=cosB tan a,
sin $'= sin < cot A tan #,
cos b = cos a cos <' sec <,
c = ' and
is determined by its
sine, it may be in either the first
or the second quadrant; and c,
therefore, has two values, as noted.
The corresponding two values of
b are determined from the two
values of 4,'.
The angle, C, if required, is determined from the formulae :
cot 6 = cos a tan B, (5)
cot & - cot 6 tan cot ', ( 6 )
C=(0+tf) or 0+(180-0'). (7)
This problem does not occur in navigation, and is of rare occur-
rence elsewhere.
8
Since
98 SPHERICAL TRIGONOMETRY.
Examples.
1. Given 4 = 115 36' 45"
B= 80 19 12
6= 84 21 56 find a, c, and (7.
Ans. 0=114 26' 50"
c= 82 33 31
C= 79 10 30
2. Given a=53 18' 20"
5 = 46 15 15
4 = 79 30 45 find &, c, and (7.
Ans. c = 50 24' 57"
(7=70 55 35
& = 36 5 34
3. Given B = 76 41' 13"
4 = 117 44 36
a=126 17 22 find I, c, and C.
Ans. 6 = 117 35' 35"
c= 24 16 50
C= 26 50 24
&' = 62 24 25
c' = 120 53 50
(7 = 109 34 34
115. Case V: Given the Three Sides, a, b and c
4t this problem is most conveniently and accurately solved by
means of the tangents of the half -angles (Art. W-), especially
when all three angles are required.
It is one of the most frequent problems of navigation (where,
however, only the angles 4 and B are required) and, for reasons
peculiar to the problem, 4 is found from sin 2 4 (Art. 88), and
B from cos 2 # (Art. 89).
The tangent formulae may be put in a more convenient form by
multiplying the expressions of Art. 90 by
sin(s a] sin(s b) -, sin(s c)
sin(s a) ' sin(s b) sin(s c} '
SPHERICAL TRIGONOMETRY. 99
respectively, giving
tan B=
, ~
sm(s-a) sm(s-b)
P
sin (5 c) 9
where
p /sin (s a] sin (s b ) sin (s c)
V sins
Example: Given a=56 37', 6 = 108 14', c=75 29'.
a = 56
& =108
c =75
37
14
29
0,
cosec 0.30933 . 6 33.0
sin 9.95198
sin 9.31549
sin 9.84707
s =120
10
56=
63
11
44
33
56
41
P . .
2)9.17774
9.58887
U =
i* =
io =
23
61
28
25
56
53
55
54
28
tan
tan
tan
9.63689
0.27338
9.74180
These results should be checked by the sine formula.
100 STEREOGRAPHIC PROJECTIONS.
STEREOGRAPHIC PROJECTIONS.
THE FOLLOWING CHAPTERS, TOGETHER WITH THE PRECEDING METHODS OF SPHER-
ICAL TRIGONOMETRY ARE ESSENTIAL TO THE STUDY OF NAVIGATION; AND,
AS A PROFESSIONAL SUBJECT, THEY SHOULD BE MOST THOROUGHLY UNDER-
STOOD BY ALL MIDSHIPMEN.
CHAPTEE XII.
116. Definitions. The projection of the sphere on the plane of
one of its great circles, when the point of sight is taken at one of
the poles of the great circle on which the projection is to be made,
is called a stereo graphic projection.
The plane on which the projection is made is called the primi-
tive plane, and the circle the primitive circle.
The polar distance of a point on the sphere is its angular dis-
tance on the surface of the sphere from one of the poles of the
primitive circle.
The polar distance of a circle is the angular distance of any
point of its circumference from either of its own poles.
117. The inclination of a circle is the angle between its plane
and the primitive plane.
Let WBED (Fig. 59) be a great circle cut from the sphere by
the primitive plane, N and 8 its poles, the pole 8 being taken as
the point of sight.
Let P! be any point of the sphere, and NP^BS a great circle
cut from the sphere by a plane through P intersecting the primi-
tive plane in the straight line BOD. Then NP t is the polar dis-
tance of P 1 ; and p ly the point in which SP 1 pierces the primitive,
is the stereographic projection of the point P x .
STEREOGRAPHIC PROJECTIONS',
101
The angle NSP^ is measured by one-half the polar distance of
P l ; hence Op^ the distance of p from the center of the primitive
circle, is
Op^-rtan^p,
where p is the polar distance, and r the radius of the primitive
circle.
N
^ .
\ 1 S~~ -~
! \ !//' "^
\ i 7 ^/
A. j^:
<
.\ /<|0
X. ^
\l
118. Let Cd-R (Fig. 59) be a small circle of the sphere, its
plane making an angle with the primitive plane. Its pole P
is the extremity of the diameter of the sphere, OP, passing
through the center of the circle perpendicular to its plane.
