1^3 315 
 
 ummms km calgui.ations in 
 
 HYGIENE AND VITAL STATISTICS 
 
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METHODS AND CALCULATIONS 
 
 IN 
 
 HYGIENE AND VITAL STATISTICS 
 
 INCLUDING THE USE OF LOGARITHMS AND 
 LOGARITHMIC TABLES 
 
 WITH EXAMPLES FULLY WORKED OUT 
 
 BY 
 
 HERBERT W. G. MACLEOD. 
 
 M.D., M.S. EDINBUE(4H ; D.P.H. CAMBRIDGE ; D.P.II. LONDON 
 
 First Class Honours in Public Health; Gold ^fedal, Senior, and Classical Schohtr- 
 
 ships in Arts; Fellow of the Incorporated Society of Medical Officers of 
 
 Health; Member of the Sanitary Institute; Late Medical Officer of Health 
 
 and Captain H.M.'s I.M.S.; Assistant to the Medical Oficer of 
 
 Health of the County of Essex and the Lecturer on Public Health, 
 
 London Hospital ; Demonstrator State Medicine Laboratories, 
 
 King's College, London, ij'c. cj-c. 
 
 ^UustrateO^ 
 
 LONDON : 
 CHARLES GRIFFIN AND COMPANY, LiMrrEn 
 EXETER STREET, STRAND. 
 1904. 
 
 [All ri'jhts reserved.] 
 
t^3 
 
 UBRJUO- HEALTH 
 UBRARY 
 
 <?.'2- 
 
PEEFACE 
 
 Jn the following pages the most hnportant Calculations and 
 Formulae connected with Hygiene and Vital Statistics are ex- 
 plained, and are illustrated by numerous Examples fully worked 
 out. 
 
 An explanation is also given of Logarithms and Logarithmic 
 Tables and of their use. 
 
 The methods usually adopted are shortly described, so that the 
 Calculations which follow may more readily be understood. 
 
 In the Analysis of Foods, &c., details of laboratory technique 
 are omitted, as being outside the scope of the book. 
 
 The Examples are taken chiefly from records of practical 
 work, and some from Examination-papers. 
 
 Medical men and' others working at Public Health, find it 
 inconvenient to consult several books on Mathematics, Chemistry' 
 &c., for Calculations which it is necessary for them to know, and 
 the details of which are not given in standard Text-books on 
 Hygiene. To them in particular I hope these pages will be 
 found useful. 
 
 Work done in various Laboratories — at Netley (honoured by 
 the names of Parkesand de Chaumont); King's College, London ; 
 the Jenner Institute ; the University of Edinburgh ; &c., and 
 experience gained in teaching Hygiene, and as a Medical Officer 
 
 211138 
 
vi PREFACE 
 
 of Health, have acquainted me wilh the requirements of 
 Candidates for QualijScations in Public Health. 
 
 My thanks are due to Mr. G. Maxwell Lawford, M.Inst.C.E., 
 for his Formula} ; to Mr. J. E. Mackenzie, D.Sc, Ph.D., for kindly 
 revising some of the proof-sheets ; and to Dr. Glaisher, of 
 Cambiidge, for permission to reproduce the Table of Glaisher's 
 Factors. 
 
 I am obliged to Messrs. Baird & Tatlock, Casella & Co., A. 
 Gallencamp tt Co., J. J. Griffin & Sons, Mr. J. J. Hicks, Messrs. 
 Negretti tfe Zambra, and Town son &, Mercer, for blocks specially 
 made, or lent, for illustrating the text. 
 
