against a retardation ft, which brings the body to rest in T seconds, the
body will go a distance
T
| t ft, dt.
0
But if a is a given function of 8, then, since
#v°/ds = a,
v% = }V* + |ads:
J 0
*
& dt 1 2
— F — = 2 — #
and since ds v (V* + ſº 3.
therefore t = ſºv. + 3|ado- *ds
giving t as a function of s.
Thus if a is constant,
= W -- at . (1.)
s = Wi + \at” (2.)
}v% = *W* + QS e (3.)
2 e-º-
and, as before, t = v/ (W **) V
11. Again, since—
=* = +
T di Tài,
ds
dv d ()- d º ds
dt T dº | * | T d \*) dº
( ds ds ds
_ dºt _ dºt
tº-º ds” ds ds?
‘dt 2 dź = dt 3"
(#) (#)
d2t (#) d°t ...a .
- - -— I — F - –- ?)
ds” \dt ds? ’
a simple example of change of the variable in Differentiation, of great practical use in the reduction
of Bashforth’s experiments.
In these experiments the time t in decimals of a second is recorded by a Chronograph at
which each of a series of equidistant screens, l feet apart, were cut by a shot; and t is given as a
© ë tº dt (!?t te
function of s, and not s as a function of t; so that we obtain ds and . from the screen records,
S *4
by differencing or interpolation.
(6873) D
• *, *, * : *%
“. . . ; 32
*
_* .# f
f
.*'.g.
*
º:
*", º',
{
IO
Then the velocity v at any distance 8 ft is the reciprocal of dt/ds, while the retardation is
dºt
ds?
from which we see the convenience of Bashforth’s method of dividing the retardation due to the
resistance of the air into two factors, one of them being the cube of the velocity, and the other
3 .
Q) y
º dt'
b therefore *-.
eing therefore is
Now, in Bashforth’s notation, reckoning the velocity in thousands of f/s, and assuming that
the resistance of the air varies as the cross-section or as d”, where d is the calibre in inches,
d’s d? v \8 a dºtſ v ) W d2t
* † = — Kſ = — 109° iſ — ). so that K = 10° -º-º:
dº? *(iv) 1() #(º , SO tha. d;2 dº?”
and K is Bashforth's constant of resistance, which is tabulated from his experiments.
If the shot weighs W lb., then as shown subsequently, the resistance of the air is a
force of *
y º vº pounds, or W #º poundals.
Suppose that three consecutive screens are cut by the shot at instants of time
t
$ 77.3 'mºn,
m-1}
as registered by the Bashforth chronograph; and we thence determine the velocity of the shot and
the resistance of the air at the middle or mth screen. *
We assume that the time t can be represented by a quadratic function of the distance s; so
that, measuring s backwards and forwards from the middle screen where s = 0.
t = tº + as + ba”,
and the graph of t is a parabola.
For the determination of a, b,
tº-1 = t, — bl + blº, t,41 = a + bl + blº,
so that 2a l = tººl - tº-1, 201* = tº-1 – 2t, + tail.
Now at the middle screen, where s = 0, and the velocity is V,
dt 1 — ... dºt —
2b;
iſ a y = *, + =
giving W = 21ſ (t,x, — tº 1),
the average velocity from the 1st to the 3rd screen.
The resistance of the air is—
y #v. - y tº-1 – 2# + ºnt, Vs poundals.
Thus in Round 479 (Report of Experiments, 1879), at the 5th screen,
2al = 3439 – 2031 = 1408;
2bl” = 3439 – 5458 + 2031 = 0012;
so that log 2al = 1.1485,
log 21 = 2.4771.
log W = log 1/a = 3:3286, W = 2131 f/s;
while log 2bl” = 3.0792.
log l’ = 4:3522.
log 2b = 8-72.70.
The resistance R of the air, in pounds, is given by—
R = W.2% V3:
9
and with W = 50 ;
log W = 1.6990,
log G = 1.5077,
T
log y = 0-1913,
log 2b = 8.7270,
log W* = 9-9858,
log R = 2.9041, R = 802,
about 16 times the weight of the shot.
Ill
When the time t is always a quadratic function of the distance 8, then—
#: = 2b, a constant, which is 10 º K, in Bashforth’s notation;
and the retardation of the shot and resistance of the air vary throughout as the cube of the
velocity.
Then, by integration,
= & + 2bs,
and
supposing s = 0, and v = V when t = 0.
The average velocity U over s is given by U = s/t f/s; and now
so that the average velocity over s is the actual velocity at the mid-point of s, and is the
Harmonic Mean (H.M.) of the initial velocity V, and final velocity v.
In a similar manner, taking five screen records, we shall find at the middle screen, the mth
SCI’626. Il-
dt
l d = # (tºti - tº-) - Hº (t,nt, - n-s),
* = 1 (t 2t I 2 --
alsº – 3. ( m+1 ~ **m + ºn-1) - Hºg (ºr, &º-sº ºn + ‘,-) 3
and taking the middle of seven screen records,
dº
ds F # (*#1 º ºn-s) tº- +% (**** Gº- ºn-s) + ºw (?,*s tº- tº-3),
d2?
12 dº? == # (?,*1 tº- 2t, + tº-1) tº- #, (**** tºmºse 2t, + ºn-s) + sº (**** + 2t, + ºn-s);
2
giving closer approximations to the values of # and #: at the middle screen.
(Proc. R. A. Institution, Vol. XV, 1886.)
12
WEIGHT AND FORCE.
12. We must now determine the motion produced by a given force acting upon a body of
given weight, so that the third Fundamental Unit, the Unit of Weight, must be introduced.
We use the word weight advisedly, so as to agree with the precise language of the successive
Acts of Parliament.
The weight (poids) of a body is the quantity which is determined by the operation of weighing;
to weigh (poise) #. body, we place it in one of the scales of a balance, and equilibrate it by certain
lumps of metal called weights, placed in the other scale; and the sum of these weights is called
the weight of the body.
Where great precision is required, the operation of double weighing is employed to eliminate
any possible inaccuracy in the balance, and a correction for the buoyancy of the air is als
made. r
As the field of force in which we live is that due to the attraction of the Earth, it is.
convenient to take the attraction of the Earth on a lb weight as the unit of force, and to call it
the force of a pound; this is the British unit of force in universal use in all practical problems of
architecture, engineering, mechanics, and artillery.
A force of P pounds acting on a weight of W lb will now produce acceleration a celoes, given
by the equation—
P
e-
&
g - W’
This equation is deduced from a fresh aviom, Newton's Second Law of Motion, which asserts that
“Change of motion is proportional to the impressed force, and takes place in the direction of, the
straight line in which the force is impressed.” (Principia Mathematica, Philosophiae Naturalis, p. 12).
Here change of motion means what we have called acceleration; so that if in a weight of W lb
an acceleration a celoes is produced by a force of P pounds, while experiment shows that a force
of W pounds always produces acceleration g celoes is a weight of W lb, therefore by Newton's
Second Law,
a P
g TW
This experiment was first carried out by Galileo, in 1590, by letting different weights fall from
the Leaning Tower at Pisa, when he found that the weights all took the same time to fall; contrary
to Aristotle's dictum, then and for long afterwards in universal acceptance, which asserted that the
time of falling was inversely as the weight, which would make g proportional to the square of the
weight. The best experimental proof of this law is the indirect one given by Newton, who found
that the motion of a simple pendulum is independent of the weight and material of the bob.
13. Now suppose a force of P pounds to act on a weight of W lb, through s feet, and for t
seconds, and to generate velocity v fjs from rest; imagine, for instance, a carriage on wheels on a
level railway, propelled by a horizontal force of P pounds.
Then, since a = PG|W,
Pgs
*** = F - –tº–
#! CMS W. '
Pg!
v = (tt = +.
- W
therefore # Wv°g = Ps,
Wvg = Pt,
the acceleration a not appearing in the last two equations.
. The product Ps of a force P pounds acting through s feet in its own direction, is called the
work done by the force, in foot-pounds; and since Ps and #Wv°/g are convertible, Wv°/g is also
met sured in foot-pounds, and is called the kinetic energy, or stored-up work of a body weighing W lb
moving with velocity v f/s.
The work required to raise W lb, from rest to rest, through h feet vertical, is Wh foot-pounds:
and now if the weight W is allowed to fall freely h feet vertical, it will acquire velocity, c f/s
given by º
#v°/g = h,
so that #Wv°/q = Wh,
thus illustrating the equivalence of the work done in raising the body, and given out again in the
shape of kinetic energy in falling freely.
The prºduct Pt of a force of P pounds acting for t seconds is called its impulse; and, by
analºgy with foot-pounds, Pt is expressed in second-pounds; and since Pt and Wv/g are con-
Vertible, Weſg, called the momentum or quantity of motion of W lb. moving with velocity e. fſs is
also expressed in second-pounds, being the mechanical equivalent of a force of P pounds acting
for t seconds. -
w * ~ * g.º. ." . º.
* “. . . ; º: * *
gº. *
*:-
... - r * : . .” 4 : ...A & $5, º.º., f.ſº
* * * I **ś ſº
* * * ... sp.: £3, ºº
13.
Thus, starting from rest,
Ps = }Wv°/g (foot-pounds),
- Pt = Wolg (second-pounds).
But if the body starts with velocity V f/s, * -->
as = }v” — $V”, at = v – W.;
and then
Ps = }Wv°/g – #WV*/g (foot-pounds).
Pt = Wv/g – WW/g (second-pounds).
If W is given in tons, then the kinetic energy $Wv°/g is given in foot-tons, and the
momentum Wv/g in second-tons. * *
14. Let us apply these formulas to the solution of an important practical problem, the
determination of the amount of work required to transport a carriage of given weight W lb. in
a given time t seconds, a distance l feet on the level, against a uniform resistance of R pounds.
Imagine a train on a railway, or an omnibus in a street, or a column of artillery on the march,
constantly starting and stopping.
The resistance R would be that due to a smooth incline a, in which sin a = R/W ; so that the
gradient is one in cosec a or W/R.
We suppose the carriage to start from rest, and to come to rest at the other end of the
journey, being constantly accelerated up to full speed and then constantly retarded. *
Let P denote the pull exerted in pounds, and let s denote the distance in feet through which
it acts, and v the velocity in f's which is generated.
Then, by the preceding principles, since the acceleration and retardation are constant,
therefore the average velocity, lſt f/s, is half the maximum velocity v fºs; or
w = 2i}t.
Again the kinetic energy of Wv°/q ft-lb is generated by the force P pounds, diminished
by the resistance R pounds acting through s feet, and dissipated by the resistance R pounds acting
through l—s feet, no breaks being employed at first; and therefore
#Wv°g = (P – R) s = R (l — s) (ft. lb).
Thence - *Wv°/g = Rl (P – R)/P;
and Ps = Rl, *
so that the work done by the pull P pounds is always the same, namely, Rl ft. lb, the work
required to transport the load of W lb. at uniform speed through l feet.
Also if the force P acts for T seconds during the motion,
T = 2s/v = st/l = Rt|P,
OI’ PT = Rt,
so that the impulse PT second-pounds of the pull P pounds acting for T seconds, is equal to the
impulse Rt second-pounds of the resistance R pounds, acting for the whole time, t seconds.
Now, supposing the time of the journey tºseconds, or the average velocity l/t f/s, is assigned;
then, since v = 2 l/t,
JWv°/g = Włºſłgº = Rl (P —R)/P,
R2
Ol' P = R = Wlºgā’
the requisite pull on the carriage ; and
-P- #gtºR
_ Rt Wl
T = p = t (1 º
* = 1, or t = V( .#) seconds; and then P is infinite,
The least value of t is given by #gtº
Tº
%
9
s and T are zero; but
Ps = Rl.
and PT = Rt = V(WRl/#g) (second-pounds).
The velocity v f/s requisite to carry the weight of W lb through l feet against a resistance
of R pounds, must now be supposed communicated instantaneously, as if W was shot from a gun,
by an infinitely great force P pounds, acting for an infinitely small time T seconds, but such that
this product PT, called the impulse, is finite and equal to v/(WRl/#g).
In this manner we analyse the motion due to so-called impulsive forces, to be discussed
hereafter. -
As R the resistance of the road is diminished, so the minimum time of the journey,
w/(W//#g|R) seconds, is measured ; until at last for the ideal case of a perfectly smooth road
offering no resistance R, the minimum time is infinite, unless we can create an artificial resistance
dº { } At
when requisite.
(6873) E
* . . . . . . ; ; ;&#
- .* * *º : *
º
#:
º
*::::
- *.*.*
s:x: ..º.
. . . . .
• • , 3. -. i.
. . ; ; , ; *.
‘l. * * * * **,
** ...< * ,-
W.
3.
* *
' ºs-
14.
This shows the necessity with the improvement of the road, of great break-power for rapid
travelling; the brake-resistance being applied, as usual in brakes acting at the periphery of the
wheels, so as to dissipate the kinetic energy of the moving body by friction; or, as recently pro-
.
*
posed for tramcars, the brake-power may consist of a powerful spring, contrived so as to store up
the energy of the carriage in stopping it, afterwards to be given out in starting.
An apparatus is now being tried which can stop and start the car in 20 yds; provided,
the weight of this apparatus can be kept within reasonable limits, some economy may be expected
from this arrangement in tramcar or train working. - w
15. Suppose then that the breaks can be made to call up an additional resistance of R' pounds;
to effect this the brake-shoes must press on the peripheries of the wheels with a total thrust of
R//p pounds, p denoting the coefficient of friction.
Then when working at full power, the pull P pounds must be taken off and the brake-resistance
R" pounds must be applied simultaneously; and now
#Wv°/g = (P – R)s = (R + R') (l − s.) (ft. lb);
while as before w = 21ſt = 2s/T;
so that, given l, W, R, R' and the time t seconds, or average velocity lſt f/s,
s _ R + R' •
T-PIR" * ~
WP (P - R) (R,+ R');
gº T' P + R' y
R’ Wl
R + Ri Rº
OT - P = W I'
* - RIRſ ºf
- * S T W l
— — — - 1 – — :
and l t R + R' #gt”’
and the work required is
R! WP
R + R' $gt”
R" WP
R + R' #gt”
the part representing the work dissipated in friction by the brakes.
For a level road, and for carriages on wheels, we may put R = 0; and the minimum time of
the journey is V ( W º seconds; and the carriage, weighing W lb, must now jump off
R-FR’ ºg
º- R -- R/
^) – v/ ( W w) f.s.,
Wv/g = V/{W(R + R')l/#g} second-pounds,
under the influence of the equivalent impulse PT, supposed to act instantaneously; but the
passengers would now suffer as if shot off from a gun. -
These preceding principles will enable us to solve any problem of railway motion in a short
run on the level (or on an incline) from one station to the next, when the pull of the engine and
the resistance of the road are taken as uniform.
16. In a long run the maximum velocity v f's would on the preceding principles become
excessive, so that we must take v as the superior limit of velocity, not to be exceeded ; and when
the velocity v is attained, it is kept uniform until it is requisite to make a stoppage.
Now the pull P pounds must be exerted through s feet, such that
(P — R)s = }Wv°/g,
W lºv?
OT S = 2 * .
PER g :
tº: which the pull must be reduced to the resistance R pounds to keep the velocity uniform at
Q) I/S,
To make a stop in sº feet,
P3 = Rl + (ft. lb),
with velocity
and momentum
(R + R')s' = }Wv2/g,
/ W #v?
OT - 8
RTR 7
and the time lost in starting and stopping is (s -- sº)/v seconds, over the time required to run at
full speed v f/s. -
15
17. Let us at once apply these principles to the investigation of the quickest possible run of
a train on the level, from one station to the next, l feet away; the train weighing W tons gross,
and the weight on the driving wheels of the engine being w tons; taking the adhesion as 4 times
the weight, and supposing the train fitted with continuous brakes, so that the maximum pull of the
engine is uw tons, and the maximum retarding force of the brakes is p.W tons; A is taken in
practice to vary from one-fourth to one-sixth, according to the state of the road; and the
greatest incline up which the train can crawl is one in W/uw. *s
In the corresponding problem of an omnibus on a road, or a battery of artillery, w must be
taken as the weight of the horses. g tº e &
Leaving out of account the resistance of the road at first, that if t seconds is the least time in
which the journey can be performed, and v f/s the greatest velocity attained, steam being shut off
and brakes applied simultaneously when the velocity v is attained,
Q) = 2l/t,
and #Wv°/g = Aws = p.W(l — s) (ft-tons),
so that we must replace the previous P and R' by pºw and pºw, and put R = 0.
Then } F. w". w’
and * = w”. 9
so that t = A/ (w + ºw #) seconds, : Asºr *
Aww #9 º
the quickest possible journey; and then the greatest velocity attained is
dººme “Az zºº º plw
w = 21ſt = V(w #20) f/s.
For instance, for a run of 1 mile, l = 5,280;
and if W = 100, w = 14, p = }, t = A/ 114 × 6 x Śwo = 127 seconds;
- 14 16 - N
e = 21ſt = 83.15 f/s; nearly 57 miles an hour.
But suppose the greatest running velocity is limited to v fls, less than the previous value
partly obtained; the time of the run, t seconds, will now be increased; for when full speed v f/s is
attained, steam must be shut off, and with the assumption of no resistance the train will proceed
with uniform velocity v f/s, to a certain distance, s' feet short of the station, at which the brakes
must be applied to make the stop.
Now we have
#Wv°/g = Aws = p.Ws' (foot-tons),
giving s and s', and the train being 2s/v seconds under steam, and 2s'ſv seconds under the brakes,
and (l — s — s')|v seconds at full speed v f/s, the whole journey will take—
r
(, ; ; ; ) = + (.
–H 1) * (seconds),
w 21g -
exceeding the time lſo seconds, required for the journey of l feet at uniform speed v f/s, by—
W );
(8 + s^)/v = (y + 1 2/19 (seconds),
so that the average speed is reduced to loſ(l-F s -- s').
For instance, if in the above numerical application of the train, the velocity is limited to 45
miles an hour, c = 66 fs, and the time of the journey of one mile will be increased to about 130
seconds.
16
18. As another application, consider the problem of an omnibus, required to slow down to a
velocity of 1 f/s every 100 yards, to pick up and set down a passenger, the maximum speed
being limited to 7% miles an hour; given that the weight of the omnibus is 25 cwt, of the driver,
conductor, and passengers is 35 cwt, and of the horses is 30 cwt, and taking an adhesion of
one-fifth, the pair of hind wheels being braked. ſº g
Here the pull of the horses P = 6 cwt., while the given weight W = 90 cwt; and the
velocity is made to grow from W = 1 f/s to v = 11 f/s; so that the horses must pull through
8 feet, given by—
Ps = }Wv°/g – WWºlg (ft-cwt);
or - s =
The omnibus may now be considered to proceed with uniform speed v = 11 f/s, the horses
doing little or no work, and to diminish speed again to W = 1 f/s at the end of 300 ft, the brakes
must be applied to the hind wheels for 8 ft, creating a resistance of 6 cwt, one-fifth of the weight
of 30 cwt on the hind wheels, acting on 60 cwt, the weight of the omnibus and passengers.
Therefore V* – wº
63' = 60
S 2g
or s' = 18.75 ft. }
There remains l — s—s' = 253.125 ft described with uniform velocity v = 11 f/s, and
therefore in time (l – 8 – 8')|v = 23:01.14 seconds; while s and 8' ft. are each described with
average velocity #(11 + 1) = 6 fls, and therefore in time (s -- sº)/6 = 7-8125 seconds, so that
the run of 100 yards takes 30-8239 seconds, giving an average speed of 300 -- 30-8239 = 9.7 f/s,
or 6.6 miles an hour.
Each horse will now do 3 × 112 × * = 9/450 ft-lb every 308239 seconds; so that he is
e 9,450
working at a useful rate of about 3TX 550 ºf '55 of a horse power; rather hard work for a horse,
as Watt's estimate of 33,000 ft. lb per minute, or 550 ft. lb per second, is about 50°) o greater than
the performance of a strong dray horse, drawing a weight out of a well.
Work out the same problem when the roads are frosty and slippery, so that the adhesion is
reduced to one-tenth; alsº when only one hind wheel is braked; also for a gun carriage and six
horses, supposing the shaft-horse only to act as a brake.
19. With a mean uniform resistance of the road of R tons, equivalent to an incline of one in
W/R, our equations become changed to— -
#Wv°/g - (uw *º R)s = (a W + R)s' (foot-tons),
and now the whole time of the journey is
P l W W
l F – -º-º-º-º-º-º-º: ºf
(l − S + 8 /v Q) + º R + AW-E R+ º 2g seconds,
For instance, if in the above numerical instance of the railway in 17, the resis
road is taken as 20 pounds a ton, then R = 20 W/2,240 = W/112 ; and º, º:
the minimum time of the journey will be 1574 seconds, the maximum velocity attained bein !
-about 67 fis; but with velocity restricted to 45 miles an hour, or to v = 66 f/s, the time of .
journey is but slightly increased. 5
Work out the problem of a battery of artillery on a road, checked from a trot to a slow walk
every 200 yards. -
17
PROBLEMS ON RECTILINEAR MOTION AND TRAIN MOTION ON THE LEVEL. Y-
1. Prove that if a weight of W lb resting on a smooth horizontal plane (e.g., a carriage on
wheels on a horizontal road or railway) is acted on by a horizontal force of P pounds for t seconds
through s fº the velocity acquired will be— -
º-ºº: P
W gt = W(& *) f/s, •.
2
the energy acquired will be * gt? = Ps fi lb,
the momentum acquired will be Pt = V/( PW º seconds-pounds,
and 8 = } & gt”.
2. Determine the length in which a barge of 50 tons, moving at 3 miles an hour, can be
brought up by a rope round a post, supposing the maximum weight the rope can support is one
ton 2 Ans, 15-125 ft.
3. Determine in tons the mean thrust on the terminus buffers, which stop in 6 ft a train of
200 tons, going at 6 miles an hour; determine also the time this takes in seconds?
Ams. 40% tons, 1++ seconds.
4. Cutting a foot off the muzzle of a 6 in. gun was found to cause the velocity to drop from
1,490 to 1,330 f/s, the shot weighing 100 lbs. Calculate in tons on the in” the pressure on the
base of the shot as it is leaving the muzzle. Ans. 11-13.
5. Prove that the shortest time from rest to rest, in which a chain, capable of bearing a safe
load of P tons, can raise a weight of W tons out of a hold h feet deep—or lower the weight into
the hold, is—
v/( P º seconds.
P – W g -
Prove also that the greatest load which can be raised or lowered in t seconds is—
h
P (l -º-º-º: #) tons.
6. Prove that a brake resistance of 385 pounds per ton will destroy a velocity of 60 miles an
hour in 704 ft, and in 16 seconds. *
With a coefficient of friction 0:16, prove that a train going 30 miles an hour can be brought
up in about 84 yards by continuous brakes, supposing them to press on the wheels with three-
quarters of the weight of the train.
Determine the shortest distance in which the train can pull up.
7. A train of 60 tons performs a journey of 20 miles in one hour, with 9 intermediate stoppages,
each of 2 min., at intervals of two miles, the resistance of the road is 10 lbs. a ton, and the brake
power of the engine and brake van, half the weight of the train, is one-sixth of the weight.
Determine the pull of the engine and its horse power at full speed,
8. Prove that a train going 45 miles an hour will be brought to rest in about 378 yards by
the brakes, supposing them to press with two-thirds of the weight on the wheels of the engine
and brake vans, which are half the weight of the train, taking a coefficient of friction 0.18.
Prove that an engine capable of exerting a uniform pull of 3 tons can take this train, weighing
120 tons, on the level from one station to stop at the next, two miles off, in about 3 min. 38; secs.,
the speed being kept uniform when it has reached 45 miles an hour. - Q
(A man alighting from a carriage going v f/s on a road, or platform, with coefficient of
adhesion pu, will come to rest when the carriage is a away; determine w.) -
9. Determine in tons the greatest train an engine capable of exerting a uniform pull of 3 tons
can take on the level, from one station to the next a mile off, in four minutes, supposing the
resistance of the road estimated at 20 lb. a ton, and the brake power at 400 lbs a ton in addition.
Ans. 203 tons.
(6873) F
18 °.
10. Prove that a train going 60 miles an hour can be brought to rest in about 313 yards by
the brakes, supposing them to press on the wheels with two-thirds of the weight of the train,
taking a co-efficient of friction 0:18, in addition to a passive resistance of 20 lbs, a ton.
Prove that the mean uniform pull to be exerted by an engine to take this train, weighing
100 tons, on the run from one station to stop at the next, two miles off, in four minutes, is about
2-6 tons.
11. Prove that the pull of a locomotive engine is pdºl -- D pounds, for a mean effective pressure
of p lb per in”, where d denotes the diameter of such of the two cylinders, l the height of stroke
and D the diameter of the driving wheels, all in inches;
and with an adhesion of one-nth, or p = 1/m, the weight on the driving wheels must be mpdºl -- D lb.
Determine the pull of an engine and the weight on the drivers, on which d = 20, l = 24,
D = 60, for a mean effective pressure p = 100 lbs. in the square inch, taking an adhesion of one-
sixth. Ans. 73 tons, 43 tons.
(This is the well-known rule for the pull of an engine, given in Molesworth's Pocket Book,
and elsewhere.
To prove this rule it is only requisite to consider the work done by P pounds, the traction
due to the adhesion of the driving wheels, during a complete revolution, when the engine
advances TD inches, and the work done is therefore TDP inch-pounds.
This must be equated to the work done by the steam in the two cylinders, each of which is
filled twice with steam of average pressure p lb. per square inch, and of volume 37td*l; and
therefore the work done by the steam is 4 x + Trdºlp inch-pounds; and therefore Tâ’lp = TDP,
and the T's cancelling— -
P = pa”l -- D.
A similar method, slightly more complicated, will apply for a compound engine.
12. In an express engine the driving wheels are 8 feet in diameter, and the load on them is
15 tons; the cylinders are 18 x 28 inches.
Find the pressure of steam which will skid the wheels, with an adhesion of one-sixth.
Determine the ratio of the velocity of the engine to the velocity of the piston at any point
of the stroke.
13. From King's Cross to Grantham is 105 miles, and there are twenty-seven intermediate
stations. The average resistance of a parliamentary train stopping at all stations is taken as
8 lbs a ton, while the resistance to an express train which runs through without stopping is taken
at 10 lbs a ton, the brake power in each case being taken at 80 lbs a ton additional.
Supposing the speed to be kept constant by reducing steam when it has reached 30 miles an
hour, find which train is most expensive in fuel, and by how much per cent.
Work out the same problem with their resistances and full speed doubled.
' ' '...' . . . .
A . -
19
MOTION ON INCLINED PLANES.—TRAIN PROBLEMS ON INCLINES:
20. When a body weighing W lb slides down a smooth incline making an angle a with the
horizon (or rolls down on wheels with smooth axles), the normal thrust on the plane is a force of
W cos a pounds; but the component of gravity, W sin a pounds, parallel to the plane is unbalanced,
and therefore produces acceleration a celoes, such that
C! W sin a
— =
9 W->
OY a = g sin &.
By means of the inclined plane we dilute the effect of gravity, and theoretically we measure
the acceleration g more conveniently.
But practically there is no such thing as a smooth plane in Nature; although by providing
the moving body with wheels we can make the body roll down the plane almost as if the plane
was smooth, the slight discrepancy being due to the friction of the axles and the rotary inertia of
the wheels.
Suppose a body is placed on an inclined plane, like the lid of a desk, and suppose the inclina-
tion is gradually increased until the body is on the point of sliding. -
It was found experimentally by Morin that, provided the surfaces in contact are kept the same,
and are carefully lubricated, this critical angle is independent of the weight W of the body, and
of the extent of surface in contact with the plane.
It was also found experimentally that on increasing the inclination still more, so as to make
the body slide, the acceleration of the body down the plane was constant, and independent of the
weight and extent of surface in contact with the plane. *
Generalizing from the experiments, we are able to enunciate Morin's Three Laws of Friction,
employed in Theoretical Mechanics.
20
LAWS OF FRICTION.
I. When the state of the surfaces in contact remains the same, the maximum or i.; •
amount of tangential friction F which can be called into play is proportional to the norm
pressure R, and *
F = R tan b,
where q is the angle of inclniation of the plane at which the body will rest without sliding.
The angle q is called the Angle of Friction, or the Angle of Repose.
II. The friction F is independent of the extent of surface in contact.
III. The friction F is independent of the velocity when sliding takes place.
It is found experimentally that the value of F/R when sliding takes place is slightly less than
when the body is just on the point of motion; we see this verified when the driving wheels
of a locomotive slip and skid; this is expressed by saying that friction is slightly less than
stiction.
The Laws of Fluid Friction are entirely different to those for the Friction between Solids.
FRuid Friction being independent of the pressure, but proportional to the area and the square
of the velocity approximately.
Incidentally also the Laws of Friction are verified from the observation that the maximum
#. of a locomotive is proportional to the weight on the drivers, independently of their
lameter.
Careful experiments by Thurston, Kennedy, Beauchamp, Tower, &c. (“Research on Friction' of
the Institution of Mechanical Engineers, 1883, 1888), have shown that these Laws of Morin
are only a rough approximation to the truth; but we shall employ these laws in the subsequent
applications, in the absence of any other simple and more accurate formulas.
These laws will be assumed to hold equally well for a body sliding down an incline, like a
ship on the launching ways, or for a railway train on wheels on an incline, q denoting in each case
the limiting angle or gradient at which motion just begins of itself, or at which the body will
move with constant velocity if once started.
In performing the experiments on Friction on a small scale, we have taken the lid of a desk
as representing the incline; but in the future applications it is better to think of phenomena on as
large a scale as possible, and to consider a railway or road incline.
21. Suppose now that a body weighing W lb is placed on an incline, and that p is the slope
when motion is on the point of taking place.
Now let the inclination be increased to a, and let us determine the acceleration a down the
plane.
Assuming Morin's experimentallaws that the resistance parallel to the plane is always R tan $
pounds in the direction opposite to motion, where R is the normal thrust on the plane in pounds,
then since no motion takes place perpendicular to the plane,
R = W cos 2,
f
and considering the motion down the plane,
a W sin a – R tan ºf
g W
= sin o. — cos & tan ºp
= sin (c. — p) sec (b.
Suppose the body starts from rest at 0, and acquires at P the velocity v f/s in s feet and t
seconds; then
#v? E C S = gs sin (c. * --> $) SéC ºb,
w = at = gt sin (2 - b) sec (b.
Through the point O draw OH horizontal, and OK. sloping downwards at the angle p, meeting
the vertical PKH through P in H and K; then, since OP = s,
PH = 8 sin &, PK = 8 sin (c. — p) sec (p;
so that #v% = g, PK,
or the velocity at P is that which would be acquired in falling under gravity g vertically from the
line OK; PK is then called the impetus of the velocity v.
Thus a velocity 57.6 f/s × m/h has an impetus 52.8 ft, which is sufficient to carry a train 1 mile
up in 1 in 100, 2 miles up in 1 in 200, &c.
Suppose now that the body is projected from Pup the plane, and that the retardation is aſ
celoes ; then
a' W sin & -- R tan ºf
g T W
= sin (2 + ſp) sec (b.
21 4. ;4
If the velocity v' at P is just sufficient to carry the body back to 0,
#
v” = a's = gs sin (2 + (p) sec p = g, PK',
if OK! is drawn from O, sloping upwards at an angle b, meeting the vertical through P in K'.
The impetus of the velocity of is now PK’; and now if this inclined plane is used as a machine,
so that the work done in raising the body from P to 0 is given out again by the body in falling
back to P (e.g., a clock) the efficiency of the machine is
PK sin (2 — b)
tº-
---
º-
PK' T sin (a + i).
We shall find the lines OK and OK' useful in solving geometrically problems of motion on
inclined planes and railways; they are maximum limiting gradients on which a carriage will
stand without running back.
When b = 0, the plane is smooth, or behaves as such ; and the lines OK, OK' coalesce with
the horizontal line OH.
If t' is the time in seconds from P back again to O,
v' = a't' = gt' sin (2 + h) sec p.
But in finding the time from 0 to P, or back from P to O, when the downward velocity v or the
upward velocity v' is supposed to be known, it is simpler to consider the average velocity from O
to P, or P to O, to be $v or #v' f/s; and the corresponding time will be s/#v or s/#v, seconds.
22. Suppose we have a number of inclined planes, passing through a fixed horizontal axis
through O, and we wish to find the locus of bodies, starting simultaneously from Q, after a given
time of t seconds, the planes all having the same limiting angle 3.
Draw OE vertically downwards, of such a length that a body would fall freely from O to E
in t seconds, so that OE = }gt”; and on OE, draw a circle touching the line K'OK, sloping down-
wards at the angle q.
Then the time of sliding down any chord OP, between OE and OK, will be t seconds; so
that the different bodies will all lie in the circular arc OPE.
For the time down OP = s/#v= OP/#v/(2g.PK)
- 2OP, º- yº = t seconds
- V apk - V - - 3.
since OP” = OE.PK.
The chords such as OP' on the other side of the vertical OE, between OE and OK', will
possess the property that a body projected up PO, with velocity just sufficient to reach O, will
take the same time t seconds to reach O for all positions of OP', 'b still denoting the angle of
friction.
As another example, suppose AO and QB are two inclines of 2 and 8, having the same limiting
angle b, and that a body starting from A slides down AO, and passes without change of velocity
up the plane OB, as in a switchback railway, a small curved part being introduced at O; suppose
it comes to rest at B, and slides back again to A', and so on ; it is required to determine the
position of B, A', &c.
Draw the line AB through A, sloping downwards at an angle b, and cutting OB in B; then,
since the velocity at any point on AQ or OB is that which would be acquired in falling vertically
from AB, it follows that the body will come to rest at B, and
OB – sin OAB – sin (a – ?)
*-T- * -——- -->
OA sin OBA. T sin (3 + (5)
If the horizontal line through A meets OB in b, then Bb is the lost space due to friction, up
which the body must be pushed to regain the level Ab.
Supposing AO is a slope of 2, greater than q, and OB a slope of 8 less than $, and that a
body starting from A and sliding down AO passes without change of velocity on to OB, and comes
to rest at B: then obviously in the same manner, the point B will be determined by drawing AB
through A, sloping downwards at the angle q, to cut OB in B.
We can prove easily that the work required to transport a body from B to Aalong the inclines BU
and OA is always the same as that required to lift the body vertically to the level of A, and then
to drag it horizontally to it.
If the plane QP meets a plane below 0, inclined at an angle b the horizon, in P, the time of
sliding down QP is proportional to the length of the incline OP; but the velocity with which the
body reaches P is independent of OP.
(6873) G
22
Š
to
i
i
28. In railway inclines the slope is generally estimated by 1 ft vertical to so many feet m
horizontal, so that if a. is the slope of the incline of one in m,
tan a = 1/m.
The friction also is generally estimated by the resistance in pounds r per ton weight of the
train, so that on the level the coefficient of friction
p = tan b = rſ2240;
sin q = rſ2220,
the resistance of the road being thus equivalent to that of a smooth incline 4.
Then when running freely down an incline 2, the acceleration
f = g sin (2 – b) sec p
= g (sin & — u cos x).
accurately
But in railway inclines m is large enough for us to replace sin a or tan a by 1/m, and cos 2
by unity, so that -
1 – aſ" – "
f = g|... – 2) sº- 9. ãº)
the formula employed in practical problems.
More accurately,
e 1. 1 1 \– # 1 1 )
M F — = — - F — 1 - 9 @ 9
Sl]] O. v/(m” + 1) #(l + #) #ſ 2m.”
770, 1 \–3 -
sº- - tº- = 1 - • . . . .
COS C& E V(mº -E 1)1) * (1 + ;) - 27m.” 5
so that the error committed in the above approximate formula is only about 1/2m” of the whole, or
50/mº per cent. ; for m = 50, only about 0:02 per cent.; and thus it is immaterial whether we take
an incline of one in m to mean 1 ft. vertical form horizontal, or for m on the slope; or whether we
call a slope of & an incline of one in cos x or cosec 2, the former being used on the plan, and the
latter on the constructed work.
With the same approximation, the resistance of the road on the level or the brake power,
estimated at r pounds per ton, may be replaced by the resistance of an uphill incline of 1 in
2240 —— r. *
According to the experiments of Smeaton and Rennie,
The resistence of Stone Tramways is to be taken as 20
*~.
* AA |
pounds a ton.
22 25 Paved Roads 22 5 x 33 39
25 22 Macadam Roads , 35 44 to 67 32
2 3 95 Gravel Roads 32 39 1 50 25
2? 39 Soft Sandy Roads , 32 210 22
#
TRAIN PROBLEMS ON INCLINES.
(1) Prove that an engine exerting a pull of P tons, starting a train weighing W tons up an
incline of 1 in m, the resistance of the road being r pounds per ton, will in the time t seconds
generate the velocity
in the distance
the energy accumulated in foot-tons being
P 1 7, \?
E = W | *- — “ — — . . ) kat”.
W (. 7%. im #9t
(2) A train of 120 tons is to be taken from one station to the next, a mile off, up an incline of
1 in 80, in four minutes without using the brakes. Prove that with no road resistance the engine
must exert a pull, until steam is turned off, of about 6,203 lbs.; and the weight on the drivers
must be 37,218 lbs. = 16*6 tons, with an adhesion of one-sixth.
(3) Prove that in order to take a train of 100 tons from rest at one station to stop at another
a mile and a-half off, up an incline of 1 in 100, in six minutes, the velocity must reach 30 miles an
hour; the steam must be shut off and the brakes applied 108 yards from the station, and the
engine must exert a pull of about 1:4 tons; supposing a brake pressure on the wheels of half the
weight of the train, and a coefficient of friction of one-sixth.
(4) Determine, in tons, the greatest load the Strong locomotive—with cylinders 20 by 24 in.,
driving wheels 5 ft in diameter, and mean effective pressure 100 lb/in”—can take up an incline
rising 96 ft in the mile, from rest at one station to stop at the next station, two miles off, in eight
minutes; taking an average resistance of the road of 12 lbs a ton, and the brake power at 400 lbs
a ton; the engine and tender weighing 106 tons. Ans. 164 tons.
23
(5) A train of W tons is to be taken from rest at one station up an incline of 1 in m to stop
at the flext station a ft off in t seconds, the resistance on the level being equivalent to an incline
of 1 in n, and the brake-power to 1 in q.
Prove that the pull of the engine, if uniform, must be :—
l
} + *(; +})(; +} +;
O. 7, m n q
W *(; +++)-1 tons,
Q. 770 70, Q
and prove that the extra expenditure of work over taking the train at uniform velocity is.
Wa2 1
#gº q
(; + 1 + } foot-tons.
&m n q
(6) Prove that the weight of the greatest train an engine with cylinders d x lin, and driving
wheels D in in diameter, with a mean effective pressure of p lb/in”, can take in the last journey of
ex. 5, is:
tons gross.
(7) Prove that the loss of time in going from A to C–for instance, from Stanstead to
Newport, on G.E.R.—two points on a railway at the same level eight miles apart, due to an
incline of 1 in 100 from A up to B, and an incline of 1 in 300 from B down to C, instead of going
on a level line from A to C at a uniform velocity of forty-five miles an hour, is 2 minutes 20
seconds, equivalent in time to a detour on the level of 1 mile 60 chains.
It is supposed that with full steam on the velocity drops from forty-five to fifteen miles an
hour at the summit B, and that in descending the incline full steam is kept on till the velocity is
again forty-five, after which the velocity is kept uniform by partly shutting off steam ; and prove
that this happens at a point Q, distant from B about 1 mile 892 yards. Kr
Prove that the train would come to a standstill, or, as the Americans say, would be “stalled
on the grade,” if the incline from A to B was a quarter of a mile longer.
Prove that if the weight of the train is 200 tons, and the resistance of the road is 14 lbs a ton,
the pull of the engine from A to Q is 245 tons, and from Q to B is #3 ton; and determine
whether there is any extra expenditure of work due to the inclines, on this and the return
Journey.
Determine the requisite weight on the driving wheels, taking an adhesion of one-sixth.
Compare the Shap and Beattock inclines, causing a loss of 9 minutes (Times, 22nd October,
1889).
(8) Determine the loss of time and the extra expenditure of work—if any-–due to crossing a
pass by a railway having an incline of 1 in m up and 1 in n down, instead of going in a level
tunnel of length l feet through the pass; supposing that W fls is the maximum speed allowed on
the line, and that the velocity of the train drops from V to v f/s at the summit of the pass;
determine the pull of the engine P in tons for a train of W tons weight, taking the resistance of
the road at r lb per ton.
Determine the length of detour of equal time.
24
IMPULSE.
24. We have already given the name impulse to the product Pt of a force of P pounds and
of the t seconds for which the force acts, and shown that the impulse Pt and the momentum Wv/g
which it generates are equal and convertible; and therefore measurable in the same unit, which
we have called the second-pound; by analogy with the unit of work, the foot-pound, in which the
work Ps and its mechanical equivalent, the kinetic energy #Wv°/g, are measured. º
In a large class of mechanical questions, such as the action of hammers, pile driving, the
collision of railway carriages, the impulse of a steamer against a pier, the motion of a pro-
jectile in the bore of a gun; or, on a small scale, in the impact of billiard balls; a very great force,
P pounds, acting for a very short time, t seconds, produces a finite change of velocity and of
momentum, so that the impulse Pt is finite. º
In such questions it is convenient to begin by supposing the impulse to act instantaneously,
so that the change of velocity takes place while the bodies have hardly moved; afterwards to
investigate more closely what really takes place. *
In the preceding examples on the dynamics of rectilinear motion, the Principle of Energy
has been employed, in which the kinetic energy generated or used up is considered as the
dynamical equivalent of the work done on the body, both being measured in foot-pounds or foot-
tons.
But now we shall employ the Principle of Momentum, which asserts that the momentum
generated—or destroyed—in any direction is the dynamical equivalent of the impulse in the
same direction, both being measured in second-pounds, or second-tons; the impulse being defined
as the product of the time and the mean force in that direction.
In the case of the impact of one body upon another, the same impulse, consisting of the same
force acting for the same time, acts on each body, but in opposite directions; so that the momentum
of the system is unchanged, and is the same just before and just after the impact, if we neglect
any extraneous forces acting on the bodies as insensible compared with the mean force of the
impulse.
p Let us apply these principles immediately to the solution of the following practical
questions:—
(1) “An inelastic pile, weighing w tons, is driven vertically a feet into the ground by n blows
of a hammer weighing W tons, falling h feet. *
nWº J.
W -- w a
drive the pile down very slowly against a uniform resistance.
If the pile is crushed at each blow a small length, measured in feet by ar, the mean
Prove that tons would have to be superposed on the pile, in addition to W, to
Ww
W –– w
thrust exerted by the hammer during a blow is
º tons, and acts for aſh of the time of
falling of the hammer.
Determine also the time of each movement of the pile.
(2) “If the resistance of the ground increases in proportion to the penetration of a pile, and
n blows of a hammer of W tons, falling h feet, drives a pile of w tons, a feet, prove that
27, W* h
-
W -- w a
ons, gradually superposed, would drive the pile the same distance.
(3) “Prove that, if the safe building load on a pile is one-math the resistance to penetration
of the ground, and the pile is intended to bear a safe load of M tons, the pile will be sufficiently
driven home when the penetration at each of the last few blows of the hammer is
W2),
-ºw I feet
(called the refusal of the pile), where W denotes the weight of the hammer, w of the pile in tons,
and h the height in feet the hammer falls.
Prove that, if the chain which raises the hammer can bear a safe load of P tons, the minimum
time of raising the hammer is w/ rºw of the time of falling.”
Considering a single blow in question 1, the hammer in falling h feet acquires Wh foot-tons
of energy, and acquires the striking velocity V f/s given by
#
V2 - gh, OI’ *WWº/g - Wh o e 6 º º & © s g & & (1)
25
The pile being supposed inelastic, the hammer does not rebound, but pile and hammer move
together with the same velocity; supposing this velocity is U f/s; immediately after the blow,
U is determined from the condition that the momentum is unchanged; and therefore, in second-
tons,
(W + wyu/g = ww WV/g,
W
Úr U = w H. W. . . . . . . . . . . . . . . . (2)
and the available energy of the hammer and pile is now
W2 kV2 W2}.
I 2 *º 2 - tº & e gº o sº e g ſe *
#(W + w)U*/g = WTº g W -- w foot-tons (3)
the energy dissipated (liberated) by the blow being
Wºh With
W H w T W -- w
so that the work lost bears the ratio w; W -- w to the whole work, and this is less the greater
the ratio W : w. wº tº
Denoting by R the mean effective resistance of the ground in tons, this energy of the
hammer and pile is sufficient to carry them a distance a/m feet against this resistance, and
therefore by the principle of energy,
Ra W2),
ºn ~ WTº
_ nW* h
Úr = wº; . . . . . . . . . . . . . . . . (6)
We have not allowed for gravity, when the pile is driven vertically, when the effective
resistance R tons must be increased by W -- w tons to obtain the total resistance to motion;
consequently R tons is required to be superposed on the hammer and pile, weighing W -- w tons,
to drive the pile down very slowly. \
The above supposes the blow instantaneous, and consequently that there is no deformation
of the pile, and that the thrust between the pile and hammer during the impact is infinite; but
in reality some deformation must take place, and supposing that the pile is crushed a small length,
denoted in feet by a, and that P. denotes the thrust in tons between the hammer and pile during
the blow, then Pa measures in foot-tons the work dissipated by the blow in crushing the pile,
and therefore—
Wh — foot-tons . . . . . . . . . . . . (4)
W2), W w/,
tº-º-º: I h — * ======s===
_ Ww h *
OT = wº; . . . . . . . . . . . . . . . . . (6)
Supposing t denotes in seconds the time the blow acts, then when a and therefore t is vely
small, P is very great, and we may neglect R in comparison with it; and by the principle of
momentum,
Wat W
* r º-º +. - — ”
Pt = W(V – U)/g = w Ug W H w g’
and therefore t = } = aſh of the time of falling of the hammer; since the hammer is
l
W/g = h/#V seconds in falling h feet . . . . . . . . . . . . . . . . . . . . (7)
After the blow, the pile describes aſm feet with average velocity #U, and therefore the time
of penetrating the ground is
* a W -- w a
|Un T W Wr,
W -- w a w *
= 'º-º ºr of the time of falling of the hammer . . . . . . . . . . . . . . . (8)
W wh
Suppose, however, we consider the case where the crushing of the pile a becomes comparable
with the penetration, b feet suppose, at each blow. Now R, the effective resistance of the ground,
becomes comparable with P; so that for the t second the blow lasts, considering the change
of momentum of the hammer,
seconds
Pt = WW/g – wu/g . . . . . . . . . . . . . . . (9)
and considering the change of momentum of the pile,
Pt – Rt = wV/g. . . . . . . . . . . . . . . . (10)
so that Rt = WW/g – (W + w}U/g . . . . . . . . . . . . . . (11)
a quantity we have previously neglected as insensible.
(6873) H
26
Pile and hammer are now, just after the blow, moving with common velocity V f's, and
will therefore penetrate a distance y feet against the mean effective resistance R tons given
by
Ry = }(W + w)U*/g. . . . . . . . . . . . . . . (12)
But while the blow lasts the pile will move bodily a distance 2 feet, such that the energy
liberated by the blow is used up in crushing the pile a ft. against a thrust of P tons, and
moving it 2 ft. against a resistance of R tons; so that:
Wh — ;(W + wyvº/g = Pa + Rz, f*
Or Wh = Pa + Rb . . . . . . . . . . . . . . . . (13)
since y + 2 = b . . . . . . . . . . . . . . . . . (14)
Knowing W, w, h, and observing b and w, we have sufficient equations for determining
P, R, U, y, and 2 ; but as these equations become rather complicated, and the results are of no
practical value, we shall not pursue them further. e
The preceding considerations, treating the pile as inelastic, are requisite for roughing out the
theory of pile driving before introducing the refinement of taking into account the elasticity of
the pile; for this further treatment the reader is referred to Rankine's Applied Mechanics, Sec-
tion 616, and to the works of Collignon and St. Venant.
25. We have supposed the resistance of the ground uniform, but if, as in Question 2, we
suppose the resistance to increase as the penetration, then the average resistance will be half the
maximum resistance: denoting this by R tons, then the work required to penetrate a feet
is #Ra foot-tons, which must be equated to the total available energy of the blows; so that
1B, nWºh
#Ra = WT º'
_ 2nW* h
OT R = wº, ,
Here again the only effect of taking into account gravity when the pile is driven vertically,
is to reckon the penetration a, not from the surface of the ground but from the depth which
the pile will sink under its own weight and the weight of the hammer superposed; and
then the additional penetration of a feet will be obtained by gradually superposing R tons.
The penetration at each blow will no longer be the same, but supposing that a, feet is the
penetration produced by r blows, then the rth blow produces penetration
a;-a, -, feet, against an average resistance of #(a, + a_i)R/a tons;
and the available energy of each blow being W*h/(W + wy foot-tons,
W2h
therefore #(a,” — a, 1%).Rſa = WI, 7 #Raſm,
so that a,” — a, 1% = a”/n,
giving a,” = a”/n,
and a, - a, –1 = [Vr – V(r – 1)]aſn
the penetration at the ºth blow; a formula similar to that giving the time it takes for a falling body
to describe equal vertical distances.
The time occupied by the pile in penetrating the ground at each blow must be investigated
subsequently.
We may, however, determine the time the pile would take to penetrate a feet under a single
blow of the hammer, falling nh feet, against a resistance of the ground increasing as the pene-
tration, by comparing the motion of the pile with that of a piston, moving from the middle to
the end of a stroke; when, as pointed out in Ex. (5), No. I, the maximum velocity is ºr times the
average velocity.
Supposing then that hammer and pile start moving together with velocity Uffs, the average
velocity of penetration is now U/#T f/s; and the time of penetration is
W + w #7ta W -- w łTra
# *Eº 2 tº-º-º-º: 4.
#Ta/U = WT V- seconds = TWT ºff,
of the time the hammer takes to fall m/, feet.
But the determination of the time occupied with successive blows of the hammer, falling h
feet, must be reserved for the present, till the circumstances of motion under a force proportional
to the displacement have been investigated.
If the resistance of the ground was really uniform, then gradual loading to drive the pile
down would be dangerous, as the pile would not move till the full load required was Superposed,
and would then, tend to start with a run; with increasing resistance, however, this method is safe,
and is used in sinking a caisson.
27
, 26. In building operations the safe load on a pile is taken as a certain fraction, one mth, of the
force required to drive the pile; if the pile is designed to bear a safe load of M tons, the force
required to drive the pile should be mM tons, and in Question 3 we wish to determine when the
pile is sufficiently driven from an observation of the penetration of the pile at each of the last few
blows. ſ
The penetration b at each blow being small compared with the total penetration, we may
consider the resistance R of the ground uniform, and replacing the resistance by mM tons,
then
W2h
mM = wº
Ol' b = W*h
-* mM(W + w)
To raise the hammer in the least possible time, notice that the extreme tension of the chain,
*P tons, must be exerted vertically through a feet until sufficient velocity v f/s is imparted
to carry the hammer freely through h-a feet to the full height required, h feet; and therefore, in
foot-tons, -
#Wv°/g = (P – W)w = W(h — a ;
so that Pa = Wh,
and *g – ºr "h, veg – i.
Ol' - W = Wºr").
Therefore the average velocity of ascent, #v, is v/ { (P-W)/P } times the average velocity of
falling #V ; and therefore the least time of ascent is vſ. ſ P/(P-W) - of the time of falling ; and in
this way we gain an idea of the greatest speed at which the pile driving can be carried on with
given appliances.
(4) An inelastic pile weighing half a ton is driven 12 ft into the ground by thirty blows of
a hammer weighing 2 tons falling 30 ft.
Prove that it would require 120 tons in addition to the hammer to be superposed on the pile
to drive it down slowly, supposing the resistance of the ground uniform ; but 240 tons if the
resistance increases as the penetration. *:
Prove that with uniform resistance, each movement of the pile takes 0.0228 sécond.
(5) If the resistance of the ground to the penetration of an inelastic pile is 60 tons, prove that
15 blows of a hammer weighing 1 ton falling 20 ft will drive the pile 4 ft into the ground, the
pile weighing 3 ton.
Prove also that the time of each movement of the pile is 0.01863 second.
6) A pile is driven a feet vertically into the ground by n blows of a steam hammer fastened to
the head of the pile. Prove that, if p is the mean pressure of the steam in pounds per in?, d the
diameter of the piston in ins, l the length of stroke in ft, W the weight in lb of the moving parts
of the hammer, w lb the weight of the pile and the fixed parts of the hammer attached to it; then
the mean resistance of the ground in pounds is—
mW !
- 47td?n) –,
W –– w (W + +7tdºp) C!
Putting #Tilºp = P, the total thrust of the steam on the piston, prove that the whole time
of an ascent and descent of the hammer is, in seconds— -
Vº A/ 2PW ) + V. W }
| (p. TW3) (F + W
and prove that the volume of steam used at uniform pressure p is 1 + W/P times the volume of
the cylinder.
Discuss the best rate of working of a steam pump, required to pump a given quantity of
water against a given head and frictional resistances.
(7) Why, when we wish to drive a hammer shaft home into the hammer head, do we hit on the
butt end of the shaft, and not on the hammer head? Compare the distances the shaft is driven in
the two cases. .
(8) Prove that if a hammer head weighing 2 lbs striking with a velocity of 50 f/s a nail, ºr in
in diameter and weighing 1 oz, drives the nail 1 in into a plank of wood, then a bullet 0-5 in in
diameter and weighing 1 oz striking with velocity 1,500 fſs, will penetrate 1:16 in of the wood;
supposing the resistance uniform and proportional to the sectional area of the hole.
Determine also the penetration of the bullet, supposing the resistance proportional to the
penetration.
Determine also the time of movement in each case.
Answer.—When the resistance is uniform, the nail is 0.0034375 second, and the bullet is
0.00013 second in penetrating the wood; but when the resistance increases as the penetration,
the nail takes 0-0027 second, and the bullet penetrates 1.077 in, and takes 0.000094 second.
28
(9) A hammer head weighing W lb, moving with velocity v f/s, strikes a nail weighing
* lbs fixed in a block of wood weighing M lbs, which is free to move; prove that if the mean
resistance of the block to penetration is R pounds, the nail will penetrate each blow a distance
of—
MW2 #ºf
* - feet
(MTWTº) (W Ivy gF “
taking wºrd jº seconds,
during which the block M will move
MW* ** feet,
(MTWTºjº gR
MW
(M + W -- w (W + wy
W -- wi
But if the block is fixed, the penetration will be increased to (l + W; ) times this penetration,
and will take Wv/gE seconds. (Compare setting a plane or holding up in rivetting.)
so that a fraction of the total work is utilized in penetration.
(10) Prove that the mean resistance of the wood is 204 pounds to a nail weighing 1 oz; sup-
posing a hammer weighing 1 lb striking it with a velocity of 34 f/s drives the nail 1 in into a fixed
block of wood.
If the block is free to move and weighs 68 lbs, prove that the hammer will drive the nail only
£4 in.
"Prove that the nail is 00052 and 0005128 second in penetrating the wood in the two cases;
during which the block if free will move 0.015 in.
(11) A rifle weighing 10 lbs is suspended horizontally by two equal parallel threads, and is
observed after firing a bullet weighing 1 oz to recoil so as to ascend one foot.
Calculate the velocity of the bullet; and prove that if the length of the barrel traversed by
the bullet is 3 ft, the bullet is about 0-00466 second traversing the barrel, and that the rifle will
have moved about 2-118 in when the bullet is leaving the muzzle.
(12) Prove that if a Ballistic Pendulum, made of a box filled with sand weighing W lb, and
suspended by equal parallel vertical cords of length L ft, is found to recoil through an are whose
chord is c ft as measured on a tape drawn out in the recoil, when struck in a horizontal line
passing through the centre of gravity of the box by a shot weighing w lb ; then the striking
velocity of the shot and the energy liberated by the impact are respectively—
W –– w g st l(W7 W c2 sº
Wi e V (£), f's, and (W4 w foot-lb.
Graduate the tape.
Prove that if the shot penetrates into the sand a distance a feet against a uniform resistance
2w a ^ ; seconds, during which the box of sand will have moved
W -- w e
R pounds, this will take
about aw/(W + w) feet.
Thus if W = 2,000, L = 8, c = 6, and w = 20, then v = 1,212 fis, and if the penetration of the
shot into the sand is 2 ft, a = 2; and then R = 227,250 pounds; the time of penetration being
0-0033 second, during which the box will have moved about 0.2376 in. -
(13) Prove that when the weight of the powder charge, Plb, is taken into account, and it is
supposed that the density of the powder gas is always uniform, the velocity V f/s of the shot
weighing W lb, and the velocity of recoil U f/s of the gun and carriage weighing M lb are, with
no elevation, connected at any point by the equation—
(W + P)W = (M + P) U,
to that the weight of the charge may be supposed shared equally between the shot and the gun.
Prove that if the bore is lift long, and the powder pressure is supposed constant, the shot is
–– - 2M + P l seconds
#(V + U) M + W. H. PW
te g * * e e º * W + |P
traversing the bore, during which the gun and carriage will have recoiled TM-H wº P
and the thrust of the powder on the base of the shot and on the base of the bore of the gun is
respectively,
l feet;
M + W. H. P. J.WW2 and M + W H P }MU’
M + , P gl W -- $P gl
Work out the same problem when the carriage sides on a plane, having a given coefficient of
friction; also when the carriage slides on a railway truck of given weight.
The time occupied by the shot in the bore of the gun is important when Colonel Sébert's
Velocimeters are used.
Show how to represent in a graphic diagram by curves the velocity and energy of the shot,
and the developed work and pressure of the powder in the bore of a gun.
pounds.
29
(14) A coal train consists of an engine and tender of weight M tons and r wagons, each of
weight m tons, with buffers in contact, and a foot slack of chain.
If the tractive force of the engine is P tons, then when the last wagon is started, the velocity
U in f's is given by *
#U*/g = Pa (rM + 4*(r. — 1)m}/(M + rm)”, Pa (r.M + š, (r-1) m}/(M + rm)”,
and the engine will have advanced ra feet, taking *
v/{rM + #r(, — 1)m} V(2a/Pg) seconds.
Discuss the motion when steam is shut off and brakes applied to the engine and tender.
(15) Supposing that the penetration of a body by a projectile weighing m lb moving with
velocity v f/s is a feet when the body is fixed, prove that if the body is free to move on a smooth
horizontal plane, and weighs M lb, and is b feet thick, the body will be perforated if
{) i Ma/(M + m); and that after passing through the body the projectile will retain the
velocity
3% + Mv4 + º + ")/Mał, ſº,
the resistance being supposed uniform; while the body will have communicated to it the velocity
m – m V/{1 — (M + n)b/Ma} v f/s
M + m º
It is supposed that the projectile passes through the centre of gravity of the body, so that no
rotation is communicated to the body.
Prove that the time occupied in perforation is
1. b seconds
1 + VH1 – (M + m)b/Ma} }v * >
(16) If the body is approaching with velocity V f/s, the bullet will be imbedded if
& S- wº (1 + y) and the body will retain the velocity (MV — mo)/(M + m) f/s, being stopped
7?? g -
M
if MV = mv ; and the time occupied in penetration will be ºr tº - (l + ¥) * seconds.
h M + m v/ #v
But if the body is perforated, it will retain the velocity
MV — mo + V/{m”(1 + V/v)” – (M + m) mºb/Małv f/s
M + m, 2
while the bullet will retain the velocity
me — MV + V (M*(1 + V/v)” – (M + m)Mb/a}v f/s
M + m 2
and the time of perforating the body will be
1. b
II vſ IV[IIvºy-OMHz)/Ma, in "
This is the dynamical problem required in shooting big game.
(17) If water flowing in a main or pipe l feet long, with velocity V f/s, is shut off quickly and
uniformly in t seconds by a stop-valve, the pressure of water in the pipe near the valve is increased
by Gly/1444t pounds on the square inch, G denoting the density of water in pounds per cubic
foot, about 62°4. -
, Thus if l = 50 ft, W = 24 fis, t = 0.1 sec, the pressure is increased by 162.5 lb per square
11) CI).
(18) Show how to determine the velocities after impact of two elastic spheres from Newton's
experimental law of rebound, that in the direct impact the velocityof separation bears a constant
ratio e to the velocity of approach, where e is a proper fraction, called the coefficient of restitution.
Prove that the energy lost in the impact is #(1 — e”) Cvº/g ft-lb, where v is the relative
velocity before impact, and C is the harmonic mean of the weights of the spheres.
Conversely, prove that if the molar Kinetic Energy lost in the direct collision of two spheres
is to depend only on their relative motion, and not on the actual velocities in space, the law of
rebound must be the one assumed by Newton.
(6873) I
30
WORK, POWER, AND HORSE-POWER.
27. The preceding illustrations have considered the problems arising in the varied motion of
bodies, starting and stopping, and moving with variable velocity; but to the engineer a more
general subject of practical interest is the consideration of the power, amount of fuel, and running
expenses required in the uniform motion for a long period of a railway train, of a steamer at sea,
or of a mill or machine. -
We shall require three definitions (i) of Work; (ii) Power; (iii) Horse-power. .
(i) Work is measured by resistance overcome, that is, by the product of the resistance and of
the distance it has been overcome; thus, if a resistance of R pounds is overcome through s feet,
the work done is Rs foot-pounds, as in Section IV. (e.g., in grinding corn or boring a gun). . .
(ii) The Power of an agentin uniform motion is measured by the work done in a given time,
say in the unit of time; if the Resistance of R pounds is overcome at a velocity of v f/s, then Rw
will measure the power of the agent in fſs-pounds.
(iii) The horse-power is the arbitrary unit of power employed in practice by engineers; it was
defined by Watt as a power which performed 33,000 ft-lb. per minute, or 550 ft-lb. per second.
Thus, if the resistance R pounds is overcome at a velocity offs, the H.P., (horse-power) of the
agent is Rw/550. Watt's earliest engines being constructed to clear mines of water, his estimate of
power was apparently based upon a rule easily applicable to the case of pumping, or “forking”
water; one horse-power forking a hogshead of water per second one foot high, or 10 hogsheads of
water per minute one fathom high ; taking a gallon of water as weighing 10 lbs., and a hogshead
as 55 gallons, the mean of 54 and 56 gallons, by which the hogshead is variously defined 4
hogsheads of 56 gallons going to the ton.
Watt derived his horse-power from actual experiments carried out at Messrs. Barclay and
Perkins' brewery, with certain of the very heavy horses belonging to the firm. These horses
... were caused to raise a weight from the bottom of a deep well, by pulling horizontally on a rope
assing over a pulley. He found that the maximum work on which it was safe to reckon was
22,000 ft-lbs per minute. By adding 50 per cent. to this, he obtained 33,000. The margin,
11,000 ft-lbs, he allowed for loss in engine friction, &c.
The French cheval vapeur, defined as doing 75 kilogramme-metres—542-5 ft-lb.-per second,
is about 1% per cent. less than our horse-power; and recently it has been proposed that French
tº engineers should use the poncelet as unit of power, defined as raising 100 kg (1 quintal) per
second 1 metre high. The poncelet would thus be about 30 per cent. greater than 1 horse-
power,
QUESTIONS ON POWER AND HORSE-POWER.
(1) Find the horse-power of Niagara, where 18,000,000 cubic feet or 562,500 tons passes per
minute, falling 162 ft. -
(2) A cistern on Cooper's Hill, at a height of 250 ft above the river, is to be filled with 20,000
gallons of water every 24 hours by pumps, worked at Old Windsor Lock, by a turbine with a fall
of 4 ft. Determine how much water the turbine will require, and its horse-power with an efficiency
of 70 per cent.
(3) Find the horse-power of an engine which pumps up water from a depth of 50 ft and
delivers it at the rate of 1,000 gallons per minute through a pipe whose cross-section is 1 square
foot. A gallon of water weighs 10 lbs, and its volume is 0.16 cubic ft. Answer.—15; horse-
power.
. (4) Given, that the coal consumption of a locomotive engine is 2 oz per ton per mile at 50
miles an hour (Findlay's Railways) determine (i), the coal and water required to take a train
of 200 tons at this speed from London to Edinburgh, a distance of 400 miles; (ii) the coal burnt
... per mile ; (iii) the horse-power of the engine; (iv) the average resistance of the road in pounds
per ton, taking a consumption of 24 lbs of coal and 25 lbs of water per horse-power hour.
Answer.—(i) 4-5 tons of coal, 44-6 tons of water; (ii) 2.5 lbs; (iii) 500 horse-power; (iv)
1875 lbs per ton.
This implies a pull by the engine of 3,750 lbs, and a load on the drivers of over 10 tons, with
an adhesion of one-sixth ; and with cylinders 18 in by 24 in and driving wheels 7 ft in diameter,
a mean effective pressure of over 40 lbs per square in. ,
(5) An engine of 400 horse-power can draw a train of 200 tons gross up an incline of 1 in 280
at 30 miles an hour. Determine the resistance of the road in pounds per ton. Answer.—17 lbs
per ton. -
(6) Prove that the horse-power of an engine drawing a train of 120 tons up an incline of 1 in
224 at 30 miles an hour is 336, taking the resistance of the road on the level at this speed at 25 lbs
a ton.
Determine also the horse-power of an engine drawing a train of 200 tons up an incline of 1 in
140 at 30 miles an hour, resistance of road 18 lbs a ton; or of the engine drawing the train down
an incline of 1 in 140 at 50 miles an hour, resistance at this speed 50 lbs a ton.
Determine the pull of the engine, and supposing it constant, find how far the train would go
up an incline of 1 in 100, before the velocity dropped from 50 to 30 miles an hour, taking an
average uniform resistance of the road of 20 lbs a ton.
31
(7) Prove that generally the horse-power required to draw a train of W tons, up or down, an
incline of one in m, at a speed of s miles an hour, against an average resistance of the road on the
level at this speed of r lb per ton is -
2240\ Ws
( + == ) 375
On a very steep incline, this formula must be replaced by the accurate formula,
(r cos x + 2240 sin 2) Ws/375,
where a is the slope of the incline; this is derived from the well-known theorem, implied in
Section W, that the work required to draw the body up a rough inclined plane is the same as that
required to draw the body on a similar level road, and afterwards to lift the body vertically to the
requisite height on the incline, assuming Morin's laws of friction to hold.
(8) If a bicyclist always works at the rate of one-tenth of a horse-power, and goes twelve
miles an hour on the level, prove that the resistance of the road is 3-125 pounds.
Prove that up an incline of 1 in 50 the speed will be reduced to 5.8 miles per hour, supposing
a man and machine to weigh 12 stone; and that the mean vertical thrust on each pedal is about
72 lbs, supposing the wheel of the bicycle 56 in in diameter, and the cranks 4 in long.
(9) Supposing a tricyclist and rider, weighing together 200 lbs, to run uniformly at 8 miles
an hour down an incline of 1 in 100 against the resistance of the air and of the road, without
working the pedals; prove that to go up an incline of 1 in 200 at the same speed, the rider must
be working at the rate of -064 of a horse-power; and that the mean thrust on each pedal
will then be about 12-672 lbs, supposing the cranks 5 in long, and making 100 revolutions a
minute.
(10) Define Work, Potential Energy, Horse Power, Efficiency, and Mechanical Advantage.
A Machine of mechanical advantage 25-6 is of such design that the power is applied at the
teeth on the rim of a wheel of one foot radius. The machine is employed to raise vertically a
weight of one ton, and is worked steading without friction by an engine of unit horse-power.
Shew that the wheel makes one turn per second. - -
It is needless to multiply numerical applications of these principles; the theoretical student
will, however, find collections of examples on this subject in Twisden's Practical Mechanics,
chap. ii. -
*2. In railway problems the speed is always given in miles an hour; and then the horse-
power required to overcome a resistance of R pounds at a speed of s m/h (miles an hour) is
R s x 5280 -- (3600 x 550) = R s/375. - -
Various empirical formulas, depending on the state of the road and of the rolling stock, are
given for the resistance of a train on the level, expressed in pounds per ton r, at a speed of s miles
an hour, such as
* = 3 + s^/217. . . . . . . . . . . . . . . . . . (1)
2 = 8 + 8°/171. . . . . . . . . . . . . . . . . (2)
* = 12 + s”/114. . . . . . . . . . . . . . . . . (3)
all of the form r = A + Bs" . . . . . . . . . . . . . . . . . (4)
consisting of a constant term A, giving the resistance at starting, and a term B s”, increasing as
the square of the speed 8, and tending to impose a finite limit to the running velocity.
So far as the resistance of the air is concerned, it is assumed to vary as the square of the
velocity, and taken by Barlow as 106 lbs per square foot at 45 miles an hour.
To determine the coefficients A and B experimentally, the gradient is observed, say 1 in m, on
which the train will just start running down of itself; and then the terminal velocity, say Sm/h,
is observed which the train acquires on a steeper gradient of say, 1 in n. Then
A = 2240/m, A + BS* = 2240/n,
whence A and B can be determined.
Such experiments were carried out at a very early period of railway history by Barlow,
Transactions of Institution of Civil Engineers, Vol. III, and Strength of Materials, Appendie
—Lardner, and Scott Russell, British Association Reports, 1838–9, and others; but no recent
experiments appear to have been made, at least in this country. f S.’
(Engineering News, p. 584, 9th June, 1892; “New Facts as to High Speed Train Resistance.”)
The horse-power then required to draw a train of W tons at a speed of s miles an hour up or
down an incline of one in m, against a resistance of the road given by formula (4) is :-
2240) Ws,
772 );
(A + Bs” +
by which examples 4, 5, 6, can be re-calculated, using the numbers in formulas (1), (2), (3).
Gradients are measured by 1 in cot 2 on plan, 1 in cosec 2 on the ground in railways,
Gradients on High Peak Road, of 1 in 7,8}, and 10%, are worked by stationary engines, and 1 in 14
by locomotives; load, 1 truck. On Khojak there is a gradient of 1 in 2}.
32
At sea the speed is always measured in knots, a knot being a speed of one nautical mile, or
mean minute of ſatitude per hour; roughly speaking, a knot is a speed of 100 ft a minute–more
accurately, 101 ft, or 101.4 ft —equivalent to taking the nautical mile—of 1,000 fathoms—as
6,000 ft or 2,000 yards; and then the horse-power required to overcome a resistance of R pounds
at a speed of V knots is RV/330; but more accurately, RV/325.65, taking the nautical mile to be
6,080 ft, as on the Admiralty measured mile, so that one horse-power, 550 ft-lb per second, is
about 326 knot-pounds (Rankine). e ge e -
Dr. Troubridge finds for an 8-oared boat, over a 4 mile course in 21 minutes, speed 1000 flm
or 10 knots, a resistance 75 lbs; and therefore I.H.P. per man 0-28, about seven times that of a
labouring man in a day's work. - % & y
A steamer differs from a locomotive in being practically always on a level road. Granted
fair weather, the coal consumption and the economical conditions of a voyage can be accurately
calculated. Allow 1.4 lb of coal per horse-power-hour; thus the City of Paris burns 300 tons of
coal a day, at 20,000 H.P.
ADDITIONAL QUESTIONS ON H.P. OF STEAMERS, &c.
(10) Determine the longest paying voyage of a steamer—e.g., the Sirius of 1838—of 700
tons capacity for coal and cargo, when freight is a penny a mile for 10 tons, supposing the
steamer to go 8 knots with a consumption of 20 tons of coal a day, the coal costing 12s, a
ton, and the steamer £20 a day for wages, repair, and deterioration of capital. Answer.—
2,880 miles, a voyage of fifteen days.
Compare this with a modern steamer of 5,600 tons, going 12 knots on 50 tons of coal a
day.
y (11) Determine the most economical speed of a passenger steamer or troopship, costing
#400 a day for provisions, wages, &c., given that the speed is 8 knots on a daily consump-
tion of 50 tons of coal, costing 10s. a ton, and that the daily consumption of coal varies as
the cube of the speed. Answer.—16 knots. h
The amount of fuel consumed per hour by a steamer varies as the cube of the velocity. She
consumes 1-5 tons of coal per hour at 18s. per ton, when her speed is 15 knots. She costs for
other expenses 16s. per hour. What would be the least cost for a voyage of 2,000 miles?
The wages of the crew of a steamer are £22 per day, and when the steamer performs a
voyage between two given ports in 9 days it consumes 550 tons of coal at 16s. 8d. per ton;
assuming that the total amount of coal used varies as the square of the speed. Prove that the
most economical voyage will cost £495. -
(12) Determine the resistance of a steamer in tons, when 8,000 effective horse-power is
required to drive it at 174 knots; and generally when N indicated horse-power is required
to drive it at a speed of n knots, the efficiency of the propelling machinery being e per cent.
Answer.--67-5 tons; 326 Ne/224,000m tons.
(13) In a steamer of 5,000 tons displacement and 400 ft long, when driven at 15 miles an
hour, the draught of water appeared to be increased 1 ft forward and diminished 1 ft aft. Prove
that the effective horse-power required in consequence was 2,240.
(14) Given that 50 horse-power is required to drive a steamer 100 ft long at a speed of
8 knots, prove that 6,400 horse-power will be required to drive a similar steamer 400 ft long,
similarly immersed, at 16 knots, assuming that the resistance is proportional to the wetted surface
and the square of the velocity through the water.
Prove that in a given voyage from one port to another, the cost of coal per ton of cargo will
be the same in the two steamers. -
(15) Given that 100 horse-power is required to drive a steamer of 120 tons at a speed of 9
knots on a daily consumption of three tons of coal, prove that 12,800 horse-power will drive
a similar steamer of 7,680 tons at 18 knots on a daily consumption of 384 tons of coal, the
resistance varying as the wetted surface and the square of the speed; and prove that for a voyage
of given length the coal required will be proportional to the tonnage of the steamer.
(16) Suppose it is found by experiment that the resistance of similar vessels is propor-
tional to the wetted surface and the square of the speed, and that a model steamer, 6 ft long,
and displacing +}t; th of a ton of water, experiences a resistance of 0-198 lbs at a speed of 2 knots.
Prove that a similar steamer 600 ft long, and displacing 10,000 tons, will require 20,000 horse-
power to drive her at a speed of 20 knots, assuming an efficiency of the propulsive machinery of
60 per cent. With a coal consumption of 1.4 lb per H.P. hour, this steamer will burn 300 tons a
day. -
Design similar 22 and 24-knot steamers to cross the Atlantic, 2,800 miles, in 5% and five days.
Calculate the grate areas at 22 I.H.P. per square foot of grate area; also the boiler volumes,
at a ſt” per I.H.P.
(By similar experiments on the resistance of models in an experimental tank, Mr. R. E. Froude
is able to calculate by Froude's law the horse-power required for a given speed of the full-sized
steamer; the model being made on a scale of one to a, and run at a speed on a scale of one to v/a.)
(17) Prove that if the linear dimensions of two similar steamers are in the ratio of a to 1, then
when their speeds are in the ratio of V/a to 1, the resistances will be found to be as a” to 1,
(Froude's Law of Similitude.”) and the horse-powers (and grate areas) as a” to 1, the revolutions
of the engines as 1 to v/a, the piston speeds as Va to 1, and the steam pressures as a to 1; also
the resistance of the water will vary as the wetted surface and the square of the speed conjointly,
33
(For denoting by l, s, v, r, h.p.) n, p, the length, wetted surface, speed, resistance, horse-power
revolutions, and steam pressure of the smaller vessel, and by capital letters the corresponding
quantities in a similar larger vessel, a times longer, then
and = +*-* = a&º, so that P F Q.
For a voyage of given length the coal consumption varies as the horse-power directly
and the time directly or the velocity inversely, so that the coal consumption in the two steamers
is as a” to 1, that is, as the tonnage; so that the cost of fuel per ton of cargo is the same for the
same voyage in the two steamers; but the larger steamer can earn more money in a given
time by reason of her greater speed.
If the grate area is designed to vary as the horse-power, the larger steamer requires more
grate area in proportion to her size in the ratio of a” to 1, instead of a” to 1, as in geometrical
similarity; while the volume of steam required is greater in the ratio of a” to 1, and at a pressure
greater in the ratio of a to 1. With similar engines the scale of size of the boilers must be
considerably increased in the larger steamer; and in recent cases of disappointment in expected
speed of large steamers, the cause can generally be traced to insufficient boiler power.
(18) Explain how the horse-power of an engine is determined experimentally by means of
a friction brake. Suppose that the band of the friction brake extends over the upper half of
a fly-wheel of radius r feet, and that the band is kept tight by means of a weight of W lb
hung at One end, and a spring balance at the other end. Prove that if the spring balance registers
a tension of W’ lb. when the engine is making N revolutions a minute, the horse-power of
the engine is—
2TNr(W – W’) -- 33,000.
(For practical details and description of friction brake dynamometers, a paper by W. W.
Beaumont, M.I.C.E., in the Proceedings Institution of Civil Engineers, Vol. XCV., should be
consulted.) º
(19) Determine the horse-power transmitted by a belt passing round two pulleys, and
running at a speed of 600 ft per minute, supposing the difference of tension of the two parts is
1,650 lbs. Answer.—30 horse-power.
Prove that for a rope round pulleys, running at 3,000 ft a minute, to transmit 40 horse-
power, the tension on the driving side must be 880 lbs, supposing it double the tension on
the slack side.
(20) Prove that the towing horse-power of a tug going W knots is
WV #!” – a”
326 2al
supposing it is, found that the tow rope, weighing W Ibs and l feet long, sags a feet in the
middle, when the ends of the rope are at the same level.
(6873) K
34
II. THE ABSOLUTE UNIT OF FORCE.
29. We have shown that a force of P pounds acting on a weight of W lb will produce
acceleration a celoes, given by the equation
Q, P _ W. .
g - wº or P = #º
and that if the force, acting through s feet for t seconds, causes the velocity to grow from V to
v f/s. o
Ps = }W* — #WV. (foot-pounds);
g 9
Pt = y º y (second-pounds).
In these equations W always appears qualified by the divisor g; not because W is an invari-
able quantity, as is generally taught; but because our unit of force, the force of a pound, is a vari-
able unit, and depends in magnitude on the local value of g. -
To avoid the trouble and waste of space in writing and printing y in the equations of
motion, it was customary formerly in treatises on Dynamics to replace y by a single letter M,
thereby saving a line in printing; and then M was called the mass of the body; but we shall avoid
this usage, as being in our opinion the source of great confusion in Dynamical teaching.
But we can get rid of g in our equations of motion in a different manner, by changing the unit
of force. -
Suppose we write the equations
Pgs = }Wv? – WW2,
Pgt = Wv — WW,
and now put Pg = Q ; this is equivalent to taking a new unit of force one-gth part of our former
unit, in accordance with a suggestion due to Gauss for the purpose of the comparison of the
magnetic measurements made in gravitation measure all over the world; this new unit of force is
an invariable unit, the same for the whole universe; we call it the poundal, a name due to Professor
James Thomson. -
Conversely a force of Q poundals will be a force of Q/g (= P) pounds; and if Q poundals acts
upon a weight of W Ib and produces acceleration a, f/s”,
then P = ** = . . . a,
9 9
Or Q = Wa,
Or a = Q/W,
from which g has disappeared.
The poundal is now the force which produces unit acceleration, one celo or flsº, in a lb weight,
or makes its velocity grow one fls per second.
*- Now if Q poundals, acting through s feet for t seconds, causes the velocity of W lb to grow
from Wto v f/s (imagine a carriage on a level road, or a shot in the bore of a gun).
Qs = }Wv° — $WV* (foot-poundals),
Qt = W v — WW (second-poundals).
The words foot-poundal and second-poundal are framed by analogy with foot-pound and second-
pound; but we can also say the momentum is Wv lb-veloes.
We are obliged to coin these new words, because, as De Morgan says (Calculus, p. 66), “we
must not wait for words because Cicero did not know the Differential Calculus (or Dynamics).”
With g = 32, the poundal is the force of attraction of the Earth on a half-ounce weight, so
that the poundal is rather a small unit for practical questions.
*
35
We have defined the weight of a body as the quantity measured out by the operation ef
weighing in the balance, expressed in lb so that the weight is the same as what is often called
the mass, if also expressed in lb. -
But when the gravitation unit of force (the force of a pound) is employed, the weight of a
body also gives the force in pounds with which the body is attracted by the Earth; and as this
attraction of gravity is inseparable from matter, the word weight is sometimes used in a secondary
sense as meaning this force of attraction; but we must not believe the books which maintain that
this secondary meaning of the word weight is the only true meaning, as this is in opposition to the
usage of the Acts of Parliament, the Bible, Shakespeare, &c. -
The usual definition given that “the weight of a body is the force with which it is attracted
by the Earth’ is, with gravitation units, merely an inexact truism, inexact because it neglects
the discount of g due to the rotation of the Earth.
With absolute units this definition leads to such perplexing statements as “the weight of a
lb weight is roughly the weight of half an ounce.”
The poverty of the English language, in having only one word weight to bear these two
distinct meanings, is unfortunate; the French language is better off, as it translates by poids the
primary sense of weight, as meaning quantity of matter (mass); and by pesanteur the secondary
sense of a force of attraction by the Earth (gravitation). --
With the poundal as unit of force, we never use the secondary meaning of weight as gravi-
tation; it would be absurd to speak of the weight of a body as so many poundals.
The absolute unit of force was proposed originally by Gauss for the purpose of the com-
parison of the different measurements, observed in gravitation measure, of the intensity of
magnetism at different parts of the Earth's surface. +
The attraction of the Earth in any locality provides such a convenient and invariable measure
of force that all instruments, great and small, for measuring force and work are calculated and
graduated originally in gravitation measure; the reduction to absolute measure if required being
made subsequently by means of the local value of g; presumably determined previously with the
greatest attainable accuracy by means of pendulum experiments. z
In any dynamical problem at a given place on the Earth's surface, it is immaterial whether we
employ the gravitation unit of force, as formerly, or whether we change to the absolute unit of
force; the only apparent alteration in the dynamical equations consisting in the removal of g as
a divisor on one side of an equation to become a multiplier on the other side.
To test this, the preceding collection of examples can be reworked with the absolute unit of
force, the poundal. º
The force of a pound is a gravitation unit, and depends on the local value of g; so that it
varies slightly for different parts of the Earth's surface accessible to us; the variation is, however,
so slight as to be unimportant in practical engineering and gunnery problems.
But the force of a pound becomes completely changed, when as in Astronomy we take
a voyage into space, and consider dynamical phenomena on the surface of the Sun, Moon, or
lanets.
p Again in the comparison of minute scientific and electrical experiments at different parts of
the Earth's surface we require a cosmopolitan unit of force, and this is secured, according to the
suggestion of Gauss, by making the unit of force one gth part of the local force of a pound ;
following Prof. James Thomson, we have called this force the poundal. \
Thus Joule found by experiment that the Mechanical Equivalent of Heat, measured by the
amount of work which is capable of raising the temperature of 1 lb of water from 50° to 51°F. is
772-55 ft-lb at Manchester, where g = 32.2;
= 24876 cosmopolitan ft-poundals,
= 772.86 ft-lb at Paris, where g = 32.18.
36
THE CENTIMETRE-GRAMME-SECOND (C.G.S.) SYSTEM OF UNITS.
30. In the C.G.S. system a complete dynamical terminology has been made. o
We have already introduced the kine as the unit of velocity, and the spoud as the unit of
acceleration. º
The absolute unit of force is called the dyne, defined as giving to one gramme an acceleration
of one spoud, or as making its velocity grow one kine per second; so that, by the Second Law of
Motion, F dynes acting on m grammes will produce a spouds of acceleration, given by
a = F/m, or F = ma.
If a constant force of F dynes acting on mg through s cn for t seconds generates yelocity
v kines, then #mvº is called the energy in ergs, and mv is called the momentum, in boles; and
thus
Fs = #mvº (ergs),
Ft = mv (boles);
while F's = #mv° — ; mV* (ergs),
Ft = mv — m V (boles),
if the velocity grows from V to v kines.
Since 1 ft = 30:48 cm, 1 lb = 453-59 g,
therefore 1 poundal = 30:48 x 458-59 = 13,825.4 dynes;
1 ft-poundal = (30.48) × 453.59 = 10° x 4217 ergs,
Therefore, with g = 32-19 celoes,
the force of 1 pound = 32.19 x 3048 x 753.59 = 105 x 4:45 dynes,
the work of 1 foot-pound = 32.19 × (30:78)” x 753.59 = 107 x 1:326 ergs.
This shows the minute size of the C.G.S units, and their unsuitability for practical questions;
although they are now used universally in recording scientific experiments, and in electrical
calculations.
In recording very large numbers for which the language of numeration fails, we adopt the
system shown above of exhibiting the number as the product of the two factors, one being the
power of 10, of which the index is the characteristic of the logarithm of the number; a similar
system, with negative power of 10, being used for very small decimals.
Numbers in Astronomy run so large that all our units appear equally small, so that C.G.S.
units are not unsuitable for Astronomical constant, and we shall generally employ them for that
purpose. (Everett, Units and Physical Constants).
The M.K.S. System is sometimes employed; then the energy of M kg moving with velocity
v m/s (metres per second) is #Mv° joules, the unit of energy or work being called the joule, of
10° x (10)* = 107 ergs, since M kg = 1000 M g, and v m/s = 100 v kines.
The power which does a joule per second is called a watt; and in electricity a watt is the
equivalent of a volt-ampère ; but power is usually reckoned in kilowatts, each of 1,000 watts.
The absolute unit of force in the M.K.S. system is 10° dynes; and thus the force of a pound
= 4:45 M.K.S. units; and a foot-pound = 136 joules; one horse-power = 550 x 1:356 = 746
watts; one kilowatt = 1,000 -- 746 = 1-34 H.P. ; the kilowatt and poncelet being very nearly
equal, and about 34 per cent. greater than a H.P.
For the poncelet does 100 kilogramme-metres of work per second = 100 x 9.81 joules per
second, or watts = '981 kilowatts, so that the kilowatt = 102 poncelets, and is thus about
2 per cent. greater than the poncelet.
An incandescent electric lamp requires on the average 14 watts per candle power; so that a
lamp of 2,000 candle power would require 8 kilowatts, or about 11 H.P. to drive it. -
The Mechanical Equivalent of Heat, in metric units, measure by the work required to raise
1 kg of water 1°C, is
426.5 kilogrammetres at Pars, where g = 980-94 sponds:
= 426-3 kilogrammeters at Manchester, where g = 981-34 sponds:
= 4.184 cosmopolitan joules.
EXAMPLES.
1. Prove that the K.E. (Kinetic Energy) of translation of the Earth, relative to the Sun, is
10° x 2.687 joules; and determine the K.E. of the Moon relative to the earth.
2. Calculate in spouds the local value of g (i) on the surface of the Moon; (ii) on the surface
of the Sun, taking the mean density of one-eighth that of the earth.
Prove that the Sun's weight is then about 330,000 times the Earth's weight.
3. Two bodies, weighing m and m'g, are in contact on a smooth horizontal plane; and when
the first is struck by an impulse of B boles, the second is acted upon by a force of F dynes in the
same direction, and overtaking the first with velocity v kines in t seconds and in s em.
Prove that
v = B/m, t = 2Bm"/Fm, s = 2Bºm"/Fm’.
37
3* With given adhesion p (say one-sixth) when must a man start running so as catch a
carriage going at velocity V fis and jump up without shock. -
In alighting, how far off is the carriage when the man has come to rest on the ground 2
4. Supposing a lb of air to occupy 13 ft“, and to behave like a cloud of inelastic particles,
prove that the pressure of wind against a plane perpendicular to its direction should be given by
0.077 tº or 0.165 Vº (poundals ft), and by 0.0024 vº or 0.0052 W* (pounds per ft2), supposing
the velocity of the wind is given as v f/s or V miles an hour. t
This gives a pressure of 10.5 lb/ft” for W = 45, about one-tenth the pressure observed by
Barlow with railway carriages. With a velocity of wind 78 miles an hour, a pressure of 70.2 lb/ft”
was observed at the Forth Bridge.
- 5. Determine the resistance in dynes/cm2 (barads) to a celestial body moving with velocity
v kines through space which is filled uniformly with meteoric dust to the extent of 1g per a cm3.
6. Prove that if a. inches of rain falls uniformly in one hour with average velocity due to falling
'h ft, the impacts will produce a steady pressure of 0-00288 gha poundals/ft”, or 0-00288 ha.
pounds/ft”.
# the pressure due to a fall of 3rd inch of rain in 10 minutes, striking with velocity 20 f.s.
Explain why a waterfall h ft high can support a column of water 2h ft high.
7. Prove by Torricelli's theorem that if liquid of density w lb. per ft” is issuing from a vessel in a
jet of which the cross section is A ft” at the vena contracta, and that if p denotes the hydrostatic
pressure in pounds per ft” in the vessel when the jet is stopped, then
(i) the quantity which issues in t secs. is At v'(2gpw) lb ;
(ii) the momentum which issues in t secs. is 2Apt second-pounds;
(iii) the H.P. of the jet is 2°g’Apºw” -- 550.
Prove that if the jet impinges normally on a plane area at rest, the thrust is 2Ap pounds,
but if it impinges on a cup-shaped vane, the thrust can be made 4Ap pounds; and if the jet is
received by a number of these cup-vanes fixed on the circumference of a revolving wheel—the
Pelton wheel—the mechanical efficiency of the wheel is theoretically perfect when the vanes
revolve at half the velocity of the jet; but the efficiency is halved if the vanes are flat. -
For instance, with a head of 2100 ft of water, or a pressure of 910 lb/in", a Pelton wheel at the
Comstock mine making 1150 revolutions a minute, with a peripheral speed of 1100 ft/m or 12.5m/h,
half the velocity of the jet, gave 5 H.P. for every miners inch, of 1.6 ft/m of water.
8. Assuming that the velocity of a steam jet follows Torricelli's law, and the steam and water
are issuing by equal pipes from a boiler at a pressure of p lb/in above the atmospheric pressure,
and that a and p denote the density of the steam and water in lb per ft“, prove that
(i) the velocity of the steam jet v/ p
OT
23 59 water , 2
(ii) the quantity of the steam jet A/.
3 * 25 water , p
..., the momentum of the steam iet e
(iii) J* = 1;
33 3 9 Water 35
(iv) the energy of the steam jet p
59 , Twater , T a
Explain the action of Gifford's injector by this theory; and denoting the cross section of the
jet in in” by A, prove that the theoretical quantity of water which can be fed into the boiler in t
seconds is
+'. At{ V(2gpp) – V(2gpa)} lb.
Compare the efficiencies of the injector, and of the steam pump, required to feed water at the
same rate. -
Practically, to overcome friction, the cross section of the steam jet must be increased to
B in” suppose, so that the quantity fed in is
+'. Atv/(2gpp) – 4's Btv/(2gpa) lb.
9. Prove that the thrust of a pair of paddle wheels, driving a vessel at velocity u f/s with n
revolution a second is pau (np - w poundals, a denoting the area in ft” of a pair of floats, and
p the circumference in ft of the circle passing through the centre of the floats.
Prove that the efficiency is n/mp or 1 — s, s denoting the slip.
*10. Give a geometrical construction for the determination of the direction of projection of a
billiard ball so as to strike another ball after a determinate number of impacts with the cushions,
taking a given coefficient of restitution e. *
11. Prove that if a stream of small particles moving in parallel directions with velocity u f/s,
impinges successively on two smooth fixed planes of elasticity e at right angles to one another,
the average resultant thrust is w u” (1 + e) poundals, where w is the weight in pounds per ft3 of
the particles.
12. Prove that, on the supposition that the wind acts like a stream of particles, an ordinary
windmill will be doing most work when going at one-third of its speed when unloaded, but a
Canadian windmill should run at one-half the unloaded speed. .*
Determine, by means of ex 4, H.P. of the windmill, or of the sails of a ship, as in fig. 5.
13. Discuss the theory of the Screw Propeller. *
14. An overshot water wheel of 12 ft radius takes it supply from an unlimited quantity of
still water, which leaves it at the extremity of a horizontal diameter. -
Show that, in order that it may be as effective as possible, it must make approximately one
revolution in 4-712 seconds. º
(6873) L
38
THE COMPOSITION AND RESOLUTION OF VELOCITY AND ACCELERATION.
CURVILINEAR MOTION.
31. So far we have treated of the Rectilinear Dynamics of the motion of a body in a straight
line; but now we proceed to discuss the motion of a body due to two or more coexistent com-
pound velocities, as for instance the motion of a man walking across the deck of a steamer in
motion, the motion of a steamer across a tide-way, the motion of a shot fired from a gun on a
steamer in motion, or at another vessel in motion, the vertical and horizontal rise of the tide on
a sloping shore, &c. *. -
TRIANGLE AND PARALLELOGRAM OF VELOCITY AND ACCELERATION.
Velocities and Accelerations are compounded and resolved according to the rules of the
Parallelogram or Triangle of Velocities and Accelerations, in exactly the same manner as Forces
are compounded and resolved according to the Parallelogram or Triangle of Forces.
For if a body moves from O to P, for instance, a steamer, while at the same time a body is
carried from O to Q, across the deck, then the body will have been moved from O to R, where
OR is the diagonal of the parallelogram OPRQ.
These displacements may be supposed to take place simultaneously, in the same time, or
consecutively in any time whatever.
If they take place simultaneously, in the same time, t seconds, then OP/t and OQſt represent the
compound average velocities, and ORſt the result and average velocity, so that we may take OP
and OQ to represent to scale the compound velocities, and then OR will represent to the same
scale the resultant (average) velocity; and thus the Parallelogram of Velocities is established.
Thus if a steamer is going through the water with a speed of V knots, represented by the
vector OP, while the water is setting in the direction OQ with velocity v knots, represented
by the vector OQ, or PR, then the true course and velocity of the steamer is represented by
OR.
It is usual to employ the Triangle of Velocities, as requiring only three lines, in preference to
the Parallelogram, requiring five.
Example 1. Determine the true course and velocity of a steamer steering due north by com-
pass at 10 knots through a 4 knot tide setting south-east, and determine the alteration of course
in order that the steamer should make a true northerly coast.
2. A ship sailing north-east by compass through a tide running 4 knots, finds after 2 hours
she . made good 4 miles south east; determine the direction of the current, and the speed of
the ship.
3. A headland bears S. 59° E., and a 44 knot tide sets past it S. by W.; find the course to
fetch the headland, with a speed of 93, knots.
By taking OP and OR to represent the growth of component velocities, we prove in the same
way that OR represents the growth of resultant velocity; and thus the Parallelogram of Accelera-
tion is established.
All motion in Nature is relative motion; we have no anchorage in space, and thus can
have no idea of absolute motion in space. -
But in motion on the Earth's surface, we consider motion as relative to the land, and investigate
all other motion as relative to it.
A body is sometimes said to be at rest when it has no relative motion to the surrounding
objects, say the deck of a steamer, or a railway carriage, or even the surface of the Earth. (Max-
well, Matter and Motion, chap. ii.) -
In Astronomy, we refer all motion either to the centre of the Earth, or else to the centre of
the Sun, or else to the C.G. of the Solar System; and the whole science of Astronomy consists in
disentangling the various elements of the relative motion of the celestial bodies.
32. Knowing the velocities of two bodies A and B, relative to a third body C, the relative
velocity of A and B is obtained by compounding the velocity of one body A with that of the
revised velocity of B. *
Take for instance the case of an ice-boat ABC, with a mast at O, mounted on runners
at * C, which permits motion on the ice only in the direction CO, so that no lee-way is
OSS10 162.
We suppose that the ice offers no appreciable resistance to the motion in the direction CO;
and now if OV represents the velocity of the boat, and OW the true velocity of the wind, as
shown by the drift of the clouds and waves, or by a vane on a lighthouse; the apparent wind, as
shown in direction by the vane on the mast, will have the direction WW ; and when the boat is
going at full speed, this direction will be parallel to the sail OD. b
For if the velocity OV drops to OV', the apparent wind V’W will fill the sail, and urge the
i. faster; if the wind falls to OW", the apparent wind WW’ will back the sail and retard the
Oat.
At full speed then we may consider the apparent wind appreciably parallel to the sail, and
thus the vane on the mast will also set itself parallel to the sail.
For given velocity of wind W, and given angle 2 of sail OD with OC, the velocity V of the
boat will be greatest when OW is at right angles to VW, and thus the true wind will be an angle
o, abaft the beam, which -
W = W cosec 2.
39
When the boat is sailing at an angle 8 with the wind, the component velocity to wind-
ward, -
U = V cos Á
W sin (8 - 2) cos Á
sin &
y
= | W sin (28 – 2) — sin o.
2 Sil] &
and this is a maximum, and equal to #W(cosec 2 – 1), when sin (28 – 2) = 1,
Or B = #7t + § 2.
Suppose, for instance, we can set the sail so that sin & = }, or less; then the maximum
velocity with the wind nearly abeam, is 3W or more; while the maximum wind and velocity is W,
or more; so that it is possible to beat to windward faster than the wind is blowing in the opposite
direction.
For a similar reason it is found quicker to beat to leeward instead of sailing dead before the
wind.
In a sailing vessel, the resistance of the water increases with the speed, so that at full speed
the apparent wind WW or OW' must fill the sails, and produce a normal pressure OP', the forward
component of which, OQ' parallel to the keel, must equal the resistance of the water; and thus
we see how it is possible for a ship to sail to windward ; OW representing the true wind; OW the
velocity of the ship (allowing for leeway); OW' the apparent wind, OP' its component perpen-
dicular to the sail, considered as the effective component, and OQ' the component of OP along the
keel CO, urging the ship forward, Q'P' being the component causing leeway.
When the ship is just gathering way after tacking, the apparent wind is the true wind OW;
and its component OP perpendicular to the sail has a forward component OQ, and a transverse
component QP.
33. In a steamer the individual particles of smoke are carried down by the true wind OW,
but the aggregation of particles of smoke makes a long trail which takes the direction OW’ of
the apparent wind. --
A steamer, passing between two other steamers going the opposite way, will see the wind-
ward steamer blotted out by its own smoke, and will blot out the leeward steamer, in different
direction, except for a true beam wind.
Bradley observed in 1729 a similar phenomenon with the fixed trail of smoke from a chimney,
when rowing in a boat, and was thus led to the discovery of Aberration in Astronomy.
He also noticed the difference in direction of the vane on the mast of a vessel in motion, and
of the vane on shore or on a vessel at anchor. -
As another illustration of Aberration, consider the theory of the speed sights on a gun (or
torpedo carriage).
When the gun is fired from a fort at a moving target P, the hind sight is moved from A to a,
so that Aa is parallel to PP', the course of the target, and the ratio Aa/AB is made equal to the
ratio of the velocity of the target to the average horizontal velocity of the shot. *
Then if the gun is fired when a, B, P are in line, the target P and the shot will arrive simul-
taneously at P', in AB produced, taking AB as the axis of the gun.
If the gun itself is mounted on a vessel, the fore sight is moved from B to b in a line parallel
to the keel, through a distance Bb proportional to the speed of the vessel; and now Ab will
represent in direction and magnitude the velocity of the shot over the water; so that if the gun
is fired when a, b, R are in a line, the target and the shot will arrive simultaneously at Q' in the
line Ab produced. *
If Ab meets the axis of the line in D, the shot must be supposed to start off from D with full
average velocity, in order to strike the target exactly at the point aimed at.
But the shot starts from rest in the bore, and may be taken as having an average velocity in
the bore of one half the muzzle velocity; so that the point O from which the shot must be
supposed to start with full velocity is about at double the length of the gun, is distant from the
muzzle M ; and now if Og' is drawn parallel to Ab, the shot will strike the small distance Q'g' on
one side of the mark. -
If, for instance, the distance AB is 6 ft, and the average velocity of the shot is 1200 f/s, then to
graduate the scales Aa, Bb for speed in knots, taking the knot as 100 ft a minute, the graduation
must be 6 × 100 + 72000 = 1/120 ft, or 0.1 inch apart. -
A similar phenomenon of aberration occurs in consequence of the velocity of recoil of the
gun; thus if V is the velocity of the shot in the bore as it is leaving the muzzle, and U is the
velocity of recoil, 2 the tangent angle of elevation, and 8 the angle of departure, then
- V sin &
t = --——
f an 8 = y.ºu.
= tan & (1 +\see 2),
when U/V is small.
In this way part of the phenomenon called jump in Artillery may be explained.
40
31. In Aberration in Astronomy, the velocity of light replaces the velocity of the shot, the
telescope replaces the gun, and the earth and the heavenly body observed replace the steamer
and the target.
When an observer on the earth at E, points his telescope to observe a star S, in the true
direction ES, the telescope must be turned through the angle SES', called the aberration, to point
in the direction ES', where Ess’ is the triangle of velocities, compounded of the velocity of light
V in the direction SE, and of the reversed velocity of the Earth in the direction 8's.
- sin sEs’ v
Then sin Es’s T W
and since the fraction v/V is small, we may replace sin sEs' by the c.m. of ses', and Es's by sRQ,
O being the point of the sky towards which the Earth is moving; and then the angle 8'EO is
called the earth's way. - -
Q)
W
or multiplying by 206265 to convert c.m. into seconds, the aberration in seconds = k sin (earth's
way), -
where k or 206265 (v/W) is called the constant of (annual) aberration.
With C.G.S. units, W = 3 × 1019 kines, or 30 quads; while with a solar parallax of 8”8, and
treating the Earth's orbit relative to the Sun as a circle, of a crn, *
Thus the c.m. of aberration = sin (earth's way),
27ta = 4 × 10° x cosec 8”8
= 4 × 1010 × 200,265 –– 8:8 -
and w = 27ta/T, where T = 365 x 24 × 60 x 60.
By logarithmic calculation—
log 206265 = 5:31:44
log 8-8 = '9445 -
log cosec 8”:8 = 4:3699, cosec 8":8 = 23440.
log 4 x 10° – 9: 6021
log 27ta – 13-9.720
log 365 = 2.5623
log 24 = 1.3802
log 602 = 3'5564
log v = 6-4731, v = 29.73000.
log V = 10.4771
log v/V = 5-9960
log 206265 = 5:3144
log k = 1.3104, k = 20°46
When the orbit of the Earth is taken as a circle, the point O is 90° ahead of the Sun in the
ecliptic, and k or 20’’:46 is the aberration of the Sun; that is, the Sun's true place is 20°46 ahead
of the apparent place. As the Sun advances about 1° degree in the ecliptic every day, this shows
that its light takes 8 m. 18 s. to reach the Earth.
Aberration will make any fixed star in the ecliptic perform a S.H.M. of angular amplitude
20”:46, and of period one year; a star at the pole of the ecliptic will describe a circle of angular
radius 20”.46 in the same period' one year; so also will a star in celestial latitude A, but this
circle will be seen projected into an ellipse in the celestial sphere, of excentricity cos \, the
major and minor semi-axes being 20°46 and 20%-46 sin X; and the angular velocity will be
271- sin X.
- T 1 — cos”) sin”(nt — l)'
where nt and l denote the longitude of the Sun and of the star.
The rotation of the Earth, causes an aberration called diurnal aberration, in which the point
O is the east point, and v is replaced by the velocity due to the rotation of the Earth which in
latitude 6 is
4 × 10° x cos 6 -– 24 × 60 × 60 kines;
so that we may put the diurnal aberration in seconds = %' cos 6 sin (angle between star and east
point), while k', called the constant of diurnal aberration, is given by
k' = 206265 × 4 × 109 − 24 × 60 × 60 = 0”32.
Thus k'ſk = 365 sin (solar parallax) = 365 sin 8”8,
log 365 – 2:56:23
log sin 8”8 = 56301
log k"/k = 2.1924
log k = 1.3104
log k' = I-5028, k = 0°32; 2 . "
and thus diurnal aberration is practically insensible,
41
In the case of lunar and planetary aberration, the velocity of light must be compounded with
the velocity of the body, so that the aberration is similar to that investigated for firing at a moving
target. - e e
. Other familiar illustrations of aberration are seen in the deflection of the direction of falling
raindrops, when we are carried in a train through a shower.
ExAMPLEs.
1. State and explain the proposition called the parallelogram of velocities?
Distinguish between uniform motion and uniform acceleration ?
When is a train said to be going at full speed P ſº * A &
How can you determine, by means of a pendulum inside a railway carriage, whether the train
is going at full speed? . :* §. e
A train travels at the rate of 45 miles an hour. Rain is falling vertically, but owing to the
motion of the train, the drops appear to fall past the window at an angle tan-1 1-5 with the vertical.
Find the velocity of the rain-drops.
If the rain-drop were divided into 1000% equal spherical portions, prove that the cloud so
formed would have a velocity of -044 feet per second. e
The velocities are supposed to be full speed velocities, and the resistance of the air to vary as
the square of the product of the velocity into the diameter of the drop. - -
2. Define linear, angular, and areal velocity: and explain how they are measured.
What is meant by relative velocity ? -
A person travelling in a railway train observes that external objects appear to be moving in
a direction contrary to that of the train, and with velocities which depend upon their distances
from the observer.
Explain this phenomenon.
EXAMPLE.—Determine the angular velocity of the Earth, of the Moon and Sun round the
Earth; of the hour, minute, and second hands of a watch; of the driving wheels of an engine,
7 feet in diameter, when going 60 miles an hour; of a projectile fired with a velocity of V f/s
from a gun in which the rifling makes one turn in n calibres; of a bullet fired with velocity
1,850 f/s from the 0:303 inch rifle, the grooves making one turn in 10 inches.
3. Prove that the time in which it is possible to cross a road, of breadth c feet, in a straight
line with the least velocity, between a stream of vehicles of breadth b feet, following at intervals
a feet, with velocity W f/s, is
c /a . b
w (. + #) seconds.
4. Determine the quickest course to take from one bank to another, in two parallel currents,
separated by a straight edge; or in going from one point to another of two vessels passing along-
side; determine also the direction to take in the jump.
Discuss the quickest course of a steamer across the Channel, allowing for the setting up and
down of the tide.
5. A body is floating down the middle of a stream with velocity V; determine the direction
and Velocity with which a stone must be thrown so as to hit the body, the point of projection
being h feet further up the stream, whose breadth is Qe.
6. A sportsman moves his gun round his shoulder as a fixed point so as to cover accurately a
small bird which is moving uniformly, and fires when it is nearest him. Prove that he will miss,
unless the scattering angle of his gun be greater than Q sin-" where v, W are the velocities of
the bird and shot, the firing being supposed point—blank, and n the ratio of the distances of the
bird from the muzzle of the gun and the shoulder. --- -
7. The velocity of a swarm of meteors is n times the Earth's orbital velocity; show how to
find the true direction in which the meteors are moving, having given the right ascension and
declination of the radiant point, and those of the Sun's centre. [Neglect the eccentricity of the
Earth's orbit.] -
* Compare the curvature of the track of a steamer with the curvature of the trail of the
SIO Oke.
(6873) M
42
THE HODOGRAPH.
35. When a body P is describing a curved path PQ, the velocity is always changing, and an
acceleration measured by the rate of change of velocity is always in operation.
If a vector OW is drawn from a fixed point O to represent in magnitude and direction the
velocity of P, then as P moves along its path PQ, the point V will trace out a curve called the
Hodograph by Sir W. R. Hamilton, the inventor.
If OW changes to OW' as P moves to an adjacent point Q, then VV’ represents the change
of velocity, or the velocity to be added or compounded with OV to change the velocity to OV’.
If T seconds is occupied in going from P to Q, the average acceleration is WW’/T, and there-
fore the acceleration of P is lt WW’/T, the velocity 2V in the hodograph.
The velocity of V in the hodograph represents, therefore, in direction and magnitude, the
acceleration of P. co
Sometimes it is convenient to draw the vector OV perpendicular to the velocity of P; and
then the velocity of V will be perpendicular to the acceleration of P. -
For instance, the effect of aberration considered above is to make a star describe about its
real position a curve similar and similarly situated to the hodograph of the Earth's orbit.
We shall prove subsequently that the Earth's orbit is an ellipse about the Sun in a focus, and
that the hodograph is a circle. Aberration therefore causes the stars to describe circles about
their real position in planes parallel to the ecliptic. We see these circles projected on the
celestial sphere as ellipses, of excentricity equal to the sine of the celestial latitude.
CIRCULAR MOTION.
36. When a body P describes a circle of radius h ft, with constant velocity V f/s, the
hodograph described by the vector OV is another circle of radius V, described in the same period,
T = 2arr/V seconds, or with the same angular velocity n = 27/T ; the angular velocity being
defined by the number of radians which the radius OP turns through in one second.
The acceleration a will be given by the vector Ou of the hodograph of W, and therefore Oa
is in the direction PO, and
a = m\V,
the product of the angular and the linear velocity.
Also, since - n = V/r,
therefore - - a = V2/r,
or Oa is the third proportional to OP and OV. (Maxwell, Matter and Motion, chap. vii.)
Again a = nºr = 4trºr/Tº :
or if N, the number of revolutions per second is given, then
N = 1/T, and a = 47°N°r.
If R is the number of revolutions per minute,
R
0, E *(j), = "PR*n/900.
When the period T is small, a fraction of a second, it is usual to employ its reciprocal N, the
number of revolutions per second, and vice versä. -
In consequence of the importance of this theorem, it is advisable to give another demonstration.
If P was free to move, it would proceed in the tangent PT with velocity V, and therefore,
after t seconds would have reached a point T, where PT = Wt.
But in consequence of the constraint of being compelled to move in the circle PQ, the body
will be found at Q; so that if a is the average acceleration in celoes,
TQ - #at”.
Produce TQ to meet the circle PQ again in Q', then TQ, TQ = TP*, or lt'TP*/TQ = ltTQ',
TP2 1, Vºtº 2W2
But It is – it # = +
while lt TQ' = 2r ;
so that V* = ar, or a = W*/r.
Thus in crossing the Channel at a speed of W knots, the cliffs appear to descend or ascend with
constant acceleration V*/R knots an hour, but with constant angular velocity V/R, a sea mile
and an hour being the units, and R denoting the radius of the Earth in sea miles, 5400 + 4 +.
So also in the case of a man who has fallen overboard; thus, if the vessel is going 15 knots
the man will lose sight of the masts, 100 ft high from the water, in about 21 minutes.
Notice the analogy with Ohm's law, which asserts that the current A amperes through
resistance O ohms under an electromotive pressmre V volts is given by A = V/O ; and then the
power W in watts or volt-ampères is given by 4
W = WA = v2/O = A*O.
43
Suppose the body P to weigh W lb., and moves on a smooth horizontal table, attached to Q
by a rope of length r, then W and therefore n will be constant, while the tension of the rope will
be
Wnºr, or WW/r, or WnV, poundals,
Or Wn°r/g, or WW*/g”, or Wn W/g, pounds.
When the body makes N revolutions a second, or R a minute, then R = 60N, and 27taN = v
27 N = w ; and the tension of the rope is -
47°WN’a, or trºWRºa/900 poundals.
This tension is felt at the fixed point, O, as a pull away from the centre, and from the point
of view of O is called the centrifugal force. -
But P feels the pull of the rope as a centripetal force.
The centripetal force at P and the centrifugal force at O constitute the two opposite aspectſ
of a stress, the tension in the rope OP, now called the central tension.
Even when no rope or connexion is visible, as between the Earth and the Moon, or the Earth
and the Sun, still a central tension must exist, due to the propagation of the attraction of gravi-
tation through space.
37. In the first investigation of the motion of the Moon round the Earth, or of the Earth and
planets round the Sun, it is sufficient to consider these orbits as circles; and now if the periodic
time is T seconds, and r is the radius of the orbit in feet, v the velocity in the orbit, and n the
angular velocity,
n = 27/T, v = nr = 27tr/T;
so that the central acceleration in the circular orbit is 47°r/T".
Kepler's Third Law asserts that “the squares of the periodic times are as the cubes of the
mean distances from the Sun, or that Tº cº r"; and therefore 47°r/Tº varies inversely as r*”
and thus Newton's Law of Gravitation is verified among the planets, as a corollary on Kepler's
Third Law. r º
It was from an arithmetical calculation on the orbit of the Moon that Newton first verified
absolutely his law of gravitation.
Tradition asserts that Newton, in 1665, having observed the fall of an apple, and looking up
to the Moon, proceeded to calculate how far the Moon—or the apple if carried up so far—would
fall in the same time.
The Moon's parallax being about 57, the distance of the Moon from the centre of the Earth
is cosec 57' = 60 times the Earth's radius; so that, assuming that the acceleration of gravity varies
inversely as the square of the distance, the acceleration of the Moon should be 32-2 + (60)*
celoes, or that the Moon would fall 16:1 feet in the first minute, or that the acceleration is 32.2
when a foot and minute are units of length and time. N.
Taking the Moon's period as 27 days, T = 27 × 24 × 60 minutes, while r = 60R, R denoting
the radius of the Earth in feet, and R = 6,080 x 180 × 60 -- T, taking the geographical mile as
6,080 feet; then
log 6,080 = 3.7839
log 60 = 1.7782
log 180 = 2.2553
7.8174
log Tr = -4971
log 60 – 1.7782
log r – 9-0926
log 27 = 1°4314
log 24 = 1.3802
log 60 = 1.7782
log T = 4-5898
log 27t = •7981
log 27/T = 4,2083
log 47°/T* = 8.4166
log r = 9-0926
log 47°r/T* = 1.5092
47°r/Tº = 32.3,
a very close agreement.
44
Unfortunately Newton took the geographical mile as the same as the statute mile of
1,760 yards, or 5,280 feet; so that he found at first
4trºr/Tº = 28;
and the discrepancy was so large that he laid aside his calculations for 19 years till 1864, when the
correct length of the geographical mile was brought to his notice.
If we calculate the acceleration of the Earth relative to the Sun, we must now, with the
foot and seconds as units, put
T = 365 x 24 × 60 × 60, log T = 7-4989,
r = R cosec II = 23440R, log r = 11.6843,
with a parallax II = 8”8; so that 47°r/Tº = 0.0186 celoes; and the Earth would fall 0-009 feet, or
about 0.1 inch in the first second towards the Sun, if suddenly stopped in its orbit.
The attraction of the Earth on a body on its surface is then more than 1920 times the
attraction of the Sun on the same body, placed at the mean distance of the Earth. .."
'The acceleration of the Moon to the Earth being about 32°3 –– 3,600 or 1 + 112 = 0:009
celoes, and of the Earth to the Sun being 0-0186 celoes, we see that the acceleration of the Moon
is always towards the Sun, so that the Moon's orbit relative to the Sun is always concave to the
Sun. -
In C.G.S. units, 47°r/T* = 0-6 spouds; while the weight of the Earth E = 10” x 6.134g; so
that the average attraction between the Sun and the Earth is 10° x 3-68 dynes.
CENTRIFUGAL FORCE.
38. As the speed V is constant, it is generally a difficulty to the beginner to understand how
there can be any acceleration; but the acceleration V*/r, acts continually in changing the direction
of the velocity V without changing its magnitude.
Experimentally we feel this acceleration when we whirl a body round in a circle at the end
of a rope, or when we have to exert ourselves to sit up when running round a railway curve, or
in the thrust asserted against the outer rail by a railway carriage running round a curve; also in
some kinds of fuze set in action by the rotation of the shell.
To the person holding the fixed end of the rope, the tension pulls away from the centre, so
that the force which reacts on the constraint came to be called the centrifugal force. But the body
which describes the circle feels the tension pulling towards the centre, the Moon for instance,
feeling the centripetal force of the Earth's attraction.
The centrifugal and the centripetal force are, however, only the two aspects of the stress, in
this case a central tension, which is the mutual action of the moving body and the constraint
which makes it describe a circle (Matter and Motion, xxxvii).
- There will be no need of confusion in the use of the word Centrifugal or Centripetal Force
if we attend carefully to the fact that they are the opposite aspects of a central stress,
which may be a tension, as in the case of a whirling body at the end of a rope, or of the
attraction between the Earth and the Moon, or may be a pressure, as in the case of a train
running round a curve, and bearing against the outer rail.
The force Wa poundals, causing the actual acceleration a celoes of W lb, is called the
effective force of W; and it is convenient to adopt D'Alembert's principle of considering “the
reversed affected forces in equilibrium with the impressed forces.”
The reversed effective force of a body describing a circle is now the same as the centrifugal force,
and it is therefore in equilibrium with the impressed forces; so that the employment of the
centrifugal force is useful as reducing questions of circular motion to statical problems.
39. We can realise this circular motion practically by the
45
CONICAL PENDULUM,
in which the plummet P at the end of a thread OP of length l feet, fixed at 0, describes a hori-
zontal axle of radius r with velocity v, and angular velocity n, in T seconds, making N revolutions
3, second ; so that -
w = nr = 2+r/T = 2+Nr, n = 2+\.
Supposing the thread OP makes a constant angle a with the vertical,
then sin a = rſl; { {
and, in gravitation measure, denoting the tension of the thread by R pounds, and resolving vertically
R cos a = W . . . . . . . . . . . . . . . . (1)
while resolving horizontally -
Rsin a = "it . . . . . . . . . . . . . . . . (2)
g -
so that tan ox = * = 47-2N2/ sin a,
gr g
= –0– = *
Or cos & = ſārūj = #7.
The period of revolution in seconds of the plummet
1 ! cos & OC
T = ± = 27t = 27t — ;
N A/ 9 nº A/ g
so that, when & = 6, the period is 2T v/(l/g) seconds, the same as for the plane oscillations of a
pendulum; and conversely, if the period is T seconds, a convenient mode of determining g.
g = 4trºl/T",
Instead of being attached to a thread, the plummet may be supposed carried round in the
interior of a spherical bowl of radius lift and centre O, rotating N times a second about a vertical
diameter; and now the plummet will be carried round steadily by the bowl at a depth g/4T*N*ft
º i. ºte: but this must be less than the radius for the plummet to leave the lowest point
Of the bowl. -
If the length l is comparable with R, the radius of the Earth, for instance if the conical
pendulum swings in the shaft of a deep mine, being suspended from the surface of the Earth in a
shaft l feet deep, then if the period of revolution is T seconds, the value of g at the bottom of the
mine is *
_ 47°l (1 – )
*- 9 = T, \* - B).
For, employing the gravitation unit of force,
R cos & = W cos Á
Arwr = R sin a + W sin 8;
gT"
and when & and 8 are small, we may put R = W, and sin a = rſl, sin 6 = r)(R – l).
\ 47°r — ” r rR
Thus † = i + R-7 = ºrity
47-?! !
Or g = TT 1 — #):
If suspended from a point l feet high outside the Earth, we must write R + l for R in the
last expression, and now g at the surface of the Earth is given by
_4trºl *—
g = Tº THE
– 9- ! 1
Or T = ºv/(; Hºº)
the same as the period of a very long pendulum making plane oscillations.
Thus if l = R, we shall find that the pendulum makes about 24 revolutions or complete
oscillations, to and fro, in a day; or one revolution in an hour.
But if l is infinite, the pendulum makes 24 + V2 = 17 oscillations in a day; as for instance
a railway train on a perfectly straight railway, the Channel Tunnel for instance (§ 84). -
(6873) N
§3.
46
40. A plumb-line may be considered as a conical pendulum, whirled round by the rotation of
the Earth. -
We suppose that the Earth is a perfect sphere, and that it does not change its shape
and assume a figure of equilibrium, as the level of the ocean, in consequence of the rotation.
Denoting by G the acceleration of gravity at the pole, and in latitude l by g, then a plummet
weighing W lb., supported by a plumb-line will be in equilibrium under three forces:–
(i) WG poundals tending to O, the centre of the Earth ;
(ii) Wg poundals, the tension of the plumb-line; n
(iii) Wn”R cos l poundals, the reversed effective force, or the centrifugal force due to the
rotation of the Earth, tending away from the Earth's polar axis, R. denoting the radius of the
Earth in feet, and 27t/n the sidereal day in seconds, equal to 86,164 seconds, the mean solar day
being 86,400 seconds.
Producing the plumb line to meet the equator in Q, then PQa' will be the true latitude; PEa.
being the Geocentric latitude, and EPQ the angle of the centre.
We must take EPQ as the triangle of forces, so that
WR cosi T J T G. :
and we may take EQ, PQ, and PE to represent n°R cos l, g and G to scale. .
Drawing QFL parallel to the polar axis EN, then EF represents n°R, so that PF represents
G — n°R, the value of gravity at the equator, which we may represent by H.
Then g = PQ = y/(PM2 + PL”)
= V/(G° sin *l + H2 cosº l);
while sin EPQ = EQ sin l = m*R sin l cos l – (G — H) sin l cos l,
PQ 9 g
_ (G — H) sin l cos l
tan EPQ = G sin” l + H cos” I
These formulas are exact on the supposition that the surface of the Earth is a perfect sphere;
but it is found by geodatic measurement that the Earth has the shape it would assume, if fluid,
being slightly distorted from the spherical shape assumed under the gravitation of its parts by
the whirling effect of its rotation. ºf
There is therefore no real loss of accuracy in employing approximate formulas;
and since the angle EPQ is small, we may suppose that PR = PQ, where OR is the perpendicular
from Q on PO ; *.
so that PQ = PE – ER = PE – OQ cos l,
Or g = G – n°R cos”!
– G (1 — "Roose;
= g (1 ** cosº).
Taking G = 32, #TR = 90 × 60 × 6,080, and 2 T/m = 86,167, the number of seconds in the
siderial day, we shall find G|n^R = 289 = 17*; so that if the Earth was made to rotate 17 times
faster without change of shape, g would vanish at the equator.
The angle of the centre EPQ is the apparent gradient of the sea with respect to the vertical
EP to the centre of the Earth; and the c.m. is approximately n°R sin l cos l/G, or a gradient of one
in G|}n’B sin 21; in latitude 45° this would be one in 578, an angle of about 6'. -
To make this angle of the centre 1°, the Earth must spin about three times faster.
A falling body at P has acceleration G in the direction PE; but the direction of the plumb-
line is PQ, and its tension is Wg poundals. - -
41. The energy of the rotating body, in foot-pounds,
#WV*/g = }Rr.
But if the tension R is expressed in poundals, then
R = Wa = WV*/r,
and & #WW* = }Rr,
the kinetic energy in foot-poundals.
º The tension of the rope, or constraint required to make a body describe a horizontal circle, is
therefore to the pressure on the ground as V*/r to g, whether expressed in absolute or gravitation
units. -
Take the case of a railway engine or carriage, running round a curve of radius r ft., with
velocity V f/s. ; -
The push of the flanges against the outer rail is a force of WV*/gr tons, if the engine or
carriage weighs W tons. .. S
If the gauge of the line is a ft, and the height of the C.G. is h ft, the load on the outer rail,
obtained by taking moments round the inner rail, will be
2
aw WVºl.
gra
2
Iw WVºl.
gra
2
while the load on the inner rail will be
47
This load on the inner rail will vanish, and the engine will be on the point of capsizing
when
W = V/(gra/2h).
To obviate this, and to equalise the load on the rails, the sleepers on a curve are sometimes
inclined; and now the load on the rails will be equal when the inclination & of the sleepers is such
that the resultant of W vertical, and WW’ſgr horizontal, is perpendicular to the sleepers, or ,’
tan c. = Vºſgr.
This angle & is the average slope of the surface of water in the boiler, or in a tumbler on the
engine, or it is the average deviation from the vertical of the thread of a plumb-line, the engine
running steadily. -
If, however, the tumbler of water was placed on a smooth table, then so long as the engine is
running in a straight line with constant velocity, the tumbler will remain at rest on the table; but
on entering a curve, the tumbler will continue in its former straight course, and will therefore
describe on the table the involute of the curve.
ExAMPLEs. TWISDEN, CHAP. IV., 728-735.
(1) Find in pounds the horizontal thrust on the rails of an engine weighing 20 tons, going
round a curve of 600 yds radius at 30 miles an hour.
(2) Determine the side-long thrust on the rail of a train going north and south in a given
latitude.
(3) Determine the radius of the circle on which the sand will lie, flowing from the sand-pipe
of an engine at a given distance from the rails. - -
(4) Prove that the rim of a flywheel will burst if the circumferential velocity is vſ(gT/w) f/s,
where w denotes the density of the material in lb/ft”, and T the tenacity in lb/ft”.
(5) Prove that the indiarubber band of a bicycle will become slack when running at more
than (TgdT/W) f/s, where W denotes the weight of the band in lb, T the tension in pounds, and
d the diameter of the wheel in feet.
Find when the copper driving band of a rifled projectile will become slack.
(6) Determine the diminution in the tension of a belt running round pullies, due to centrifugal
whirling; and find the speed at which the greatest horse power is transmitted.
Power is communicated to a pulley by means of belting. T is the greatest tension commu-
nicable to the belting in pounds. Determine the maximum amount of power transferable when
the pulley makes n revolutions per minute, taking account of the centripetal acceleration at the
pulley. (Weight per foot of belting = m lb, radius of pulley = a ft, coefficient of friction = u, and
arc in contact with belt = a).
Hence show that the H.P. transferred is greatest when the pulley is making
120 , /2T
7TC! 3r.
revolutions per minute.
(7) If B is describing a circle round. A fixed, and A is released, show that A will describe a
cycloid, while their C.G. will describe a straight line, and B will describe a trochoid; and determine
the tension of the thread connecting A and B. -
(8) Compare the attraction between the Sun and the Earth with the tension of a string
º to a pound weight which is swung round horizontally at a distance of three feet twice
{l, SGCOI). Cl. -
(9) Show how to determine the resistance of the air to a sphere by means of a Whirling
Machine, consisting of a horizontal beam made to revolve about a fixed vertical axis with con-
stant angular velocity, and carrying at the other end the sphere suspended by a fine thread; by
observation of the angle which the thread makes with the beam. -
(10) “We use Hydro extractors to dry cloth; the cloth is rolled tightly round a horizontal
cylinder which is made to rotate and the water flies out of the cloth in straight lines. For a
cylinder of 18 inches diameter we find 1,200 revolutions a minute the best speed. A 36-inch
cylinder at 800 revolutions we found not quite so effective, and a 72 inch at 500 less effective than
the 36 inch. We are going to erect an 18-foot cylinder; which velocity must be given to produce
the same effect as the 18 inch.”” Comment and calculate.
(11) Describe Newton's method of determining the Laws of Rebound of elastic spheres.
Two equal ivory billiard balls are suspended in contact by equal parallel vertical threads, so
that the line joining the centres of the balls is horizontal and two feet below the points of attach-
ment of the threads.
Determine the co-efficient of restitution between the balls when it is found that allowing one
ball to start from the position when the supporting thread makes an angle of 60° with the vertical
causes the other ball after impact to come to rest at a point where its centre is 1 foot 8 inches
from its original position of equilibrium.
Solve the same question when the threads are 4 feet long, also when the balls are loaded so
as to weigh 12 lb and 8 lb.
48
HARMONIC VIBRATION.
42. Suppose P is a describing a circle with centre O and radius r feet with velocity V f/s, and
angular velocity n so that W = mr.
From P drop the perpendicular PM, PN on Oa, Oy two straight lines at right angles; then as
P describes the circle with constant velocity, the points M and N which we may look upon as the
shadows of P, will vibrate backwards and forwards on the diameters AoA' Bob', as Jupiter's
satellites when observed in the plane of their orbits, in almost the same manner as the piston of a
steam engine of which OP is the crank revolving uniformly, provided the length of the connecting
rod is considerable, and in exactly the same manner as the piston of a donkey-pump without a
connecting rod. - -
If OP coincides with OA at the time t = O, then at any subsequent time t seconds, the circular
measure of the angle AOP is nt, so that
OM = r cos nt, ON = r sin nt,
nt being called the phase of the vibration of M.
The velocity of P being n.OP and perpendicular to OP, it follows from the resolution of
velocities that the velocities of M and N are n.MP and n.NP respectively; and the acceleration of
P being n°.PO, it follows, from the resolution of accelerations, that the accelerations of M are n°. MO
and m”.NO respectively, the order of the letters indicating the direction of the acceleration.
Vibrations such as M and N describe are called Simple Harmonic Vibrations (S.H.M.) because
all small vibrations, of pendulums, of springs, and of bodies sounding musical notes, can be ulti-
mately analysed into vibrations of this nature. -
Suppose the weight of the piston M is W lb and actuated horizontally at M ; then the force
required to overcome the inertia is Wm”.MO poundals, a maximum at A and A', the ends of the
stroke, and vanishing at O, the middle. -. * -
We may suppose N to represent the piston of a vertical engine, and now at a certain point C
above O, such that OC = g/n”, gravity and the effective force of the piston are equal, and elsewhere
the combination of gravity and reversed effective force will be Wn”.CN poundals, in equilibrium
with the pressure of the steam and the stress in the piston rod. -
The crank describes a circle 27tr feet in circumference in T. seconds, while the piston moves
backwards and forwards through 4r ft. ; the velocity of the crank, or the maximum piston velocity,
is therefore #7 times the average piston velocity.
We can now determine the time occupied by the pile in a previous question, driven into the
ground against a resistance which increases proportionally as the penetration.
Representing to scale by OA, the penetration at the first blow, then A1A3, AAAs, . . . will
represent the penetrations at the 2nd, 3rd, . . . blow, obtained by drawing the arcs UAG, U.A.,
U30A, . . . . with centre Q ; and the times occupied will be represented by the angles UOA,
U10As, U20A3, . . . ; so that the time occupied after the rth blow, is to the time after the first
blow as sin-1 (1/w/r) to #7t.
49
III. MOTION UNDER AN ATTRACTION TO A FIXED POINT O,
PROPORTIONAL TO THE DISTANCE FROM O.
2
43. If OM = a, then the acceleration of M isº, in the direction Ow.
Then if M is attracted to O, with intensity proportional to OM (as if controlled by a spring
which keeps M in equilibrium at O), the equation of motion is
d?a,
# = - ºr,
when p denotes a constant,
It is convenient to replace p by n” (and then n is called the speed of the S.H.M.); so that
d?a,
d; - - m*a, o o e e e º º e º © º º & º ſº ſº (1)
representing oscillations about a position of stable equilibrium given by a = 0.
It is easily verified by differentiation that, A, B, a, b, representing arbitrary constants,
this differential equation (1) is satisfied by a = A cos nt, or B sin mt, or A cos nt + B sin nt, or
A cos (nt + t); all representing periodic vibrations, performed in the period 27t/n; and if we take
the last expression.
a = a cos (nt + e),
then a is called the amplitude of the vibration, and e is called the epoch.
Writing it -
w = a cos n(t – T),
then T is called the phase; and when t = T, the point M is at A, where OA = a.
To integrate (1) directly, we multiply both sides by º then
da d’a — mºa. da: .
dt dº T dº ’
and integrating with respect to t,
2
#. = C — $nºw”.
#
Denoting the amplitude by a, so that # = 0, when a = a, then
C = }n”a”;
2
and # = *(*-*),
Ol' # = — my/(a” — w”),
the negative sign being taken because the motion begins at A towards O.
Therefore **=––––.
da: (a” — a”)
and integrating again
Q,
—*— = cos- 4,
(a” — wº) 0, .
n(t — t) = |
•S
Or w = a cos n(t —T).
(6873) r O
| da, •7,
When t = T, aſ = a, i = 0;
t = 1 + \rh, a = 0, *. E – 720, .
* ºf 5
da,
t = = — a. º. = 0 :
T + T/m, a *Tºt y
t = 1 + ºrn, a = 0, . = 7ta ;
da:
t = 1 + 2*h, * = a, , = 0;
and M is again at A ; so that the period is 2T/m, as already stated; and since the period 27t/n is
independent of the amplitude a, the oscillations are isochronous, that is, are always performed in
the same time, e.g., a mixed note of a pianoforte wire.
The point M on Oa, follows the point P on the circle AP, moving with constant velocity na, so
that M is the foot of the indicate of P; and the angle A0P = n(t — T); since
a = a cos n(t — T).
The point N also performs a S.H.M. in the same period 27t/n, given by
3) = a sin n(t — T);
and the composition of the two S.H.M.'s of M and N gives the motion of P on the circle
w” + y^ = a”.
We may suppose P to represent the crank of a steam engine, and M a point on the piston rod
produced, provided the connecting rod is sufficiently long for its obliquity to be Leglected, or, as
in a steam pump, where the connecting rod is suppressed.
51.
THE SIMPLE PENDULUM.
44. The small oscillations of a simple pendulum are approximately S.H.M. , , ,
For if l denotes the length of the thread in feet, and W the weight of the plummet in lb,
then the tangential acceleration is d; so that
dt
2
y º = — W sin 6
or - # = – g in 6 = -gsin;
and when s is so small that sin º can be replaced by its cm. º
d’s – 98
iſ, T ~ 7”
representing S.H.M., in which nº = g/l; so that the period of the pendulum is 27tv/(l/g/), and the
time of a single swing is TV/(l/g/) seconds.
If, as the plummet passes through the lowest point A, it is struck by a horizontal impulse
across the plane of vibration, this plane will assume a new position, depending on the magnitude
and direction of the impulse; but the period of oscillation will be unaltered if the vibrations
remain small.
For instance, the impulse may be so applied perpendicular to the plane of vibration, so as to
shift the plane through 45°; and now if this impulse is applied in the same direction at the end of
a swing, it will make the plummet move in a circle, as in a Conical Pendulum. -
Let a body weighing W lb be suspended from an elastic rod or spring CD, attached at D
to a fixed point.
Then if allowed to descend slowly, W will come to rest at O, where P, the pull of the spring
in pounds is equal to W lb., the weight of the body. w
Considering CD as an elastic rod, of cross section Kin”, then Hooke's law asserts that if L is the
unstretched length CD, and if the elongation CO is denoted by l,
P !
* = E 1.
where E is a constant, called the modulus of elasticity of the substance of the rod.
Here P/K is the tension in pounds per in.” in the rod, and l/L is the eatension produced; so
that Hooke's law asserts that t -
tension
extension
(Twisden, chap. i., § 6).
Iu the position of equilibrium of the weight P = W, or KEl/L = W.
But now suppose P is pulled down further to B', and let go; the weight will oscillate in a
S.H.M. about O.
For if we denote OM by w, where M is the position of the weight below 0, after t seconds,
then the equation of motion is
= E, the modulus of elasticity (in lb/in”)
W d24,
# = W-P,
where P, the pull of the rod, is now given by
-* ! + æ ( 4 ).
dža,
so that dº? T l
where m” = g/l ; so that the weight will perform a vertical S.H.M., synchronizing with the
oscillation of a simple pendulum of length l = CO, the permanent average set of the rod or spring
OC, due to the weight W lb.
If OB' = OC = OC, the weight W will rise to C, and if originally let fall from C, it will pass
through O and come to rest again at C at double the depth of O, in time T/n = Tv/(l/g), and will
continue to oscillate between C and C’.
But whatever the original distance OB from O, the weight will continue to oscillate to equal
distances, OB and OB' above and below () in a S.H.M. of period 27tv/(l/g) seconds.
The rod CD may be inverted and made into a column, and the weight W placed upon it;
if suddenly inverted, the shortening of the rod or spring will be 31.
The weight W may be supposed to be the body of a carriage on springs; and in all these
cases the vertical oscillation of W will constitute a S.H.M., synchronizing with the small oscillation
of a pendulum of length l, the permanent average vertical set of the springs. -
52
Next consider the small vertical oscillation of a floating body, a ship, for instance.
Denote the displacement of the ship by W ft” or W tons, so that
W = wV, or W = nW,
where w denotes the weight in tons of a ft” of water, or n denotes the number of ft” of water to
the ton (n = 35, for sea water).
Now let the ship receive a small vertical displacement of a feet.
The upward buoyancy of the water is now w(W + Aa) tons, where A denotes the (average)
water line area of the ship in ft”, so that the equation of motion is
}* = — w(V -- Aw) = — waſ ;
d’a _ gwa — — 9A e
Ol' d; T *w- * = ºv : ;
so that l, the length of the simple equivalent pendulum, is given by l = V/A.
If the ship is floating in a dock of area B ft”, then the small vertical displacement a ft. of the
ship will make the water rise Aw/(B–A) ft on the dock wall, and therefore
Aa: Ba:
a; + B – A Or B – A feet
on the ship's side; so that the extra buoyancy will be
ABæ
'M) B – A tons;
and therefore the equation of motion is
W. d’a _ 2D ABæ
g dø T ~ * B = At
Or dº — gABæ .
dº? W(B — A)”
B – A 1 1
t – — = — - — ; 2
so tha ! = W AB (, #).
reducing as before to W/A, when B is infinite.
53
FORCED WIBRATIONS.
45. Sometimes a body is forced up and down in a vertical oscillation, which may be con-
sidered as a S.H.M., for instance the piston of an engine (or the screw propeller by the pitching of
the steamer), and it is required to determine the maximum stress called into play at the attachment.
Suppose a body N, weighing W tons, is forced up and down from B to B' and back again in
a period T seconds in a S.H.M.
The force which drives N up and down may be supposed due to a vertical spring, reaching
to C, and of such a strength that the weight W tons will deflect the spring through l = CO,
where O is the middle point of the stroke, and where the length l is given by the equation
T = 27tv/(l/g), or l = gT*/47°.
N is a piston, moving in a vertical cylinder as in a marine engine, the stress in the piston T
rod at N will be proportional to CM, and equal to W.CN/CO tons, being a thrust when the
piston is below C, changing to a pull above C.; so that above C the downward acceleration of N
will be greater than g, and water will not lie on the top of the piston.
At the bottom of the stroke the thrust is a maximum, and equal to
w(1 + #) where 2a denotes the length of stroke, - w( 1 + º) == w(l + ;) tons.
* ga
where v = 27ta/T denotes the velocity of the crank-pin.
At the top of the stroke the pull in the piston rod is a maximum, and equal to
w(;- 1)= W. – 1).
In a horizontal engine the maximum thrust and pull are each equal to
Wa/l = Wv°/ga tons.
If L = 2a in the height of stroke in feet, and R the number of revolutions per minute,
Waſl = tº R*LW/1800g ;
also the maximum piston speed is ºr times the average speed.
(6873) P
54
:
i
.
DECAY OF WIBRATIONS IDUE TO FRICTION.
46. We observe that vibrations die away unless periodically renewed; as, for instance, in a
clock pendulum. . . .
The decay of the motion is due to the frictional resistances, and it is usual in small oscillations
to suppose these resistances to vary as the velocity, so that the equation of motion now
becomes
d?a: da:
= — m”a — k ‘
d? d;’
d?a,
da: 2 on —
Or # 4. 2n cos 3; + n’a = 0,
in writing, for convenience, 2n cos Á for k.
We can verify by differentiation that the solution of this differential equation is
a = ae-" ºf cos (nt sin 8 + e),
representing a S.H.M. of period 27/n sin 6, of which the amplitude dies away at compound
discount at the rate n cos Á, called the modulus of decay.
The motion of M is now represented graphically by supposing it to follow P, as before, but P
now describes the equiangular spiral -
* = ae - 99988 when 6 = mä.
When the vibrations of this system are maintained by a periodic applied force, represented by
E cos pt, the differential equation of motion is
d°4, b da, 2 o' —
d;2 + 2n cos 3; + ºr = E cos pt,
the particular solution of which is obtained symbolically as
- E cos pt
& E 3.
D* + 2n cos Á. D + n”
where D represents the operator d/dt, and this can be written
D* + m” — 2n cos & . D
& E (D* + n”)” – 4n” cos” (8. D2 E cos pt
_ (n° – p”) E. cos pt + 2np cos Á. E sin pt
(n° – pº)” + 4n°p” cos” & g
since D* cospt = — pº cos pt;
2mp cos Á
77° — p” 3.
v/(n” + 2nºp” cos 28 + p")
E cos (pt – tan-"
Ol' (pt – tan
& E
representing a S.H.M. of period 2 Tlp, of amplitude E (nº 4. 2n"pº cos 23 + p.)- and of change
2np cos - -
**, 3, this is called the forced vibration, due to the impressed vibration
of phase tan-" *—p” ”
E cos pt. - -
To this must be added the free vibrations ae -nt coss cos (nt sin 3 + e), to complete the solution;
but the free vibrations die out rapidly.
Taking the ease of free vibrations with frictional decay given by
w = ae-”** sin (nt sin 3),
&nd supposing the amplitude of oscillation maintained constant by successive impulses as the body
passes through O, its position of equilibrium, then since
# = nae-"e"sin (3 – nº sing).
dt
therefore the velocity V with which the body must start from O is given by
W = n.a sin &.
55
The velocity with which the body returns to O is obtained by putting nt sin & = T, and is
therefore We-T coº, so that an impulse, changing the velocity by V (1 — e-" º is required each
time the body passes through O, to keep up a constant amplitude of vibration, as in a clock.
To maintain a forced vibration of amplitude a and period 27t/p by an applied periodic force
E cos pt,
E = avy (nº + 2 n°p” cos 2 & 4-pº).
If the period of the forced vibration is equal to the free period of the system,
p = n sin 3. -
and E = a n” cos Áv/(1 + 3 sin “8).
But if p = n,
E = 2 a m” cos Á.
*
Suppose for instance that the point D in fig. 15 is made to vibrate vertically according to the
law,
= b cos pt,
and that the inertia of the spring CD is neglected, so that the pull
l + a - y & — º
P = KE * T * * = W | 1 - * u, e.
L (1 +*zº)
then the equation of oscillation of the weight W, supposing a frictional resistance proportional to
the velocity, is now -
y
W /d? 0. 4) #) sº- W -ºms. P * = r a – y
g ( dź” + 2n cos Á oft / T = — W –7–,
or with * = g/l,
dt”
so that E = n°b, and
+ 2n cos Á # + n’a = n°y = n°b cos pt,
12mp cos e)
- 2b t tº- -
J} = 37. (ºr tan-º-º- p”
v/(n+ + 2nºp” cos 28 + p")
The amplitude of oscillation is therefore small when the ration/p is small, that is when the
free period is long compared with the period of the forced vibration ; this is the theory of Yarrow's
Vibrometer, an instrument for measuring the vertical vibrations of a torpedo boat ; the point D is
attached to the vessel and compelled to vibrate with it, while the body at N is practically at rest
in the air; so that a pencil attached to N will record on a revolving drum the vertical oscillations
Of D.
3. Two weights W. and W, are connected by a spring of such a strength that when W, is
held fixed W., performs N complete vibrations per second. Shew that if W, be held, W, will make
NA/W 2/W, and if both be free they will make NV(W. H. W.)/W1, vibrations per second, the
vibrations in each case being in the line of the spring. *
4. A spherical shell contains a sphere of equal weight, supported by springs of equal length
and strength, attached at opposite ends of a horizontal diameter. Prove that, if it is fired at a
vertical elastic plane, the shell will strike the plane again after a time equal to
half the period of free oscillation.
5. An endless elastic band, of natural unstretched length 27ta ft and of weight W lb., is
stretched round a wheel of radius b, which is made to revolve with angular velocity n ; prove
that, if the pull required to stretch the band to double its natural length is E pounds, the pressure
of the band on the wheel, in pounds per unit length, is
1 1 Why?
E (. - }) --> 2gtº
Supposing n is now increased so that this pressure vanishes, and that the band leaves the
wheel, but remains always circular and concentric in position, prove that it will oscillate about a
mean radius c, given by
1 1 W7,2
Tc - a T2gtE,
º - 27t C \
in a period ºn (; —l)
Discuss the case of
E 3 Wall”/2gt.
47 . The solution of
#. = + wº
is given by * = A cosh nt + B sinh nt, or a cosh (nt + e) or a sinh (nt + e),
representing unstable motion of M, as if repelled from O with intensity proportional to its distance
from O.
* A particle P will describe this motion if placed on an upright cycloid; and approxi-
mately, if placed near the highest point on a vertical circle, the equation of motion now being
d?6/dt = nº sin 6 = n°6,
when 6 is small. -
We see the same-kind of motion when a window blind runs down.
With a frictional resistance varying as the velocity we may write the equation of motion,
2
º: = m^a — 2n sinh & º
dža, da
OT º, - nºw – 2n cot 8.5
the solution of which is
* = aente-" + be-nt."
OI’ w = aent tan #8 + be-n' cot 38.
48. With high velocities through a resisting medium it is usual to assume that the resistance
varies as the square of the velocity, as in the case of the motion of trains, steamers, or projectiles. .
Take the case of a train weighing W tons, starting from rest under a constant pull of the
engine of P tons; R tons denoting the constant part of the resistance of the road, due to inclines,
and the resistance of road and axles; and Xvº tons denoting the variable part of the resistance, at
velocity v f/s, *
*
Supposing the train to have gone a feet from rest in t seconds, the equation of motion is
W 024, W do or W dº
cº-º ºmºmºs--º * * = P – tºº 2
7 dB, * g ºf * , # = P R – Av’.
The train will start with acceleration f, given by
_ . P – R.
f = g —w >
and the full speed being denoted by V, then
XV* = P – R
so that we may write the equation of motion
dv 7,2\
# = f (1 T wº
Integrating this equation
dv V ty
t = | —“— = -- tanh T'
|za ºwn =}***
so that * = ft
a. V tanh W .
Again, since v = da/dt, integrating again,
tºº , ft
= Y or cosh ſº.
- ? log cosh &.
57
If a spherical body is let fall in a medium in which the resistance varies as the square of the
velocity, as in Newton's experiments carried out on falling bodies in St. Paul's, then in these
equations f must be replaced by g, the acceleration of gravity, corrected for buoyancy, and V will
be the terminal velocity in the medium, that is, the velocity with which the body will fall vertically
when the upward resistance of the medium balanced the downward pull of gravity.
Newton's experiments have been recently repeated in the Eiffel Tower (Comptes Rendus, July,
1892). -
h terminal velocity is observable in sparks from fireworks, raindrops, hailstones, meteorites,
and in a balloon or parachute.
At any other velocity v, the resistance of the air to the body, weighing W lb, will be a force
of W (v/V)* pounds, causing acceleration g (v/W)”.
A steamer may be supposed to be always moving on a smooth level road, so that we may
put R = 0, and take P tons to represent the constant thrust of the engine, so that 2,240 PV/550
is the effective h-p of the engines at full speed V f/s.
When the body is projected vertically upwards, or when the engines of the steamer are
reversed, the equation of motion becomes
dv v/2
— - 1. -
dt' ( + V.)
so that | w -: tan',
Y2 y
42 W Og secº,
= , , W
the time t' being now measured backwards from the instant at which the velocity is zero.
Suppose, for instance, the engines are reversed when the steamer is going full speed; then
r' = W, so that
V V
t' = } T = 0-7854 -.,
• " ? f
& E (log v/2) y - 0-6931 y *
Take the case of a steamer in which the horse-power and the tonnage are each equal 10,000,
going full speed at 18 knots; then W = 18 × 100+ 60 = 30 fs, taking the knot as a speed of
HO0 ft. per minute, while
2240 PV/550 = W
f = ′ = ** = }}
J T W T 2240W T 42
Then t = 0-7854 x 30 × 72 –– 11 = 90 seconds;
a = 0-6931 x 750 x 72 —– 11 = 1,200 feet.
(6873) Q
58
EXAMPLES.
(1) Prove that if an engine can pull a train of W tons at a velocity V on the level, against
resistances varying as the square of the velocity, the engine exerting a constant pull of P tons,
then up an incline a the velocity will fall to WA/(1 – W sin aſ P), and that down the incline with-
out steam, the velocity will reach Wv/(W sin &/P).
Prove that if the actual velocity v varies from the average velocity V on a long railway
journey periodically according to the law v = V + U sin nt, or W -- U sin ns, the average
horse-power and consumption of fuel is increased by # U*/V* or #U*/V* of that required for
uniform velocity W.
2) Work out the same questions when the resistance varies as the first, third, or fourth
powers of the velocity.
59
MOTION UNDER AN ATTRACTION TO A FIXED POINT, INVERSELY PROPORTIONAL
TO THE SQUARE OF THE DISTANCE FROM THE POINT.
49. When a body is carried to a great distance from the Earth, say to the distance of the
Moon, the attraction of gravity is, by Newton's law of Universal Gravitation, found to vary
inversely as the square of the distance from E, the centre of the Earth. -
The parallax of the Moon being 57, the distance of the Moon is cosec 57' = 60 times
the Earth's radius; so that the acceleration of gravity is reduced from 32 to 32/60°, and the
Moon, or a body at the same distance, will fall 16 ft in the first minute towards the Earth.
Suppose the body is carried up to a distance 2a from (E the centre of the Earth), and let a
denote its distance after t secs., and let R ft. denote the radius of the Earth, g the acceleration
of gravity at the Earth's surface.
Then at a distance EM = a from E, we must take the acceleration as gB*/a", so that our
equation of motion is now -
d’a R2
*\. IF = - 9 a.
Multiplying both sides by #.
da d’a _gBº da ,
dº dº? T 22 dº ’
and integrating, with respect to t,
da,” gR” '1 1
l — - *- = 2 - * -----
# E = ** + constant = gr: (: #)
da,
since = 0, when a = 2a.
dt
To integrate this equation further it is convenient to use an eccentric angle 6 indicated
on the figure, such that a = EM = 2a cos”;6, AM = 2a – a = 2a sin”;6 = a vers 0; -
and then
day? * d62 |R2 R2
§ da - 202 º cos”;6 if " *. (secº;6 – 1) = ‘. tan *# 6,
- d62 R2
Cr d;2 = %. SéC *#6.
We put n*a* = gF", in accordance with Kepler's Third Law (§ 61); and now
da: -
º: = ma tan #6 = #n AT;
dt
wº dt
while n; = 2 cosº = 1 + cos 0;
and integrating nt = 0 + sin 6,
giving the time t (or mean anomaly nt) in terms of the eacentric angle 6; so that the time t is
proportional to the area AEP, and P therefore moves on the circle AP as if attracted to E. -
Expressed as a function of w,
?vt = sin- ve:-- + ve:-2. *
60
Similarly, if the body was repelled from O, we should have
d’a: R2 day? </1 1. 1. l
# = g . . . . =9R'ſ. – ) = *(; – )
t? 2a as 2a w
and nt = sinh u + w,
on putting a = a(cosh u + 1);
so that 7tt = ve-sº + ºn-vº. N
The time of falling from A to Eis obtained by putting 6 = T, and is therefore T/n = TV (a"/gB");
this is ºv/2 or 0.177 of the period of A in a circular orbit round E. tº
Thus the Moon, if suddenly stopped in her orbit, would fall into the Earth in less than
27 x ºv/2 = 4.8 days; while the Earth, if suddenly stopped, would fall into the Sun in between
3 * e e le
64-5 (1 + 3°) days, e denoting the excentricity of the Earth's present orbit, about 1/60, so that
the greatest difference of time would be about 3% days.
For if V denotes the velocity of A in its circular orbit of radius 2a,
V*_ gR. or U = gº
za 4 a” 2a
and the period of A is
4~a 4 2aº
-U = 47 A gR”
= 4\/2 times the time of falling from rest at A to E.
At E the velocity is theoretically infinite; but an infinite velocity is not possible in Nature
without supposing some other quantity is also infinite; in this case the Earth would have to be
condensed into a particle of infinite density at E, for the law of the inverse square to hold close up
to E.
We may suppose the body to reach E by falling through a diametral shaft cut through the
Earth; and now, on the assumption that the Earth is homogeneous, the attraction of gravity
changes to being proportional to the distance from E.
This depends on two theorems of Attraction.
I. The attraction of a homogeneous spherical shell on a point in its interior cavity in zero.
II. The attraction of the shell or of a solid homogeneous sphere on an external point is the same
as if its weight was condensed into a particle at its centre, and is therefore CW/r”, at distance r.
Denoting by p the mean density of the Earth, and a the distance of a point in the diametral
tunnel from the centre, the acceleration of gravity at this point is—
47pCº/* = {T/Cr = gº/R,
g denoting the acceleration at the surface when a = R ; so that
g = 4TpCR = #pC × 10°, +
gF2 = 4"rp.CR* = CE,
E denoting the weight of the Earth (in grammes) and C the gravitation constant (§ 51).
The equation of motion in the diametral tunnel is therefore—
d’a _ 9°
# = -R = – pºa,
giving a S.H.M. in which the speed p = y/(g/R).
Draw the ordinate RQQ' meeting the circle APE in Q, such that V , the velocity at R, is equal to
p. RQ'.
Then if 8 denotes the cin. of the angle Q’ER, the time from R to E in the diametral tunnel is
(#7 – 8)|p = (#7 – 3) vſ(R/g).
º
61.
But if V, the velocity at R, is acquired in falling freely in outside space from A, a distance 2a
from E,
1 1
W* = * ! :- — it
2gR (i. #)
1 1
= 2n?R3 | * – it
2p*R (i. })
_ 2p*R*(2a – R)
2a
R
= p?...". *;
p Q. RQ
so that we must make
RQ” R
RQ. T a .
The whole time of reaching the centre is now t + tº seconds, where
at = sin-1 v/(2aR – R*) + veh- Rº, pt' = }T – 8
Q,
To shoot a body weighing W tons away from the Earth vertically upwards to a distance 2a
ft ſrom the centre, say the distance of the Moon, requires energy,
f R
l 2 cº- sº - sº-as tºº *
*WW*/g = WR (1 ...) foot-tons;
and with a parallax of 57, 2aſ R = cosec 57' = 60, while R, the radius of the Earth,
= 90 x 60 x 6,080 –- #7 ft.
Thus if, as in Jules Werne's story, “Le Voyage dans la Lune,” W = 10,
then #WW*/g = 10 x 90 x 60 x 6,080 x ## -- #7t
= 205,000,000 foot-tons.
If the strength of the powder is taken as 50 ft-tons per lb of powder, this would require
a charge of 4,100,000 lbs., or about 1,800 tons of powder.
The time occupied in reaching the Moon would be nearly ºv/2 or 0.177 of the period of the
Moon, or 27 x 0.177 = 4.78 days. t -
50. To shoot the body away to infinity requires a velocity v/(2gB) f/s; but to shoot it up to
a height R at the North Pole requires a velocity v/(gR) f/s; and then if fired horizontally, the
body would, in the absence of resistance, skim the Earth like a grazing satellite in a period
2T/n = 27 v/(R/g) seconds,
the saune period as the oscillations in a diametral tunnel, on the supposition that the Earth is
homogeneous. - - - -
The period of the Moon being 27 days, the period of a satellite at one-ninth the distance of
the Moon would, by Kepler's Third Law (§61), be one day, and the satellite would appear stationary
in the sky; and this would make the period of a grazing satellite 27 + (60); or about one-17th of
a day, so that a body in a diametral tunnel, or in any straight tunnel, would make 17 complete
oscillations in the day.
Then, if V denotes the velocity of the grazing satallite of the Earth,
W = 17 × 4 × 10' + 24 × 60 × 60 m/s,
= 17 × 4 × 5400 –– 24 = 15300 knots;
&
while a velocity 15300V/2 = 21634 knots would send the body to infinity; this is a velocity of
about 6 sea miles or 7 land miles per second.
In the plane of the equator such a satellite would fly past a point 16 or 18 times a day, or
every 90 or 80 minutes, according as it moved eastward or westward, and the velocity over the
ground would be 744 or 84 kilometres per second.
Gravity at the equator is thus diminished by 1/17° or one 289th part in consequence of the
centrifugal whirling of the Earth's rotation; and if the Earth were to rotate 17 times faster without
change of shape, bodies at the equator would be on the point of flying off into space, and else-
where the plumb line would point to the Pole Star, and the surface of water would be paral’el to
the equator.
(6873) R
62
So also the period of the Earth round the Sun being 365 days, and the Sun's angular radius
being 16', the period of a satellite grazing round the Sun would be 365 (sin 16')3 days, or between
one-8th and one-9th of a day, say three hours.
The Sun's angular radius as seen from the Earth being 16 or 960", while the Earth's radius as
seen from the Sun is 8”8, the Sun's parallax, it follows that the velocity of a grazing satellite of
the Sun with a period of 3 hours is
* * * * * = m/s,
8.8 × 3 × 602 T 5
and V2 times this velocity, or a velocity of m/s will send a body to an infinite distance from the ‘’
surface of the Sun; conversely this will be the velocity with which particles entangled by the
Sun's attraction will strike the surface.
The energy liberated by the impact or combustion in the Sun's atmosphere of these particles
is joules, or calorics per kilogramme, or 82,340,00 British Heat Units (B.T.U.)
per pound; and it has been calculated that the heat radiated by the Sun could be developed by
the annual impact or bombardment of a quantity of matter, one-quarter of the weight of the Earth,
falling on the Sun from an infinite distance.
63
THE GRAVITATION CONSTANT.
51. Theorizing on the observed phenomena of the motion of the Moon, Sun, and Planets,
compared with the motion of bodies, on the surface of the Earth, Newton was led to enunciate his
general law of gravitation – s - º
“The attraction between any two particles in Nature is proportional to the product of their
weights, and inversely proportional to the square of their distance.”
Thence he proved that the attraction between two spherical bodies, for instance the Sun and
the Earth, weighing S and Eg, when their centres are w cm apart, will be given by the expression
CSEw-” dynes; where C is a constant, called the gravitation constant, being the attraction between
two small spheres, each weighing one g, when their centres are one cm apart. * * *
We shall employ the C.G.S. units, and now for a body on the surface of the Earth, weighing
Wg, the attraction of the Earth is a force Wg dynes, where g = 981 spouds, the acceleration of
gravity. - • . . . . *
Then if R denotes the radius of the Earth in cm, 10° -- $7, V its volume in cm°, E its weight
ing, and p its mean density,
CWE/R* = Wg. *
Or CE = gR*,
. Gl’ #TpCR = 9 ; §pC X 109 = 9, pC - # X 10-9:
so that C must be known to determine E the weight of the Earth, or p its mean density, and
19006 Q)07"SO,
It was conjectured by Newton (Principia, III, prop. x) that p is between 5 and 6, or that the
mean density of the Earth is about half the density of lead, an extraordinary close guess.
But C has been measured directly by the Cavendish experiment, recently repeated by
Poynting and C. W. Boys, in which the attraction between given weights is directly measured,
and it is found that a mean value of C is 10–9 × 6-597.
Now, with R = 10° -- #7t, g = 981,
log R – 8:8038
log R* = 17.6076
logg = 2.99.17
loggR* or log CE = 20.5993
log E = 27-7799, E = 1097 x 6.134g.
Also log R* = 26.4114
log #7t F •622 |
log W = 27.0335, W = 1027 × 1.031 cmº,
and - log p, or log E/V ‘7464, p = 5:577,
-:
Or immediately, since pC = #g x 10−",
log pC = 7×5657
log C = 88194,
SO that log p F. ‘7463, p -: 5'576.
The attraction of gravitation between two spheres, each weighing W grammes, will be one
degree when their centres are one centimetre apart, if -
Wºe = 1, or W = 1/M/c ;
and log 1/c = 7.1806
log 1/v/c = 3.5903
1/v/c = 3893.
But the relative acceleration of the speres will be one spoud, if
W = 1/c = 107 ×
this quantity of matter is sometimes called the astronomical unit of mass.
On the M.K.S. system, R = 107 -- $7, E = 1024 x 6'134, g = 9.81 : so that C = 10-11 x 6'597
- log C = TI-8194
log 1/C = 10-1806 -
log v/(1/C) = 5.0903, V(1/C) = 10° x 1231,
so that two spheres, each weighing 123 kg, will attract with 10-12 units of force when their centres
are 2 km apart; this unit of force it is proposed to call the gauss, being 10° dynes,
64
On the F.P.S. system of units, the equation—
F = C, WW’ſ.”
gives F the attraction in poundals between two spheres, weighing W and Wºlb, when the centres
are a ft apart; and C is given by the equation—
g = 47twQR,
where g = 32.2 f/s”, TR = 180 × 60 x 6,080 ft;
and w = 5.577 × 1,000 + 16, the mean density, in lb/ft";
log w = 2.5423
log #TR = 7.9423
log 47twR = 10:4846
logg = 1.5079
log C = 9-0233, C = 10–9 × 1-055.
Ҽ
The attraction of gravitation between two spheres, each weighing one lb, when their centres
are one foot apart, is thus about 10-9 poundals.
When the weights W and W’ are abandoned in free space to their mutual attraction, their
C.G. remains at rest; and therefore the equation of motion of W is
W d”
diº
2
W'a, ) CWW’
W -- W’
2?
dº C(W + W’).
Or &
dt? a;2
and we should have obtained the same equation of relative motion if we had reduced W’ to rest
by applying to W and W’ the reversed acceleration of W’, namely CW/w”.
- If the bodies start from a distance 2a cm from centre to centre, then after tseconds, we find as
before
a' = a(1 + cos 6),
where mt = 6 + sin 6,
and *a* = C(W + W’).
Thus if the sum of the radii of the spheres is 25 cm, the time for the two spheres to come into
contact is
2a3 b b b°)
- *-* - *-* - seconds.
v/{c(W + Wi)} {sº I A/. + V(. #)}~ $
When the diameters of the bodies are small compared with 2a, we may put a = 0 or 6 = ºr
when the bodies are in contact; and then mt =T.
Thus for instance, if the bodies each weigh one metric tonne, and their centres are initially
one kilometre apart, then with C.G.S. units,
2a = 10°, C = 10-8 x 6-597, W = W’ = 10%;
so that n° x , x 10* = 10-8 x 6-597 × 2 × 10",
#):p. An° = 10 – 17 × 6-597 x 16
log n° = 15,0235
log n = 8:51.18
log mt = 4971
log t = 7-9853
log 24 x 60 × 60 = 4.9366
Iog tſ(24 x 60 × 60) = 3.0487;
on the time is 1,120 days, for the bodies to come into contact.
65
The period of a grazing satellite round a spherical planet, of density p and radius R, being
2Tſp, when p” = g/R = #TpC,
depends only on p and is independent of the radius R; Maxwell proposed as a cosmopolitan unit
of time the period of a grazing satallite round a planet whose mean density is that of water, so
that p = 1, as is approximately the case with Jupiter.
With C.G.S. units ...--
C = 10-8 x 6'597, 1/v/c = 3893,
and this unit of time is
–%– F. seconds.
v/(#7tc)
The period of a grazing satallite round a planet of mean density p will be v/p of these units.
Thus for the Earth, p = 5:576, and the period of the Moon would be
(60) (5.576)} = units.
EXAMPLES.
1. Prove that two spheres, each 50 cm in diameter and weighing one tonne, will take about
2,525 seconds to come into contact, if their centres are initially one metre apart. *
2. Calculate the number of seconds which two spheres, each weighing 750 lbs, and one foot
in diameter, will take to come into contact, the centres being initially two feet apart.
3. A sphere weighing 450 grammes is suspended from one arm of a balance, and counter-
poised. A weight of density 12, in the form of a circular cylinder, with its axis vertical, is
brought exactly under the sphere, and it is found that the apparent weight of the latter is
increased by 0:36 gramme. The radius of the cylinder is 25 cm, and its height 50 cm, while the
distance between the top of the cylinder and the centre of the sphere is 50 cm; find the mean
density of the earth, neglecting the action between the cylinder and the counterpoise.
4. Having given that the Earth's mean density is 53, that the density of iron is 7-5, and that the
earth's quadrant is 10 million metres, find the thrust in dynes due to the mutual attraction between
two spheres of iron one meter in diameter, the spheres being in contact and stupported on a smooth
horizontal table.
5. Two small equal spheres are in contact, one with the upper and the other with the lower
surface of a large horizontal slab of lead; compare the difference in the apparent weights of the
two spheres, due to the attraction of the lead, with that due to difference in altitude. (Take the
mean density of the earth to be half that of lead.)
6. Two spheres of lead, each a meter in diameter, are placed a kilometre apart; prove that if
abandoned to their mutual attraction they will come together in about 463 days. Assume the
mean density of the Earth to be half that of lead, and the value of g on its surface to be 980 cm/s”,
or spouds.
7. The diameter of one of the minor planets is 100 miles; assuming that it is spherical and has
the same mean density as the earth, prove that a particle projected upwards from its surface with
a velocity exceeding 460 feet per second will never return.
*
Jº;
EQUATIONS OF MOTION OF A BODY REFERRED TO CO-ORDINATE AXES.
52. The motion of a point P (a projectile for instance), is given by ar, y, its co-ordinates, at
the time t seconds, such that (Fig. 19), OM = a, ON = y; we suppose On horizontal, and Oy vertical,
and OP the vertical elevation of the trajectory.
tº a we W d tº a tº
Then the velocities of M and N are . and %; and these are the component velocities of P
2 2 -
The acceleration of M and N are # and % the component accelerations of P
If the curve OP is a twisted curve (as happens in Gunnery in consequence of drift) we take
a third co-ordinate axis Oz, perpendicular to the plane woy, and denote the distance of P
from the plane w0y by : ; then the component velocity and acceleration of P parallel to Oz
&l'é
* and dº
dt (lt.”
(6873) S
66
UNRESISTED MOTION OF A PROJECTILE.
53. Take for instance the motion of a projectile, under gravity, when the resistance of the,
air is left out of account ; the equations of motion are now (Bashforth, Motion of Projectiles
Chap. II)
d’a, dº/
= 0, ± = - g.
Integrating these equations with respect to t, supposing the body projected from the origin
O with velocity W at an elevation 2,
da:
º: = a constant = V cos a,
dt
# = a constant — gt = V sin o. — gt;
and integrating again
a = Wit cos &
3y = Wit sin & — gt”.
Therefore t = aſ V cos & ; and
ga,”
z E & tan Cº - —“ —
3/ 2W2 cos^2°
the equation of a parabola as is seen by writing it in the form.
a.” — ** sin a cosa imº - 2Vºy cos” &
W2 . * 2W2 COS2 & 7 V2 Sin? & )
OI’ Jº — SII). C. COS 6. E - y F,
( g ) 9 ( 2g 3/
so that the co-ordinate of the vertex are (#Vºsin 22/g, #V* sin ºaſg);
and the latus rectum is 2W cos” aſg.
The time of flight T is obtained by putting y = 0; and then
T = 2W sin &/g;
and thus the range
R = 2W" sin & cos &/g = W* sin 22/g.
Therefore, for given V, Ris a maximum when sin 22 = 1, or a = }ºr, or 45°.
The velocity V required for a given range R with given elevation 2 is given by
W = V/(gR coses 2 x);
thus, for example, the velocity required with an elevation of 1° to strike an object at a range of
1,000 yards is f/s.
We may write the equation of the parabola
so that tan 2 = 4 + = 4
&
= tan 6 + tan b,
f 6, 4 denote the angular elevation of the top of the ordinate y at the beginning and end of the
range; this enables us to calculate the requisite elevation of a mortar so as to graze the top of s
parapet y feet at a distance a feet, so as to strike a magazine R — a feet bºi
The velocity v at any point is given by
da,” 2 g
#4 ## = (Vcoe y + (v in 2-gly
los?
#v
#V* – g(Vt sin & – #gtº)
I
#V* — gy.
-67
If we denote T — t by t, then since V sin & = gT,
y = }gTt — $gt” = }gtt', Col. Sladen's formula.
At the vertex A, t = t = }T, and the height of the vertex
k = \gT* = 4Tº = (2T)*,
taking g = 32, hence the practical rule—“The square of twice the time of flight in seconds is the
height of the vertex of the trajectory in ſeet.”
Firing up a slope, making an angle 8 with the horizon, and taking Oa in the slope,
a = Wt cos & – #gt” sin 3,
sy = Wt sin & – #gt” cos Á,
derivable from the equations of motion
and a now denotes the tangent elevation of the gun; the quadrant elevation being a + 8.
Then with the same notation as before
T = 2W sin & sec 8/g,
and denoting the impetus #V*/g by h,
R = 4h sin a cos (2 + 8) sec 3
= 2h{sin (22 + 3) — sin &} sec 6,
a maximum for given impetush and slope 8, when sin (2 x + 3) = 1, 2a + 8 =#T, a = }T – 8
and as before e &
= }gtt' cos Á,
so that the distance from the plane measured vertically is # gtt'.
ExAMPLE.-Prove that if E is the elevation required for a range R on the level, then firing
up or down a slope of angle A with tangent elevation given by
3. sin 6 = cos A sin E,
(or tan 6 = cos. A tan E, approximately so that the back sight is depressed to cos A times its
original height for firing on the level).
the time of flight is unaltered, but the range is changed to
cos (A + 6)
cos Acos E” (or by R sin A tan E approximately).
R
68
THE PARABOLIC TRAJECTORY.
54. A purely geometrical proof of the principal properties of the trajectory, when there is no
resistance, may be given here.
Suppose the body is projected from O in the direction OT with velocity V f/s; then in
the absence of resistance and gravity, the body will in t seconds reach a point T, such that
OT = Wit feet. * , - -
But in the same time t seconds Galileo asserts that gravity will have caused the body to a
point Q vertically below T, at a depth $gt” feet, the same depth OW the body would have fallen
in t seconds if let drop from 0. -
According to ancient writers on Gunnery, OT was called the motus violentus, and TP the
motus naturalis, while the trajectory OP formed by their combination was called the motus naturalis;
and Galileo was the first to show that the motus naturalis may be supposed resolved into the
motus violentus and motus maturalis, considered as simultaneous, or successive in time.
o
Having now OT = Wi, TP = }gt”,
the elimination of t lends to the invariable relation for all points on the trajectory OP,
OT = W** = *W* = 40H
- TP - ºf T
if ÖH is measured vertically upwards equal to #V*/g feet; this is the vertical distance the body
would have to fall to acquire the velocity of W, and it is called the impetus of the velocity W.
We have now—
OT2 = 40H. TP, or PW2 = 4HO . OV,
and therefore the curve OP is a parabola, from a fundamental property of the curve.
If OT is horizontal, we obtain the usual relation
y” = 4ha,
connecting OH = h, OT = y, and OV = w ; from which the curve derives its name parabola, from
having been used by the Greeks for the graphic comparison (Tropafloxi) of squares and square
roots.
Thus, the wheels of a train going 30 m/h, on coming to a drop of 0.1-in in the rails, will go
1 foot before touching the next rail (Barlow).
Again, a rifle bullet which, if fired horizontally at a height of 9 ft above the surface of a
lake, strikes the water at a range of 400 yards, will, in the absence of resistance, have an initial
velocity of 1600 f/s, and a time of flight of three-quarters of a second. *.
We proceed, however, to translate this relation into one proving that the curve OP is such
that the distance SP of any point P from a certain fixed point S, called the focus, is equal
to the distance PK of P from a horizontal straight line HK, called the directria ; this being the
usual definition given of a parabola.
The focus S is determined by drawing HYS perpendicular to OT to meet the circle described
with centre O and radius OH in S; and then SO = OH, SY = YFI.
Draw PN horizontal to meet OH in N, and let SH meet PW, drawn parallel to OT, in Z.
Then, since PW2 = 4HO OV,
and by similar triangles,
therefore PN* = 4HY.YZ -
= (HY + YZ)” – (HY — YZ)?
– HZ3 — SZ”
= HP2 – SP3,
()]." SP% = HP2 – PN2 = HN* = PK2,
and SP = PK,
the fundamental property of the parabola. -
To describe the parabola mechanically, take a straight edge OR, a set square PM, and a
thread of length SO, its ends attached to S and M. -
Then if the thread is kept stretched by a pencil at P, a parabola will be traced out as the set
square is moved along the straight edge OR.
For SP + PM = SO = OH = MK, and therefore SP = PK.
If a circle is drawn with centre S and radius SO, and SP produced meets the circle in Q
then º SP + PM = SO = SP + PQ,
Ol' PM = PQ,
so that the point P on the parabola OP moves so as to be equidistant from the circle OQR, and the
horizontal straight line OMR. -
69
Another demonstration may be given of this proposition, by bisecting OT in F, and producing
HF to meet KPM in L. ;:
Then since - OT2 = 40H. TP,
therefore TF? = }OTº = TL. TP,
so that the circle described round the triangle FPL touches TF at F. -, *-y
Also SF = FH = FK = FL,
so that FPS = FLS = FSL = FPK,
and therefore SP – PR.
We notice also that SL is parallel to OT, and that the circle round FPL passes through S.
Si FT Vt V
In Ce TP T iſ, T gº -
and that the velocity v at P is compounded of the original velocity of projection V, and ºf
the vertical velocity gt (which has been poured into the body at a uniform rate g, to use Clifford's
expression), we may take FTP as the triangle of velocities; FP is thus the tangent at P, and
**
Now OSF = OHF = FLK = FSP,
and OFS = FPS;
therefore the triangles OFS, FSP are similar, and therefore, the tangents OF, FP subtend equal
angles at the focus S, and intersect in a point F midway between OH and PK, two well known
proporties of the parabola.
Also vº FP” – SP PK,
and since #V*/g = OH, therefore #v°/g = PK, so that PK is the impetus of the velocity v at P.
The triangles HSK, TFP being similar,
SK FP v.
SH T FT T V
the directrix HK of the parabola may therefore be taken as the hodograph of the motion of P (§ 35)
SK being perpendicular and proportional to the velocity at P; and the velocity of K being
horizontal and constant, shows that the acceleration of P is vertical and constant.
Produce OS to meet the parabola OP in D, and draw DE perpendicular to the directrix.
Then - OD = OS + SD = OH + DE,
so that D lies on the parabola HD, having focus at O and vertex at H, and the parabola OPD lies
inside this parabola HD, touching it at D, since their common tangent bisects the angle SDE.
The parabola HD is therefore the envelope of all the parabolas, such as OP, and D is the
extreme range attainable on the straight line OD, with the given impetus OH from O, the requisite
direction of projection OF bisecting the angle DOH, and the focus S of the corresponding
parabola lying in OD. -
If OT meet, the direction HE in Xand DE produced in e, then X is the middle point of HE,
and OH = Ee; the direction of the enveloping parabola HD thus passes throngh e, so that
OD = De
and OD is therefore the vertical distance a body would fall from rest in the time of flight from
O to D.
Also XD is the tangent to the parabolas at D, and perpendicular to OX; so that the tangents
at the ends of a maximum range on a plane through the point of projection are at right angles.
If the parabola HD revolves round the vertical straight line OH, the inside of the paraboloid
will be the space covered from O, with the given impetus OH.
Suppose, for instance, that OD is the trace of an inclined plane through O ; this plane will cut
the paraboloid in an ellipse with focus at O, and this ellipse will be the area covered by a gun at .
O on the inclined plane. -
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70
The section of the paraboloid made by a vertical plane whose trace is PK, will be a parabola,
which will bound the area covered on a vertical wall PK by a fire engine at O, supposing O
is the greatest height the engine can send the jet vertically. Determine the locus of the points
on the wall at which the jet must be directed so as to attain this limiting area.
To determine the two directions of projection from O with given velocity, an impetus OH,
to strike a given point P, describe a circle with centre P and radius PK, cutting the former circle
with centre O and radius OH, in S and S'; then the required directions are perpendicular to
HS and HS', or bisect the angles HOS and HOS', and are therefore equally inclined to the bisector
of the angle POH, the direction of projection for maximum range on the plane OP. * *
The lower parabola with the smaller angle of projection is that required for direct fire; and
the higher parabola for high angle or mortar fire. ++ -
If the circles do not cut, the point P is out of range (outside the paraboloid HD), and if the
circles touch, the point of contact lies on OP, so that the two parabolas coalesce for the maximum
range on OP; and OP is a focus chord, and the tangents at O and P are at right angles.
But if PO is given and we have to determine the smallest velocity which will send a body
from 0 to P, we bisect the angles POH, OPK by OT, PT, meeting in T, and draw HTK
horizontal; then OH is the required minimum impulse; or otherwise we describe a circle on OP
as diameter, and then draw the horizontal tangent HTK at the highest point.
Suppose, for instance, that MPP'M' is the end elevation of a building, and it is required to
throw a body (e.g., the caber) over it with the least exertion, standing on the ground OM, we
describe a circle on PP' as diameter, and draw the horizontal tangent at the highest point, and
this will be the direction of the parabolic trajectory.
If PP', the top of the building, is horizontal, then PP' will be the latus rectum of the
parabola. . . . - -
Having determined the directrix HK of the parabola, the point O is determined by describing
a circle with centre P, and radius equal to the height of HK above the ground, to cut at the
ground in O.
Similar geometrical considerations will show that a ship will come under the fire of the
guns of an elevated fort at a range 2h -- a, but is helpless to return the fire until the range is
reduced to 2h – a (the Helpless Zone); a denoting the altitude of the fort above sea-level, and h
the impetus, or 2 h the maximum range of the guns on the level; also that a range V(4ah—3a*)
is most suitable for the ship to batter the walls of the fort.
Another construction is given by Thomas Gray in his Theory of Gunnery, 1730, for the direc-
tions of projection requisite with given impetus OH to strike a given point P from O ; the circle
OHC is drawn through H and O, touching OP at O, and then the vertical line ptt' is drawn at a
distance from O one quarter of the distance of PK, cutting the circle OHC in t, t'; then Ot, Ot’
will be the required directions of projection. *
For Ot” – OH. {p,
so that OT* = 16.OH. to
or TP – 4tp,
4. TP tp
so that PO T pO’
and therefore Ot'T is a straight line.
If the vertical tangent to the circle cuts OP in q, then P will be out of range if OP is greater
than 404.
The same construction will hold for ranges on the horizontal plane OR, or on a descending
plane OD.
The time of flight from 0 to P, when Ot is the direction of projection will be twice the time
of falling freely down tp, or V (8tp/g) seconds.
Then if t, tº are the times of flight from O to P, when Ot, Ot’ are the directions of projection
OP’ = 16Op” = 16pt.pt' = 16. Agtº. §gt”,
Ol' OP = #gtt'.
Or we may strike the circle, centre C, touching OP at O, and draw the vertical line FF, at
half the distance from O of PM ; and then OF, OF' will be the requisite direction of projection
with the given impetus OH, to hit the point P. 5
Or again, if OH is produced vertically upwards to h so that Oh is four times the impetus OH
and the circle is drawn touching OP at O and passing through h, the direction of projection OT
will cut this circle T in a point vertically above P, thus giving the range on OP.
Produce OP, OT to meet the vertical line through any point R in p, t
7
then TP - OT:
th. O#2 *
while TP - OT 9
tp Ot
SO that fp - OT - OM
tR T Of T OR’
or Mp is parallel to Ot.
71.
This gives a method of construction of the parabola by points; for if M, T, p divide OR, Qt,
and tR in the same ratio, that is, if Mp is parallel to Ot Op and TP drawn parallel to tº will
intersect in a point P on the parabola touching Ot at O, and passing through R. *- e
Also, if the parallelogram MpHp' is completed, then pp' is parallel to TV; so that, if the
parallelogram OtBK is completed, and any straight line pp' drawn parallel to its diagonal tº, then
Op and p"T drawn parallel to OK will intersect in P on the parabola OPR, and if t + tº = T', the
time of flight from O to D,
t t” T'
Om mT) T OD’
so that Om, mID
so 8. Pm = }gT1? OL)2 - :
which shows that the time of flight from O to D varies as the square root of Prm, if m is a fixed
point on the line OD.
A similar method will apply for the construction of the hyperbola or ellipse; but now we
must draw KO in the direction of the diameter through O'; and then if O’ is the other end of the
diameter, Op and O'p will intersect in P on the hyperbola or ellipse OPR.
This follows because -
PV RK
OW = Win'
PW p'K tp. RK
so that, by multiplication,
PV2 RK2
OW OFW = OK. O'K.
the property of a point P on the hyperbola or ellipse OPR.
Also the velocities of T and p are in the constant ratio Ottº ; and since the velocity of T is
constant in a parabolic trajectory, it follows that the velocity of p is constant; and this explains
why, when we watch the body P from the point of projection (), the body, referred by the eye to
a constant distance OR, appears to descend with constant velocity.
Conversely the body P will appear at R to rise with constant velocity, as is noticeable in
catching a cricket ball. -
Denoting by tº the number of seconds the body takes to move from P to R, then
MP Rh Tt º'
PT - a - OT - z
and since PT = #gt”, therefore MP = #gtt', Col. Sladen's formula; and pP = #gTt, where T = t +t',
the time of flight from Ob' to R.
More generally Prm = #gtt" if tº is the time from P to D.
Assuming, according to Robins and Hutton, that the energy imparted to a projectile in a gun
is proportional to the charge of powder, and measuring the strength of the powder by E, the
available energy in foot-tons per lb of powder, thus if P lb of powder gives muzzle velocity
V f/s to a projectile weighing W lb,
#WV*/g = 2,240EP (foot-pounds),
or the impetus
h = }Vºlg = 2,240EP/W.
Supposing then that R = V/g- 2h is the range in ft obtained from a proof mortar elevated
at 45°, then E, the strength of the powder, is given by—
E = RW/4,480P.
With an elevation of 15°, R = h ; and
t = RW/2,240P.
For example, if it was found that 60; lbs. of powder could send a shot weighing 385 lbs in the
absence of resistance to a range of 12 miles, then E = 90 (foot-tons per lb of powder).
But the resistance of the air is found to modify considerably the above theory in practical
applications of gunnery; although with heavy projectiles and low velocities, as in howitzer fire,
the parabolic theory is still very useful to the gunner as a first approximation. -
As numerical calculations, calculate range tables for the 8 in. howitzer, fired with a projectile
of 185 lbs., with charges of 10, 7, 6, 5, 4, 3, 2 lbs. of powder, giving the velocity, elevation, and time
of flight for 200, 500, and 1,000 yards. Use Bashforth's X, Y, Tºtables for y = 0.
A jet of Water, or a stream of bullets from a Maxim gun, will form an apparently continuous
parabola in the air, like an inverted catenary; and would stand as an arch in the air if suddenly
solidified. -
{A}
72
The horizontal component of the velocity being constant, equidistant vertical planes will cut
aff equal quantities of matter, as the horizontal distribution of weight is uniform; and forming the
jet into a chain, and inverting it, it will serve as the chain of a suspension bridge, in which all the
weight is supposed concentrated in a uniform roadway. *
The C.G. of the jet will therefore coincide with the C.G. of the parabolic area. OPR, and will
therefore be at a height two-thirds the height of the vertex of the jet.
This shows that the average height of a projectile in a parabolic trajectory is two-thirds of
the height of the vertex; this result is practically useful in allowing for the tenuity of the air in a
$ong trajectory, as showing that a good approximation is obtained to the average density of the
air traversed by the projectile at a height in the atmosphere of two-thirds of the estimated height
of the vertex. Thus in a range of 12 miles, with an estimated height of vertex of 3 miles, assume
as a first approximation the mean density of the air as the density 2 miles high; this is about O of .
the density at the ground.
A jet of water or stream of bullets may be directed vertically upwards, and the C.G. of the
jet will be at two-thirds of the height; conversely in a stream of corn running through an opening
to a floor at a depth h below the C.G. of the falling corn will be at a depth #h below the opening.
According to ancient theories of Gunnery (Regiomontanus, Muller, and Sandbach, 1471;
Tartaglia, Nova Scientia, 1550; Quesiti et Inventioni diversi, 1554) the trajectory was composed of
(i) the motus violentus, where the velocity was so great that the shot flew in a straight line, the
extent of the motus violentus being the point blank range, an error which prevails to this day; (ii) the
motus miatus, the curvilinear part of the trajectory; (iii) the motus naturalis, in which the body fell
vertically downwards, as it tends to in reality in the neighbourhood of the vertical asymptote,
when the resistance of the air is taken into account, so that after all the ancient theory was a fair
imitation of the true trajectory.
- Galileo combined the motus violentus PT, and the motus maturalis TQ into the motus miatus PQ
throughout the trajectory, and thus obtained a parabola; but when this theory was accepted by
artillerists it was necessary to suppose that the velocity of the shot was very much less than its
real value, a discrepaney that could not be detected till Robins invented his Ballistic Pendulum,
1743. ſ
Robins was also able to detect with this instrument the resistance of the air, which he found
enormously greater than was suspected; his results naturally met with considerable criti-
cisms from the teachers of the ancient theory, such as Prof. Müller, in his Treatise of Artillery,
Supplement, 1768, p. 108. -
When the resistance of the air is taken into account, all the simplicity of the preceding theory
disappears, so that in the following collection of examples we neglect this resistance, and employ
the geometrical constructions explained above.
73
EXAMPLES ON PARABOLIC MOTION.
(1) If a man can throw a cricket ball 50 yards vertically, how far can he throw it; and if he
can throw 90 yards, how high can he send it. Determine the velocity and time of flight.
Determine the angle of elevation for maximum range with bowling in cricket.
(2) Prove that, if R is the maximum range of a shot, when the weight of the charge is equal to
that of the shot, then a charge of P lb. of powder will send a shot weighing W lb with elevation
P tº
o, to a range ºr R. Sin 22.
W
(3) Determine the charge of powder required to send a 32 lb shot to a range of 2,500 yards
with 15° elevation, supposing the muzzle velocity is 1,600 f/s, when the powder is half the weight
of the shot; and determine the time of flight.
What is the strength of the powder, in ft-toms per lb, implied in this (Hutton's) rule?
What area could be covered on the Earth? and what area on the Moon, whose g is about one-
sixth of our terrestrial g”; or on the Sun, whose g is about thirty times our g?
(4) Determine the charge of powder required to send a 68 lb. shot with 15° elevation to a
range of 3,000 yards, given that the velocity communicated by a charge of 10 lbs of powder is
1,600 f/s. - - -
6, Determine the velocity, elevation, and time of flight for a projectile to attain a range of 12
miles, clearing a mountain 3 miles high, half way.
If the projectile weighs 380 lbs, determine the charge of powder, of the same quality as in the
last example. Determine also the least charge of this powder which will send a projectile weighing
400 lb to a range of 20 miles; and determine the time of flight.
(6) Determine how a perfectly elastic ball can be thrown from the foot of one telegraph post
so as to graze the tops of the next two, rebounding from the ground between.
A perfectly elastic ball is thrown with given velocity so as to rebound from a horizontal pave-
ment and a vertical wall, and to be caught again by the hand at the same position. Determine the
direction of projection.
(7) Prove that a shot fired with velocity V and elevation a, ricochetting from a smooth hori-
zontal plane, with a coefficient of restitution e, will describe a series of portions of equal parabolas,
. .. 2W si ſº tº
T-I in time Tg. **, and describe the subsequent motion.
* 2W2 sin 22
In a range — ---
g
(8) A gun and carriage, of weight W lb., free to recoil on a plane, fires a shot weighing v lb.
at elevation 2. The energy E given out by the powder charge being constant, determine the
range, and the elevation for maximum range. -
(9) A gun, suspended at elevation 2 by equal parallel ropes, will send a shot of one nth its
weight to a range—
4n (1 + n) h tan 2,
if h is the height ascended in the recoil.
(10) Determine the percentage of change in the strength of the charge and in the gain or
loss of muzzle velocity when the point of mean impact at a proof range of 500 yards is 4 ft
higher or lower than ordinary.
(11) Prove that a piece of mud thrown from the top of a hansom cab wheel, of diameter d ft,
the cab moving with velocity v f/s, will, when it strikes the ground, be at a distance #vv/d in front
of the point of contact of the wheel with the ground.
Prove that the velocity v must exceed 4yd for the piece of mud to clear the wheel.
(12) Determine the envelope of the sparks thrown off by a Catharine wheel; or the maximum
range of the fragments of a bursting flywheel or shell.
(13) Prove that the boundary of the splashes of mud on the side of an omnibus moving with
velocity V f/s coming from a wheel of radius a ft is the parabola -
a’ — a” = 4h(y -- h),
- *
referred to horizontal and vertical co-ordinate axes Oa and Oy drawn through the centre of the
wheel, h denoting the impetus #V*/g of the velocity V. w -
Determine the height to which the mud splashes from the hind wheel will reach on the back
of a bicyclist.
(14) A and B stand on a rough inelastic plane, and A wishes to send a cricket ball towards B;
for what value of the coefficient of fiction will it be immaterial whether he throws the ball
towards B as far as possible or bowls it along the plane towards A, the initial velocity being the
same in either case.
(6873) U
74
(15) The Stairease Problem. A man running along the level with uniform velocity v f/s, comes
to a smooth marble staircase, which he mounts by a periodic series of regular bounds. Prove that
if the pressure of his foot during its contact with the ground and steps is constant, his C.G. will
describe an undulating curve composed of arcs of parabolas; and that if each step is b ft broad
and h high, and the man's weight is W lb, the pressure of his foot must be— -
b | /b 202
W } |(} tºº †) pounds,
(16) Deflection due to inequality of ground. Prove that if 2 is the elevation required for a
range R, then if the wheels of the gun carriage are at different levels, so that the axle makes an
angle 8 with the horizon, the shot will strike R tan & sin & to one side, and R tan & vers 8 below
the mark aimed at.
(17) Prove that in the pneumatic dynamite gun of the Vesuvius, fired at constant elevation 2,
the pressure of air required for a range of WR yards is 18WR cosee 22/C lb in”, C denoting the
volume of the bore in in", and W the weight of the projectile in lbs.
Determine the change of pressure required at the same range, when overhauling the object at
given speed.
(18) The Hiawatha question. Determine how fast he can run, from the following data–
“Strong of arm was Hiawatha ;
He could shoot ten arrows upwards,
And the tenth had left the bowstring,
Ere the first to earth had fallen.
Swift of foot was Hiawatha ;
He could shoot an arrow from him,
And run forward with such swiftness
That the arrow fell behind him.”—Longfellow ;
and also supposing that he could shoot an arrow once a second,
TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION.
55. Supposing the body P to move to the adjacent position p in the time At, where the
are Pp = As, so that w = It (As/At) = ds/dt; and supposing the tangent TP has turned through
an angle Asſº, then the radius of curvature p = — lt (As/Ala) = — ds/day, the minus sign being
introduced because iſ is diminishing along the trajectory.
The acceleration of P in the direction of the tangent TP, is the rate of growth of the velocity
in the same direction, and is therefore
du . . d’s dv d;v”
—H-, Ol' –H.
dt dt
as in rectilinear motion.
But the acceleration in the direction of the normal QP, being the rate of change of velocity
in that direction
it ". sin Ay
t At
is (where v' denotes the velocity at p)
=lt aſsin AkAile Avº = Q) dy
Anle At dt
But since p = — ds/day, the normal acceleration along PQ
a dº dº vº
ds dt T p
dale
° di.
_ _ | dolº ds /d\,\*
()]? *= (#) dº, T (#).
Take, for instance, the motion of a projectile P in a resisting medium, in which the resistance
R acts in the direction PT.9pposite to the motion, and is of magnitude to produce acceleration,
in the body; so that R = Wr poundals, or Wr/g pounds, if W lb is the weight of the projectile.
Then, resolving normally, the resistance R disappears, and
dº .
* † = g cos / g ſº te g © e º tº • 0 e e (1)
75
Again, resolving horizontally, g disappears, and
* = -r cos y e e s tº 0 & 0 & © tº dº º e (2)
dt
w denoting da/dt, the horizontal component of the velocity.
Equation (1) may be variously written,
u% = – g cos’ \',
or, writing p for tan ºk,
# = - 9.
so that # = -; . . . . . . . . . . . . (6)
# = - . . . . . . . . . . . . . ()
# = -º, . . . . . . . . . . (5)
while from (1) and (2),
du r
vdA), -ºs g’ e vº *> ve º º º e we o © e * (6)
equations theoretically sufficient to determine the trajectory, when r is a given function of the
velocity v. •
As an exercise on these equations, prove that at a point of minimum velocity
* +g sin J = O
and, at a point of maximum curvature,
* + 3 g sin * = O.
19. A tennis ball is served from a height of 8 feet; it just touches the net at a point where the
net is 3 ft 6 ins high and hits the service line, 21 feet from the net; the horizontal distance of the
server from the foot of the net is 39 feet; prove that the horizontal velocity of the ball is about
169 º per second, and find the angle which the direction of projection makes with the hori-
ZOntal. t
20. A lawn-tennis ground AGBCLD is rectangular in shape, and the net GHKL is stretched.
across the court at a uniform height of 3 ft 6 in, while the service line is nearer the base line than
the net. If the server stands at the corner A, and holding the ball at the height of the net’ hits
it so that after rising 1 ft 2 ins it just clears the net, show that it will be a fault unless the ball
pass over HK, where GH = HK = KL,
21. It is required to throw a ball from a given point below a plane, inclined at an angle & to
the horizon, so as to strike the plane. Prove that the impetus must exceed C cos a, where c is the
distance of the point from the plane.
22. It is required to project a ball from a given point with given velocity V so as to strike a
vertical wall above a horizontal line on the wall. It is found that when projected on the vertical
plane at right angles to the wall the elevation must lie between 6, and 6. Prove that the points
on the wall towards which the ball may be directly projected lie within a circle of radius
V* sin (6, -6.)
g sin (6, + 6,)
23. A shot, fired with impetus h at elevation 2, strikes the top of a target; and when a
vertical plate of thickness a, and resistance n times the weight of the shot is placed in front of
the muzzle, the shot, after perforation' strikes the bottom of the target. Prove that the height of
the target is -
700,
4 na tan” c. (cos x — 7,
Q
76
24. A man throws a stone from the edge of a cliff with velocity U at an elevation a, and r
seconds later he throws another stone with velocity V so as to hit the first stone. Prove that he
must employ an elevation 3, given by
, UV sin (a - 8)
9 TV cos & EU cos a
#
and determine the coordinates of the point of meeting, and the time of flight. Discuss the limits
within which the problem is possible. * ~~
25. From a fort a buoy was observed at a depression i below the horizon. A gun was fired
at it at an elevation a, but the shot was observed to strike the water at a point whose depression
was '. Show that in order to strike the buoy the gun must be placed at an elevation 6, where
cos 6 sin (6 + i) cos” i sin i*
cos a sin (a + i) cos” tº sin i
26. A gun is mounted at a given spot so as to command the horizontal plane on which it
stands. Its mounting is such that the direction in which it is pointed, must lie in a given plane
inclined to the horizontal at angle a. Prove that the points on the horizontal plane, which its
projectiles with given muzzle velocity can hit, lie on an ellipse of eccentricity sin a.
27. Prove that if the sole effect of a wind on a projectile is to produce an acceleration g/n
in a horizontal direction, the area which can be covered from a given point on the ground with
given impetus of projection h in an ellipse of excentricity (n° + 1)- and area
_4 Th” v( + #) º
772
28. If jets of a fountain proceed from a very large number of small holes equally distributed
over a small hemispherical surface, whose axis is vertical and vertex upwards, with equal velocities
along the radii of the hemisphere, prove that the form of the fountain is a paraboloid of revolu-
tion. Prove also that the average pressure per unit area produced by the falling drops on the
horizontal plane through the hemisphere varies inversely as the height to which the velocity of
effius is due. . Prove also that the pressure at any point of the circular area covered by the falling
drops varies inversely as the length of the chord through the point, at right angles to the diameter
through the point.
29. A sphere falls vertically and strikes a smooth wedge that can move on a smooth table,
prove that the latus rectum of the parabola subsequently described by the ball is
2V1/(1 + e) M sin a cos a Nº
g M + m sin” a
where e denotes the coefficient of restitution, m the weight of the sphere, M of the wedge, and a
the angle of the wedge. tº
30. Prove that if the time of flight from O on a horizontal range through O is constant, the
focus of the parabolic trajectory lies on a fixed parabola.
Prove also that if a body is fired from O so as to strike a fixed horizontal plane above or
below. O at a constant angle, the focus of the parabolic trajectory lies on a fixed hyperbola.
Prove that the focus of the parabolic trajectory described by a body projected from a fixed
point O so as to strike another fixed point P lies on a fixed hyperbola.
Resolving horizontally and vertically,
d’a , da;
d;2 T ds’
*! — — , dy
i = - r + - g;
and eliminating r,
da, dºy dy dº , da:
dº dº? dº dº? d;"
an equation equivalent to (1).
Writing p for dy/da,
t du , da;
thus - ay jaw
ll] p = #/.
% = (##–%;)(#)
dt T \d; dº? dº dº? ...) %
dp da
that — — — — “ U ,
§6) tºka, dt dº g
77
Take for instance the case of a resistance proportional to the velocity; with a terminal
velocity w, so that
r = g(v/v);
d’a g , dº – 9 dº
then d;2 T ... " I = w dt
* = – 2% – g
d;2 T w dº º
To integrate these equations, write them in the form
dº ſº – 9.
d; I dº. T Tºw’
d”y //dy – – 9.
#(; ; )= w
and then, with given elevation 2 and velocity of projection V, the first integral of these equations is
log # = log V cos 2 – gt
ap 2
log (#4 •) = log (W sin & + w) — #:
º = V cos 2e-9/*,
% = (W sin a + wye-gº” – w.
dt
Another integration gives
* = N* * * (1 – e-ºe),
9
*@ 2
$y = Ww sin a + 'w (1 — e-vº) – wi;
9
and the elimination of t gives the equation of the trajectory
Q0.2%
w” &
== vº. 4 log ( – viºr. )
3/ * tan • 4 vº. 4 ; Og 1 Vw cos &
Therefore if R denotes the range on Ox, T the time of flight, w the angle of descent, v the
striking velocity, H the height of the vertex of the trajectory, and t the time of flight to the
vertex, t
gT/w
R _ Vw cos &
T T W sin z`-E wº
sº t /
= 1/+ V sin a , ”
9
V sin a + v sin w = gT,
V cos 2
V sin & + w
log (V cos &/V cos w) = gT/w,
H = were =* log (1 + W sin •) etc.
9
When w is infinite, the resistance is zero, and by expanding the logarithm we obtain the
equation of the parabola as a limiting form.
V cos & — v cos w =
gT,
w
Since (V sin a + wº. – V cosa (; + w)=0.
the hodograph is a straight line parallel to the tangent at the point of infinite velocity; the
tangent at the point of minimum velocity is therefore at right angles to this line.
Determine the envelope of these trajectories, and prove that the angles of elevation and descent
are complementary in a maximum range on an inclined plane through the point of projection,
(6873) - X
78
56. In the Principia, Lib. II, Prop. X, Newton has investigated the resistance of the air. . .
which is called into play when the trajectory is such that TP varies as OT", instead of as OT”,
the case considered by Galileo. . . -
We may take O T and the vertical O V as co-ordinate axes, putting O T = a, T P = y, and
thus we may write the equation of the trajectory
n—l
SO that p = dy = n”. - "(...) y
& Q? C/
Or OT = n.FT,
if the tangent at P meets OT in F. 1.4
Denoting by R the resistance of the air in pounds, and by r the retardation which would be
produced by it in a projectile weighing W lb, so that ; =#.
then the equations of motion are, as with rectangular axes,
* d’a da:
— E - ?” -F 3
cº ds
dºy — — ... dy
# = - ri, + g,
- da, d” dy d”a, da:
SO at di. T â, d; T * dº
Or * dp da =
dt dt
Denoting the component velocity da/dt by u, then
dp g
t w"
da, w” ”
so that w” = g|% =—*— º
a n(n − 1) \a.
dy? . 2n-2 n–2
There * = pºu? = (:) 9° (...)
# = pºu nºt. n(n − 1) \a.
= —“– aeſt) = * m.
**H*(·) in Li 99;
so that if v denotes the velocity ds/dt,
FP3 dy? ng FP2
* = * * *** = 'º' *-*
- dt” da;
_ _ dºw” FP
da, FT
m – 2 FP
2n − 2 * TF 7
and then R = r = * ~ * 2 FP.
W g T 2n − 2 TP
Also Z – º – 2 1.
q2 7, FP’
to that if we assume Newton's law of the resistance of the air, varvi
-- ~! ~~: º º g º , varying as the square of the
velocity and the density of the medium, this density must vary inversely as FP, Q.
79
:
:
When n is negative, the trajectory is hyperbolic in shape, and CV is taken as the vertical
asymptote; and writing — n for n,
ng FP"
7, I. 1 TP
2
while R – * n + 2 FP
With n = 1, the trajectory is an ordinary hyperbola, and
R – 3 FP
W – 4 TP -
The equation of the trajectory, when the resistance varies as the velocity, referred to oblique
co-ordinate axes, one CW the vertical asymptote, and the other CT parallel to the tangent at the
point of infinite velocity, may now be written in the form, with CT = a, CW = y, * *
% = log ga, e
w? w w/(V* + 2 V w cos & -- w”)
This investigation by Newton of the resistance required for the trajectory to be a hyperbola
appears undertaken in consequence of conventional motions extant on the subject, as the following
quotation from an Elizabethan writer will show :— -
Thomas Digges in his treatise on the Newe Science of Great Artillerie (Pantometria, 1591),
a sort of translation and commentary on Tartaglia's Nova Scientia, appears to have made a
random guess that the trajectory is a parabola or hyperbola, as he remarks that “The bullet
violently thrown out of the peece by the furie of the poulder hath two motions, the one violent,
which endeavoureth to carry the bullet right out of his line diagonall, according to the direction
of the peece's axis, from whence the violent motion proceedeth; the other naturall in the bullet
itselfe, which endeavoureth still to carrye the same directelye downeward by a right line perpen-
dicular to the horizon, and which dooth, though insensiblye, euen from the beginning of little and
little draw it from that direct and diagonall course. - . . .
“These middle curve arkes of the bullet's circuite, compounded of the violent and the natural
motions of the bullet, albeit they be indeed mere helicall, yet have they a very great resemblance
to the Arkes Conical. And in randoms above 45° they doe much resemble the hyperbola, and in
all under the ellipsis. But exactly they neuer accorde, being indeed spirall mixte and helicall.”
Digges appears to be endeavouring to explain Tartaglia's refutation of the fallacy, not yet
extinct, that for a certain distance at the outset, called the point-blank range (bout-portant, but-
blanc) the trajectory is a straight line. A similar explanation is due to a Busca, a contemporary
Spanish Officer (Memorial de Artilleria, January, 1889, p. 33).
Denote by a and w the angles of departure and descent, and by V and v the initial and final
velocities in a real trajectory of range R, height of vertex H, and time of flight T. This
trajectory will obviously lie between the two parabolic trajectories of the same range R and
angles of projection a and w; and the heights of the vertices of these parabolas being #R tan a
and #R tan w, their velocities of projection being v/(gR cosec 2a) and V/(gR cosec 200), and the
times of flight being -
A/(2R tan aſg) and V (2R tan w/g),
therefore
#R tan a < H. 3 #R tan w,
V > V(gR cosec 2a) > V(gR cosec 2w) > r,
v/(2R tan aſg) < T : V (2R tan w/g),
formular useful in a first approximation. *-
Lieut. Wolley, Dod, R.A., found by calculation (Proceedings R.A. Institution, Vol. XVI) for a
projectile weighing 380lb, fired at 40° from the 9.2-in wire gun with a velocity of .2375 f/s, a
range of 20765 yards, a height of vertex of 17110 ft, an angle of descent of 53°50', with a time of
flight 63.8% and a striking velocity 1090 f/s.
Here R = 62295, H = 17111,
2 = 40°, w = 53° 50', T = 63.8.
Also, by numerical calculation,
#R tan 2 = 13068, +R tan w = 21305;
v/(gR cosec 2 z) = 1427, V(gR cosec 20) = 1451 ;
v/(2R tan 2/g) = 57, M(2R tan w/g) = 73;
thus exhibiting the scale of these limits of approximation.
... • ‘. 8 O
MoTION OF A BODY ON A SMOOTH PLANE CURVE, UNDER GRAVITY, OR
GIVEN FORCES.
57. Denoting by R the normal reaction of the smooth curve, and X, Y, the component
impressed forces on the body P, weighing W lb, then, with the gravitation unit of force, the
equations of motion are - -
}# =x-minº . . . . . . . . . . . (1)
} = x + Rºr . . . . . . . . . . (2)
or revolving along the tangent and normal,
W dº — v. de v dy - w
— — — — — - Y _*_ º e G º º •. …” e * º 3
g ds X ds + ds ' (3)
W v2
"" = – x 4 + y + + R,. . . . . . . (4)
9 p ds als
Integrating equation (3),
weg-ſex is + Yay) + H . . . . . . . . . . ()
the Equation of Energy, giving the velocity v at any point; and then equation (4) determines R,
the pressure on the curve. -
Suppose the body to move under gravity on a smooth curve, e.g., a switch-back railway,
then X = 0, Y = — W, and equation (5) becomes
#Wv°/g = H – Wy,
ÖI’ #v°/g = h – y,
if the velocity v = 0 when y = h; and then the velocity at any other point is that which would
be acquired in falling freely from the level of the line y = h, the distance h — y being the
impetus of the velocity v.
Then the reaction
W 12 2
R = W cos – º – w(co- -*).
Ale 9 p z */ p
so that the carriage would leave the rails if at any point the level of the carriage above the centre
of curvature of the track is less than twice the depth below the starting point.
If the track is continued round in a loop, so that the people in the carriage are for a time
head downwards, then for the carriage to keep on the track, and for the people to keep in the
carriage, the vertical height Pºg' of the carriage P’ above the centre of curvature Q' of the track
must be less than twice the depth P'M' below the level AMM’ of the starting point A.
In consequence of friction and other passive resistances, the carriage will not reach its
original level at R, but will be a certain distance below, up which distance the carriage must be
pushed to make a fresh start. Drawing the sloping line AR, we may, as in $21, suppose that the
velocity at any point P to be due to the impetus PM, the vertical distance of P below AR.
- Writing the equation -
4Wv°/g -- Wy = Wh,
then Wv°/g is called the Kinetic energy, and Wy the potential energy; their sum Wh, the total
energy, being constant; all these energies being measured in foot-pounds. -
In a swinging pendulum we see the interchange of kinetic and potential energy in constant
operation.
81
COMPONENT WHELOCITIES AND ACCELERATIONS WITH POLAR CO-ORDINATES.
58. Referred to polar co-ordinates (r. 6),
a = r cos 0, y = r sin 6 ;
and if OP cuts the curve AP at the angle p, called the radial angle,
tan (p = tº. cost =}| sing = }
denoting the arc A.P.
Therefore the component velocity along OP, called the radial velocity,
ocosº-º-º:
T dºg's T ºf ’
and the component velocity perpendicular to OP, called the transversal velocity,
..., , d8 ral6 rd6
a sin b = |* = #.
Also d6/dt is called the angular velocity of P round O ; angular velocity being always estimated
in radians per second ; so that the transversal velocity at a distance r feet or centimetres should be
drð/dt celoes or kines.
Independently.
da, dr 2-d6
ºf T ... cos 6 +cos 6,
dy dº 7:00 .
dt =# sing + cost; sin 6;
so that the radial velocity
*cos 0 + 4 sin 9 = %
(lt dt d;
and the transversal velocity , .
da: #. dy *º- zdó
... sin 9 + #, cos 6 = dº ’
Differentiating again with respect to t,
d’a (d. — ?” #) cos 6 — (? dy d6 + 7° º sin 6,
d? *E=º J;2 (7.2 dź dº dº?
d°y d?), d62", . ( dr d6 626
—- F — - ?? --— 2* ** * — *
Gltº # - rºº) sin 6 + (*, *, +, #) cos 6;
and therefore the radial acceleration
d24, dºy sº, a d” d6°.
# cos 0 + ... sin 9– d;2 * if: ;
and the transversal acceleration
d’a . 0.2, dr d6 0.26
Eºs º, sin 6 + %cos 0 =2% # * * dº
usually written # # 7.2 # )
to which it is equivalent, as is seen by performing the differentiation.
(6873) Y
82
CENTRAL ORBITS.
59. Polar co-ordinates are useful in Astronomy, when we have to investigate the motion of a
planet with respect to the Sun as origin, under a radial force R, and sometimes also a transversal
force T, estimated in dynes per gramme.
The equations of motion are then—
2 2
d?” , d6 = R,
alſº d?
1 d *) = Y
}}(ºft = T.
The quantity "dē/dt is denoted by h; and then dA/dt = }h, the rate at which the sectoreal
area A is being swept out by the vector OP revolving about 0. 14
We also denote 1/r by w; so that
& = u_* = u%h ‘’’
and now T = *i; w”h 06'
{)T - —H = — e
- d6 Q43
& dr – dr d6 — 1 du 7,5 – du
Again d; T dé da T wº jºu tºº j%
d°r '#|-} )= - ‘(; du dh)
# = hu' (-; )= - hu # ***)
so that R = – 2 -
so that ++, + \l = - - - -—#–. -->
_ R _ T du * . . . s
72n.2 Å2,3 d6'
the differential equation of the orbit.
a ds” dr” + ,406” * *(i. º Ji?
Also 77" F — = — *-* + 2, *
d;2 Gli2 Glt? d62 p?
*ºms
In a central orbit, in which the attraction is always directed to the origin O, T = 0, and h is
constant.
Denoting the central force towards the centre, in dynes per gramme, by P, so that P = — R,
2
P = *(; + •)
whence the value of P is found when the equation of the orbit is given.
ExAMPLE. —Prove that P varies as wº or r-8 in the orbits (Cotes's spirals),
au = sin nô, nô, sinh m0, exp nô, cosh mó.
60. The fact that h is constant or that the radius vector of the body from the origin sweeps
out equal areas in equal times in an orbit described under a force tending to the origin, goes by
the name of Kepler's Second Law.
The following geometrical proof is given by Newton —
Let the time t be divided into a number of small equal intervals, T seconds; and in the first
º let the body describe the straight line AB, with uniform velocity, being acted on by no
OTC 62.
In the second interval the body, if no force acted upon it, would proceed to c, where Bc = AB,
but if an impulse is applied at B in the direction BS, the body will proceed in a new direction BC,
and by the triangle of velocities will be found at the end of the interval at C, when co is
parallel to SB, and therefore the triangles SBC, S.Bc, SAB are equal and in the same plane.
N
83
In the same way, if centripetal impulses act at the end of each interval T, causing the body
todescribe the straight lines CD, DE, . . . . the triangles SAB, SBC, SCD, SDE, . . . .
will all be equal, and will lie in the same plane.
Now when these impulses act at indefinitely small intervals, they become equivalent to an
incessant force, tending always to S, while ABCD. . . . becomes a continuous curve; and the
radius vector from S sweeps out equal areas in equal times. -
Conversely if equal areas are described in equal times, the body is acted upon by a force
tending to S.
Denoting the area swept out in a second by #h, then in the very small interval of time T,
we may take the area SAB = }ht, while we may take AB = vT, v denoting the velocity at A; also
AB is ultimately the tangent at A, and SY, the perpendicular AB is ultimately the perpendicular
on the tangent, which we have denoted by p.
But the triangle SAB = }SY. AB,
Ol' - #ht = }SY. vºt, ultimately;
so that h = pu.
Denoting the radial angle SAY by p,
te _ , d6 .#
then sin q = 7 ds
& d6
2 - ?? & - .2 ºv
#9 r sin (p = 7 ds
while Q) = %
so that J. = ?” d6.
dt
Since v = h/p, it follows that the Hodograph of a central orbit is similar to the inverse of the
pedal of the orbit, the vectors of the Hodograph being now perpendicular to the velocities which
they represent ; and the velocity in the hodograph will be perpendicular and proportional to the
acceleration in the orbit.
61. The most important case in Astronomy is the elliptic orbits described, according to Kepler's
Second Law, about a centre of force in the focus at S; and now the polar equation of the ellipse
being,
# = n = 1 + ecos 0.
du g
! : = —
d6 e sin 6,
2
1%. = — e cos 6,
so that (; + w) = 1,
and P = hºu?/l,
so that P varies as w” or r-”, and thus Newton’s Law of Gravitation is deduced from Kepler's First
Law, that the planets describe ellipses of which the Sun occupies a focus.
Denoting by T the periodic time in seconds, and n the mean motion or mean angular velocity,
so that
n = 27/T,
then #hT = the area of the ellipse = Tab,
Or h = nab, h” = nºa. lº, since l = bºſa;
so that P = n”aº-
84
f{EPLER'S LAWS OF PILANETARY MOTION.
* * Kepler, by long continued observation and measurement of the Sun's diameter and motion in
longitude, noticed that (i) the variation of the Sun's apparent diameter d could be expressed by
the formula D (1 + e cos 6), where D denotes the mean angular diameter (about 1920') and
6 the Sun's longitude from perihelion, e being a small constant, about 1/60; (ii) that the Sun's
daily motion in longitude was proportional to the apparent area or square of the diameter.
Since the apparent diameter is inversely proportional to the distance, he deduced the laws
called Kepler's Laws—
(i) That the relative orbit of the Earth (or any planet) and of the sun is an ellipse, with a
focus at the Sun, if the Sun is supposed fixed, given in polar co-ordinates by l/r = d/D = 1 + ecos 6.
(ii) That r*.d6/dt = h is constant, and that the Earth sweeps out by its radius vector from the
Sun equal areas in equal times. -
The third law of Kepler—(iii) The squares of the periodic times of the planets round the
Sun are proportional to the cubes of their mean distances—was easily inferred by arithmetical
calculation, when once the distances of the planets from the Sun were measured, in terms of the
Sun's distance from the Earth. ** ; :
Newton, theorizing on these laws, inferred from law (ii) that the Earth is attracted by the
Sun, and from law (i) that the attraction must be inversely proportional to the square of the
distance, while law (iii) showed that the attraction of the Sun was of the same nature on all the
planets (§ 84), and generally between any two particles of matter.
We shall employ the C.G.S. system of units, and measure a in centimetres, T in seconds, and
S and P in grammes (§ 142); the unit of velocity is now called the kine, of acceleration the spoud,
of force the dyne, and of work the erg.
Now if we consider two bodies, the Sun weighing Sg, and the Earth or planet weighing Pg,
at a distance r cm, the c denoting the gravitation constant (§ 51), the attraction between them will
be CSPr-* dynes, causing acceleration in P and S of CSr-” and CPR-4 spouds respectively, and
therefore a relative acceleration of C(S + P)r-* Spouds; so that the relative orbit of S and P is a
conic section, in which -
m*a* = C(S + P), Or 47%a” — C(S +P)T*.
Suppose, for instance, that the two bodies S and P were originally at a very great distance
apart, the Sun and a connet for instance, and approaching with relative velocity V kines so as to
pass, in the absence of gravitation, at a minimum distance of d em.
In consequence of gravitation the relative orbit will be a hyperbola, and the direction of
relative velocity will be ultimately turned through the angle between the asymptotes, an angle
2a = 2 tan - {C(S + P)/dV*}:
while the nearest approach of S and P will be changed from d to
d(Sec a) tan a.
Taking the Earth and the Moon as the two bodies, and denoting their weights by E and M, and
comparing their relative motion with that of the Sun and the Earth, considered as the two bodies,
then from Kepler's Third Law,
S + E mºa'
ETM T nº.'
in which we may put n'/m = 13, with 13 lunations in the year; and also
a' cosec 8”.8 23546
--- amamº sº------------, sº
C!, cosec 57.1 69 °
ve S + E Fºr ºf , a *
4. r – 10% g © ...}
and thence we find E + MI ()* x 3.576 :
Writing this,
S + E s 1 + E/S
ETM T E . IIME:
then since M/E and E/S are small, we may put
S/E = 105 x 3.58.
The Sun's mean angular semi-diameter being taken as 16', or 16' being the parallax of the Earth
as seen from the Sun, the Sun's radius is therefore sin 16’ cosec 8”8 = 960 –– 8:8 = 110 times the
Earth's radius, nearly double the distance of the Moon; and thence we find that the mean density
of the Sun is only about one-third of that of the Earth; but the acceleration of gravity on the surface
of the Sun is about 30 times that on the surface of the Earth.
85
. In a similar manner, from the astronomical observations on Jupiter, that the fourth satellite is
at a distance of 25 radii of Jupiter, and that its period is 16 days 18 hours, and that Jupiter's
radius is 11 times the Earth's radius, we can infer that (i) the weight of Jupiter is about 270 E.;
(ii) that the mean density of Jupiter is a little greater than that of water; (iii) that g on Jupiter
is about 2170 spouds, or 2-2 times terrestrial gravity. -
So also the periodic time of Saturn round the Sun being 10,759 days and the periodic time
of one of Saturn's satellites being 16 days, while the apparent semi-diameter of the satellite's
orbit as seen from the Sun is 176"-25 we find that the weight of Saturn is about one-3,490 of the
weight of the Sun.
62. As the point P moves on the ellipse A P under a force to the focus S, the corresponding
point p on the auxiliary circle Ap, called the eacentric follower, moves so that Sp also sweeps out equal
areas in equal times, for the &
area ASp a
area. ASP T b . sº
The excentric follower p therefore moves as if attracted to S, and so also does any other
excentric follower on any ellipse on the same axis A A'; and it will be found that this force varies
directly as the distance Sp, and inversely as the cube of the distance pn from nNX the polar of S.
It can be proved as an exercise that this force can always be resolved into two forces;
tending to two fixed points F, F in the axis AA", varying inversely as the fifth power of the
distance. (Sylvester, Phil. Mag., 1866).
The two points F, F, are situated in CA, such that CF. CF1 = CA”; and then
Fp : Flp = FA : FA = CF, CA = CA; CF.
The force to S which makes the excentric follower p describe the circle Ap in the periodic
time T seconds is
47-? C. × 3. Sp.
T3 pn”
and this force can be replaced by the two components,
1672 CF2 . CA6
T. C.S. Fº
16trº CFix. CAs
Tº CŞ. Firs' "
a long pH,
long pH";
components, which are always in a constant ratio. - f
If these components were equal, the points S, F, F, would coalesce in A, and the force to A
would vary inversely as the fifth power of the distance from A. -
In the particular case where S is the focus of the ellipse, the polar of S is the directrix NX,
and then PN, the distance from the directrix is proportional to SP; so that the force varies
inversely as SP”, as shown previously. -
Now if we denote the angle ACp by v, where p is the excentric follower on the auxiliary
circle, then v is called the eacentric anomaly, and :
the area ASp = º: ACp — triangle CSp
= }a”(v — e sin v),
so that
the area. ASP = }ab(v – e sin o);
and, by Kepler's Second Law,
t area. ASP v — e sin v
T area of ellipse 27ſ
so that mt = v – e sin v,
and nt is called the mean anomaly.
Expressed in terms of the true anomaly 6, the angle ASP from perihelion A,
sino - Mp a MP SP sin 6 l sin 6
Op T 5 a T b b(1 + ecos 6)
V(1 – e’) sin 6.
1 + e cos 6
e -- cos 6
COS Q) =
- IT ecos 6
6% Z
86
_ sin-1 vſ (1 – e’) sin 6 ev'(1 – e’) sin 6
so that nt = sin 1 + e cos 6 1 + ecos 6
Denoting by v the velocity in the elliptic orbit
#v% = #h” h” HZ #h” HP
2a – 7. 1 1 l 1
— lon2a2 = m^a" | f – – ) = m^a” | – — — ) :
= sna' –F we(; ) *(i. st)
so that v is the velocity which would be acquired by the body under the central force to S, if it
moved to P in the straight line UP, starting from rest at U, as in § 49.
In a parabolic orbit, as of a comet, a is infinite and n is zero, but
h?
ZŽ ISP'
m*a* = Z2/Z”, and #v% =
Suppose, for instance, the relative orbit of the Sun S, and a planet P was a circle of radius
a cm; then
4Tºaº = c(S + P)T*,
\ 27ta} 27taş a;
Or T = McCŞTP)} = 38.93 V(STP) = 24.460 V(SIP) seconds; *
and the relative velocity W is given by
W2 = 47%a” c(S + P.
* T2 O),
But now suppose the square of this relative velocity is doubled, so that
| W2 = c(S + P)
2 Cº, 2
the relative orbit becomes changed to a parabola of semi-latus-rectum, l = 2a, and
h” = 4c(S + P)a? = c(S + P)Z.
The distance will now become doubled after
$v/2 Q} seconds.
v/ v/{c(S + P)}
63. If particles are projected from a given point O with given velocity, they will, if subject
to an attraction to a fixed point, varying inversely as the square of the distance, proceed to
describe ellipses with one focus at the fixed point, and with major axes of the same length, so that
the other focus of an elliptic orbit will lie on a circle (or sphere) with centre at O. x
We may suppose this realized with bodies projected with given velocity in different direc-
tions from a gun or volcano on the surface of the Earth, the velocity being sufficiently large for
the variations in the direction and magnitude of gravity to become sensible.
In the absence of resistance the trajectories will be ellipses with one focus at E, the centre of
the Earth, and the other focus S on a circle (or sphere) described with the point of projection O
as centre with radius R/n, supposing the velocity of projection is sufficient to send a body vertically
upwards to a height O H, one nth of R, the radius of the Earth.
We use the letter S now to denote the empty focus of the elliptic orbits, and E to denote the
common focus at the centre of force, the centre of the Earth.
Drawing any straight line ESS' to meet this circle (or sphere) in S and S', then bodies pro-
jected in the directions perpendicular to HS and HS' will have the same range OR on the Earth’s
surface, obtained by making the angle OER equal to twice the angle OES ; and then OHS, OHS'
will be the angles of projection with the horizon at O.
The maximum range OA is obtained by making the angle OEA double the angle OEC,
where EC is the tangent to the circle OH.
The two elliptic trajectories OPR, OP R now coalesce into a single ellipse, and their foci S, S'
coalesce at C ; so that the chord of the maximum range oA is the latus rectum of the correspond-
ing elliptic trajectory. --- -
Now if the angle OHC is denoted by 2,
cos 22=1/n,
giving the angle of elevation for maximum range, and then if a. is given in minutes of angle, the
maximum range will be 180 × 60 — 42 geographical miles.
Produce OS to meet the corresponding elliptic trajectory in D; then
OD + DE = OS + SD + DE = 20S + OE = OH -- HE,
so that the trajectory touches a fixed ellipse, with foci at E and Q, and vertex at H.
If we revolve this ellipse about E0, it sweeps out a prolate spheroid; and the interior of
this prolate spheroid is the space covered from the point O. -
s
..-
-.
º
t
87
Since OA = 200 = 20H = 2R/n,
therefore, if AB is drawn perpendicular to EO,
OB = OA*/2R = 2R/n”;
so that the area covered on the Earth's surface from O is 1/n” of the whole surface. tº e *
Thus if nº = 2, a hemisphere can be covered from the point Q, and the angle of projection for
maximum range is 22°. Therefore the least velocity with which a body can be projected from
the Pole at O, so as to reach the equator, is that sufficient to send the body to a height #y/2R, at
the Pole, a velocity vſ(2x/2 − 2) or about 0.9 times the yelocity of a grazing satallite, and there-
fore about 13,800 knots (§ 50); and the angle of projection is 22°4. e e
If n = 1, and OH = R, bodies projected horizontally will skim round the Earth like a satellite,
and if projected at an angle 8 minutes with the vertical will have a range of OR of 28 geographical
miles, the points O and R being the ends of the minor axis of the trajectory.
If OH is greater than R, the two trajectories for the same range OR will go round the Earth
opposite ways. º e ©
If OH is infinite, the trajectories will be parabolas with a common focus at E, while for still
greater yelocities of projection the trajectories will be hyperbolas. tº a ge © (a
As in the former parabolic trajectories we can prove that in these elliptic trajectories such as
OPR, P moves so as always to be equidistant from the ground at M, and from the circle through
O and R, whose centre is at S, so that in the figure 27.A
PM = PQ and also SP = PK.
For EP + PS = EO + OS = EM -- SQ,
so that EP – EM = SQ – PS,
Or - PM = PQ.
Also - SP + PE = SO + O.E = EH = EK,
so that SP = PK.
Bisect the angle OEP by EF, then FO = FM;
also SF = FH = FK = FL,
if SL, drawn at right angles to HS, meets HF produced in L.
Then since SP = PK and SF = FK, SPF = FPK,
or FP bisects the angle SPK, and is therefore the tangent at P.
Also FSP = FKP = FHO = FSO,
and therefore the tangents FP, FO subtend equal angles at the focus S, as well as at the other
focus E.
All the 9ther letters in the figure will be found to correspond to points denoted by the same
letter in the fig. 20A of the parabolic trajectories, and to indicate corresponding properties.
If O is at one ºf the poles of the Earth, the vertical plane OP of the trajectory will have a
fixed direction, while the Earth turns round, so that the body will meet the Earth again at a point
R not in the original meridian of projection.
Thus, a body at the North Pole, sliding at 5 m/h on a sheet of smooth ice 60 miles in radius,
Will reach the circumference at the opposite end of the diameter through the point arrived at
(Academy, 22nd April, 1892).
For any other position of 0 on the Earth, the plane OR will have an aberration due to the
motiºn of O. and R will be deflected from the original vertical plane of projection.
It is sufficient however for ordinary dynamical and ballistic problems to take the surface of
the Earth as at rest, as great complication arises from considering it a moving platform. -
The circle HK is the hodograph of all the elliptic orbits, since the velocity in any ellipse OP
is proportional, and perpendicular to SK.
The acceleration of P is proportional and perpendicular to the velocity of K in the hodograph;
it therefore passes through E, and is proportional to the angular velocity of EP.
So also in Fig. 26, the circle UV is the hodograph of the elliptic orbit AP, described round the
Sun Ş. as a focus, since the velocity is perpendicular and inversely proportioned to SY, and there-
fore directly proportional to HZ or HU.
The acceleration of P is therefore perpendicular and proportional to the velocity of U, and
therefore passes through S and is proportional to the angular velocity of SP.
88
But by Kepler's Second Law, the angular velocity of SP varies inversely as the square of SP;
so that in addition Kepler's First Law, “The orbit of a planet with respect to the Sun is an
ellipse, the Sun being in a focus,” leads to Newton's Law of Gravitation, that “the planet is
attracted by the Sun with a force varying inversely as the square of the distance.”
This geometrical demonstration is taken from Maxwell's Matter and Motion, § 132. -
We notice that if UH is produced to meet the hodograph circle UW again in T, and if the
tangent at P is produced to meet the hodograph circle again in T. and T," then TT. T"
will be a triangle, inscribed in the circle UV, and circumscribed to the ellipse AP, the empty focus
H being the orthocentre of this triangle.
If T'H, T"H produced meet the hodograph circle again in U", U", then SU’, SU” and the
ellipse in P', P’, the points of contact of the sides T^T, TT" of the triangle TT Tw.
The hodograph of the Earth's elliptic orbit round the Sun being a circle, it follows that the
stars, in consequence of aberration, will appear to describe circles in planes parallel to the ecliptic
(§ 35) even when the excentricity of the orbit is taken into account; but now the true position of
a star will not be at the centre of the circle.
The vector HU of the hodograph may be resolved into two constant component vectors SW
and HS; so that the velocity at P,
= *.
- SY -
Q)
. HV,
%
2
–
#
}
T
; HV =# .
;
may be resolved into two constant components, perpendicular to the radius rector SP and the
major axis AA', of magnitudes
#
;
_ 2+ a” T a Trs 27 a”é
SV = + , and #. . HS = + , ,
in the ratio of one to e.
This theorem is useful in the exact determination of the aberration of the Sun, or of a star on
the ecliptic; for if G) and l denotes the longitudes of the Sun and of the Star, and to the longitude
of perihelion A, the aberration of the star will be proportional to the component velocity of the
Earth perpendicular to the direction of the star; and will therefore be represented by
X { cos (G) — l) + e cos (as — l)}
when X is a certain constant, such that k = Av/(1 — e”), if k denotes the constant of aberration at
the Sun's mean distance (§ 34).
EXAMPLES.
1. Prove that the latus rectum of the Earth's orbit divides the orbit into two parts described
in the ratio of 1 to 1043, or in 0.489 and 0.511 of the year, taking e = 1/60, -
2. If the moon had a satellite whose weight is the same fraction of the weight of the moon that
the moon is of the Earth, and whose distance from the moon is the same fraction of the moon’s
distance from us, find the period of the satellite.
3. A spherical shell of Small radius describes a circle of radius R with velocity V about a centre
of gravitation at S; and when the shell is at P, it bursts with an explosion which generates
velocity v away from its centre. Prove that the fragments will all pass through the line SP
within a length
8 W3 v R.
V4 – 6 Wºvº -- v 4
and that if p is small, the stream of fragments will form a complete ring after a time grkſ,
approximately. 2
4. A stream of meteorites, originally moving in a straight line K with velocity V, comes under
the influence of the attraction of the Earth, whose centre moves with velocity v in a straight line
which intersects the line K, and is inclined to this line at an angle a. Prove that, if the distance
of the Earth from the line K is originally very large, a length -
*R_{v: — 2Wv cos a + v -- 29R}”
v sin a
of the line of particles will fall upon the Earth.
89
4. DYNAMICS OF A RIGID BODY.
MOMENT OF INERTIA AND RADIUS OF GYRATION.
64. In problems on the Dynamics of Rotating Rigid Bodies, a quantity called the Moment of
Inertia about the axis of rotation is required. -
DEFINITION.—The Moment of Inertia of a body about an axis is the sum of the products of the
weight (in lbs) of each particle of the body and of the square of the distance (in feet) of the
particle from the axis. ra
Thus if Oz is the axis, and m lb is the weight of a particle at the point a, y, z, then
Xm (w" -- y”) is the M.I. (Moment of Inertia) of the body about Oz, the summation extending to
all particles of the body. - tº-
If W denotes the total weight of the body in lb, then
W = Xm ;
and now if we write
>m (wº + y”) = Włº,
then k is a certain length in feet, called the Radius of Gyration of the body about Oz ; and we may
suppose the whole weight of the body condensed in a ring at a radial distance k from the
axis O2, without altering its M.I. about Oz.
Take, for instance, a fly wheel of weight W lb, rotating with angular velocity m = d6/dt
about a fixed axis Oz; or a projectile rotating about its axis. -
Then a particle of the wheel, weighing m lb, at a distance r feet from the axis O2, will hav
velocity v = rm f/s, and therefore K.E. = #mrºn”/g ft-lb ; so that the total K.E. of the fly wheel is
>4mrºn'/g = }W ºn’ſgft-lb.
The energy of rotation of a projectile weighing W lb fired with velocity V fis from a d inch
gun, in which the pitch of the rifling is n calibres, will thus be -
... 47-2%2 WW2 ft-lb, or 47%2
n°d” 2g 102d?
For instance the work stored up in the rotation of a solid cylindrical proof projectile
weighing 2048 lb and fired with velocity 2100 f/s from a 16-inch gun in which the rifling at the
muzzle makes one turn in 30 calibres, is about 345 ft-tons. Compare this with the striking
energy.
So also in slamming a door; to overcome the work required to turn the spring lock, an equal
amount of work must be stored up in the angular motion of the door, represented by
#WKºn?/g (ft-lb). -
h In the notation of the Integral Calculus, if w denotes the density at any point, in lb ft“,
then
of the striking energy.
W/2 =| ſº + y”) daydydz,
or in other words, the M.I. about an axis is the space integral of the product of the density and the
square of the distance from the aaris.
When the body is homogeneous and of volume V ft“, then W = wV, and
Wk? = | (a” + y”)dardydz,
the space integral of the square of the distance from the avis.
Similarly for an area. A ft” about an axis Oz perpendicular to its plane
Ak.” = | (a" + y”)dady,
* *.
while about axes Oa, Oy in the plane
Ak.” = ||rºwdy,
Ak,” = ſeded,
the surface integrals ºf the square of the distance from the awes Oz, Oa, Oy; using k, k, k, to denote
the Radii of Gyration of the area about the axes Oa, Oy, Oz. -
(6873) 2 A.
90
For a plane lamina, of weight W lb., and uniform or variable superficial density a lb per ft.
W = || orda:dy,
Wk,” = ||award,
Wh,” - |award,
tº
Wk 2 = || a (a.” + y”)dardy,
the surface integrals of the product of the superficial density and the square of the distance from the aris.
We notice that, for a lamina,
Wk.” – Whº -- Wh,”,
Or k” = k.” + k,” . . . . . . . . . . . . . . . (I)
an important Theorem for Moments of Inertia.
65. About an axis in the plane of the lamina, making an angle 6 with Oa',
Whº? = | ſo (y cos 0– a sin 6) dady,
= Wk,” cosº 6 – 2F cos 6 sin 6 + Wh,” sin” 6,
where F denotes the quantityſºzyded, called the product of inertia of the lamina with respect to
the axes Oa, Oy.
By turning the axes through a suitable angle, F can be made to vanish; and these new axes
are called the principal awes of the lamina at O; and now
K* = k,” cos’ 6 + k,” sin” 6;
and if the ellipse
2
3/.
+ £- = 1
ap?
2 zº
ky ,”
k
is drawn, then the k about any axis through its centre is the length of the perpendicular from the
centre in the parallel tangent; and this ellipse is called the momental ellipse of the lamina at O.
For a line or wire, of variable linear density a lb per foot run,
Wk? = ſº + y”)ds,
when W = | ords, the weight of the rod in lb. ; so that Wk” is the line integral of the product of
the linear density and the square of the distance from the aais.
, We notice that in finding the M.I. of compound bodies round the same axis, we can add or
subtract M.I., but not k”.
66. Another Theorem (II) connects Wk”, the M.I. about any axis, with Whº, the M.I. about a
parallel axis through the C.G. at a distance h.
For taking the axis O2, -
Wk,” = Sw(w? -- y”),
Wk” = Xw ((a — v.)* + (y – y)*}
= X w(w” + y”)
– 2a2wa — 23/Xwy
+ W(z” + y”)
= Wł,” – Whº, -
since >mw = Waſ, Xmy = Wy;
and therefore Wk,” = W(k? -- hº)
QI" ** = }* + h^ . . . . . . . . . . . . . (II)
The value of k” need therefore be calculated only for axes through the C.G. of a body,
* - - ~ * - , “... . . . . .
- - .*
- * :
- * - - *~ ****
91 y ... • •.
ºl. - -
EXAMPLES.
For a straight rod OA, of length l, in Oa,
- ^!
W = |ad. Wa: = |ards, Wk? = | oražda.
() 0 J0
If the rod is of uniform density, we replace W by l and a by unity; and
l/c2 = | wºda = }lº, k” = }l”.
0
About an axis through the middle point of the rod,
#!
//52 =| w”da = -lºlº. k” = -lºr!”.
1 2 * > 1 2
Similarly for a rectangle, of sides a and b feet, about axes Ow, Oy through its centre O parallel
to the sides a, b, -
$0.
Ak.” =| a"bda = #a’b, k,” = Tºra”;
1 2 3.
-a%
treating the rectangle as an assemblage of lines of length b and breadth da ; and by symmetry
- k,” = +0°.
About Oz, perpendicular to the plane of the rectangle,
le.” = k.” + k,” = Tr(a” + b%) = Tºr (diagonal)*
by Theorem I. .”
About the edge a of the rectangle k” = }b”; and about the edge b, k” = }a’; so that about
an axis through a corner of the rectangle, perpendicular to its plane,
k” = }(a” + b%).
By superposing a number of such rectangles we build up a book, or brick-shaped right solid;
and denoting its height by c, then, about an axis through the centre, parallel to the edges c,
}* = Tir (a” + b%);
and about an edge c k” = (#a” + b%).
with similar expressions for k” about the other edges and principal axes.
Cut the solid in two by a plane CD through two opposite edges c.; then the M.I. about Oz of
each half is halved, while its volume is also halved; so that k” is unaltered, and
/* = +3 (a” + b%),
for either half, a right prism on a right-angled triangular base.
Changing to the parallel axis through G, the C.G. of the triangular prism,
/...” = Tºr (a” + b”) — OG”,
— Tº (a” + b%) — alº (a” + b%)
= +'s (a” + b%),
Changing to the axis coinciding with the edge through E,
#, = +ls (a” + b”) + GE”
= 's (a” + b%) + + (a” + bº) = } (a” + bº).
Placing two such triangular prisms back to back, we form a triangular prism on an isosceles
base; and now the M.I. being doubled by doubling the volume, the k” about E will be, as before,
k.” = } (a” + b%),
where b denotes the height and 2a the base of the isosceles triangle.
Changing to the parallel axis through H, the C.G.,
K.” = } (a” + b%) — ºbº = }a” + 4's b%;
and changing to the edge through C,
2 – 4 ~8 2 tº- 2
K.* = }a” + +'s b% + 4b* = } a” + #b%.