The -polar distance of the circle is PQ PR. Its inclination,
(f>, to the primitive WEED is measured by the arc NP, which
measures the angle between their poles, since the angle between
the planes is equal to the angle between the normals ON and OP
drawn from to the two planes.
102
STEREOGPA^HIC PROJECTIONS.
The line WE, the intersection of the plane passed through NO
and OP with the primitive plane, is called the line of measures of
the circle QQ^R.
119. Let NSWE (Fig. 60) be a circle cut from the sphere by a
plane through NS and OP, the axes of the primitive circle and the
small circle QR, respectively, and WE the straight line in which
it intersects the primitive plane. Then WE is the line of measures
of the circle QR.
N
P being the pole of the circle QR, PQ = PR is its polar distance,
and PN its inclination (since the angle between two circles is
equal to the distance between their poles) . 8 is the point of sight,
and q and r the projections of the extreme or principal elements
of the oblique circular cone SQR, which is formed by the pro-
jecting lines of all points of the circle QR.
Denoting the polar distance of the circle QR by p, and its
inclination by <, we have, in the two right triangles SOq and SOr
STEEEOGRAPHIC PROJECTIONS.
103
-at" the distances on the line of measures from the center of the
primitive circle to the projected extremities of the diameter of QR.
120. The Stereographic Projection of a Circle is a Circle.
Let QR (Fig. 61) be any circle, P the vertex of a cone tangent to
the sphere along QR, and Q any point in QR. Pass a plane
through SQP, cutting from the primitive a line, pq, which is the
projection of the element PQ; cutting a tangent plane at 8 in ST,
which is parallel to pq; and from the sphere a circle (not shown
in the figure) to which PQ and ST are tangents, 8Q being the
chord of contact. Fig. 62 represents the circle omitted in Fig. 61,
rotated into the plane of the paper.
104 STEREOGRAPHIC PROJECTIONS.
Draw PK parallel to pq (Fig. 62) ; it is also parallel to TS,
and the alternate interior angles TSQ and QKP are equal. But
as T8 and TQ are tangents from a common point T, they are
equal, and the angle TSQ = angle TQS Bangle KQP. Hence
PK=PQ.
K-, .,?
FIG. 62.
The triangles Sqp and SKP are similar, hence
or pq = PQ - , which is equal to a constant, since PQ,
pS and PS are all fixed lengths.
Since pq is constant, the point p being fixed, it follows that the
locus of q is a circle whose center is the projection of the vertex
of the tangent cone.
STEREOGRAPH: c PROJECTIONS.
105
We shall see hereafter that this gives us a simple construction,
to find the center of the projected circle.
121. Angles on the Sphere are Unaltered in the Stereographic
Projection. Let MR and MT (Fig. 63) be tangents to two
curves on the sphere at the common point M. (These curves are
S "*--..
FIG. 63. T
not shown in the figure.) Let these tangents be projected on the
primitive plane by the planes RIMS and TMS, respectively, in the
lines mr and mi, and let MaS and Mis be the circles cut from the
sphere by these planes; also let the lines cut from the tangent
plane at 8 be SR and ST. Since the tangent plane is parallel to
the primitive, the lines SR and ST will be parallel to mr and mt,
and the angle RST-rmt.
106
STEREOGRAPHIC PROJECTIONS.
Join RT. Now, since R8 and RM are tangents to the same
circle, they are equal; and for the same reason TM=TS; hence
the triangles RMT and R8T are equal, and the angle RMT=
RST=rmt. That is, the angle between two lines drawn tangent
to the sphere at a common point is equal to the angle between their
projections.
Y
FIG. 64.
/
As the angle between two curves on the sphere is measured by
the angle between the tangents at the common point, it follows
that angles on the sphere are unaltered by 'the stereographic pro-
jection.
122. The principal properties of the stereographic projection
are those just proved; viz.:
1. That all circles are projected as circles.
2. That angles are not altered.
STKREOGRAPHIC PROJECTIONS.
107
These properties enable us to construct the projections of
spherical triangles by convenient and simple methods.
123. To Project Any Circle of the Sphere, Given its Polar Dis-
tance and the Projection of its Pole. First Method. Let NESW
be the primitive circle (Fig. 64), and Z the projection of the
point diametrically opposite the point of sight, the latter lying
in the axis of the primitive circle at a perpendicular distance r
from Z, below the primitive plane.
Let P be the projection of the pole of the given circle ; draw the
diameters NS through P, and WE at right angles. The diameter
NPS is the projection of the great circle through P; and we re-
volve it about the diameter N8 into the plane of the primitive,
bringing tlie point of sight to W. NS is the line of measures, and
108
STEREOGRAPH: c PROJECTIONS.
Q } the intersection of WP (produced] with NQE, is the revolved
position of the pole P.
Lay off from Q, Qr and Qt equal to the given polar distance,
and draw Wr and Wt, cutting the line of measures in r and t 19
the extremities of the diameter of the required projection (Art.