 HERBERT W. G. MACLEOD. 
 November 1903 
 
TABLE OF CONTENTS 
 
 CHAPTER I 
 Chemistry 
 
 To determine the true Position of Equilibrium of a Chemical Balance : 
 Examples. Tempeiature Scales : Examples. Mass. Weight. Density: 
 Absolute and Eelative. Normal Temperature and Pressure. Units of 
 Weight and Volume. Avogadro's law. Gramme-Molecular Weight. 
 Density of Air and Gases : Examples. Changes of Volume and Tem- 
 perature with Constant Pressure. Charles' law. The Absolute Scale. 
 Formula : Examples. Correction of Volume for Normal Temperature. 
 Formula : Examples. Changes of Volume and Pressure with Constant 
 Temperature. Formula : Examples. Correction of Volume for Normal 
 Pressure. Formula? : Examples. Density under pressure. Formula? : 
 Examples. Boyle's and Mariotte's law. Formula : Example. Change 
 of Volume with Changes of Temperature and Pressure. Formula? : 
 Examples. Weight of a given Volume of Gas at a given Temperature 
 and Pressure. Formuhe : Examples. Mixtures of Gases and Vapours. 
 Dalton's law. Tension. Saturation. Maximum Density. Partial 
 pressures. Formula : Examples. Absorption of Gases. Dalton's and 
 Henry's law. Coefficients of Absorption and Solubility. Normal 
 Solutions. Valency : Examples. Parts per 100,000. Grains per 
 gallon pp. I -18 
 
 CHAPTER II 
 
 Specific G-ravity 
 
 Principle of Archimedes. Specific Gravity : Of a Solid ; by the Balance. 
 Formula : Example. By Nicholson's Hydrometer : Examples. Of a 
 body lighter than and insoluble in Water. Formula : Example. Of a 
 substance heavier than and insoluble in Water. Formula : Example. 
 The Specific Gravity tube : Example. Hare's Apparatus. Westphal's 
 and Sartorius' Balances : Examples. Twaddell's Hydrometer : Examples. 
 Beaume's Hydrometer and Scale. Specific Gravity of a Liquid relative 
 to Water : Examples VV- 19-25 
 
viii TABLE OF CONTENTS 
 
 CHAPTER III 
 Meteorology- 
 Weight of Aqueous Vapour : Examples. Vapour Tension Tables : Examples. 
 Weight of Air saturated with Moisture. Formuhe : Examples. Dry 
 Air. The Dew-point. Absolute and Relative Humidity. Formuhe : 
 Examples. Hygrometers : Daniell's, Regnault's, Dines'. Dry- and Wet- 
 Bulb Thermometers. Glaisher's and Apjohn's Formulae. Table of 
 Glaisher's Factors. The Baroscope : Example. Density of Air relative 
 to Water, to Mercury, and to Glycerine. Barometers. Barometric 
 Corrections for (i) Capacity. Formula?. (2) Capillarity : Table of Cor- 
 rections. (3) Index-Errors. {4) Temperature : {a) Mercurial-column. 
 Formula? : Examples, (h) Mercurial-column and Brass-scale. Formuhr : 
 Examples. Coefficients of Expansion of Mercury and Brass. Schu- 
 macher's Formula. Table of Corrections for Brass-scales. (5) Reduc- 
 tion to Sea-level. Formula : Example. Ordnance datum. Correction 
 for Gravity. Formula. Inches into mm. Formula : Example. Estima- 
 tion of Heights by the Barometer : Example. Strachan's Formula. The 
 Vernier : Divisions and Use. Temperature Records. Radiation : Solar 
 and Terrestrial. Graduation of Rain-gauges. Formula : Examples. 
 Wind-Velocity : Robinson's Anemometer. Formula : Examples. Wind- 
 Pressure. James' Formula : Examples . . . .pp. 26-50 
 
 CHAPTER lY 
 Ventilation 
 
 Pure and Expired Air. Velocity of Air-currents. Formula : Examples. Air- 
 Supply : de Chaumont's and Carnelley's Formula' : Examples. Air- 
 Supply for Horses and Cattle. De Chaumont's Formula for Size of 
 Inlets and Outlets : Examples. Diffusion. Friction. Dimension and 
 Shape of Apertures : Examples. Ventilation by Fireplaces and Circular 
 Fans : Examples. Velocity and Volume of Air-currents. Anemometers : 
 Examples. Estimation of Superficial and Cubic Space in Houses and 
 Hospitals: Examples. COo in Air : Pettenkofer's Methods PP- 51-75 
 
 CHAPTER V 
 Water 
 
 Unit of Heat. Specific Heat: Examples. Latent Heat. Water-Supply: 
 Hawksley's, Symons', and Pole's Formula^ : Examples. Yield of a 
 Stream. Formuhe : Examples. The Hydraulic Ram. The Suction 
 Pump : Example. Velocity of Efflux : Example. Tlie Syphon. The 
 Bramah Press ; Example. Chemical Calculations : Total Solids, 
 