92
A right prism on a regular polygonal base can be divided into a number of such prisms on
isosceles triangles as base; and for each of these prisms, as well as for the whole, about its axis
of figure,
Å;2 - #7° + #a”,
r denoting the radius of the circle inscribed in the polygon, and 2a the length of a side of the
polygon. º e
By keeping r fixed, and making a zero, the polygon is changed into a circle of radius r ; and
therefore, for a right circular cylinder about its axis,
}* = }r°.
This is independent of the height of the cylinder; so that making it indefinitely thin, we
obtain k” = }r” for a circular disc of radius r, about an axis through the centre perpendicular to
its plane; and therefore, by Theory I, about a diameter of the circular lamina,
}* = }r°.
EXAMPLE.
Determine the number of ft-lb of work stored up in the rim of a fly wheel making 40
revolutions a minute, the inner and outer radius of the rim being 9 and 10 feet, and the weight of
the rim 10 tons.
The M.I. of the circular bodies can also be determined by integration.
The k” of a circular line or wire, about an axis through its centre, perpendicular to its plane,
is obviously r*, r denoting the radius of the circle.
About a diameter, the M.I.
f*27r
270°].” – | zdó. 7" sin” 6 = Tr"; or k” = }r”.
a' ()
About a tangent line, k” = }r” + 7% = },”, by Theorem II.
-Considering the circular disc of radius a as built up of a series of concentrig circular filaments,
the M.I. about its axis of figure,
Ta2/c2 =| 27trár. r" = }Tra*;
J 0
}* = }a”, as before.
About a diameter, cutting the disc up into filaments perpendicular to the diameter.
Q. &
Ta”k? = | 2yda . 49* = 4| (a” — wº)}da,
–6. 0
and substituting a = a cos 6,
}ºr
27.2 – 4 ~4 ; , , 4 •y — 4 24 3 - 4.
Troºk:” – *|| sin' 6d.6 = }aºr = }ºra",
}* = }a”, as before.
2
Similarly for the ellipse; +} = 1, ... = 3, # = 32.
Treating the cylinder as built up of a number of coaxial elementary discs, of radius a and
thickness da, (fig. 31), then the k” about the axis Oa is still #a”; but about the axis Oy through
the centre perpendicular to the axis of figure, the k” of the elementary disc Pp being 4a2 + 2*,
by Theorem II., the M.I. of the cylinder (denoting its height by h)
Wk,” = | ăſt Traºda (#a” + æ")
—%h
= }Traºh -- Tºrtraºh’;
and W = Tra”h,
so that k,” = }a} + Tºhº.
For a thin rod, a = 0, and ſº = Tºſh”, and for a thin disc, h = 0, and k” = }a”, as before; the
k” for the cylinder being the sum of these two.
Generally, for any prismatic bar,
ky” = k” + +'s h”,
* k is the radius of gyration of the cross sectional area about the axis in plane parallel to
$/.
For a cone, of height h, and radius of base a, about the axis of figure Oa,
h
Wk,” = | Try”da: , ;",
0
and % =
&\}
and V - 47ta”h, *...
so that /* = **a*.
93
About the axis Oy, perpendicular to the axis of figure,
h
Vk,” = |ayasay + a”)
0
= gºtraºh -- 4tra*h”
k,” = #: a* + #h”.
For a sphere about a diameter
Wk? – ſ Try”da. . ; ſ’’,
- +. (a” — wº)"da.
JQ
- ** — 2a3a* + æ")da:
O
= Tra”(1 — 3 + +) = +*Tra";
and W = 4tra”, so that k” = #a”.
Similarly, for the ellipsoid
&? y” 22 - 1
h;2 = }(bº + c2), k,” = }(cº + a”), k” = }(a” + bº).
(For other examples, consult Twisden, Chap. V, and Calculus, Chap. II.)
67. Taking any solid body, reférred to three rectangular axes Oa, Oy, Oz, then its M.I.
about an axis OP through O, whose direction cosines are l, m, n, will be,
WAE2 = || | p8°dardydz,
when 8, the distance of the points a, y, z, from the line OP, is given by
8* = a + y” + 2* – (la + my + m2)*
= (m^+ n”)a? -- (nº + lº),” + (* + m”)2”
— 2mmy2 — 2nlza – 21may ;
so that denoting the M.I. of the body about Oa, Oy, Oz by A, B, C, and the products of inertia (P.I.)
with respect to the axis (Oy, Oz) (Oz, Oa) (Oa, Oy) by D, E, F, then
Wk” = AP + Bm2 + Cm” – 2Dmm – 2Eml — 2Flm.
Now draw the quadric surface Aw” + By” + Cz” – 2Dyz – 2E2a – 2Fay = K,
where K denotes a constant.
Then if the axis OP meets this surface in P,
Al” + Bm2 + Cº.” – 2Dmn – 2Enl – 2Flm = K/OP*,
so that WAE2 = K. OP”,
or k” varies inversely as OP”. º
Since k” is obviously always finite, on the supposition that the density of matter is always
positive, this surface is an ellipsoid, called the momental ellipsoid.
The axes of this ellipsoid are called the principal awes of the body at O; and by taking these
as co-ordinate axes, the products of inertia, D, E, F, are made to vanish; and thus
WAE2 = Al” + Bm2 + Cm”,
Or ** = lºk,” + mºk,” + n°kz”.
As an exercise, construct the momental ellipsoids of the preceding figures. e
Principal axes are axes of permanent rotation; for suppose a body is rotating about Oz with
angular velocity 7, the centrifugal force of a particle at (t, y, z), weighing m lb, being tº mV(w”-- ź
poundals directed away from the axis Oz, will have components rºma and r*ny parallel to Oa and
Ov.
3/ These components of the centrifugal force have moments
rºmyz and — rºmaz about Oa and Oy,
and therefore the resultant couple of constraint of the
axis Oz, required to keep the body rotating about O2 will have components