119), and c, the middle point of r 1? is its center.
f
Second Method. Find Q, as in the first method, take Qr (Fig.
64) equal to the given polar distance, and draw Wrjr, giving one
extremity of the diameter at r v . Draw rV tangent at r, meeting
ZQV in V, the vertex of the tangent cone, and join VW, cutting
NS in c, the center of the required circle (see Art. 120, Fig. 61).
Draw the circle with radius c/v
This method is most convenient when the pole is on the circum-
ference of the primitive circle, as in Fig. 65. Here we simply
draw the diameter NS through P and WE at right angles, laying
off the polar distance Pr, and drawing the tangent at r to intersect
NS in the center c.
STEREOGRAPHIC PROJECTIONS. 109
124. To Project a Great Circle. When the polar distance is
90, the circle is a great circle and passes through W and E (Fig.
66), which are each 90 from P.
Proceed as for a small circle to find Q, lay off 0r = 90, and
draw Wr t r. The tangent cone to the sphere becomes in this case
a tangent cylinder, of which ZQ is the axis ; and the line from W,
parallel to ZQ, cuts the line of measures at c, the required center.
Draw the circle with radius cr^ cW.
Since W, r and E are three points of the required circle, a per-
pendicular to the middle point of Wr^ will intersect N8 in c, a
much less convenient way of getting c.
125. To Find the Locus of the Centers of the Projections of
All Great Circles Passing Through a Given Point. Let P (Fig.
67) be the given point through which the projections of great
circles are to pass; draw the diameters NFS and WE at right
angles, and draw WPQ to Q.
1. The projections of all great circles through P must also pass
through a point 180 from P; hence draw the diameter Qr, and
draw Wr, cutting NS, the line of measures, in T; then T is the
projection of the point 180 from P.
Since all the required circles pass through P and T f their center
must lie on the straight line perpendicular to PT at its middle
point c; the line is called the line of centers.
2. Since a great circle may always be drawn through the points
W, P and E (Art. 124), the point c may be found by drawing a
perpendicular to WP at its middle point, intersecting N8 in c.
3. The triangle WcP is isosceles, and the angle PWs=ihe angle
WPS, which is measured by J(90 +QN) =%QNW; i. e., the arc
QEs = QNW. Hence lay off QEs=QNW, and draw Wcs. This
is equivalent to laying off a polar distance QNW from Q; and
thus the line of centers is the projection of a small circle passing
through the iwe of sight and having the polar distance QNW
= 180 <, where denotes its inclination.
110
STEREOGRAPHIC PROJECTIONS.
4. From the figure, Wr = QE, and rSs=lSO -QEs = 180
-QNW=QE. Hence lay oft*WSs=2QE f and draw Wcs.
5. Since the angle WEs=QZE=WZr, a line joining E and s
is parallel to Qr, which gives a fifth method.
Of these methods the most convenient are :
( 3 ) Lay off QEs =QNW and draw Ws f
(4) l&joftWS8=2QE.
126. To Draw a Great Circle Through P Making a Given Angle
with NS. The tangent to the required circle at P (Fig. 67)
makes the required angle (x) with PZS (Art. 121) ; the per-
pendicular to the tangent (i. e., the radius), makes with PZS the
angle 90 x. Hence construct ZPC = 9Qx, intersecting the
line of centers in C, the center of the required circle.
STEREOGRAPHIC PROJECTIONS.
Ill
Note that the projection of a great circle always meets the
primitive circle at the extremities of a diameter, LK, of the
primitive circle.
127. To Find the Pole of a Given Circle. 1. Let Wr^E (see
Fig. 66) be a given great circle. Draw the diameters WE and NS
at right angles, and draw Wr r. Lay off rQ = 9Q and join WQ,
cutting NS in P, the required pole.
2. Let r^ (Fig. 64) be a given small circle; through its center
c draw the diameter NS, and WE at right angles. Draw Wr^ and
Wt 19 intersecting the primitive circle in r and t. Bisect the arc
tQr in Q, and draw WQ intersecting NS in P, the required pole.
128. Given the Projected Arc of a Great Circle, to Find its
True Length. Let NM (Fig. 68) be the arc to be measured.
Fing the pole X of this arc, and draw the straight line XMm;
then Nm is the true length required. For XMm is the projection
112 STEREOGRAPHIC PROJECTIONS.
of a small circle which passes through the point of sight; i. e.,
through a pole of the primitive ; and since it passes through the
poles of two circles, NM and the primitive, it makes equal angles
with them, the triangle NMm is isosceles, and NM=Nm.
The distance Nm might also be found by drawing the tangent
Me to meet NS at c; then a tangent to the primitive through c
meets the primitive at ra. In other words, c is the vertex of a
tangent cone of which a small circle passing through M and m
(pole at N) is the base. But this method would generally be in-
convenient.