TABLE OF CONTENTS ix 
 
 Chlorine. Hardness : Total, Permanent, Temporary : Examples. 
 Magnesia. Free and Albuminoid Ammonia. Wanklyn's Process : 
 Examples. Dissolved Oxygen. Thresh's and Winckler's Methods : 
 Examples. Oxidisable Organic Matter. Tidy's Process : Examples. 
 Nitrites and Nitrates : Griess' Test. Phenol-Sulphonic Method : 
 Examples. Standard Solutions for Estimating Poisonous Metals in 
 Solution 76-95 
 
 CHAPTER VI 
 
 Soil 
 
 Percentage of Air and Moisture : Examples. Specific Gravity : Apparent 
 and Real : Examples. Pore-Volume : Example. Water-Capacity : 
 Example pp. 96-98 
 
 CHAPTER VII 
 Sewerage 
 
 Circular Pipes. Wetted Perimeter. Hydraulic Mean Depth. Formula\ 
 Velocity of Flow: Eytelwein's Formula: Examples. Maguire's Formula. 
 Data for Separate and Combined Systems. Lawford's Formulae for 
 Velocity and Discharge. Egg-shaped Sewers : Examples . pp. 99-104 
 
 CHAPTER VIII 
 
 Diet and Energy 
 
 Standard Diets : Percentage Composition of Foods : Examples. Energy 
 evolved. De Chaumont's Formula : Examples. Energy of a Food 
 in Calories : Examples. Mechanical Work in Foot-tons : Examples 
 
 pp. 105 -1 10 
 
 CHAPTER IX 
 
 Foods 
 
 Milk. Correction for Temperature. Total Solids. Adam's, Werner-Schmidt's, 
 Hoppe-Seyler's, and Ritthausen's Processes : Examples. Richmond's 
 Formula and Slide-Scale : Examples. Milk Standards. Estimation of 
 Fat abstracted and of Water added : Examples. Butter : Moisture, 
 Volatile and Fixed Acids. Albumen in Foods : Kjeldahl's Method : 
 Example. Alcohol : Proof-Spirit ; Over- and Under-Proof. Alcohol 
 in Beer : Mulder's Formula : Example. Acidity in Beer : Example 
 
 pp. 111-120 
 
X TABLE OF CONTENTS 
 
 CHAPTER X 
 Logarithms and Logarithmic Tables 
 
 Definitions. Method of using Tables. To find the Logarithm of n Number. 
 Negative Characteristics : Examples. To find the Number from the 
 Logarithm : Examples. Multiplication, Division, liaising to a Power 
 and Extraction of a Eoot by Logarithms : Examples . pp. 121-129 
 
 CHAPTER XI 
 Population 
 
 Estimation by Logarithms. Registrar-General's Method. Increasing and 
 Decreasing Populations : Examples worked out by Geometrical and 
 Arithmetical Progression. Newsholme's Method : Example. Marriage- 
 Rate : Example. Birth Rates : Annual, Quarterly, and Weekly : 
 Examples. Death-Rates : Crude, Corrected, and Standard. Registrar- 
 General's Factor. Correction for Non-Residents. Comparative Mor- 
 tality Figure. Infantile Mortality : Examples. Zymotic Death-Rate. 
 Case Mortality. Incidence of Disease. Combined Death-Rates. Density 
 of Population. Formula : Examples .... pp. 130-138 
 
 CHAPTER XII 
 Life-Tables 
 
 Data required. Formula^ : Examples. Expectation of Life : Examples 
 worked out. Mean Duration of Life : Example. Farr's, Willich's and 
 Poisson's Formula^ : Exami)les PP. 139-143 
 
 Appendix 144 
 
 Index 14S 
 
METHODS AND CALCULATIONS IN HYGIENE 
 AND VITAL STATISTICS 
 
 COREIGENDA. 
 
 Page 25, line i, for Hygrometers read Hydrometers. 
 ,, 51, footnote, for 0.3 per cent, read 0.03 per cent. 
 
 ,, 52, line 4 Pt xeq. for — ^ =4000 to end of (i). 
 40 
 144 
 
 600 
 
 , 12000 , 36000 ^ 
 
 read — 77- = 36000 -^-^ — = 600 , 
 48 ^ 60 60 
 
 10- — • = 7-5 feet per second. 
 4 
 59, line 19, for A — 1 1. 25 square inches read A = 20 square inches . 
 61, ,, 6, for greater than read less than. 
 