> *my2 = 1°D about Oa,
—X rºmaz = — ya?E about Oz.
These couples vanish, and the axis of rotation Oz has no tendency to become displaced, if
D = 0 and E = 0; and then Oz is called a principal avis at O.
Rotating parts of machinery should therefore be so shaped that the axis of rotation is a
principal axis of the rotating piece. a- tº
The superior steadiness of running of the inside cylinder locomotive over the outside cylinder
type is to be attributed to the nearer coincidence of the driving axle with a principal axis of the
system
(6873) 2 B
94
D’ALEMBERT'S PRINCIPLE.
68. This principle is the application of the Third Law of Motion,
“Action and Reaction are equal and opposite ”
to the formation of the analytical equations of the motion of a Material System.
First consider the plane motion of a body considered as an assemblage of particles; and let
w, y, denote the co-ordinates of a particle, m, lb its weight, X, Y poundals the component
impressed forces, and P, Q, poundals the component internal molecular forces, due to the
reaction and constraint of the adjacent particles of the body.
Then the equations of motion of the particle are (§ 52).
d?a,
— F P
* IE X + P,
d”y ---
mº = Y +Q.
so that we deduce
d? dža,\ *
m(*# sº yº.) = (a:Y — yx) + (a:Q — yP).
Denoting summation throughout the body by 3, then
sm; = XX + XP,
sm; = XY + XQ,
- A dºz dža,
sm(s . sºme vº) = X(w) — yx)
+ X(a:Q — yP).
Now by D'Alembert's Principle, or rather by the Third Law of Motion, the system of forces
represented by P, Q form a system in equilibrium, since to any one force of them there always
corresponds an equal reaction or force.
Therefore, by the Principles of Statics,
XP = 0, XQ = 0,
and X(a.0 — yP) = 0;
expressing the fact that the sum of the components in any direction is zero, and the sum of the
moments about any axis is zero, when the system of forces is in equilibrium.
Therefore, for the motion of the body or system,
dža,
>m. -: XX 6 f © Ç g © gº g e g g & * (1)
d?
Smº. ra. XY © d g g * º g ſº Q & tº o & (2)
*(#-º)=sey --> . . . . . . . . . (3)
the three equations of motion, deduced from D’Alembert's Principle,
95
If we had considered motion in space generally, referred to three rectangular co-ordinates
axes Ow, Oy, Oz, we should obtain in exactly the same manner the six equations of
motion—
d24,
*... = XX,
777, (l/? X.
d?y
m; = XY,
dź2.
"if - XZ,
(J22 (/27 ',
Xmſ ºf – 2" ") = X(7 – 2 Y),
"(; %) */
ſ/24: (7°2 -
sm(.. — , , ) = X(2X – w/),
dº (/#2
d?, (/?a,
Xmſ tº — y = X(a Y — y X).
Clt? (ſt?
Forces such as X, Y, are called impressed forces, and such as P, Q, molecular or cohesive
forces, while the name effective forces is given to the forces of inertia of the particles, such as
dža, d”y } 3. * - & º
dº ’ ” Iñ poundals; and now D'Alembert's Principle can be stated concisely in words as
“ the impressed forces and the reversed effective forces from a system in equilibrium ; the molecular
or cohesive forces forting a system in equilibrium among themselves;”
and thus by D'Alembert's Principle, all Dynamical theorems can be stated in a statical form.
These equations of motion can be simplified by the introduction of the motion of the Centre of
Gravity or of Inertia.
777,
INDEPENDENCE OF THE MOTION OF TRANSIATION OF THE CENTRE OF
GRAVITY, AND OF ROTATION ABOUT THE CENTRE OF GRAVITY, OF A BODY
OR MATERIAL SYSTEM, ACTED UPON BY FINITE FORCES.
69. The co-ordinates ºn, y, of the centre of gravity of the body or system are defined by the
equations— - -
Xma = Xma, Xmy = 2my,
()]" Wv = Xma, Wy = Xmy,
W or Xm lb denoting the whole weight of the body.
Differentiating with respect to t,
da, de d/ dy
> - - - W Tº 9 t!/ = W a/ &
” di dt >mº alſ'
d” – w dºc d”y d°y
Xm 7/2 T W d?” >m. - Wºź.
"I'l, r. v. A £2. al”, p. —º (ſº
Therefore, Wºº. – SX, W ... = XX . . . . . . . . (A).
or the C.G. (Centre of gravity) moves like a particle, at which the whole weight W Ib is collected,
under the action of all forces applied parallel to themselves at the C.G.; or in other words, all the
weight and all the forces may be supposed collected at the C.G., and the motion of the C.G. will
be unaltered; so that so far as the motion of translation of the C.G. of a body is concerned, the
body may be considered as a particle concentrated at its C.G. -
We are now able to restate Newton's First and Second Laws of Motion.
I. The C.G. of a system perseveres in its state of rest, or of uniform motion in a straight
". except in Bo far as it is made to change that state by forces acting on the system from
without.
II. The change of motion of the C.G. of the system is proportional to all the impressed forces,
and is in the direction of the resultant of these forces, applied at the C.G. 4-
Thus the C.G. of the Solar System is either at rest or moving in a straight line with
uniform velocity, and any change in this motion would be due to the attraction of the stars.
Again, when we investigate the trajectory of a projectile, we follow the path of its C.G.
THE THIRD LAW-Action and Reaction are equal and opposite;
now merely amounts to the definition of a stress; and does not hold at a distance when the stress
is propagated through ponderable matter, which is being accelerated or retarded.