The solution of the reverse problem is obvious, to lay off the pro-
jected length NM on a given circle Nr^S, from its true length Nm,
by simply drawing from the pole X, Xm to m, cutting the given
circle in M .
129. Spherical Triangles on the Earth Regarded as a Sphere.
Definitions. The axis of the earth is the diameter about which
the earth revolves.
The north and south poles are the extremities of the earth's axis.
The equator is a great circle of the earth perpendicular to its
axis.
Meridians are great circles of the earth passing through its
poles, and are therefore perpendicular to the equator.
The latitude of a point on the earth's surface is its angular dis-
tance north or south of the equator (measured on the meridian
passing through the point) .
The longitude of a point on the earth's surface is the angle at
the poles between the meridian of the point, and a fixed meridian
taken as the origin of longitudes, as the meridian of Greenwich,
or of Washington Longitude is reckoned from the zero meridian,
positive to the west, negative to the east, from O h to 12 h , or from
to 180.
130. To Find the Great Circle Distance between Two Given
Points. The angle between the meridians of the two planes is
STEREOGRAPHIC PROJECTIONS.
113
obviously the algebraic difference of their longitudes. The two
adjacent sides are the co-latitudes of the two places ; and we have
two sides and the included angle (Fig. 69) to project the triangle,
measure, and compute the desired parts.
FIG. 69.
Example : Find the great circle distance from Cape Flattery,
Long. (AJ =8 h 20 m W., Lat. (LJ =48 N., to Java, Long. (X 2 )
= 7 h 28 m E., Lat. L 2 = 9 S.
Take the meridian of Cape Flattery as the primitive circle (Fig.
69) and P as the North Pole of the earth. Lay off PM 1 = 90 L^
9
114
STEREOGRAPHIC PROJECTIONS.
= 42. Draw through P a great circle making the angle 123
with PMi (or 33 with PP') . See Art. 126. This is conveniently
done by laying off P'Q'c 114, twice the angle cfP', then draw-
ing PCC-L to c the center of the required circle PM 2 P'. Find its
pole p by Art. 127. Make PQs=99 (the north polar distance
of Java, 9 S.), and draw p s cutting off PM 2 , the projected length
of 99. Draw the diameter M^m^ and the diameter perpendicular
to it; then find the center of the great circle through MJl z by
perpendicular bisector of chord M^M Z) meeting line of centers at
c 2 , and draw MJtt z m^ The triangle PM^M 2 is the projection of
the given triangle.
After finding the pole, p 2 , of M^M 2 m^ the true length of M^M.,
is given on the primitive circle (Art. 128) by the arc D intercepted
between p^M^ and p 2 M 2 (produced), about 118 30' . The angle
PM^M 2 is measured by the projected arc mp of the great circle
90 from M 19 the true length of which is marked M^ on the primi-
tive circle, about 70.
131. Numerical Computation. From Art. 105, Spherical
Trigonometry, calling
\ 2 -Xi=A, 6 = 90-L 2 , c=90-L 1 ,
we get
tan (f> = cos ( A 2 Aj ) cot L 2 ,
$' = (90 -LJ -$ =
cosZ> = sinL 2 sec < sin(
cot(A 2 Ajcosec
L! 48 00' 00"
X 2 \,=123 00' 00"
^2
4
_9 00' 00"
= 73 47' 8"
cos
cot
tan
9.7361 In
0.80029w
cot 9.81252n
D
M,
D
0.53640
47' 8"
=118 30' 35"
= 70 23' 30"
=7110.6 geographical miles.
sin 9.19433w
sec 0.55404
sin 9.92943
cos 9.67880w
esc
cos
0.01763
9.72160n
cot 9.55175
STBREOGRAPHIC PROJECTIONS. 115
The values found from the projection agree as closely with the
computed values as could be expected from the scale of the pro-
jections (Fig. 69).
Examples.
Project the triangle and compute the great circle course
between :
(1) San Francisco, L 1 = 37 47' 48" N., A 1 = 8 h 9 ra 43 s W. ;
Manila, L 2 = 14 35' 25" N., A^=8 h 3 m 50 s E.
Ans. = 100 '50' 55".
M^N. 61 45' 34" W.
(2) New York, L=40 40' N., A= 4 h 55 m 54 s AY.;
Cape Good Hope, L = 33 56' S., X= l h 13 m 55 s E.
(3) San Francisco, L = 37 47' N., X= 8 h 09 m 43 s W.;
Sydney, L = 33 52' S., A=10 h 04 m 50 s E.
132. The spherical coordinates of a point on the earth, latitude
and longitude, as defined in Art. 129, can be determined only by
astronomical observations of the heavenly bodies, the sun, moon,
planets and fixed stars. ^ \ (fc^ftf
^In such observations these 01103 cotes are seen projected on the
background of the sky, and appear to have a revolution about an
axis from east to west. This apparent revolution is due to the
rotation of the earth on its axis in the opposite direction, but we
shall regard the apparent revolution of the heavenly bodies as real.