 , ^ ,. J. 1000 , 1000 
 
 91, last line, for -\ = i read .'. — — — i . 
 
 4/ 4/ 
 
 82, lines 10 and 11, for x = 666,468 cubic feet x 6.25 
 
 = 4,165,425 gallons 7-ead : 515=170,615,808 cubic feet x 
 
 6.25 — 1066348800 gallons. 
 
 95, line 6, for i c.c. =0.1 grm. Pb read i c.c. =0.1 mgr. Pb. 
 
 ,, 7 from bottom, for i c.c. = 1 mgr. Pb read i c.c. =0.1 
 
 mgr. Pb. 
 
 ,, ,, 12, for I c.c. =0.1 grm. Cu read i c.c. =0.1 mgr. Cu. 
 
 103, Fig. 24 (Note), /o?- radii of the sides is 1.50 read are 1.50. 
 
 107, line 5, /or 1.3 read 1.2. 
 
 113, ,, 21, for in 100 c.c. read in 10 c.c. 
 
 115, ,, 10, for total solids read non-fatty solids. 
 
 ,, ,, I ^, for y -.6 read o.y -.0.6. 
 
 1 26, last line, for 7 + 6=1 read 7+6=1. 
 
 135, line 3, omit (or crude death-rate), 
 ,, ,,8 from bottom, omit (crude). 
 
 ,, ,, II ,, ,, „ {i.e., the crude death-rate). 
 
 136, ,, 14 ,, „ /or notified cases rectd^ notified deaths. 
 
 ,43,/o.A/3!5?„„rfA/93» 
 V 1000 ▼ 1000 
 
X TABLE OF CONTENTS 
 
 CHAPTER X 
 
 Logarithms and Logarithmic Tables 
 
 Definitions. Method of using Tables. To find the Logarithm of a Number. 
 Negative Characteristics : Examples. To find the Number from the 
 Logarithm : Examples. Multiplication, Division, Raising to a Tower 
 and Extraction of a Eoot by Logarithms : Examples . pp. 121-129 
 
 nFTAPTKR, XT 
 
METHODS AND CALCULATIONS IN HYGIENE 
 AND VITAL STATISTICS 
 
 CHAPTER I. 
 
 CHEMISTRY. 
 
 To Determine the true Position of Equilibrium in a 
 Chemical Balance. — Raise the beam and free the pans by turn- 
 ing the handle. If the oscillations of the pointer are of equal length 
 to the right and left of the scale the instrument is correct. If of 
 unequal lengths, correct as follows by taking an unequal number 
 of readings (say 3), commencing with the first swing of the index. 
 
 Oscillations of the pointer to the right of the zero are written 
 + ; those to the left - . 
 
 Example. 
 
 ' ^ ist to right 8.6 ^ oscillations ( f */" '^^^ 5-3 
 
 3 Oscillations - 2iid „ 7.8 L 2nd „ ^.5 
 
 I Z^^ „ 6.2 
 
 22.6 9-8 
 
 22.6 9.8 
 
 7-5-4-9== +2.6 
 
 The balance points 2.6 in excess to the right. 
 
 2 6 
 .'. -^-= 1.3 to the right is the zero-point. 
 
 (2) + 
 
 ^, .„ ^. f ist to right 4.3 f isttoleft 7.2 
 
 2 Oscillations I ^^_^ „ 3.2 3 Oscillations J 2nd „ 63 
 
 I 31-d n 55 
 
 7.5 I9-0 
 
 + 
 
 2 ^ 3 
 
 A 
 
2 CALCULATIONS IN HYGIENE 
 
 Difference = 2.5 to the left in excess. 
 
 .-. _l5.= 1.2 to the left is the zero-point. 
 
 2 
 
 Temperature Scales. 
 
 The three scales in use are (i) the Centigrade or Celsius — univer- 
 sally used for scientific work ; (2) the Fahrenheit, adopted in Great 
 Britain for Meteorological observations and largely used in Hygiene; 
 (3) the Reaumur, wiiich is used in Russia chiefly, and may be 
 neglected. 
 