93
EXAMPLES.
(1) Prove that the latus rectum of the Earth's orbit divides the orbit into two parts described
in the ratio of 1 to 1043, or in 0.489 and 0.511 of the year, taking e = 1/60. ”
(2) If the moon had a satellite whose weight is the same fraction of the weight of the moon that
the moon is of the Earth, and whose distance from the moon is the same fraction of the moon's
distance from us, find the period of the satellite.
(3) A spherical shell of small radius describes a circle of radius R with velocity V about a centre
of gravitation at S; and when the shell is at P, it bursts with an explosion which generates
velocity v away from its centre. * - r
Prove that the fragments will all pass through the line SP within a length -
*
8 V8 v R
V4 – 6 W292 +63)
and that if v is small, the stream of fragments will form a complete ring after a time ºr R/v,
approximately. - * ~ -
(4) A stream of meteorites, originally moving in a straight line K with velocity W, comes under
the influence of the attraction of the Earth, whose centre moves with velocity v in a straight line
which intersects the line K, and is inclined to this line at an angle 2.
Prove that, if the distance of the Earth from the line K is originally very large, a length
*R_{v: – 2Wv cos a + v" + 2gR}}
v sin a
of the line of particles will fall upon the Earth. - - . . .
(5) Prove that if a volcano can send a stone to a vertical height R/n, where R denotes the
Earth's radius, then at elevation a the range will be 2R/8, where
cot #8 = } (n − 1) cot a + 3 (n + 1) tan a.
_ – T - * —-
º * ~ + & Vºx :
% 6......., & wº. 2 # * , ** = & Kºx /
_ſ , Yaº
A 4X > : /vae A x Vºx- *a***)
4 ºra, 4 ſºrº, º żº :*
Qº0. -*- ( )
W - * - ſ
F £ - e t-e”— »."2 * = o, o/6 Cy ~, 4 -é6 ~~ Yazº. A cº-c = e, e z-e = e,
a 2 –4 =
g+, Vºz º'e ' y -
4, Z-24 2.9708 – 4:03.3% =G4% ºfte -4--,-
. W. a. 7 * & .6– - /.3°70 y + d > 0 33% 7:572) /, a y 3 -
~ £4–?
(6873) 2 B