133. The Celestial Sphere. The celestial sphere is an imagi-
nary spherical surface of indefinite radius, having the eye of the
observer at its center, and upon which the heavenly bodies are
projected.
The poles of the sphere are the points in which the earth's axis
produced meets the celestial sphere, and about which the celestial
sphere apparently revolves once in 24 hours.
The equinoctial (or celestial equator) is the great circle 90
from the poles obviously the same as the great circle formed by
the indefinite extension of the earth's equator.
116 STEREOGRAPHIC PROJECTIONS.
Hour circles are great circles through the poles or the great
circles produced by the indefinite extension of the earth's
meridians.
The zenith and nadir are the points of the celestial sphere
vertically overhead and vertically underneath the observer. They
may be conceived to be the points in which the celestial sphere is
pierced by the indefinite extension of the radius of the terrestrial
sphere -through the position of the observer.
The celestial horizon is the great circle of which the zenith and
nadir are the poles ; or it is the great circle cut from the celestial
sphere by the tangent plane to the earth at the observer's position.
Since the radius of the celestial sphere is infinite, we may sub-
stitute for the tangent plane, a parallel plane through the center
of the earth, for these two planes will meet the celestial sphere
in the same great circle, which is the celestial horizon. (See Fig.
71, Art. 135.)
Vertical circles are great circles passing through the zenith and
nadir.
The celestial meridian (of a place) is the hour circle passing
through the zenith and nadir ; it may also be defined as the vertical
circle passing through the poles. The celestial meridian is ob-
viously the great circle formed by the indefinite extension of the
plane of the terrestrial meridian of a given place.
The prime vertical is the vertical circle passing through the east
and west points of the horizon, and is perpendicular to the
meridian and to the horizon.
The declination (d) of a point of the celestial sphere is its
angular distance north or south of the equinoctial, measured on
the hour circle passing through the point. It is marked north or
south like latitude on the earth.
The polar distance (p) of a point of the celestial sphere is its
angular distance from the pole; p = 9Q d if the declination is of
STEREOGRAPHIC PROJECTIONS. 117
the same name as the pole from which it is reckoned, otherwise
p = 90 + d.
The altitude (h) of a body is its angular distance above the
horizon.
The zenith distance (z) of a body is its angular distance from
the zenith; z = 90 -h.
The hour angle (t) of a point of the celestial sphere is the angle
at the pole between the meridian of a place and the hour circle
passing through the point.
The azimuth (Z) of a body is the angle at the zenith between
the meridian of a place and the vertical circle passing through the
body.
The amplitude of a body is the arc of the horizon between the
east or west point and the body when in the horizon.
The ecliptic is the great circle of the celestial sphere which is
the apparent path of the sun, due to the real motion of the earth
around the sun.
The vernal equinox, or the first point of Aries, is the point in
which the ecliptic cuts the equinoctial, as the sun passes from
south tojnorth declination.
The right ascension of a heavenly body is the angle at the pole
between the hour circle passing through the body and the hour
circle through the vernal equinox.
134. The Astronomical Triangle. The spherical triangle
whose vertices are the pole, the zenith and any point M on the
celestial sphere is called the astronomical triangle. The angle at
the pole P is the hour angle (t), the angle at Z is the azimuth
(Z), the angle at the point M is the position angle. The side
PZ is the co-latitude (90 -L) ; the side PM is the polar distance
from the elevated pole, and is equal to 90 d when the latitude
and declination have the same name, to 90 +d when they have
different names ; the side ZM is the zenith distance of the point,
or the co-altitude, 90 h.
118
STEREOGRAPH ic PROJECTIONS.
The relation of the astronomical triangle to the definitions of
Art. 133 is shown by the diagram of Fig. 70 :
NWS = Horizon.
Mrrii Vertical circle of M.
Mm, = Altitude of M.
= Azimuth of M.
PQP'Q' = Meridian.
(Q'WQ = Equator.
I PMP' = Hour circle of M .
mM Declination of M .
[ MPZ = Hom angle of M.
Either of these two groups forms &' system of spherical co-
ordinates, by which the position of a body on the celestial sphere
is known with reference to the observer's position :
1. When its hour angle and declination (or polar distance) are
known.
2. When its azimuth and altitude (or zenith distance) are
known.
135. The latitude of a place on the earth is equal to the declina-
tion of its zenith or to the altitude of the elevated pole.
STEREOGRAPHIC PROJECTIONS.
119
Let pzq (Fig. 71) be the terrestrial meridian of a place z, pp^
the axis of the earth, and q^Oq the equator.
Extend the axis of the earth, the equator, and the radius (or
vertical) of the place, to meet the celestial sphere in P, Q and Z.