3.
T(3) DYNAMICS OF A RIGID BODY.
MOMENT OF INERTIA AND RADIUS OF GYRATION.
64. In problems on the Dynamics of Rotating Rigid Bodies, a quantity called the Moment of
Inertia about the axis of rotation is required.
* *
ON.—The Moment of Inertia of a body about an axis is the s e products of the
weight (in lbs) of each particle of the body and of the square of the distance (in feet) of the
particle from the axis. **
Thus if Oz is the axis, and m lb is the weight of 3/particle at the point w, y, z, then
Xm (a" -- y”) is the M.I. (Moment of Inertia) of the bodyAbout Oz, the summation extending to
all particles of the body. 3 -
If W denotes the total weight of the body in lb, then
W = Xm;
and now if we write •
Xm (w' + y”) = Włº,
then k is a certain length in feet, called the Radius of Gyration of the body about Oz; and we may
suppose the whole weight of the body condensed in a ring at a radial distance k from the
axis O2, without altering its M.I. about Oz.
Take, for instance, a fly wheel of weight W lb, rotating with angular velocity n = d6/dt
about a fixed axis Oz; or a projectile rotating about its axis.
Then a particle of the wheel, weighing m lb, at a distance r feet from the axis Oz, will have
velocity v = rm f/s, and therefore K.E. = }mrºn*/g ft-lb ; so that the total K.E. of the fly wheel is
>}m,”n”/g = }W Kºn”/g ft-Ib, e
the same as that of W lb moving with the velocity of the points on the circle of radius k.
The energy of rotation of a projectile weighing W lb fired with velocity V fls from a d inch
gun, in which the pitch of the rifling is n calibres, will thus be
47.2%2 WW2 , 47%
# *; fºlb, or ;
of the striking energy.
For instance the work stored up in the rotation of a solid cylindrical proof projectile
weighing 2048 lb and fired with velocity 2100 f/s from a 16-inch gun in which the rifling at the
muzzle makes one turn in 30 calibres, is about 345 ft-tons. Compare this with the striking
energy. f
So also in slamming a door; to overcome the work required to turn the spring lock, an equal
amount of work must be stored up in the angular motion of the door, represented by
#WKºn*/g (ft-lb),
In the notation of the Integral Calculus, if w denotes the density at any point, in Ibjft",
then
W/2 =|wº + y”)aladydz,
*.
or in other words, the M.I. about an axis is the space integral of the product of the density and the
square of the distance from the avis. ---
When the body is homogeneous and of volume V ft“, then W = wV, and
a fe
Wk? = | | (a” + y”)dardydz,
the space integral of the square of the distance from the aris.
Similarly for an area. A ft” about an axis O2 perpendicular to its plane
Ak.” = |G: + y”)dardy,
{