The plane NOS, parallel to the tangent plane nzs at z, is the
celestial horizon.
The arc qz, which is by definition the latitude of z, is obviously
equal to QZ, which by definition is the declination of the point Z,
the zenith.
FIG. 71.
The elevated pole (P) is the pole visible above the observer's
horizon (the north pole for places in north latitude, the south pole
for places in south latitude) .
The arc NP, the altitude of the elevated pole P, is equal to QZ,
the latitude of z, since each is equal to 90 less the arc PZ. The
arc PZ, included between the zenith and the pole, is therefore
90 L, where L is the latitude of a place.
136. To Find the Latitude of a Given Place. The latitude of
a given place (i. e., the declination of its zenith) is most easily
found from the observation of the altitude of the sun or other
120
STEREOGRAPHIC PROJECTIONS.
celestial body when it is just on the meridian of the place. Sup-
pose the meridian altitude of such a body to have been measured
and reduced by various small corrections to give the true altitude
above the celestial horizon. The declinations of the brighter
celestial bodies are tabulated in the American Nautical Almanac.
With these two data the method of finding the latitude is shown
in Fig. 72, where the celestial sphere is projected on the plane of
the meridian of the given place. ZSNaN is the meridian, Z the
zenith, NWS the horizon, P the elevated pole, Q^Q the equator,
and QZ the latitude. If m is the observed body, then QM is its
declination and m^Z its zenith distance = 90 h : . We have from
the figure
If the body is in south declination, as at ra 2 , then
L=ZM z -QM 2 =Z 2 +(-d 2 ),
STEREOGRAPHIC PROJECTIONS.
121
south declination in this case being negative (the latitude being
north).
If the body is north of the zenith, as at M 3 , then
FIG. 73.
137. Projection of the Astronomical Triangle. Case I : Given
L = 20 N., d=32 N., h = 4:Q, body bearing W., to Project the
Triangle on the Plane of the Horizon. The three sides of the
given triangle are PZ = 70, PM=58, ZM = 50.
Let NE8W be the horizon, then Z will be the center of the
primitive circle (Fig. 73). Draw the diameters NZ8 and WZE
122 STEREOGRAPHIC PROJECTIONS.
at right angles, and take NZ8 as the line of measures. The point
of sight (Na) lies below Z perpendicular to the primitive plane
and at a distance r. The projected pole P lies between Z and N f
and for the purpose of finding this, the point of sight is revolved
about NS into the position W.
Lay off EA = 90 - L, and draw WA cutting N8 in P, the pro-
jected pole. Construct the small circle of 58 polar distance from
P by laying off Ar and Ar' equal to the polar distance 58, and
joining Wr f intersecting NS in r one extremity of the diameter
(Art. 123). The center c is found by the second method of Art.
123, the tangent at / intersecting ZA (produced) in V, and join-
ing WV, cutting NS in c; cr^ is the radius of the required circle.
Lay off EH=9Q -h = 5Q, and draw WH cutting N8 in h;
then Zh is the polar distance 50 from Z. With Zh as a radius
(and Z as center) draw a circle cutting Mr in M. This point
being at the given distance from both P and Z respectively, and
lying west of the meridian NS, is the required point.
Lay off As=ANW, and join Ws> cutting NS in c ; draw the
line of centers of great circles through P, perpendicular to NS
at c . Bisect the chord PM, and draw a perpendicular at its
middle point meeting the line of centers in c, the center of the
required great circle PM.
Draw the great circle ZM, passing through the point of sight,
as a straight line, completing the construction of PZM. The
values of the required quantities t and Z are indicated on the
projection ; the measure of the angle Z by the arc of the primitive
circle included between ZPN and ZM produced, approximately
64 ; the value of the hour angle ZPM=t is measured by the arc
of the primitive circle marked t, approximately 53 45', which
is the true length of Qm the projected arc of the equator, WQE,
intercepted between the two hour circles PZQ and PMm. The
angle PcC l is also a measure of the angle ZPN. (See Art. 126.)
STEREOGRAPHIC PROJECTIONS.
123
138. To Project the Same Triangle on the Meridian. Let
PZSN be the meridian (Fig. 74), and draw the diameters ZWNa
and NWS at right angles. Lay off PZ&3Q-L and draw the
diameters PP' and QQ' at right angles. Since the object is west
of the meridian, the center of the primitive circle is W.
FIG. 74.
From Z, lay off ZH = 90 h, and draw a tangent at H inter-
secting WZ in c, the center of the small circle having a polar dis-
tance of 90 -h (Art. 123, Second Method); with radius cH
draw the small circle HM.
From P lay off Pr=90 -d = 58, and draw a tangent at r,
intersecting PP^ in c x , the center of the small circle of 58.
With radius c^r draw the small circle rM, intersecting HM in M,
which has the given polar distance from both Z and P, re-
spectively.
124
STEREOGRAPHIC PROJECTIONS.
The perpendicular at the middle point of the straight line ZM
intersects NS, the line of centers of Z, in the center for the great
circle ZM. Construct the great circle PM in a similar way, giving
PZM, the projection of the given triangle.
The value of the angle PZM is given by the arc of the primitive
marked Z (about 64), which measures the projected arc of the
horizon Nm, which is in turn the measure of the azimuth. The
hour angle t is measured by the arc of the primitive marked t, the
true length of the projected arc Q'm^ of the equator, about 53 50'.
139. To Project the Same Triangle on the Equator. In the
projection (Fig. 75) the primitive circle is the celestial equator or
equinoctial, and the pole P is its center. " Call the vertical diam-
eter QQ' the projection of the meridian and lay off WA = 9Q L
^70, drawing EA cutting QQ' in Z, the projected zenith. Find
STEREOGRAPHIC PROJECTIONS. 125
c l9 the center of the small circle of (90 ft) =50 polar dis-
tance, and construct the circle; find Pb, the projected length of
the polar distance PM = 58, and draw the small circle inter-
secting the first one in M. Find the line of centers of all great
circles through Z, and the center c of that one which passes
through M. The construction is obvious. The value of t, 54 30',
is measured by the arc of the primitive circle as indicated. The
azimuth Z is measured by the true length of the intercepted arc
Nm of the horizon ENW, indicated by Z on the primitive circle,
about 64.
140. Numerical Solution of the Triangle. This problem is the
time sight, of almost daily occurrence in the navigation of a ship
at sea. The altitude of the observed body is measured with the
sextant, the latitude is assumed as known, and the declination
of the observed body is taken from the tables of the Nautical
Almanac.
To derive the formulae, used in the solution of the triangle, let
t = A, Z = B, in the formulae of Arts. 88 and 89 (Sph. Trig.),
sin o sin c
sin a sin c
Then
c=90-L
s c = %(L + p h)=s f h
where s 1 ' = %(h + L + p) ; substituting these values in the above
formulae,
sin 2 \t cos s' sin (s' h) sec L cosec p.
cos 2 \Z = cos s' cos (s' p)secL sec h.
126 STEREOGRAPHIC PROJECTIONS.
Note that p, the polar distance, is used, and not d.
In making the solution, always use the form given below.
h= 40 00' sec 0.11575
L= 20 00' sec 0.02701 sec .02701
p= 58 00' esc .07158
2s' = 118 00'
s'= 59 00' cos 9.71184 cos 9.71184
s'-h= 19 00' sin 9.51264
s'-p= 1 00' cos 9.99993
2)9.32307
*= 27 18' 12" sin 9.66153
t= 54 36' 24" 2(9.85453
3 h 38 m 25.6 s \Z 32 14' 34" cos 9.92726
Z 64 29' 8"
141. Case II : Given t, d and L, or Two Sides and the Included
Angle. Take L = 39 K, =2 h 20 m =35, d=S K
1. On the Horizon : Let NESW be the horizon (Fig. 76), and
draw NZS and WZE at right angles.
Find ZP, the projected length of 90 -L=EA = 51 ; construct
the projection of the equator, Ws'E, by drawing We parallel to
ZA (Art. 124), intersecting NS at c, the center, giving We, the
radius.
The pole p of the hour circle PM, making an angle of 35 with
PZ, must lie on the equator 35 from the pole of PZ or E; hence
lay off Ee = 35 and draw Pe, cutting the equator at p. Since the
poles and center of any great circle are in the same straight line
as the center of the primitive, draw Zpc^ meeting the line of
centers at c lf the center of the required circle, and construct the
great circle PM. From its pole p draw pm, making the true
length of PM, measured by p'm = 90 -d = 82.
STEREOGRAPHIC PROJECTIONS,
127
Draw ZMriL i the great circle through Z and M, completing the
construction of PZM, and find its pole p 19 laying off m 1 p 1 = 90.
To find the altitude h, draw pji^ giving mji l} the true length
of the projected h (m 1 7lf)=45 50'. The azimuth Z is given
directly by the arc NWm 1 = l2Q.
2. The same triangle projected on the meridian in Fig. 77,
where PZ 90 L as before. Draw the equator Q'WQ, which is
FIG. 76.
the line of centers, as also the locus of the poles, of all great circles
through P. Find the pole of PM making an angle of 35 with
the meridian by taking P'a=35, giving Wp the projected length
of the distance between the poles of the two circles ; find its center
c by taking P'& = 70, twice the angle P'Pc=35. From the
center c, draw PtP' and lay off Pm = 9Q -d = 82. Draw pm
cutting the great circle at M; PM projected length of Pm = 82.
128
STEREOGRAPH ic PROJECTIONS.