t
* * .
º t
*
$4
A w ~ * ''',
9 1.
yº.
l,
* v. ...
--
while about axes Oz, Oy in the plane - J * ...t
Ak.” = |Parly ' ', º
* ~ *
Ak,” = ||w”dady,
the surface integrals of the square of the distance from the awes Oz, Oa, Oy; using k, ky, k, to denote
the Radii of Gyration of the area about the axes Oa, Oy, Oz. -
For a plane lamina, of weight W lb, and uniform or variable superficial density a lb/ft",
W = |oded,
Wk,” = ſ| ory”dardy,
Wk,” = |ſ a'a"dardy,
Wk,” = |a (* + y^)dedy,
the surface integrals of the product of the superficial density and the square of the distance from the azis.
We notice that, for a lamina,
Wk,” = Wik,” + Wk,”,
Ur k” - k.” + k,” º b e te k b & O tº " is e • © tº t (i)
an important Theorem for Moments of Inertia.
65. About an axis in the plane of the lamina, making an angle 6 with Oz,
WAE2 =||a (y cos 6 — a sin 6)*dardy,
= Wk,” cos’ 6 – 2F cos 6 sin 6 + WK,” sin” 6,
where F denotes the quantityſſavard, called the product of inertia of the lamina with respect to
the axes Oa, Oy.
By turning the axes through a suitable angle, F can be made to vanish; and these new axes
are called the principal awes of the lamina at O; and now -
K* = k,” cosº. 6 + k,” sin” 6;
and if the ellipse “
2 *.
= 1
y
+
-*
a;2
7, 2
hy
k
9.
2
is drawn, then the k about any axis through its centre is the length of the perpendicular from the
centre on the parallel tangent; and this ellipse is called the momental ellipse of the lamina at O.
For a line or wire, of variable linear density a lb/ft,
Wł? = |a(r. + y”)ds,
where W =ſ ords, the weight of the rod in lb ; so that WK4 is the line integral of the product of ‘A
the linear density and the square of the distance from the awis, -
We notice that in finding the M.I. of compound bodies round the same axis, we can add or
subtract M.I., but not k”,
96
66. Another Theorem (II) connects Wk,”, the M.I. about any axis O., with Włº, the M.I. about
a parallel axis through the C.G. at a distance h. º *
* For taking the axis Oz, d
Wk,” - Xw(wº + y”),
WK = Xw{(a – vy" + (y – y)*}
= X w(w” + y”)
— 232 wa — 2y2wy
+ WG'+ yº)
= Wk,” – Whº,
since Xma: = War, Xmy = Wy;
and therefore Wł,” – W(k" + h^)
Or # = P + k . . . . . . . . . . . . . (II)
The value of k” need therefore be calculated only for axes through the C.G. of a body.
EXAMPLES.
For a straight rod OA, of length l, in Oa,
W = |ad, War = |ards, Wk? = ſard.
0. 0 J 0
If the rod is of uniform density, we replace W by l and a by unity ; and
_ ſ! -
is = ºrds = }!", a = }l;
()
le=|ra. = P, e = p.
0. *
About an axis through the middle point of the rod,
§
-- \ lk” = ſº = +!?, k” = +ºl’.
/ A 2-V -
Similarly for a rectangle, of sides a and b feet, about axes Oa, Oy through its centre O parallel
__ to the sides a, b (fig. 28),
treating the rectangle as an assemblage of lines of length b and breadth da ; and by symmetry
k.” = #0°. /* rº
About Oz, perpendicular to the plane of the rectangle,
k.” = k,” + k,” = +3(a” + b%) = +), (diagonal)*,
by Theorem I. 2/ 2 1
About the edge a of the rectangle k = 35°; and about the edge b, k = 3a'; so that about
an axis through a corner of the rectangle, perpendicular to its plane,
k” = }(a” + b”) = } Gagº
By superposing a number of such rectangles we build up a book, or brick-shaped right solid;
and denoting its height by c, then, about an axis through the centre, parallel to the edges c,
k” = 4; (a” + b%);
and about an edge c k” = } (a” + b”).
...”
... with similar expressions for k" about the other edges and principal axes.
Cut the solid in two by a plane CD through two opposite edges c (fig. 29); then the M.I.
| about Oz of each half is halved, while its volume is also halved; so that k,” is unaltered, and
k” = + (a” + b%),
for either half, a right prism on a right-angled triangular base.

* 97
Changing to the parallel axis through G, the C.G. of the triangular prism,
k = A, (a + bº) – OG'.
– H —G ºr (a” + b%) — sº (a” + bº)
= +'s (a” + bº),
Changing to the axis coinciding with the edge through E,
k, = 1s (a” + b%) + GE”
= lºs (a” + b”) + + (a” + b”) = # (a” + b”).
Placing two such triangular prisms back to back, we form a triangular prism on an isosceles
base (fig. 30); and now the M.I. being doubled by doubling the volume, the k” about E will be; as
before, d
k.” - # (a? + b°),
where b denotes the height and 2a the base of the isosceles triangle
Changing to the parallel axis through H, the C.G.,
*...* = } (a” + b%) — ºb’ = }a” + +'s bº;
and changing to the edge through C,
( ~~~~º &
A right prism on a regular polygonal base can be divided into a number of such prisms on
isosceles triangles as base; and for each of these prisms, as well as for the whole, about its axis
of figure,
à
22 – lo.2 .2
k = #7 + #a”,
'r denoting the radius of the circle inscribed in the polygon, and 2a the length of a side of the
polygon. &
By keeping r fixed, and making a zero, the polygon is changed into a circle of radius r ; and
therefore, for a right circular cylinder about its axis,
k” = }r”.
*
This is independent of the height of the cylinder; so that making it indefinitely thin, we
obtain k” = }r” for a circular disc of radius r, about an axis through the centre perpendicular to
its plane; and therefore, by Theory I, about a diameter of the circular laminá,
k? =
Thus the work stored up in the rim of a fly wheel making 40 revolutions a minute, the
inner and outer radius of the rim being 9 and 10 feet, and the weight of the rim 10 tons, is
248 ft-tons.
2
}r
The M.I. of the circular bodies can also be determined by integration. y
The k” of a circular line or wire, about an axis through its centre, perpendicular to its plane,
is obviously r*, r denoting the radius of the circle.
About a diameter, the M.I.
^2n:
27f7;k? – | rdó. " sin” 6 = ºrrº; or k” = }r”.
• O
About a tangent line, k” = }r” + 1" = }r”, by Theorem II. *
Considering the circular disc of radius a as built up of a series of concentric circular filaments,
the M.I. about its axis of figure,
Q.
Tra?/c2 =| 27rrdr. r* = }ºra*;
J 0
k* = #a”, as before.
About a diameter, cutting the disc up into filaments perpendicular to the diameter
)
Tra*k” – ſ & 2yda . 49* = sº — w”)}da,
-a, 0
and substituting a = a cos 6,
#m .
Tra*k? = *| sin' 6d6 = 4aºr = }ºra",
0.
k* = }a”, as before.
- * * 2
Similarly for the ellipse . +% = 1, k,” = }b*, k,” = }a”.
(6873) 2 C

98.
Treating the cylinder as built up of a number of coaxial elementary discs, of radius a and
thickness dr, (fig. 31), then the k” about the axis Oa is still #a”; but about the axis Oy through .
the centre perpendicular to the axis of figure, the k” of the elementary disc Pp being #a” + 4*,
by Theorem II., the M.I. of the cylinder (denoting its height by h) -
%
Wk,” = ſ Yºgº + æ”)
gº 4. I 2], 2 .
= }raºh + +"raºh’;
and W = Traºh,
so that k,” = }a” + Tºhº.
For a thin rod, a = 0, and k” = Tºhº, and for a thin disc, h = 0, and k = }a’, as before; the
k” for the cylinder being the sum of these two.
Generally, for any prismatic bar,
k,” = k” + +'s h”,
where k is the radius of gyration of the cross sectional area about the axis in its plane parallel
to Oy. - • - - ,
For a cone, of height h, and radius of base a, about the axis of figure Oz (fig. 32),
h -
Wk,” = |ayas. #y”,
2 0.
* * 0.
and * = }, ~,
a h
a — 1–6"|". 4.
Wk, = 37; a “da = To traºh ;
and *. W = 4traºh,
... SO that - º "k,” F. i”, a”.
About the axis Oy, perpendicular to the axis of figure,
vº – ſayasay : *
* , ... = gºtraºh -- 4traºh."
3 --
º * \, , ~,
2.0 a” + #’.
* *,
. For a sphere about a diameter
3/",
#.
Ve = ſ zyde.
2–
0. & .
*|| (a” — wº)*da.
s' 0
. I dº * .. s *
- *| (a” – 2a3a* + æ")dir
9 - -
=.g
Traft(1 tºº. # + #).– #Ta' ;
and p - sº v. - #7ta”, so that k? --. #a”.
Similarly, for the ellipsoid
Ž
- k. -- +(b% + cº, k,” = £) #(c” + a"), k” = }(a” + b’).
(For other examples, consult Twisden, Chap. V, and Calculus, Chap. II.):-


*9 9
67. Taking any solid body, referred to three rectangular axes Oa, Oy, Oz, then its M.I.
about an axis OP through O, whose direction cosines are l, m, n, will be,
Whº? = ſ |ſ68°dardydz, –7 4'-
2.
/ > 8, the distance of the points w, y, z, from the line OP, is given by
_-
8* = &º + y” + 2* – (la + my + m2)*
("(…}+-6-9.2%
— 2mmy2 — 2nlzac – 21may ;
so that denoting the M.I. of the body about Oa, Oy, Oz by A, B, C, and the products of inertia (P.I.)
with respect to the axis (Oy, Oz) (Oz, Oa) (Oa, Oy) by D, E, F, then
Wk” – At 4. Bm" + Cn” – 2Dmn – 2Enl – 2Flm.
Now draw the quadric surface
Aa' + By” + C2” – 2Dyz – 2Eza – 2Fry = K,
where K denotes a constant.
Then if the axis OP meets this surface in P,
Al” + Bm2 + Cn” – 2Dmn – 2Enl – 2Flm = K/OP”,
so that Wk? = K. OP”,
or k” varies inversely as OP”.
Since k” is obviously always finite, on the supposition that the density of matter is always
positive, this surface is an ellipsoid, called the momental ellipsoid. &
The axes of this ellipsoid are called the principal awes of the body at O; and by taking these
as co-ordinate axes, the products of inertia, D, E, F, are made to vanish; and thus
Wk? = At” + Brm” + Cm”,
Or k* = l?k,” + mºk,” + n°k2°.
As an exercise, construct the momental ellipsoids of the preceding figures.
Principal axes are axes of permanent rotation; for suppose a body is rotating about Oz with
angular velocity r, the centrifugal force of a particle at (a, y, z), weighing m lb, being r” my/(a”-- :
poundals directed away from the axis O2, will have components rºma and r*ny parallel to Oa and
Oy. Af
These components of the centrifugal force have moments
rºmyz and — rºmaz about Oa and Oy,
and therefore the resultant couple of constraint of the
axis Oz, required to keep the body rotating about oxwill have components 7 PC,
—T &m-
X rºmyz = r^D about Oa,
2,
i
/ Tèse couples vanish, and the axis of rotation Oz has no tendency to become displaced, if ///
D ---
= 0 and E = 0; and then Oz is called a principal awis at O. wº
~ Rotating parts of machinery should therefore be so shaped that the axis of rotation is a
principal axis of the rotating piece.
The superior steadiness of running of the inside cylinder locomotive over the outside cylinder
type is to be attributed to the nearer coincidence of the driving axle with a principal axis of the
system

* *
100
DALEMBERT's PRINCIPLE.
68. This principle is the application of the Third Law of Motion, 2
* “Action and Reaction are equal and opposite ” P.
to the formation of the analytical equations of the motion of a Material System.
First consider the plane motion of a body considered as an assemblage of particles; and let
r, y, denote the co-ordinates of a particle, m lb its weight, X, Y, poundals the component
impressed forces, and P, Q poundals the component internal molecular forces, due to the
reaction and constraint of the adjacent particles of the body.
Then the equations of motion of the particle are (§ 52).
dža.
* if: = X + P,
d”y
m; = Y + Q.
§O that wV W. Wu .V. V. Clºv wº
d? d? Åse
p(-; — y #) = (a:Y — yx) + (20 — yº).
;
Denoting summation throughout the body by X, then
! 3. snº = xx + XP,
t xm. = xy + x9.
; : A
f º f Xm{a, º gº yº) = X(w) — yx
E. gº + X(a.Q — yP).
| 3. . * , * ..., s
! sº Principle, or rather by the Third Law of Motion, the system of forcés
represent;d by P, Q form a system in equilibrium, since to any one force of them there always
corresponds an equal reaction or force.
Therefore, by the Principles of Statics, .
1 * , . .
#. §
#! t xP = 0, XQ = 0,
and X(a:Q — yP) = 0;
expressing the fact that the sum of the components in any direction is zero, and the sum of the
‘moments about any axis is zero, when the system of forces is in equilibrium.
Therefore, for the motion of the body or system,
dža. ... -
*~ * * > * = xx • * * * * & e s º ºſ e º s (1)
d w
>m; = &Y . . . . . . . . . . . . . .(?)
2. ..
*(; sº yº) = X(w) — a X) . . . . . . . . . (3)
*
dº? dº?
the three equations of motion, deduced from D'Alembert's Principle.
101
/
If we had considered motion in space generally, referred to three rectangular co-ordinates
axes Oa, Oy, Oz, we should obtain in exactly the same manner the six equations of
motion— & A
m; = XX,
mº = xy. ‘. .
sm; = XZ,
*(; tººs #) = X(9Z – “Y) = T, J
*( 4. – ...) - sex-º- 41,
*(;- yº) ==(-Y-29– –Zv, *-*...
Forces such as X, Y, are called impressed forces, and such as P, Q, molecular or cohesive
forces, while the name effective forces is given to the forces of inertia of the particles, such as
d? • * * te *
%, #. 770, # poundals; and now D'Alembert's Principle can be stated concisely in words as
“the impressed forces and the reversed effective forces from a system in equilibrium ; the molecular

or cohesive forces forcing a system in equilibrium among themselves;”
and thus by D'Alembert's Principle, all Dynamical theorems can be stated in a Statical form.
> → These equations of motion can be simplified by the introduction of the motion of the Centre of
Gravity or of Inertia.
(6873) 2 D
102.
INDEPENDENCE OF THE MOTION OF -TRANSLATION OF THE CENTRE bF
GRAVITY AND OF ROTATION ABOUT THE CENTRE OF GRAVITY, OF A BODY
OR MATERIAI, SYSTEM, ACTED UPON BY FINITE FORCES. .
69. The co-ordinates z, y, of the centre of gravity of the body or system are defined by the
equations— --- -*.
Xma = Xma. Xmy = Xmy,
‘Or Wr = Xma, Wy = Xmy,
W or Xm lb denoting the whole weight of the body.
Differentiating with respect to t,
da: dº dy dy.
— = W. “ *— = W “& ;
>m. dt' Sm"; dº ’
d’a: dºc d2 dºy
—7–- F WW –F-53 * Z – W. Z .
Xm # = W. *-i. dº?
:- :- ( *
Therefore, wº. – XX, W # = XY . . . . . . . . (A.)
or the C.G. (centre of gravity) moves like a particle, at which the whole weight W lb is collected,
under the action of all forces applied parallel to themselves at the C.G.; or in other words, all the
weight and all the forces may be supposed collected at the C.G., and the motion of the C.G. will
be unaltered; so that, so far as the motion of translation of the C.G. of a body is concerned, the
body may be considered as a particle concentrated at its C.G. i
We are now able to restate Newton's First and Second Laws of Motion- -->
I. The C.G. of a system perseveres in its state of rest, or of uniform motion in a straight
line, except in so far as it is made to change that state by forces acting on the system from
without.
II. The change of motion of the C.G. of the system is proportional to all the impressed forces.
and is in the direction of the resultant of these forces, applied at the C.G.
Thus the C.G. of the Solar System is either at rest or moving in a straight line with
uniform velocity, and any change in this motion would be due to the attraction of the stars.
Again, when we investigate the trajectory of a projectile, we follow the path of its C.G.
THE THIRD [..A.W.-Action and Reaction are equal and opposite;
now merely amounts to the definition of a stress; and does not hold at a distance when the stress
is propagated through ponderable matter, which is being accelerated or retarded.
Now, if we change from our fixed origin Q to a moving origin G, the C.G. of the body or
system/by taking up our position at G and investigating the relative surrounding motion, we
put
* = r + w', y = y + y’;
da da: , da' dy dy + dy'
dº T di " dº de T di " dº ’
2 2. 22, ' , 72 2. / 2.
dº dº I dº dº – dº I dº', / /
dº º dº. " dº dº T. J. " dº ’
and then
so that equation (3) becomes
xmſ (2 + 2)(*, *, *y) – (n + y^(8* * *
{ (; +% (y | W)(; +%
= XH (a + æ')Y — (y 4 y')x},
=dºy , s.a.dº' , dºy dºy'
©r Xma. dº? + Xma: dº?" + Xma: dº? + Xma.' dº
—dºr — dºc' dža. , dºz/
= X-Y + Xz'Y — Xyx — Xy'X.
163
Now, from the definition of the C.G.
- -- 2 2./
Xma or Xm(a + æ") = Wa, so that Xma' = 0, Xm # = 0,
- * -- 22,’
Xmy, or Xm(y -- y’) = Wy, so that Xmy' = 0, xm. = 0.
Therefore, ma. # = Wa: º and this cancels with Smir Y or 2×my;
– d2 / * dºy' *
smº. - #3m, = 0 ;
– 22- –22- * - - – -- .
Xmy #. = Wy º which cancels with Xy X, or ySX;
—dºz' dža'
>my. = y2m d; T 0;
, dºr dºa. * — ſh
Smy"; H . Smy' = 0;
‘and we are left with
( , d”y' , d’a'
sm(, , -yº) = x(x,y-yx) • * * * * * * * * (B.)
an equation of the same form as (3), with the moving point G for origin, instead of the fixed
oint O. * ~
P This proves that we may take moments about the C.G. as if it were a fixed pointſor that
the motion of a system relative to its C.G. is independent of the motion of the C.G., and therefore
the same as if the C.G. were a fixed point.
These two principles, expressed by (A) and (B), constitute the principle of the Conserva-
tion of the Motion of the C.G. of a System; they are employed in nearly all dynamical
problems.
70. In reading the ordinary treatises on Dynamics, it will be noticed that the letters m and M
are employed, and the quantities they denote are called masses, instead of the letter W
employed here, denoting quantities called weights; but it is important to notice that if measured in
lb or g, m M, or W denote the same quantity, and are numerically the same.
Our object here is to avoid the use of the confusing abbreviations of M for W/g ; we call the
quantity of matter in a body, as measured in lb or g by the operation of weighing, the
weight of the body; and it is unnecessary to call this quantity the mass, and to denote it by a
different letter M.
W.
Q
104
ExAMPLES.
..") A person being placed on a sheet of perfectly smooth ice, explain how he may reach the
bank. -
2) A man walks from one end of a boat to the other; determine how far the boat has moved,
and determine the motion of the boat when the man jumps out.
(3) Explain how, sitting in a chair, a man can advance by a series of jerks. *
(4) Prove that when a shell bursts, the C.G. of the fragments continues to describe the former
parabolic path.
(5) A rod, revolving about an axis, suddenly snaps at a given point; determine the subsequent
motion of the parts, the motion taking place in a horizontal plane.
(6) A man steps on to a turntable and walks round the rim with given velocity; determine the
motion.
(7) Find the angle at which the rod of Watt's governor will stand permanently with the
vertical, when making a given number of revolutions about the vertical spindle.
(8) Show how a stick may be balanced on the finger, at a constant inclination to the vertical,
by making the finger describe a horizontal circle with given velocity.
(9) Explain the difficulty of walking rapidly on a curve on th ice, and show how it is
ible-for a man in skating to describe a circle on one foot. KA skater, whose weight is 12 stone,
Gºon the outside edge a circle-of-10-feet-radius with unifermly decreasing velocity, just coming
to rest after completing the circle in 5 seconds; find the direction and magnitude, when he is
half way round, of his pressure on the ice.
(10) A particle is moving round a circular tube, which is placed on a smooth horizontal plane,
Prove that if the tube is initially at rest, its centre will proceed to describe a cycloid.