Draw a perpendicular to middle point of the chord Z'fiQ inter-
secting the line of centers of great circles through Z at c 1? the
center of ZMm^Na. Find its pole p i by taking ,sz t 90. Draw
FIG. 77.
p^M, giving the true length of the projected altitude Mm^ as
indicated approximately 45 50'. The azimuth PZM is measured
by the true length of the projected arc of the horizon N Wm^
that is, by the arc NQ'z, about 125.
3. The same triangle is projected in the plane of the equator in
Fig. 78, where SPN' is taken as the projection of the meridian,
and the projection of the zenith found by making PZ = 9Q
-L. Draw PMt, making the hour angle ZPM = 35, and find
the point M by drawing pM from pole p of the hour circle PM,
cutting off Mt = S projective length.
Bisect chord ZM as before by a perpendicular meeting the line
of centers at c; then construct ZM meeting horizon produced at
m, and find its pole p by joining cP.
The altitude Mm is measured by A-^ according to usual
STEREOGRAPHIC PROJECTIONS.
method, and is about 45 45'. The azimuth, MZP, is measured by
the true length of the arc of the horizon N Wm, or N'Wz
nearly.
142. Numerical Solution of Case II. By Napier's rules, drop-
ping the perpendicular from M on PZ :
tan = cos t cot d,
sin/t=:sin^ sin(Z/ + <)sec ,
cot Z = cot t cos (L + ) cosec <.
Note that when d has a different name from L, d is negative.
L = 39 00' 00"
cos 9.91336 cot 0.15477
cot 0.85220 sin 9.14356
= 35 00' 00"
= 8 00' 00"
= 80 15 53"
=119 15' 53"
= 45 53' 00"
=125 18' 40"
10
tan 0.76556
sec 0.77178
sin 9.94070
esc 0.00630
cos 9.68917n
sin 9.85604
cot 9.85024w
130 STEREOGRAPHIO PROJECTIONS.
143. Method of St. Hilaire. In the practice of navigation the
value of Z is taken from Azimuth Tables, and h is derived from
the formula of Art. 107 (Sph. Trig.),
sin h = sin d sin L+ cos d cos L cos i,
=cos(Ld) 2 cos d cos L sin 2 1/.
log cos L 9.89050 cos (L-d) 0.85717 (Table 41)
log cos d 9.99575 No. 0.13917
log 2 0.30103
0.18728 sinfc 0.71800 (Table 41)
log sin 2 $t 8.95628 A- 45 53' 23"
No. 0.13917 log 9.14356
The computed value of h is compared with the observed value,
thus furnishing a method of establishing the position of the ship,
instead of computing t by Art. 140.
Log sin 2 ^ may be taken directly from Table 45 of latest
edition of Bowditch's Useful Tables.
144. Case III : Given t, d and h, to Find L. We have given
two sides and an angle (t) opposite co-h, thus furnishing a double
solution. Assume t=32, d=21 N., ^ = 40 W. The projection
is made on the equator, on account of the simplicity of the con-
struction.
Let EQ'WQ be the primitive circle, the equator, which has the
pole P at the center. Draw the diameter QPQ', and EW at right
angles (Fig. 79). Construct the angle QPM =2=32, and PM
as shown in the figure equal to 90 d=89. Prom M as a pole
construct the small circle having a polar distance (90 h) =50,
on which lie all points having the given zenith distance of 50.
Then the two points Z^ and Z 2 , in which this circle of equal alti-
tudes intersects PQ, satisfy the conditions of the problem. Find
the line of centers for M and construct the great circles Z^M and
Z 2 M, completing the two triangles.
STEEEOGRAPHIC PROJECTIONS.
131
Since P is the projection of the North Pole, Z 1? lying between
it and the equator, is the projected zenith of a place in north lati-
tude, while Z 2 , lying at a greater polar distance than 90, is the
projected zenith of a place in south latitude. The true length of
Z^Q is measured by the arc marked L 1 = 66 50' N.; the true
length of QZ 2 , by the arc marked L 2 = 18 S.
FIG. 79.
145. Numerical Solution of Case III. Draw the perpendicular
from M to the side PZ Z 2) and by Napier's rules derive :
tan (j> = cos t cot d,
cos <' = cos sin h coseo d,
/, = 90 -(<#>: ' tan h.
132
STEREOGRAPHIC PROJECTIONS.
Note that ', being found by its cosine, may be either a positive
a negative arc.
or
h = 40 00' 00"
t = 32 00' 00"
d = 21 00' 00" N.
= 65 38' 46"
0'= (42 17' 55")
Z l= 40 13' 38"
Zj= 139 46' 22"
L 1= : 17 56' 41" S.
L,= 66 39' 9" N.
cos 9.92842
cot 0.41582
tan 0.34424
sin 9.80807 tan
esc 0.44567
cos 9.61529
9.92381
cos 9.86903 tan () 9.95899
cos () 9.88280
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