105
MOTION OF A RIGID BODY ABOUT A FIXED AXIS:
71. If the fixed axis is the axis of a perpendicular to the plane of the paper through O, and if
we employ polar co-ordinates, * * * , -,
- - w = r cos 6, y = r sin 9;
differentiating with respect to t, r being constant for a particle of a rigid body,
da: = — r sin 62%,
dt * dt’,
* = r cos 6 d6, " .
dt dt
dža. . ... a dº ..., a dº
dº = — r COS 9. }* SII). º
d”y . . a d6% d°6 .
Tº T —r sin 6: + rcos 0%;
g dy , de , dē
and therefore, * j, - 9 i = "i,
4. - º t ----------- - - ac d”y — dºw = ?” dºg -
é ſ? – T *--- is T 9 iſ: d? "
ſ § D Therefore, equation (3)/becomes
r - sm,” – N,
dt?
Assº- when N denotes X w(x Y — y X), the sum of the moments of the impressed forces about the axis Oz.
2
Now in a rigid body, the angular velocity # and the angular acceleration:
IF are the same
for all particles of the body, when rotating about a fixed axis; so that
d26 , , ,d26
Xmr” dº T W12 dº?”
where Wk” denotes the M.L. of the body about the axis Oz; and thus the angular acceleration.
d°6 N moment of forces about the axis
dº? Whº? 3 * inertia º
25 33
The moment of inertia thus represents mechanically the angular inertia or resistance of inertia
to angular motion, as experienced for instance in spinning a wheel or projectile, or opening and
shutting a door, while the weight of a body represents the ris inertiae of its C.G. to motion in a
straight line.
(6873)
.i
2 E
106
. - ATWOOD'S MACHINE.
72. We are now prepared to make a complete investigation of the theory of this machine,
when the angular inertia of the pulley and the friction of the axle are taken into account. *
Two equal weights of M lb are suspended by a fine thread over a pulley, weighing W lb ;
and equilibrium is destroyed by the addition of a small rider weight of m lb. -
Denoting by a ft the distance the weights have moved in t seconds, and 6 radians the angle
the wheel has turned through; then 6 = aſa, a ft denoting the radius of the wheel, or rather of the
groove in which the thread runs. -
ing the motion the wheel will ride up, and bear at a point P, at an angular distance q, from
the lowest point & of the bearing, q, denoting the angle of friction (fig. 33).
The equation of motion of the wheel will be, employing the gravitation unit of force,
Wk;2 d20 r . .
Tº # = (T-T), - (T + T +W) in . . . . . . . (1)
p ft denoting the radius of the axle.
The equations of motion of the descending and ascending weights will be
M + m d’a: !-- -- s
–7– d? F M + 772. — Ti * o o e gº g * > *º • tº º " (2)
M dº * *
; # = T, - M Q { } e e e te o ſº ſº * Qe * > ſº (3)
Eliminating T, and T,
W1.2 * @ = T, º T, tºº (T, + T, + w); sin ºp
gaſ dº
º 2M + m d’a:
-(2M + m, –? &#4-w)? sin q,
g di” a ~ *
dža, m – (2M + m + W) sin &
Or * T ºn ma–Pºnd) I wº
(1–5 ºn 4 + Wii,
If w lb is the weight of the rider which will just start the motion,
P si = 2D * F
sin & 2M + w -- W '
d’a g 700 – QU * •
and dº? T r 2M + w -- W k2
2M + m + w -- W — 2M + W. w( – )
Thus if the groove in the pulley is cut down to a depth such that a = },
d’a 7?? — Q0)
dº? T 93 MI'm Hºw I W’
Suppose the rider m is detached when it has descended h ft, and acquired velocity v f/s;
th 1779 = 200 — QU ... •
€0. $v *3MT. H. Hºw:
and now the retardation f" is given b º 25a- O 1-0 */~4.
f' is gi y A-4: 3– * + ) .**
, (2M + W) sing
and the weights come to rest in h" ft, where
h' = }*/f/



107
* EXAMPLES.
Q Prove that the horse power consumed in keeping a flywheel weighing W tons rotating
with R. revolutions a minute is 27trºW b sin q, x 2240 + 33000, the radius of the axle being b ft.
Prove that if left to itself, the wheel will come to rest after
27F/32R,
3600gb sin ºp
Tº R2
3600gb sin (p
revolutions, in minutes,
& denoting the radius of gyration of the wheel and shaft.
(2) A uniform circular disc, whose radius is 4, inches and weight 2 lbs, can turn about a .
horizontal axis through its centre at right angles to its plane. When started at the rate of 200
revolutions per minute, and left to itself, it is observed to come to rest in one minute.
Prove that the retarding couple (supposed constant) has the same moment as a weight of
25% grains attached to the rim at the extremity of a horizontal radius. -
Also find how many revolutions it makes,
(3) Supposing the driving wheels of an engine are slipping with a given number of revolu-
tions per secor steam is shut off, determine the number of revolutions the wheels will make
before slipping ceases, given tº ... e ls (and reciprocating parts), the load on the
wheels, And the adhesion with the rails.
(4) An a machine without friction and inertia, of mechani
supports n times its weight, both hanging by vertical threads
replaced by P lb and W lb, their vertical accelerations will be
al advantage n, so that a weight
Prove that\if these weights are
m” P – ºn W d mP — W
º PTW 9 and ºp TW 9,
in opposite directions; but their C.G. will descend with acceleration
(nP – W)?
(P + W.) (n°P + W) 9.
(5) Prove that in Weston's differential pulley, if friction is neglected, the lower block and
weight weighing W lb, will descend with acceleration
W (a — b)” – 2a P (a — b) Aº
W (a — b)* + 4, a”P + 4 (mk” + MK*) §ſ,
raising a weight P. lb on the free part of the chain, a and b denoting the radii of the upper pullies,
and MK*, mk”, the M.I.'s of the upper and lower pullies.
Prove that the radius of the axle, so that, with given angle of friction p, the machine will
require overhauling to make the weight descend, is 4(a — b) cosec p ; and determine the me-
chanical efficiency.
(6) Determine the acceleration of a system of pullies when the weights are unbalanced,
taking into account the inertia of the pullies. F
Prove that the directions of motion of the parts of a rope round a pulley are at any instant
tangents to a conic.




108
73. By Theorem (B) we still have the angular acceleration equal to the moment of the forces
divided by the moment of inertia about an axis, when the axis passes through the C.G. of the body
and moves with the body, instead of being fixed in space. -
Consider, for instance, the motion of a cylinder rolling down an inclined plane ; and suppose
the body has advanced a ft, and turned through 6 radians in t seconds(fig. 34). - - ºr
Then 6 = w/a, a denoting the radius of the cylinder. . . . . . . . . . . . …— l IZ:7
Supposing the cylinder to weigh W lb, and denoting by R and Öſpoundals the normal / “ *.
reaction and frictional constraint of the inclined plane, the equations of motion are
, , , , , ' '
W. - We sin a – F . . . . . . . . . . . (1).
0 = R — Wycos. s . . . . . . . . . . (2).
d”9
dº?
Wk? F. . . . . . . . . . . . . . . . (8).
Equation (2) determines R; while eliminating F between (1) and (3); ' '.
d’a: iº lº” d20
W . * = Wa sin & — W " " ".
; dt” g ' ' ' a dº?’ ſ
2. in .
or dº = .9 °º
- diº 1 + kºſa”
the acceleration down the plane.
In a hoop k* = a”, and the acceleration is #g sin a ;
in a cylinder k* = }a”, 3 y 99 #g sin a ;
in a sphere ** = #a”, 33 33 #g sin a.
49
Suppose for instance we were asked to distinguish between two spheres of the same diameter
*. #. ...” one a hollow sphere of gold, and the other a solid brass sphere, but gilt so as to look
lke gold. - - -
. The hollow sphere, having the greater kº, wonld roll down an incline slower than the solid
sphere, and could thus be distinguished. * *
li jº also a cask will roll differently according as it is empty, or filled solid, or filled with a
10 uld. • * * t t - ;
We notice that if a cylinder rolling on a plane is urged by a forceº poundals through its
centre parallel to the plane, then from the equations of motion, ' ' ' - . . . .
w dº . . ,
wº-x-5. y
d?6
WK2 ° º –
... = Fa,
while dº a d’é
d; T “dº”
we obtain X = W(1 + & dº,
- x a? / dt?
SO *gº: º or vis inertia is changed to W(1 + kºſa”).
*.*, *, *@99, the vis inertia is twice the weight, for a cylinder is 14 tim - has
(e.g., # billiard ball) is 13 of the weight (40°/, grea: 2 y # times, and for a sphere
stººdºº
…” 2 to COme linto Contact given there Wi vestºry irl; a , $
and will thus be increased 18.3 per cent. nere wi we to be multiplied by v/(1 + kºſa”) orv/#,
So also it is shown in Hydrodynamics that the*... a sphere moving in water is
increased by half the Weight of water displaced ; an #. rlia e i. s.
axis by the weight of water displaced. p 3 and ºf a cy linder moving perpendicular to its



{ { 109
For a carriage or train, weighing M lb, mounted on wheels weighing W lb, the effective
inertia or vis inertiae will be j
M + W(1 + kºſa”);
so that, if unchecked, it will run down an incline 2, with acceleration
M + W.
MTwº Tºº,” “
If the road is crushed uniformly by the wheels, the acceleration will be found to be
(M + W) sin (2 – 8) -
(MTW) cos & I Włºſº. 9
where 8 denotes the slope at which the carriage will just run down.
In the preceding collection of examples on the motion of trains, the effect of the rotary
inertia of the wheels may be allowed for by adding X Wkºſa” to the gross weight of the train.
A good average value of kºſa” is ; ; so that such a slight correction as the addition of about
half the weight of the wheels and axles to the gross weight of the train is sufficient to allow for
rotary inertia.
The proper height of the buffer and coupling chains of a railway carriage is affected by the
rotary inertia of the wheel; this should be at the level of the centre of effective inertia of the
carriage, composed of the body weighing M lb, with C.G. at a height h ft suppose above the
axles, and of the effective inertia W (1 + kºſa”) of the wheels, acting at their centres.
The buffer should therefore be at a height
Mh
M + W. (1 + kºſa”)
above the axles; and then the whole carriage will start off without pitching oscillations,
A photograph of a bridge accident in France, given in the Engineering News, 29 Sept., 1892,
shows the effect of buffers placed too low.
* EXAMPLES.
(1) Determine the motion of the system represented in Fig. 35, showing an arrangement
y 74.4°
devised for attaining great velocity on a rope traction railway; and illustrate experimentally by a
reel of thread on a table. 2 ex-
(2) A cylinder weighing M lb, and of M.I. Mºſests withits axis horizontal on two parallel
rails at an inclination a, and supports a weight W 15 hanging vertically by a thread wrapped on
the cylinder.
Determine the motion and the condition of equilibrium.
(3) Explain how a man can make a cylinder or sphere roll up an incline by walking on the
surface; and show how to make it roll up with constant velocity.
(4) Determine the motion of the cylinder in § 73 when the inclined plane is placed on a
railway truck.
(5) Prove that the angular acceleration of a wheel of radius a, rotating about a horizontal
axis, must be
5 a .
* 9 sin 6
O.
for a sphere in contact with the rim to remain with its centre at rest at an angular distance 6
from the highest or lowest point.
(6) If the force of restitution of a buffer spring is always e times the force of compression,
discuss the impact of two railway carriages of effective inertias M, and M, ; and prove that the
times of compression and restitution are T and T/v/e, where sº
T = - V(; + ...)(; + ...)
wher-6) and E, denote the forces required to produce compression in the buffer springs.
/*
(6873) 2 F

110
THE COMPOUND PENDULUM, AND THE EQUIVALENT SIMPLE PENDULUM.
74. Let OP be the pendulum, G its C.G., swinging about the horizontal axis through O, and at
time t seconds inclined at angle 6 radians to the vertical OA (fig. 36). ſ
Let the weight of the pendulum be W lb, and let OG = h, and Wk” (lb-ft”) the M.I. about
the axis through G, parallel to the axis through O, so that W (h” + k”) is the M.I. about the axis
of suspension.
Then the impressed forces being Wg poundals acting vertically through G, and the unknown
reaction of the axis, the equation of motion obtained by taking moments round O is
d”6 moment of forces
dt? , , inertia
– \\gh sin 6
W(ſº + k2) '
º
(the negative sign being introduced, because the moment of the impressed forces tends to decrease 6)
– – " " " — — "si
* =-jºº, = -}sin 0.
on writing l for h + kºſh.
This is the equation of oscillation of a small plummet at the end of a light thread of length
! feet; and l is therefore called the length of the equivalent simple pendulum, and if we reach
2L. OP = l, the P is called the centre of oscillation corresponding to the centre of suspension O.
& Thus a peal of bells should be hung so that all have the same l or length of equivalent simple
pendulum.
Since º OP = h -- Kºſh,
therefore º GP = }*/h,
Or OG. GP = #2;
and the symmetry of this relation proves that O is the centre of oscillation when P is made the
centre of suspension, or that “the centres of suspension and oscillation are convertible" (fig. 37).
When the angle of oscillation is small, we may replace sin 6 by its c.m. 6 radians; and now
the equation of oscillation is
** – – g – — ,
jº 7 - m*6,
on putting g/l = n”, when n is called the speed of the pendulum.
Comparing this with the equation of S.H.M. (§ 42),
d’a,
Olt?
= — mºv,
th solution of which is
A cos nt, or B sin nt, or A cos ut + B sin nt, or a cos (nf + e), or a cos m(f – T),
a being called the amplitude, e the epoch, and T the phase; we may write the pendulum oscilla-
tions as given by
6 = & cos (nt + e),
when the amplitude a is the extreme angle of deviation from the vertical.
The period of a complete double oscillation, when small, will therefore be
2T/n - 27tv/(l/g) ;
_- -
tº time of a single swing from rest to rest being TV/(l/g).
-Thus for a seconds pendulum, beating seconds,
I = Trv/(l/g),
, Of l = g|T", g = T’l,
• º <–T -
whe m e found very accurately, if l is known by experiment.
*—’
W
a'
ſ
N Ö) ©
Nºr-
Nº \!
NºnV
N
N
te
_P3 -
–
C / .
_AW / \
* /t-T’) \ |
(? AZ A 32 | E --
`A.
Fig. /2. Fig. //.
Uſ A U. C., U.
Fig. /4.
40 4, 42 A3 A4
Fig. /3. *—
3×30, 7, 92




3.530, 7, 92.

J&3 O, 7, 92
{,
*,
!,
... }
'si
* *
#|
$
i.

ºr
Aº
1.4/.
2
().
A.
( § ©
i. Fig. 5.
( Š% ')
NS 9)
2 H.
/ |
A. |
/
33'30, 7, 92


111
The period of oscillation is a minimum when l is a minimum, by variation of h; and since
};2
= h -- “.
l * + j, ’
* = 1 – “.
dh h2
which vanishes when h = k, and then *
dºl 2% J
ſº - ja ,
so that l is a minimum 2k, when h = k.
Otherwise, we may write
k \?
! = * + (v1-3)
\
so that l is greater than 2k, except when h = k. * *
We can make l very great, and the period very long, either by increasing h, or diminishing
h indefinitely, that is, by placing the centre of suspension close to the C.G. (as in a metronome,
or a ship, where the metacentre may be considered the centre of suspension).
75. When, the axis of suspension of the pendulum is horizontal, and cut into a smooth screw
of pitch p, the equation of energy gives *
#W(h” + 3 + p?) (d6/dt)* = Wg(H — h vers 0),
if the centre of gravity descends from a height H above its lowest position; so that—
(* + k + p^ (dºode) = – gh sin 0.
and therefore ! = h -- (ſº + p^)/h;
and now in addition to X and Y, the reaction of the axis exerts a horizontal longitudinal component
Z and a couple p7, given by—
* = a Piń º żºłº Fº
when expressed in gravitation measure.
Similarly the increase in l due to the pendulum being supported on friction wheels may be
investigated.
EXAMPLES,
(1) A circular hoop is freely suspended by a point in its rim ; show that the periods of its
pendular vibrations in its own plane and at right angles to its plane are as 2 to v/3.
(2) Determine the number of oscillations made in a day of 24 hours by a pendulum consisting
of a thin rod and a sphere 6 inches in diameter, the centre of the sphere being 4 ft from the point
of suspension, and the weight of the rod one half the weight of the sphere.
(3) Twisden, Practical Mechanics, exs. 812–820.
(4) Investigate the small oscillations of a system of clockwork, in which the wheels are
unbalanced about the axes, and prove that for small oscillations the length of the simple equiva-
lent pendulum is given by—
l = (Xwkºp”)/(X whp” cos 2), -
*- the weight, whithe moment, and whº the moment of inertia of a wheel about its
axis; a denoting the angle which the plane through the axis and centre of gravity makes with the
vertical in the position of equilibrium ; and p denoting the velocity ratio of the wheel.
Discuss the dynamical theory of a clock or watch. - -
(5) A pendulum, while undisturbed has a period T seconds, is deflected by a mountain
thrºugh an angle 6, and its period is changed to Tº'seconds; prove that the acceleration due to
the gravitation of the mountain is
'MºR)
__--"T"

112
FINITE OSCILLATIONS OF THE PENDULUM, ExPRESSED BY ELLIPTIC
FUNCTIONS.
76. When the pendulum swings through a considerable angle 2 a., we must resume the con-
sideration of the equation *
* – – 'sin a – – assin a.
Ž T fine- m” sin 6;
and now the complete solution of this equation introduces the Elliptic Functions into Dynamics.
Multiplying both sides by d6/dt, and integrating,
d6°
* — - tº º * vers.
* dº? C — m” vers 6
= n” (vers & — vers 6),
since d6/dt = 0, when 6 = 2;
or, since vers 6 = 2 sin” #6,
d6 e ©
iſ + 2nv/(sin” # 2 — sin” #6)
6 z/L
mt = & d:#6 * - - - - -
ov/(sin” # 2 — sin”; 6)
We now introduce a new angle q, defined by the equation
sin #6 = sin #2 sin ºp;
OT vers 6 = } vers a vers 26;
and thus b will denote the angle ADQ, where AQD is the circle on AD as diameter, touching
BB' at D, and cutting the horizontal line .N in Q (fig. 41)

For in the circle AP, –cy
* Z
A N = , AE vers 6,
and in the circle AQ,
AN = # AD vers 24 = 4 AE vers a vers 2%.
Then v/(sin” #2 – sin” #6) = sin #2 cos (b,
while #6 = sin-"(sin #2 sin (p),
dº in a cost .
dip V (1 — sin” #2 sin” (b)
Ö dºp
so that t = & == .
7? |va — sin” #2 sin” (b)
Denoting by T seconds the period of the pendulum, when swinging through a finite angle
22, and denoting the integral -
ſ dº/Ad by K,
• 6
(nº the complete elliptic integral of the first kind, with respect to the modulus k), then, when
t = }T, b = }T ; so that #m'T = K,
OT e o - T = 4K/n = 4Kv/(l/g).
113
When the oscillations are indefinitely small, we make a and therefore a zero, and now K = ºr,
and T = 3rvalg).
3.S. before.
But, as a and k increase, so also K increases; so that the period of the pendulum increases
with the angle of oscillation, at first, however, very slowly; for, expanding as far as k”,
K = }ºr (1 + 4* + . . . .);
Y
l
so that, as a correction for amplitude of swing, the period must be increased by }k” or 3 vers a of
2TA/(l/g).
Thus a pendulum which beats seconds when swinging through 6° will lose 11 to 12 beats a
day if made to swing through 8°, and 26 beats a day if it swings through 10°.
This integral for nt is called by Legendre the elliptic integral of the first kind, and, writing k for
sin #a, is denoted by F (ºb, k), while v/(1 — k” sin” b) is denoted by Aq, ; thus
F (p, k) = | *A* = ſ "(I – esinº)-dº
and q is called the amplitude, and k the modulus.
Thus, in pendulum oscillations
mt = F (b, k).
But it is convenient to represent the amplitude q, as a function of nt, instead of nt as a
function of 4 ; Jacobi proposed the notation
q = am nt, or am (nt, k),
when it is required to put the modulus k in evidence; and further in Jacobi's notation
cos p = cos am nt, sin b = sin am nt, Aq = Aam nt,
the three elliptic functions.
Jacobi's notation is rather lengthy, so now-a-days it is abbreviated, and we write cu, sn, dn
for cosam, sinam, Aam ; thus
cos b = cn nt, sin q = Sn nt, Aq = dm nt,
the three elliptic functions of nt.
According to this definition
*A, -: ſ ""(1 – “sin” by-dº;
0
yet =
|
d
b = d am nt
that -
SO Lºla, dt dt
– nAq, = n dn nt.
Since k = sin #2, therefore
c = AD/AB = AB/AE,
* = AD/AE.
. ... We denote cos #4 by k', and call it the complementary modulus; and now, if 0 denotes the
inclination of the pendulum to the vertical after t seconds from the vertical position, and a denotes
the extreme inclination,
sin #6 = k sin q = k sn nt;
cos #6 = Aq = do nt;
tan #6 = k sn nt/dn nt = tan ; a cn (K — nt);
d6/dt = 2nk cm nt;
AP = AE sin #6 = AB sn nt;
PE = AE cos #6 = AE dm nt;
AN = AD sn” nt, ND = AD cn” nt, NE = AE dm3 nt;
NQ = V/(AN. ND) = AD sn nt cn nt;
NP = A/(AN. NE) = AB Sn mt dm nt;
giving all these quantities as elliptic functions of nt, #2 being the modular angle.
(6873) 2 G

{ •w- 1.1%
º In small oscillations -> k = 0, and then
sn nt = sin nt, cn nt = cos nt, dn nt = 1;
so that the elliptic functions degenerate into circular functions, and the period is 2T/n.
With k = 1, a = ºr, and the pendulum just reaches the highest position; and now
sn nt = tanh nt, cn nt = Sech nt, dn nt = sech nt,
so that the elliptic functions degenerate into hyperbolic functions, and the period becomes infinite.
But with intermediate values of a and k, as required with the Electro-Ballistic Pendulum for
recording small intervals of time, the Elliptic Functions are required in the graduation of the
instrument.
**
...A
1.15
THE REACTION OF THE Axis OF SUSPENSION OF A PENDULUM.

77. It is important to know the magni f his rºgction in the case of a large swinging
body, like a bell in a church towerSº a ~4--4--- ºº:: º
Denote by x and Y the horizóñtāTālū vertical components–of–this reaction, considered as
acting on the swinging body; and take the gravitation unit of force, the force of a pound.
Then x, y, and W, applied at the centre of gravity G, will be the dynamical equivalents
of the motion of the body, collected as a particle at G : and since the component accelerations
of G are h(d6/dt)” in the direction GO and h(d”6/dt”) perpendicular to GO, therefore, resolving
horizontally and vertically,
Wh(d°6/dt") cos 6 – Wh(d6/dt)" sin 6 = Xg,
Wh(d.9/de) sin 6 + Wh(d6/dt)’ cos 0 = Yg — Wg; t;
while, from the pendulum motion,
I(d.6/dt) = – g sin 0.
#!”(d6/dt)* = g(2R – l vers 6).
From these equations we find—
h . 4Rh 2h
* – 1 – 2 -º- tº-
W – l 7 sin 6 + F-cos 6 + cos 6(1 — cos º
Y h 2h 4Rh 3h. 2
X 2h 4Rh) . 3h . º
W - (#– *) in 9–4 in ecos 0.
and therefore the resultant of X and Y — W(1 — h/l) is a force
T=wſ—º. 1 #1 * cos) = wº: G + R-9)
in the direction GO ; and T varies as the depth of P below the line—
y = }l + 4R,
*whence X and Y are easily constructed.
*the simple pendulum, h = l, and the tension T of the thread PO is given by—
..ºr .
T 3
W – 7
(#l + 4R — y).
At the end of a swing
y = 2R, and T/W = 1 — 2R/l;
so that, if 2R is less than l, T is always positive; but if 2R is greater than l, so that the plummet
swings through more than 180°, T changes sign, and the thread will become slack, unless replaced
by a light stiff rod.
When 2R is greater than 2', the pendulum makes complete revolutions; and now, at the top
of a revolution,
y = 21, and T/W = 4R/l – 5;
and when 2R is greater than #l, T is again always positive, and the plummet can be whirled
round at the end of a thread, without the thread becoming slack.
1íčº
THE INTERNAL STRESS OF A swinging BODY t --, -,
78. These internal stresses are most forcibly realized on board a ship rolling in the sea, not onl
in their effects as producing sea-sickness, but also in causing the cargo to shift, if not carefully
stowed, or if the cargo is grain, coal, or petroleum, in bulk. (Prof. P. Jenkins, On the Shifting of
Cargoes, “Transactions of the Institute of Nayal Architects,” 1887.) *
It is usual to consider the ship, in its rolling motion, as acted upon by two forces; ,
(i) W tons, the weight or displacement of the ship, acting vertically downwards through the
centre of gravity G; " ~~ jº f * . . . .
(ii) W tons, the buoyancy of the water, acting Vertically upwards through M, the
metacentre $
These two forces form a couple of moment W. G.M. sin 6 (foot-tons), so that the ship will roll
4. 4- t a horizontal longitudinal axis through G, like a pendulum of length GL = }*/GM ft,
tou-4 Wk"/denoting the moment of inertia of the ship about this axis of rotation (fig. 38). *
Now to find the force which acts upg, w, any infinitesimal part of the ship at P, to give it its
acceleration and to balance its weight, we refer the point P to axes Ga, and Gy, drawn upwards
through GM and perpendicular to GM. &s * ,”
This force will balance the reversed effective force of w at P, and the effect of gravity on w;
and therefore, in gravitation measure, will have components—
la (*. Y4 w cos 6, parallel to Ga.,
/ 2 º
3/ º + w sin 6, , ,, Gy.
v -- - , |
If w is suspended as a plummet by a very short thread, the thread will take the direction of
this force, and will therefore make an angle with Gar-
an—g sin 6 – c.(d’6/dt”) – y(d6/dt):
gcos 0 + y(dž0ſdº) – aſdöſdi)?"
•-5 *
Supposing the ship to roll like a pendulum of length l, through an angle 22, then—
l(d°6/dt” = – g sin 6, and #!(d6/dt)* = g(cos 6 — cosa);
and by § 76,
d”6/dt’ = — nº sin 6 = — 2n” sin # 6 cos #6 = – 2n’k sn it dm it,
(d6/dt)* = 2n” (cos 6 — cos a) = 4n” (sin” #a — sin” #6) = 40%"cnºnt.
This force will be the tension and in the direction of a short thread, from which w is suspended
as a plummet at any point P; and the deflection of this plumb line from its original mean direction
in the ship will be a measure of the tendency of a body to slide or of a grain cargo to shift; and
to a certain extent of the tendency to seasickness at this point of the ship and at this instant of its
motion.
At any instant the lines of resultant acceleration reversed will be equiangular spirals, of
radial angle p, round the centre of acceleration G as pole, the resultant acceleration at P being
7"
7
when we put GP = r, and lſd6/dt)* = g sin 6 cot ºb; so that
g; sin 6 cosec p, and the resultant effective force w 4 sin 6 cosec $,
tan b = sn nt dm nt/2k cn” nt.
Superposing the effect of gravity, the resultant lines of force or internal stress will be equi-
angular spirals of the same radial angle p, round a pole J, the position of which is obtained as
follows:–Draw LK perpendicular to GL to meet the horizontal line GK in K; deseribe the circle
in GK as diameter, and draw KJ making an angle GKJ = 4, with GK/ this will meet the
circle in J.

117
For the resultant effective force of w at P, being—
*~,
fu% sin 6 cosec q = a PG
GJ’
making an angle q, with GP, will, when ºpº. with w upwards, and taking the triangle

PGJ turned through an angle q, as the triangle of forces, have a resultant t = w . PJAGJ, making - .
'* an angle q, with JP. !
TThe tendency will clearly have its maximum value at the end of a roll, when døfdt = 0, and
= #7r, and then J coincides with K.
The plumb line at P will now set itself at right angles to JP, while the surface of water in a
tumbler at P will pass through K, and a granular substance at P will begin to slip if KP makes
with its surface an angle greater than the angle of repose of this grain. *
Thus up the mast, at a distance a feet from G, water would be spilt out of a tumbler, or sand
in a box would shift, by the rolling of the ship through an angle 22, which would not spill, or shift
if the ship heeled over steadily, until an inclination 8 (the angle of repose of the sand) was reached,
given by—
} tan 8 = (1 + aſl) tan a.
At the cent? of oscillation L, where a = — l, there is no tendency for the water to spill; and
this shows that the motion of the ship is felt least by going down below as far as possible in the
middle of the ship. '•
In a swing the body is very near the centre of oscillation, so that ordinary swinging is very
little preparation for the motion of a vessel.
A swing to act properly as a preparation for a sea voyage should be constructed as in fig. 38,
to imitate, in full size, the cross section of the ship, suspended at M ; and now the varying effect
of the motion can be experienced by taking up different positions on the deck, up the mast, and
in the cabins, constructed in this swing.
Sir W. Thomson proposes to find the axis of rotation of a ship, and the angle through which
the ship rolls by noting the direction of the plumb lines of two such plummets, suspended at two
given points across the ship; planes through the plummets perpendicular to the plumb lines at the
extreme end of a roll would intersect in J.; the horizontal plane through J would meet the median"
longitudinal plane of the ship in the axis G.; while the plane through J perpendicular to the
median plane would meet it in L, whence GL, the length of the equivalent pendulum, and there-
fore the period of small oscillations could be inferred.
The rolling motion of the ship is lesser the longer the equivalent pendulum l, or the closer
M and G are brought together by º: and arrangement of weights. t

J/
(6873) 2 H
I 18
79. The rolling and pitching oscillations of a carriage are of the same nature, so that, pro-
vided the road is good, easy motion is secured by raising the height of the C.G., as in the old
stage coaches, with the luggage piled on the top; or in walking on stilts, and generally in the
attitude of man as a biped.
Considering the rolling oscillations of a railway engine or carriage, the body may be supposed
to roll about a horizontal axis through O, midway between the tops of the springs, and now if
the body rolls about O through a small angle 6, the moment of the resilience of the springs about
O is (figs. 39, 40)
& $ w(1 + ºx a- w(1–?), a = wº
ft-tons, W denoting the weight of the body in tons, 2a the transverse distance between the
springs, and c the permanent average set of the springs due to the load W tons. *
Denoting by Wk” the M. I. in tons-ft” of the body of the carriage about the horizontal axis
through G, the C.G. of the body, and h the height of G above O, the equation of small oscillations
becomes * * *
W/1, sy d”6 — wa”
;0 + 9); = Who w: 9
* d”6 g
Or dº? T ~ 7 6,
where l = * + k, y
a” — ch Y
the length of the single equivalent pendulum; c being, as already stated, the length of the simple
equivalent pendulum for vertical oscillations. * , "
When a train pulls up quickly at a platform, a sudden jerk is felt the moment after the train
has stopped. This is due to the carriage settling back on its springs, after being inclined in
, consequence of the retardation of the train.
Denoting this retardation by ſland by 2%the longitudinal distance between the two axles,
supposing the carriage has four wheels, then equating the deflecting movement of the retarding
effective force and of the resilience of the springs KT-

Wſh b? *
''J't = W & 9,
g c –4)
so that 6 = chfſbºg, I
the slope of the carriage floor the moment.before stopping being one in 1/6.
, , Ease of motion in vehicles is secured by making l large, or a” — ch small, so that, within
limits, h should be made large or a small. " *
Brunel introduced the broad gauge with the intention of lowering the boiler of the engine
and the body of the carriage between the wheels, thinking to secure steady moving thereby;
but this we see is a fallacy, and broad gauge carriages are now carried over the wheels.
Crampton's low boiler engine is another instance of the same fallacy; and in another direction,
the new-fashioned knapsack of the soldier carried low down on the back.
The higher the knapsack, the slower and easier its swing, the best position being on the
head, where the Roman soldier carried it.
( & &ºe., A47, ºve, v_zzº-º-º-º:
(**~~9 & 4, , , S. . . .ºz. c.4: ... ?" *~…)
119
PHYSICAL APPLICATIONS OF THE PENDULUM,
* \
º
80. Having shown how the motion of the pendulum can be completely determined by the
introduction of the elliptic functions, we may proceed to give a short account of the principal
applications of the pendulum in determining the local value of g, the attraction and density
of the Earth, in preserving a standard of length, and in other important problems in Natural
Philosophy. (Mémoires sur le Pendule, publiés par la Société Francaise de Physique, 1889),
In all such applications the *::: of oscillation is generally sufficiently small for the
period of the pendulum of length l ft to be taken as 27tv/(l/g) seconds; but if necessary, a cor-
rection can be applied for the amplitude of oscillation, in the manner previously explained in § 76.
The next chief practical point is the measurement of the length l of the simple pendulum,
equivalent to a given material compound pendulum, like the pendulum of a clock.
/ This length is determined by finding the position of an axis P, such that the pendulum has
the same perio ether swung about O or P, and thus OP is the length l of the equivalent simple
pendulum (Fig. 37).
Such a pendulum is called Bohnenberger's or Kater's reversible pendulum; a large movable
weight at P is moved about till the times of oscillation about 0 and P are very nearly equal, and
their absolute equality is secured by the movement of a small weight on a screw in OP.
Supposing this small weight is w lb, and that it is placed in OP at a distance a ft from O,
and that the length of the pendulum is changed in consequence from l to l', then
and "... l = 0, if * = 0 or a = l; that is, if w is placed at O or P; and the period is increased or
diminished according as w is raoved below or above P.
But since w is small, l’ – l will have its maximum value approximately when
_ Wkº !' = Wk” + war?
l = Wh’ “ T WAT ºr "
treating the weight was a particle; so that
a:(l − 4) or #!” – (#l — a y” has its maximum v.
*/-, ..?
that is, when w = }l, and the weight w is then
Now, when a = }l,
* = -- a--, -
| Wh + 4 wil Wh’ * * @
. . . . . 4 2 4------, 4-2C
approximately, neglecting ſee Wºº'so that the length of the equivalent pendulum is inereased-
by a fraction of itself, which is equal to a quarter of the ratio of the moments round the axis of
suspension of the added weight and of the pendulum.
Knife edges are employed at 0 and P, rocking on a horizontal steel plate, with a view of
obtaining a fixed axis of suspension with as little friction as possible; this implies that the knife
edges are sharp, but if blunted so that the radius of curvature of the edge is c, the length of the
equivalent simple pendulum becomes *
!' = {(h — c)* + k}/h = l — 26, approximately.




20 *
Bessel's pendulum, constructed by Repsáld (fig. 37, A), is symmetrical in shape, but not in
density. It is swung at the same atmospheric density in its two symmetrical positions, and its
half periods (reduced to infinitely small arcs), t, and t, seconds, are noted. Supposing the
pendulum weighs W lb and displaces W’ lb of air, and that h, and h, ft are the distances of the
knife edges from the C.G. of the pendulum. On the assumption that the effect of the air being
dragged to and fro is simply to increase the angular inertia by an amount dependent on the shape
only, and on the weight of air displaced, say by W’k”, then *
l, gt. W(* + h) + W.8%
Tº TWh, E.W.A.T.) .
~
1-gº-W(eir lº) + wº
* ~ * TW. E.W.I.O.I.T.)
Eliminating Włº and W’k”, ſº. length of the seconds pendulum is
9 = W(h? --> h,”) *--
T* W(h, t,” — he t,”) — Wº #(h, + ha) (t,” — tº)
h, + h,
W^\ h, + h
lſ + 2 2 }l 1 — 2 — ; 2) "I 2-. * *.
#(t,” + t,”) + 1(l WT) (t, t,”) h, — h,
The pendulum is constructed so that ti and t, are very nearly equal, but not h, and h, ; and
then the term
/. W’ h, + h
Al ~ W )& * tº). dº i.
is small, so that h, and h, need not be known with very great accuracy, although h, + h, can be
measurel with great precision as the distance between the knife edges.
81. Taking the formula T = 27tv/(l/g), and denoting by AT, Al, Ag small finite simultaneous
increments of T, l, g; then \ sº
* →
AT - , Al Ag.
= } — #:
T - 2
g
or, in popular language, to increase the period of a pendulum one per cent. requires two per-cent.
increase in the length l, or two per cent. decrease in the value of g. * -
Thus if a clock, beating seconds, loses 9 seconds in a day, the length of the pendulum must
be shortened by 1800/(24 × 60 x 60) = 1/48 per cent., or by 0:008 of an inch if the pendulum is
39:14 inches long. **'. … *.
ts. *r j. s. --
With a view of preserving the standard of length, it was en Z Act of Parliament, 1829,
*.*.*.x hould be 39°1393 inches,
sºvº-
. … Yº 3. **-. º- C
--> Tºw
ºr:
acted b
that the length of a pendulum beating seconds at vel it.
whence the standard yard of 36 inches could be recovered, if" Wºr
When the need arose in 1834 of restoring the standard yard, destroyed by the fire at
the Houses of Parliament, it was found that the difficulty of measuring accurately the distancé
between the knife edges at O and P, on which the pendulum oscillated, was a drawback to the
practical employment of the seconds pendulum as a standard of length.
º w
:*
The length l is susceptible to changes of temperature; and to neutralize these changes, the
gridiron pendulum was devised, consisting of iron and brass or zinc rods expanding in different
directions so as to keep lunchanged; the length of each metal must therefore be inversely pro-
portional to its coefficient of linear expansion. - ***
To keep true time, two clocks may be employed with uncompéisated pendulums, or two
watches, say a Waltham and a Waterbury watch, provided their rates are always in a known
constant ratio ; a small calculation will now give the true time.
For if the two clocks or watches are set originally to true time, and if their rates are always
as m to n, then when the first has gained or lost a time t on the second, the true time will be
mtſ (m—n) behind or in advance of the first, and ntſ (m—n) of the second.
The mercurial pendulum secures the same object by having a cistern of mercury attached to
the bar OP of the pendulum, and the expansion of the mercury can be made to neutrali 49-'
effect of the expansion of the bar OP by regulating the quantity of mercur —the Surface of the 4 –
mercury is near P, the addition or removal of a sensible quantity of mercury will not sensibly
affect the period of the pendulum, by what we have just proved concerning the addition of the
weight w to the pendulum, so that it is thus possible to correct this pendulum for temperature
with great ease.




121
. From the accuracy with which the period of a pendulum cayſ be measured, by counting the
number of vibrations in a long time by the method of coincidence; the pendulum is the most accurate
instrument for comparing the value of g at different parts of the surface of the Earth, and is
universally employed for this purpose.
When we ascend from the Earth at sea level, either by going up a building or a mountain, or
by a balloon. it is generally assumed that gravity diminishes inversely as the square of the distance
from the centre of the Earth; so that one per cent. increase in the distance from the centre will
cause two per cent. diminution in g, and one per cent, increase in the period of the pendulum, or
one per cent. diminution in the number of vibrations in a day.
Thus if we take the Earth's radius as 4,000 miles, and ascend h ft from sea level, g and
the period of a pendulum will diminish 200h/(4,000 × 5,280) and 100h/(4,000:x 5,280) per cent.:
thus if the pendulum beats seconds at sea level, it will lose n beats in 24 hours, given by
ºn/(24 x 60 x 60) = h/(4,000 × 5,280); so that h|n = (4,000 × 5,280) -- 24 × 60 x 60) = 245.
82. By taking a pendulum down a deep mine and observing the change in the period of the
pendulum, a comparison can be made between the mean density of the Earth, and the density of
its crust; such experiments were carried out by Airy at the Harton coal pit, and more recently by
Sterneck in a mine 1,000 metres deep.
We shall employ the C.G.S. system of units in this, and other cosmopolitan problems; and
denote by R the radius of the Earth in centimetres, 10° -- #7 ; by p the mean density of the Earth,
and by a the density of the crust.
We assume the Earth to be spherical, and arranged in concentric spherical strata of equal
density; and now if g denotes the acceleration of gravity at sea level (in C.G.S. spouds) and F.
denotes the weight of the Earth in g— *
= 4tpKº, and g = CE/R* = 4tpCR,
C denoting the gravitation constant, that is, the attraction of gravitation in dynes between two
spheres, each weighing one gramme, when their centres are one centimetre apart (§ 51).
Since the attraction of a spherical shell on any point in its interior in zero, it follows that if we
descend a mine h cm deep, the attraction of the outer crust of the Earth, h cm thick, can be left
out of account; and gº the new value of gravity in spouds, is due to the attraction on a point on
its surface of a rºtºniº — h and Weight E — 4tra-Rºh; so that—
" – ('ſ R. — * — * - Rſ 1 – 32 h _h)-"
g’ = C(E – 4tra-Rºh)/(R – h)* = tracR(1 8. #) (l #)
= 9 { + (2 sº-ºº: *}}º. p., i.e., quam proacimé).
Therefore g increases 2 – 3a/p per cent. for one per cent. approach to the centre of the Earth:
and therefore, if the pendulum, beating seconds, gains n beats per day of 24 hours at the bºšom of
º
*=
the mine, :
*
sº==S*-* *
Thus if *
a/p = }, n = 21600 h/R.
In these experiments a pendulum beating seconds, and called a seconds pendulum, is usually
employed; the period of a seconds pendulum is therefore two seconds; and if l is the length of the
pendulum, 1 = TV (l/g), or g = trºl.
Thus if g = 981 (spouds), l = 99.4 cm ; very nearly one metre, and usually taken as such
in practical problems. §
In the Harton coal pit, 1,260 ft deep, n was found to be about 2:25, which makes alp = 0-376 :
and supposing a = 2:5, this makes p = 6-6, which is considerably above the accepted value 5:576.
But Prof. Haughton has shown (Phil. Mag, 1856) that by taking into account the mean
elevation of the continents above sea level, the value p = 5.5 can be obtained.
In a mine a kilometre deep, R/h = 10° -- #; and now if we suppose a = 2:5, p = 5.5, we
shall find n = 4:32, so that great accuracy is required in these observations. f
(6873) 2 I -

#22
The Siemen's Bathometer is an instrument for measuring the depth of the sea, depending-on
the same principles. s
Attempts have also been made (Abel Bulletin de Ferussac, 1828, p. 166) to determine, by
pendulum observations, the disturbing effect upon the magnitude of gravity, due to the position
# º º and Sun in the sky, effects which are otherwise very noticeable in the phenomena of
the Tides.
Denote by M the weight of the Moon in grammes, by r its distance from the 2entre of the
Earth in centimetres, and by 6 the Moon's angular Zenith distance.
Then the vertical component of the change in gravity will not be, as might be supposed, the
vertical component due to the Moon's attraction, but will be the difference between the accelera-
tions due to the Moon's attraction in the direction of the zenith, at the point on the Earth's surface,
and at the centre; and therefore the diminution of g will be given by -
CM (r cos 6 — R) CM coS 6
Aq = -º-
* T (=3|Rºosa IR£); 7.2
= *(oose– #) (1 — 2 R cos 6 + º'- CM cos 6
º 7” 7° 7. 7.2
- cy: (3 cos” 6 – 1), neglecting (R/r)*,
while g = CE/R^; so that
Ag M
*-*.
g T #(ºg cos” 6 – 1);
and a pendulum beating seconds will lose n seconds a day, given by m/86,400 = }Ag/g.
With E/M = 80, and rſt = 60, this makes ſ 2
77 – * (3 cos” 6 — 1), which is quite imperceptible.
PENDULUM OSCILLATIONS UNDER A FORCE TO A FIXED POINT, WARYING
AS THE DISTANCE.
83. When the pendulum OP (Fig. 36) swings inside a narrow crevasse in a gravitating sphere of
uniform density, the attraction of gravity will have a single resultant directed to Q, the centre of
the sphere, passing through G and proportional to £G ; acting as if ØG was the stretched portion
f an elastic thread obeying Hooke's law.
()
9 is follows from the fact that the attraction of a spherical shell-Qh a point in its interior in
zero, but on a erior point is the same as if the shell was gondensed inte a particlvat its centre;
so that at a distance r cm from the centre in the interior of the gravitating sphere, the acceleration.
«»n-
£2.
eaaº
of gravity g’ is given by-
g' = 4tpCrº/r? = 47tpCr,
C denoting the gravitation constant.
The resultant attraction on the pendulum OP, of weight W g is therefore a force—
#TpCW. Gayº.
and its moment about ſº
4TpCW. C.G. CO. sin OºG = 47pCWha sin 6, if CO = a.
The equation of motion cf the pendulum is thus
(h” + k”) (d”6/dt’) = — 47tpCha sin 6,
so that if we put g = 4tpCa, the value of gravity at the point of suspension O, then the length
of the simple equivalent penduluma-isT = h -- kºſh, as at first.
The crevasse must be-nārrow, and the pendulum OP must be flat and thin for gravity to vary
as the distance from 6. *: w
inside spherical cavity, the field of force would be uniform, and gravity would be equal to
the value at the centre of the cavity, and act in parallel lines, as generally assumed on the surface
•w.
of the Earth.

123
84. Outside the gravitating sphere, where its attraction varies inversely as the square of the
distance from the centre, the equation for the finite oscillations of a pendulum becomes intractable,
and we must consider only small oscillations.
Suppose the bob of a simple pendulum to swing near the ground, and that l the length of the se i
thread is comparable with R, the radius of the Earth; the deviation of gravity from absolute
. . .
parallelism now makes itself felt, and the period is not given by 27A/(l/g), *Twº * $39
G l 1 s *
271- V ( g Hº seconds,
a denoting the acceleration of grayii. e surface & Earth; this formula can be I-/
the same manner as that just employed for the pººrdulum in the crevasse, where gravity varies as
the distance from a centre Q, and is directed to (ſ; outside the Earth gravity is directed to E, the
Earth's centre, and for the very slight variations in its intensity, it may be supposed without
error to vary as the distance from E in small oscillations.
Thus the period of a pendulum of length l = R would be 27tv/(R/2g) seconds, and not
2TV/(R/g); this latter would be the period of a pendulum of infinite length, so that the bob moves
in a straight line, or slides on a horizontal plane; 27tv/(R/g) seconds would be the period of a
railway truck, oscillating in a perfectly straight tunnel connecting two adjacent points on the
Earth's surface; moreover, if the density of the Earth was uniform, the period would be the same
for any straight tunnel however long, even for a diametral tunnel.
In going through such a long straight tunnel, the Channel Tunnel, for instance, the road
would appear to be at first descending and at the end ascending; and the effect would be
mechanically the same as if descending and ascending a tight rope, like Blondin at Niagara, in
the form of a Catenary, whose directrix is the diameter of the Earth parallel to the tunnel; for s
denoting the distance from the middle of the tunnel, and y the apparent gradient of the road,
s = c tan \lº,
the intrinsic equation of the Catenary,
For a tunnel 20 miles or 32 kilometres long the gradients at the ends would appear to be about
one in 400; and water reaching to the ends of the tunnel would have a head of 800 inches or
67 ft in the middle, this being the Eßsine of a chord of 20 miles in a circle of the * / i
radius.
A satellite grazing the surface of the earth would have the same period T = 27tv/(R/g), since
g = 47°R/T', the central acceleration in a circular orbit of radius R, described in period T.
As the period of the Moon is about 28 days, and its distance from the centre of the Earth
about 60 R, according to Kepler's Third Law, “the squares of the periods are as the cubes of the
mean distances,” the period of the grazing satellite should be about 28 + (60); days, or about i
#th of a day, so that, assuming the density of the Earth uniform, a body let fall down a {
diametral tunnel, or a train oscillating freely in any straight tunnel, would make 17 complete
oscillations in the day; and in the diametral tunnel, the maximum velocity acquired at the
centre would be the velocity of the grazing satellite, that is,
27R/T = 4 x 17 x 10° -- 86,400 = 787,000 kines: -
Still considering the Earth as homogeneous, a tunnel must be cut in the form of a hypocycloid
for the oscillations to obey the same law as those investigated for the cycloid in § 10; and the
investigation will be found in the Principia, I; Sect. X. t
As an exercise the student may prove that, if the ends of this hypocycloidal tunnel on the
surface of the Earth are cut at a distance of a times the circumference measured on the great
circle on the surface, where w is a fraction, the period of oscillation in this tunnel is 2M/{a (1 — wy)
of the period in the diametral tunnel.









124
PENDULUM OSCILLATIONS ON A REVOLVING wheel. g”
*
t r
85. These oscillations can be realized practically by supposing the axis O of the pendulum
OP in Fig. 36, to be fixed to a wheel and parallel to its axis, the wheel spinning with constant
angular velocity m about a vertical axis. d
If C is the axis of the wheel, and OA the position of the pendulum when in relative stable
equilibrium, and CO = a, then with respect to co-ordinate axes Ca, Cy, through C fixed in space,
the co-ordinates of G are ***
a = a cos mt + h cos (mt + 6), y = a sin mt + h sin (mt + 6);
and if X, Y denote in poundals, the components of the reaction of the axis O, the equations of
motion are
W(dºc/dt’) = X, W(d”y/dt’) = Y;
and taking moments round G,
Wk:(dºgſdº) = Xh sin (mt + 6) – Yh cos (mt + 9);
so that, eliminating X and y,
k:(dºg/dt) = h sin (mt + 6) (dºw/dº) – h cos (mt + 6) (dyſdiº)
= ah (sin (mt + 6) (d” cos mtſdt") — cos (mt + 6) (d” sin mtſdº)}
+ hºlsin (mt + 6){d” cos (mt + 6)/dt”) — cos (mt.-- 6){d” sin (mt + 6)/dtº
= — mºah sin 6 — J.”(d°6/dt”),
C I’ (h’ + k”) (d”6/dt’) = — mºah sin 6,
representing pendulum motion with respect to the moving radius CO, with
! =, g(h’ + k})/mºah
as the length of the equivalent simple pendulum.
When the axis of the wheel is horizontal and gravity comes into play in the motion of the
pendulum OP,
W(dºw/dt’) = X + Wg,
while the other equations remain unaltered, so that
(h” + k}) (d”6/dt’) = — mºah sin 6 — gh sin (mt + 9),
and the motion is not comparable with that of a simple pendulum.
When C is moved in a straight line with constant velocity, the pendulum OP will move with
constant angular velocity; but if C moves horizontally, like a train, at starting or stopping, with
constant acceleration f, the pendulum OP, movable about a vertical axis like a carriage door, will
perform simple pendulum oscillations with f for g.
125
“86. When a cylinder rolls inside a horizontal cylinder, or a sphere inside a sphere, the line OP
º joining the centres of the spheres or the plane through the axes of the cylinder oscillates like a
- j. of a certain equivalent length l ft. - - * W.
Denoting by b the radius of the fixed cylinder or sphere, the rolling cylinder or sphere will
have turned through an angle (fig. 42) - -
*— 1) 6 (N.B. not %) -
\ a
- when the line QP has turned through an angle 6 from its lowest position.
- The equations of motion of the rolling body are therefore, employing the poundal as unit of
force, - * -
|
W(0–0). = F – We sin 9 . . . . . . . . . (1) :
- W(b - a); ~ R *s- Wg cos 6 . . • * > . . . . . . (2)
wº – i); E - Fa \ , * * . • . V - - (3) --
\a diº
Eliminating F between (1) and (3), < *.
W(b — a) Jºž T W ...(b * 4). — Wg sin 6,
• * * - *\tº - ... .
iOr - -. 0-0 +% = – g sin 6 ;
and comparing this with the pendulum equation ($ 74)
'i = – g sin 6,
- - Å;2
-o-50 ++)
the length of the simple equivalent pendulum.
Then equation (2) determines R, the pressure between the bodies in contact.
To prevent slipping, teeth may be cut on the surfaces in contact,
When the fixed cylinder or sphere is in neutral equilibrium on a horizontal axis through O,
about which it is capable of turning, this motion is slightly modified by the yielding inertia.
Denoting by MK” the M.I. about the axis through O, and q, the angle it has turned back, the
equation of motion is - º - - - -
ºv, dºd -
Mkº – Fº,
while Weſſ” – ). | **śl F
. - {{. 1). # * - Fa,
and W (b — a º = F — Wg sin 6. -
Eliminating F and dºd/dt from these equations, we find
d°9 a * /h .
... = -; sin 6,
2 D J l,
h - 1 +...+}}.
Where i = (b — a -
o a) - 1. b% W/62 -
1 + q2 MKZ
representing pen dulum motion of simple equivalent length a
(6878) - 2 K.


KINETIC ENERGY.
87. The K.E. (kinetic energy) of a single particle, weighing m lb, and having component
* ... tº da dy dz \,...;
velociti ‘..., “..., “... b
tooines in jº tº being
'da' , dy”
#m. —t-
2 : " : | Clt? + (ſt?
*
therefore the K.E. of a whole system of particles
y 'da' , dy” dzº).
T = } - —tº — ?'
smſ Glt? (ſt? + †)
= }X. ***) ". . ***). ***)
2 m{ Clt +( dt + ( dt / J
y
dt? + dt? + dź2.
*-
#
da” dy” ) dz”
l ------ ~ * - - - -ºs-
+ Psm (i. + i + #)
since Xma' = 0, Xmy' = 0, Xma' = 0.
if w, y, z denotes the C.G.
Therefore the K.E. of a material system may be divided into two parts :-
(i) The K.E. of its collective weight W or Xm, moving with the velocity of the C.G. of the
system. -
(ii) The K.E. of the system relative to the C.G.
The K.E. of a bursting shell is thus the K.E. before bursting and K.E. due to explosion.
The K.E. of a cylinder or pair of wheels rolling along a plane is - - -
da,” 062
- d?”
|W
* d??
+ VK”
- k?yday? *
Or ywſ 1 + jºin ft-poundals,
while the K.E. of a train is—
- Æ2\ \ day? -
l W - . A – – 1" * ANO ºf
(M + (1 +.) dº?” ft poundals
A locomotive engine does not therefore require a heavy fly-wheel, as the whole train serve
as a store of energy, provided however the driving wheels do not slip. * , , -
This assumes that the K.E. of a body, rotating with angular velocity d6/dt round an axis about .
which the M.I. is Wk” (lb-ft”) is—
#Wk”(d6/dt)* (ft-poundals),
or #Wk*(d6/dt)*/g (ft-pounds);
as already proved (§ 64); while the K.E. of translation of the C.G., with velocity vſ/s, is
#Wv” (ft-poundals) or #Wv°/g (ft-pounds). -

127
THE PRINCIPLE OF ENERGY,
88. “The increase of K.E. of a material system is equal to the work done by the forces of the
system, as the system passes from the initial to the final configuration.” - * " * ,
This has already been proved in § 57 for a single particle, and is therefore true for the system
of par Gles, but the principle may be proved independently.
Taking the first three equations of motion of the system—
d’a d”y d”z ſº
Xm :: = XX, Xm. dź T XY, Xm dº? T XZ;
dº?
*, * d. d’a dy d”y d2 #) * ( da: dy %)
therefore sm (#####4 ## = X. x - Y }+ zº)
Integrating with respect to t,
da,” a dy” d2? \ \
l * --- — l = cº
£3m(#. +-ji, + #) > ſexas + Yay + Zdz) + H
Then if q denotes the final velocity of a particle, and q, its initial velocity,
#2mq* – #2mgo” = 2ſ (X da + Yay + Zdz),
Or T – T = W,
the work done on the system.
This equation is called the equation of energy, and it is an integral of the equations of motion,
which in most cases can be immediately written down.
128.
MOTION OF A systEM UNDER IMPULSES.
89. Considering for simplicity the motion of a system in two dimensions (plane motion), the
equations of motion may be written - . A-
. . . --- - t d dº – sv y
... xm # => x.
º ; f .
... Sm # =>Y,
-d dy . da, — S Y — v. KY—
- . g’ g g f gy - d e is a s - t #
Integrating with respect to the time t, denoting º and # by w and v, and them initial values.
Đ () | when t = Ö), by uo and vo; then
Xmu – Xmu, = s|xa,
Xinv - Żmº = s| Ydt,
If we suppose the finite and incessant forces X, Y indefinitely increased, and the time t for
which they act indefinitely diminished, such applied forees produce what are called instantaneous
or impulsive forces, or impulses, the impulse being measured by the momentum generated or
destroyed. i
The duration of an impulse is supposed so short that although the velocities of the system,
have undergone a finite change, the configuration of the system has not sensibly ehanged: so
that we may neglect the influence of any finite forees simultaneously acting.
Now denoting the impulses
ſ Xdt and ſ Ydt by X, and Y,
30 0 -
then, supposing the body not to have sensibly changed its position,
ſey – y X).dt = a Yi — y X;
so that the equations of motion for impulses are
>mu - >mu, = XX . . . . . . . . . . . . (1)
Xme — Xmv, = 2. Y. . . . . . . . . . . . . (2)
Sm(zo – yu) — Sm(ºr, -yu.) = x(cy, — yx) . . . . . . . . (3)
As before in the equations for finite forces, equations (1) and (2) may be written
v v. Sl W. * - / -
hence the motion of the C.G. of the system is the same as if all the weight and all the impulses
were collected at the C.G.; and the motion of the C.G. is unaffected by any internal explosions,
Again, by changing to a moving origin at the C.G., equation (3) becomes transformed to
Xm(w'v' — yºu') — Xm(w'v', - y'u',) = X(a/Y, - y’X),
and hence the motion relative to the C.G. is the same as if the C.G. were fixed.



129
We notice that if no forces act in a given fixed direction, say Oa, the momentum is constant
In that direction during the motion; this is called the Principle of the Conservation of Linear
Momentum. &*.
Also if the forces have no moment about a certain fixed axis, say O2, the angular momentum
about that axis remains constant; this is called the Principle of the Conservation of Angular
Momentum. *. ~ *
These two principles, in conjunction with the Principle of the Conservation of Energy, are in
general sufficient to solve most dynamical problems.
90. The Angular Momentum of a System about an Awis, just employed, is defined to be the
sum of the moments of momentum of the particles of the system about the axis; and thus about
the axis Oz is
Xm(av — yu).
For the components of momentum of the particle at w, y, weighing m lb, being mu, no
(second-poundals) their moment about the axis Oz, perpendicular to the plane, a Oy, is
ºn v.a. — mu. y
( du #)
E 7/2. ſº -º- -
d; T & di
For a rigid body, rotating about the fixed axis Oz,
--~~~-sºº —/,

where r is constant, and d6/dt the same for all the particles; so that
QM = da, E - 7° sin 6 d6
dt dº.”
— dº — d6
d6)
d av — wu = r^ “: :
and we – yu = r is
so that the angular momentum is—
d6 d6
2 * = 2 ºz.
Xmr dt WAE dt
Denoting the angular velocity d6/dt by o, and its initial value by too, equation (3) becomes
w
Wkºo — WKºo, = Xw(a Y — yx,) = N,
, suppose, where N, denotes the moment of the applied impulses about Oz.
Hence the change in angular velocity—
tº-
Ni moment of impulses about axis
WAE2 27 inertia e
a – on =
25
This equation is sufficient to determine the motion of a rigid body, set in motion about a
fixed axis by given impulses, for instance, the
(6873) 2 L
BALLISTIC PENDULUM.
91. This was invented by Robins (1740), for determining the velocity of bullets (and cannon balls).
It consists of a large pendulum provided with an iron plate, or a box filled with sand; the bullet
strikes the iron plate, or the ball penetrates the box of sand; and the diluted velocity of the ,
pendulum and projectile carries the pendulum through a certain angle, which is measured, and %( &
thence the velocity of the projectile is inferred (fig. 43). . . . . . . . . . — 2.
Denote by W lb the weight of the pendulum; Whº (lb-ft”) its M.I. about the axis of
suspension; h ft the distance of G the C.G. from the axis O ; w lb the weight of the bullet or
cannon ball, u f/s its horizontal striking velocity, and a ft the depth below the axis at which it
strikes the pendulum, aſ the angular velocity communicated to the pendulum.
Then Wk,”o = moment of impulse about-O
= wwa,
Or • = *.
- Wk,”
If the pendulum swings back through an angle 2, then the C.G. is raised h vers a ft; and
by the Principle of Energy the work required to raise W lb through h vers 2 ft vertical,
Wh vers a foot-pounds, must be equal to the kinetic energy of the pendulum in its lowest
position, which is #Whºo”/g foot-pounds.
Therefore #Wk,”o°g = Wh vers a
Or o” = #: Vel"S 6.
l
w”u%a.” 2gh
wº T Fº VéI'S O.
2 2
#u” = W. º: gh vers &
QM, E w ºvg.)? sin #2,
so that if the point A on the pendulum pulls out a graduated tape AB through a clip at A, and
we denote OA by a, then AB = 2a sin #2, and
* - - – e.
QU) ſº Q.
so that u is proportional to AB; and thus a uniformly graduated tape will serve to read off the
striking velocities. -
Strictly speaking, when the ball remains imbedded in the box, WK,” must be increased by
wa”, but this correction is practically insensible. -
Sometimes the gun itself is slung in a pendulum, then-ealled the Gun Pendulum, and the
muzzle velocity of the shot is inferred from the angle of recoil of the Gun Pendulum ; but this
method is not susceptible of accuracy, because it is found that the blast of the powder vitiates the
records (fig. 44). - -
Denote by X, the forward horizontal impulse on the pendulum at the axis of suspension, then
X1 +. wu = momentum communicated to W lb collected at G = Who,
Who ,
(YT X, − cu ('''“ – 1
\ it'll f
wh \
Tºrº wnſ 1. ) ;
k,”
so that X) = 0, and there is no impulse on the axis, if w = kºſh = l, the length of the simple
equivalent pendulum. * •
The point P at which the ball should be aimed is thus found practically by suspending a
plummet P by a fine thread from O, and altering the length of the thread till the plummet and
the pendulum synchronize in their oscillations; in this way the centre of percussion of a sword or
of a cricket bat can be found.
The point P is now called the Centre of Percussion, while the corresponding axis throngh. O
is called the Avis of Spontaneous Rotation; and we notice that () and P are connected by ... actly
the sanne law as that for Centres of Suspension and of Oscillation in § 74. • .*N.
” ºf 131
It must be noticed, however, that although X remains the same for different positions of P,
the points aimed at in a line parallel to the axis, there is also a different impulse couple on the
axis, which may, however, sometimes be made to vanish. . . . . . . . . -
Thus if a door is brought up by a stop on the floor, the impulses on the hinges are unequal.
92. Suppose a symmetrical body, say an armour plate, suspended by chains, is struck normally
by a shot; if the shot passes through G, the C.G. of the plate, the plate will recoil without
rotation; but if the shot strikes at a point P vertically below G, the plate will revolve
instantaneously about a horizontal axis through O, the Spontaneous Axis of Rotation, such that
OG. GP = P, ... •
where # refers to the parallel axis through G (fig. 45). : . . . . .
For if U is the linear horizontal velocity communicated to G by the impulse wiſand o the
angular velocity communicated about the axis through G, then -
WU = ww,
Wk?o = ww. GP;
so that * k?a = U. G.P.”
Produce PG to 0, and suppose O is the centre of instantaneous rotation; then the velocity
of O being zero, -
(t) . OG = U = kºoſ(3P,
so that OG . GP = k?.
If the body is now free to move, and is under no forces, for instance if it is struck so that the
instantaneous axis is vertical, and that it moves on a smooth sheet of ice, G will continue to
move horizontally with velocity U, while the body will continue to revolve with constant angular
velocity of the centre of instantaneous rotation will then move along the parallel line OO', ºra-CD
the subsequent motion can be represented by rolling the circle GO fixed in the body on the i
straight line OO/fixed in space, so that the centre G moves with constant velocity U; any point
of the body thus describes a Trochoid. -
A body suspended by two or more equal, parallel, vertical chains or threads can be made to
serve as a Ballistic Pendulum, if fired at horizontally so that the shot passes through or near the
C.G.; the length of the simple equivalent pendulum is now the length of a chain or thread.
Thus an impromptu gun pendulum may be made by suspending a rifle by two equal threads
fig. 46). , -
(fig For large projectiles the Ballistic Pendulum is now superseded by electric screens and
chronographs; but the Pendulum is still useful for measuring the velocity of small arm bullets,
which are apt to miss or be deflected by the wires of a screen.
EXAMPLES.—(1.) A smooth vertical screw with flat ends is placed with its axis vertical on a
smooth horizontal table, and a nut on the screw descends from rest under gravity. Investigate
the motion, and examine what ensues when the nut impinges on the table. \
(2.) Work out the motion of a projectile in the bore of a gun with uniform rifling, powder
pressure, and friction—(i) when the gun can turn round in a collar (ii) when fixed; and find when
the copper driving bands will begin to slip.
(3.) Two equal rods are freely jointed together, and lie on a frictionless table, extended so
as to form one line; show that if an end of one of the rods is struck at right angles to its len
heir mi ints and the joint will start off with velocities in the ratios of 5: – 4: — 1.
will take up 3 times as much energy as it
if the joints were made rigid.
(4.) A freely-jointed regular hexagon lying on a smooth horizontal plane is struck at one
corner towards the centre; show that the initial angular velocities of the sides are in the ratio
10 : 22 : 1.
(5.) Determine the velocity with which a railway truck must strike a fixed obstacle so that a
rectangular body resting in it may just be turned over about an edge; and determine the
horizontal and vertical components of the reaction at the edge, both impulsive and finite
(instantaneous and incessant). z






132.
C.
THE'STEADyºmotion OF A TOP OR GYROSTAT/ANP-45
3. MQTºeN-er-in-Erie Reºrºl). EROJECT:#E.
&
º, º ... <
93. When a Top spins steadily about a fixed point O, so that the axis OC describes a right.
circular eone round the vertical OZ as axis, at a constant inclination a to the ºvertical, the instan-
taneous axis of rotation Oſ describes a right circular cone IOZ round OZ, as axis in space, and a .
right circular cone IOC round OC as axis in the body; and the motion of the Top can be imitated,”.40".
by rolling the cone IOC on the cone IOZ (fig. 47). - Tº ºa-
If we denote the angle IOC by 8, and the component angular velocity about OC by r, then the
resultant angular vettetty, which is about OI, is re
r sec & . . . . . . . . . . . . . . (1)
This depends upon a Theorem or Lemma, called the
PARALLELOGRAM OF ANGULAR WELOCITIES.
Lemma. “If OP, OQ represent the component angular velocities of a body about the axes OP,
OQ, the resultant angular velocity is represented by OR, the diagonal of the parallelogram
OPRQ” (fig. 48).
The proof is exactly the same as that required in Statics, for showing that the sum of the
moments of two forces in a plane, about any axis perpendicular to the plane, is equal to the
moment of their resultant.
For if I denotes any point in the plane POR, and Ip, Iq, Ir are drawn perpendicular to
OP, OQ, OR, then the velocity of I, upwards from the plane, due to the angular velocities OP, OQ
IS
OP. Ip + O.Q. Iq
= 2 A IOP + 2 AIOR
and this, as in Statics,
= 2 AIOR = OR. Ir,
so that this velocity of I is due to the single angular velocity OR; which is therefore the resultant
of OP and ()0.
- A machine has been devised by the Rev. F. Smith for illustrating the composition of angular
velocities, consisting of a sphere resting on two pairs of rollers, which can be turned so as to
impart the component angular velocities.
N.B.-In representing an angular velocity by a vector OR, we draw OR along the axis of
rotation, of length proportional to the angular velocity, and in the direction such that OR is the
direction of advance of a right-handed screw, due to the rotation.
The component angular velocity about the axis OA. in the vertical plane ZOC perpendicular
to the axis OC, will therefore be
ºr tan £8 . . . . . . . . . . . . . . (2)
Now OC, OA being principal aves of the Top, and C, A denoting the M.I. of the Top about
these axes, the components of Angular Momentum about OC, OA are
Cr, Artan & ; . . . . . . . . . . . . . (3)
while the resultant angular momentum will be about an axis OH in the plane AOC, making an angle
y with OC, such that
Artan B A
*::: * =% tan 8 . . . . . . . . . . (4)
tan y =
and the magnitude of the resultant angular momentum about OH will be
Cr see y . . . . . . . . . . . . . . (5)
These equations (3), (4), (5) depend on a second Lemma or Theorem called the
PARALLELOGRAM OF ANGULAR MOMENTA.
Lemma. “If OP, OQ represent the components of angular momentum of a body about the
axes OP, OQ, the resultant angular momentum is represented by OR, the diagonal of the parallelo-
gram OPRQ.” The proof is the same as for angular velocities.


, 133
Now let a denote the angular velocity of the vertical plane. ZOC about the vertical axis.
OZ in the steady motion of the Top. ‘. --- * .
The velocity of C will therefore be
OC sin a . g . . . . . . . . . . . . . (6)
But the velocity of C, considered as due to the component angular velocity r tan 8 about 0A,
is also given by
OC. r tan & . . . . . . . . . . . . . . (7)
so that
rtan 8 = p sin a . . . . . . . . . . . . (8)
Representing the resultant angular momentum Cr secºy by the vector OH, then the velocity
of H is equal to the impressed couple, due to gravity, which acts about an axis perpendicular to
the plane ZOC.
This follows from a third Lemma, which asserts that. *
Lemma, “If OH represents the angular momentum of a dynamical system, then the velocty
of H is equal to the impressed couple.” A
For the impressed couple, K suppose, generates in the element of time dt the angular
momentum Kdt about its axis; and this by the second Lemma, changes the angular momentum
from OH to OH', where HH’= Kdt; so that the *. §§
K = lt # = velocity of H,
and the axis of the couple K is parallel to the direction of motion of H.
In the motion of the Top, the axis of the impressed couple is horizontal, and H therefore
describes a horizontal plane.
y
Suppose for instance that OH represents H, the resultant angular momentum of a dynamical
system, making an angle 6 with OZ ; and suppose that L, M, N, represent the components of the
impressed couple K
(i) L about OH, * £
(ii) M about an axis perpendicular to OH in the plane ZOH,
(iii) N about an axis perpendicular to the plane ZOH.
Then the third Lemma leads to the three equations
: . . . . )
| _ d'H
J L = }
_ II d6
M = H .
N = Hsin 6%,
dt
if # denotes the angular velocity of the plane ZOH about the vertical OZ,
. In the steady motion of the Top, L and M vanish, and * = p ; so that
N = velocity of H,
= Cr sec y : usin (a — y),
= Crp (sin a – cos a tan y),
= Cr A sin a – Ar p cos a tan 8,
== Cr A sin a – A u” cos a sin a . . . . . . . . . . . (8)
by means of (4) and (8).
Now in the Top the impressed couple N is due to gravity, and is the moment about O of W
poundals acting vertically downwards through the centre of gravity G; so that, denoting Öğ
by h,
N = Wgh sin a . . . . . . . . . . . . (10)
Replacing N in equation (9) by this value, and dropping the common factor sin a, whith
equated to zero would mean that the Top was spinning with its axis vertical,
Wgh = Crg – Ap” cos a. . . * * * * , a e (u)
(6873) 2 M
ºf 34,
Solving this equation as a quadratic in p,
Wgh
Cr *
pi* — *see a - - – - Sec ox,
A
Cr \2 C24.2 4AWah *
( -āº) -: *-(º-tºo- •)
C 4AWah
•=}ºs ( ; V( – .# **) • . . . . . . . (12)
*
The minimum value of r for real values of p. given by
C*r = 4 AWGh cos a ; . . . . . . . . . . . . . . (13)
for smaller values of the angular velocity r, the Top cannot spin steadily at this inclination a to
the vertical. tºy
For instance, a Top cannot cannot spin steadily in the upright vertical position if the
angular velocity is less than
2V (AWGh/C*) . . . . . . . . . . . . . . . (14)
We notice that when (12) gives two distinct real values, p and A', and when S and S', 'y and
y' are the corresponding values of 3 and y; then
tan 8 + tan 8 = | tan a
Wgh
tan £8 tan £8' = Arº
(Sec c. — cos 2),
fan y + tan y' = tan a,
AWGh
tan y tan y' = C24.2
(sec 2 — cos a).
Thus in the limiting case when p = p/, 8 = 8', y = 'y', we find
tan y = }, tan a.
+35
94. A similar procedure will enable us to determine the steady motion of any solid of revolution,
spinning and rolling with its axis at a constant inclination a to the vertical, on a horizontal plane
or on a fixed vertical surface of revolution, while its C.G. describes a horizontal circle (fig. 49).
As examples we may take a coin, a hoop, a wheel (as in military sports), a cask, a wineglass,
and finally an elongated projectile in the air. \ * * *
Suppose the point of contact I describes in space a horizontal circle, centre O' and radius a,
and in the body a circle, centre Q and radius y, while G describes a horizontal circle, centre H and
radius b, and a denotes the distance GQ. * * *
Producing the axis CG to meet the vertical line O'H in O, then OI will be the instantaneous
axis of rotation; and the motion of the body can be produced by rolling the right circular cone
IOC, fixed in the body, on the right circular cone IOO', fixed in space.
Denoting by R the upward vertical normal reaction of the horizontal plane, and by F the
horizontal frictional constraint, expressed in poundals, then with the above notation,the dynºmical
equations of the motion of G are .#
Wbu? = F,
Wg = R;
and the impressed couple about G
N = F.O'H – R (a – b)
= Wbu” . O’H – Wg (a — b) ;
and this, by equation (9),
= Cr a sin a – A pºcos a sin à,
where A, C, now denote the moments of inertia of the body about GA, G.C.
But O'H = a cos a + y sin a,
a —b = y cos a — a sin a ;
so that this equation of steady motion becomes
{A cos a sin a + Wö (a cos 2 + y sin 2)} p.” - Cru sin a
= Wg (a — b)
= Wg (y cos 2 - a sin x) . . . . . . . . (15)
We see that the body describes a circular motion if left to itself; to keep it moving in a
straight line, a controlling force must be applied.
Thus to keep the gun-wheel moving in a straight line, a downward pressure must be exerted
on the upper side of the nave, or an upward pressure on the lower side, so adjusted as to be in
equilibrium with WQ downwards through G and R upwards through I,
So also with a hoop controlled by a stick or hooked rod,
- 136
r - l - - º º "...ºf • ". . . . . . . * * £2b -- -
THE STABILITY OF MOTION OF AN ELöNGATED PROJECTILE. . " º'
95. In the steady motion of a projectile in a medium, such as the air, the C.G. of the projectile. ,
describes a helix round an axis, when acted upon by no impressed forces; and now the couple N
is due to the effect of the displacement of the surrounding medium. *g
Suppose the projectile weighs W lb ; then when the surrounding medium is absent, the inertia
of the body is measured by W, and is the same in all directions.
, But when the medium is present, the inertia to change of motion will be different in, differen
directions, and will depend on the external shape of the projectile and upon W’, theºweight of
medium displaced. A 4."
Taking an elongated projectile, the inertia to broadside motion is obviously greater than the
inertia to motion in the direction of the axis; we may denote the inertia in these two directions by
c, and c, such that if u and w denote the corresponding component velocities of G, the total
Inomenta of body and medium in these directions GA, GC are (fig. 50)
f z
cru and cºw,
while # clu” + , cºw”
tſa. is...}otal K.E. of the motion of the body and of the medium, expressed in foot-poundals suppose,
die to the motion of translation of G, unaccompanied by rotation.
But if the Body-ha àngúlar velocities p and Tr about GA and GC, and the
corresponding components of the angular momentum of the system are c, p and cºr, the total K.E.
of the system will be * * *
l, , , ,2 l 2 l 2 2
# c,” + # cºw + # 64 p + cººr”. f
We may put *—ºr
Cl W | W/ Q,
W H W’ 'y,
va”/ ºr ºf
where a, y are certain numbers depending on the external shape of the body.
Now it is proved in Treatises on Dynamics (Thomson and Tait, Vatural Philosophy, $ 335),
that when the body has these component velocities w and w, the reaction of the surrounding
medium on the body is a couple, tending to set the axis of figure at right angles to the general -
direction of motion, of magnitude,
N == (e. * ca) 71?U.
The effect of this couple is illustrated by the tendency of a ship under insufficient weigh to
fail into the trough of the sea broadside on to the waves.
On the other hand a flattened oblate body, like a plate or piece of paper, a quoit, or a card-
board disc, in which c, - c, sets its plane horizontal in falling through water or air.
We take GZ as the direction of the initial impulse of projection ; and now if the m
taken as frictionless, and no forces are supposed to act, the Principle of the Conservation of
Momentum shows that, if the axis GC is inclined at an angle a to GZ,
c, u = F sin a,
ium is
cºw = F cos a,
where F denotes the resultant impulse;
C., 21
tan z = -li ;
Oil
%20
C
while the couple due to the medium
N = (c. – c.) tº wº tan a.
Cl
Introducing this value of N in equation (9), and dropping the factor sin a, which equated to
zero would imply perfect centering,
-\ “3 ...? :)
(c) – c.); w” sec 2 = cºrp – c.1% cos 2.
Cl __--"-
Here cº denotes the M.I. of the projectile about its axis GC, as the corresponding angular r
can exist without setting the medium in motion.
But c, denotes the M.I. of the projectile about the axis GA, increased by a certain amount
due to the stirring up of the surrounding medium; so that
ce = W.K.",
while c. - Wł,” + W’k”,
where k” depends only on the shape of the projectile.
Therefore the least admissible value of , for which this steady motion is possible is given
by
7.2 (?
- :-- 4(c. -º- ca) : - * * s • º ºf º e º º g * (16)
2
1 °6
















187
If the projectile has been projected from a gun in which the rifling at the muzzle makes one
turn in n calibres, a screw of pitch n calibres, and if the final angle of the rifling is 8, and the
călibre is d, then to secure this minimum value of r,
2 2 2
tanº & = 1. =# = *
72 20*
c. c, d”
- (e. tº-ºw es) Y3 #
C1 Cé
NW - Wy(Wk.” -- W’kº)d?
A = W( – ’) Wºwºwº,
W. k,”d”,
=w (- - )*::::. . . . . . . . . . . . . . (11)
neglecting powers of W’ſW above the first, as is allowable when the surrounding medium is.
air. . * *
If then we can determine for a body the values of a and y, or a – y, we can determine the
theoretical values of £8 and n, requisite for stability in flight. d e
But so far the only body for which the values of a and y have been determined is the
ellipsoid, values discovered by George Green (Researches in the Vibration of Pendulums in Fluid
Media, 1833); and when the ellipsoid is of revolution, Green finds that
A C
* = ATU 7 = ga.
where
O2 d’A C 1 C
2: = -> h-l - 9
A j O (a” + A)(c” + X)* a"(cº - a”) ( c” — a?); COS Q.
Gº d’A 2 2 C
- = — — — — —r -------, cosh’--,
ſ, (a' + X)(c’ +x); c(c” – a”) + (c’ – a”); COS 0.
so that
2A + C = #.
(LºC
Here 2c, 2a denote the major and the minor axes of the ellipsoid; denoting them as usual by
l and d, and putting lſd = a, the length of the projectiles in calibres, then
\
*— — 1 -l
º, , ) º 22 – 1 (º-I); cosh <!,
Q-e a"C = — —*— + –– cosh" a. ~
t = z(zº-I) ' (º-I); 9
9.
*(2A + C) = }
From these formulas the following table was calculated by Captain J. P. Cundill, R.A., and
enlarged by Mr. A. G. Hadcock, R.A. The column of values of a - y is added, as this was the

most laborious part culation. --------------— -- © º º
S en the body is a Whitehead or Howell torpedo, and the surrounding medium is water
then W = W, by the mechanical arrangement of the buoyancy chamber, and the proceeding
approximation will not hold. ſº º
It is necessary now to determine k'in addition; but this would lead us too far at present.
2
N
(6873)
138
96. TABLE of RotATION FoR STABILITY of PROJECTILEs.
A

Jength of
projectile in
calibres.
**º-º-º-º-º-º-sºmºsº
i
:
I
0
O
Minimum twist at muzzle of gun requisite to give stability of rotation
= one turn in n calibres; or a pitch of n calibres.
Value of a ~ y.
.494.18
.52032
.54431
.56643
.586.79
.60561
.62315
.63938
.65454.
.66868
.68192
.69434
.70598
.71693
.72724
.73697
.74615
.75483
.76303
.77082
,77820
.785.18
.791.94
.798.19
.80428
.80998
.81333
.82072
.82575
.83053
.83514
.83954
.8437]
.847.77
.851.65
.85536
.85898
.86241
.86571
.86887
.87211
.89723
.91543
.92.938
.93949
Cast - iron common
~7-7 v
* Solid lead and tin
shell; cavity = | Palliser shell; cavity bullets of similar
#ths vol. of shell. = $th vol. of shell. Solid steel bullet. composition to
(s. g. of iron (s. g. = 8.000.) (s. g. = 8.000.) M. H. bullets.
= 7.207.) | (s. g. = 10.9.)
%. l ſº, Ø. ſº.
63.87 71.08 72.21 84.29
59.84 66.59 67.66 78.98
56.31 62.67 63.67 74.32
53.19 59,19 60.14 70.20
50.41 56.10 57.00 66.53
47.91 53.32 54.17 63.24.
45.65 50.81 51,62 60.26
43.61 48.53 49.30 57.55
41.74 46.45 47.19 55.09
40.02 44.54 45.25 52.72
38.4% Av, 42.79 43.47 50.74
36.99 4.1.16 4.1.82 48.82
35.64 39.66 40.30 47.04
34.39 38.27 38.84 45.38
33.22 36.97 37.56 43.84
32.13 35.75 36.33 42.40
31.11 34.62 35.17 4.1.05
30.15 33.55 34.09 S v 39.79
29.25 32.55 33.07 38.61
28.40 31.61 32.1]. 37.48
27.60 30.72 31.21 36.43
26.85 29.88 30.36 35.43
26.13 29.08 29.55 34.49
25.45 • 28.33 28.78 33.59
24.81 27.61 28.05 32.74
24.20 26.93 27.36 31.94
23.65 26.32 26.74 31.21
93.06 25.66 26.08 30.44
22.53 25.08 25.48 29.74
22.03 24.51 24.91 29.07
21.56 23.98 24.36 28.44
21.08 23.46 23.84 27.83
20.64 22.97 23.34 27.24
20.22 22.50 22.86 26.68
19.81 22.05 22.40 26.14
19.42 21.61 21.96 25.63
19.04 21, 19 21.53 25.13
18.68 20.79 21.12 24.66
18.33 20.40 20.73 24.20
18.00 20.03 20.35 23.75
17.67 19.67 19.98 23.83
14.99 16.68 16.95 19.78
13.02 14.48 14.72 17.18
.50 12.80 13.00 15.18
10.31 11.47 11.65 } 3.60
Sect.
* 10.
11.
12.
13.
14.
15.
16–19.
... 20.
21–23.
24.
25–26.
Fundamental Units.
Length. -
Time. .
Weight.
Metric Units.
Density and Specific Gravity.
Linear Dynamics. -
Velocity.
Acceleration.
Variable Acceleration.
Reduction of Bashforth's Experiments.
Weight and Force.
Energy and Momentum.
Application to Problems.
Bréak/Resistance.
Train Problems.
Laws of Friction.
Motion on Inclines.
Impulse.
Pile Driving.
29. The Absolute. Unit of Force.
27. Work, Power, and Horse-power.
8.TTrain Resistance?, and examples.
30.
31,
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
46.
47.
48.
49.
#
•,§
3
The C.G.S. System of Units.
The Composition and Resolution
Velocities and Accelerations.
The Ice Boat, and Sailing Ship.
Aberration and Speed. Sights.
Aberration in Astronomy.
The Hodograph.
Circular Motion.
The Moon's Orbit.
Centrifugal Force.
The Conical Pedulum.
The Plumb Line.
Train running round a Curve.
Harmonic Vibration,
Rectilinear Motion under an Attraction pro-
portional to the Distance from a Fixed
Point.
The Simple Pendulum.
orce) Vibrations.
Frictional Decay of Vibrations.
Repulsion proportional to the Distance.
Rectilinear Motion with Resistance varying
as the Square of the Velocity.
Rectilinear Motion under an attraction or
Repulsion, varying inversely as the Square
of the Distance from a Fixed Point.
Velocity required to shoot a Body up to the
Moon, or to infinite Distance.
The Gravitation Constant.
Equations of Motion referred to Co-ordinate
Axes.
Unresisted Motion of a Projectile.
º,
of
62.
63.
64–66.
68.
69.
70.
72.
73
74–75.
76.
}-
(
78.
79
80–82.
83–84.
85.
86.
87.
88.
89.
90.
9}.
92.
94.
95
/
The Parabolic Trajectory.
Tangential
and Normal Components of Ac-
celeration * * * * * o
Resisted Motion of a Projectile,
Motion of, a body on a Smooth Pla
under Given Forces.
Compo Velocities and Accelerations with
grºw sº
Central Orbits.
The Equable Description of Areas.
Kepler's Laws of Planetary Motion.
The Excentric Follower, and the Time and
Mean Anomaly in an Elliptic Orbit.
Elliptic Trajectories. -
Moment of Inertia, and Radius of Gyra-
tion. -
Product of Inertia, and the
Ellipsoid.
Curve,
Momental
D'Alembert's Principle,
Independence of the Motion of the Centre of
Gravity and of the Motion relative to the
Centre of Gravity of a System.
Dynamical Applications. -
Motion of a Rigid Body about a Fixed Axis.
Atwood's Machine. -
Motion of a Body rolling down an Incline.
The Compound Pendulum and the Equiva-
lent Simple Pendulum.
The Finite Oscillations of a Pendulum ex-
pressed by Elliptic Functions.
. The Reaction of the Axis of Suspension.
The Internal Stress of a Swinging Body.
The Oscillations of a Carriage on Springs.
Physical Applications of the Pendulum.
Pendulum Oscillations under a Force to
a Fixed Point, varying as the Distance.
Pendulum Oscillations on a Revolving
Wheel.
Motion of a Cylinder rolling inside another
Cylinder. -
Kinetic Energy.
The Principle of Energy.
Motion of a System under Impulses.
Angular Momentum.
The Ballistic Pendulum,
Motion of a Body struck by a given
Impulse.
The Steady Motion of a Top or Gyrostat.
The Spinning and Rolling of a Body in
Steady Motion. -
. The stability of Motion of an Elongated Pros
iéctile.
96. ºf of Rotation for Stability of a Pro-
2. jectile.
-/3
/ …&









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