 The Centigrade and Reaumur scales have the zero-point at the 
 temperature of melting ice or freezing water. Fahrenheit, by using 
 a mixture of ice and salt to determine his zero temperature, obtained 
 it 32 of his degrees below the melting-point of ice. This is there- 
 fore marked 32° F. and coincides with 0° C. and 0° R. 
 
 The boiling-point of water is 100° C, 212'' F., and 80" R., the 
 respective scales being divided between the freezing- and boiling- 
 points into 100, 180, and 80 equal divisions. 
 
 As Fahrenheit's freezing-point is 2)^^ -^ ^^^^l 180 divisions separate 
 this from the boiling-point, the latter equals 32 + 180= 212''. 
 
 rni r C F - 32 R 
 
 Therefore _ o _ 
 
 Example. 
 Convert i 
 (i) 68.3^ C. 
 
 (2) -40^ C. 
 
 100 212 — ^ 
 
 33 80 
 
 C_F-32 
 
 R 
 
 10 18 
 
 8 
 
 ,r ^=.^-32. 
 
 _!'- 
 
 5 9 
 
 4 
 
 ihrenheit degrees : 
 
 68.3 F-32 
 
 9x68.3 = 5(^-32) 
 
 5 9 
 
 614.7 ="5 ^ ^ 160 
 
 
 614.7 -f 160 = 5 F 
 
 
 l^--i54-94"- 
 
 -4o_F-32 
 
 
 5 9 
 
 
 -8-^^-^^^ 
 
 -72 = F-32 
 
 9 
 
 F= -72-1-32 
 
 
 F= -40 
 
 -40' C. and -40"" F. correspond in both scales, and closely 
 approximate to the freezing-point of mercury. 
 
CHEMISTRY 
 
 (3) -27' C. -27 F-32 
 
 5 F-i6o= -243 
 5F=-83 
 
 F= - 16.6' 
 
 Convert into Centigrade degrees : 
 
 (1) 31= F. 31-3^ ^0 . -. 
 
 . . =-. — ^ — c =z Q y^ 
 
 9 5 .95 
 
 C= — ^= ""0-5^ (recurring decimal) 
 
 (2) o F. 0-32 
 
 ~9 ^5 
 
 -32_C 9C=-i6o 
 9 5- C=-x7.7'^ 
 
 (3) -19 F. -i9-32 ^C -5^^C ^eJ=-255 
 
 9 5 9 5 C=-28.3^ 
 
 Density. 
 
 Mass is the amount of matter a body contains. 
 
 Weight expresses tlie force with which the mass of a body is 
 attracted to the earth by gravitation, and represents the mass of 
 the body weighed. 
 
 Mass is an invariable quantity, but weight varies with the force 
 of attraction and difters with the latitude of a place. For con- 
 venience the mass of a body is expressed in terms of weight. 
 
 Density is " absolute "' and " relative." 
 
 Absolute density is the mass of unit volume of a substance, 
 and is estimated by weighing a measured volume. 
 
 Relative density is the ratio of the mass of any volume of a 
 substance to the mass of an equal volume of another substance 
 taken as a standard. The standard usually taken for solids and 
 liquids is pure distilled water at 4*^ C. or 39° F. — its temperature 
 of maximum density, and the density of any substance relative to 
 .water is its specific gravity. In other words, its weight in air 
 divided by the weight of an equal volume of water at 4° C. or 
 39° F. expresses its specific gravity. 
 
 In Hygiene and Physics air is used as the standard for gases 
 instead of water, and the temperature 15.5° 0. or 60° F. 
 is usually adopted for convenience, that being the degree of 
 warmth commonly met with in laboratories. It may be denoted 
 standard laboratory temj^eraturef ?a-iA i\\Q pressure of 760 mm. of 
 mercury or 30 in. is taken as the standard or normal pressure. 
 
4 CALCULATIONS IN HYGIENE 
 
 " Normal temperature and pressure," written " N.T.P.," means 
 o° C. or 32" F., and 760 mm. or 30 inches. 
 
 Tlie mass of a body is directly proportional to its density and 
 volume : Absolute density x Volume = Mass. 
 
 , , , ^ - .^ Mass Weight (expressing mass) 
 ... Absolute density ^,^^-^-^^= ^^^^^ 
 
 A gravimetric or quantitative analysis by weight estimates the 
 weight of the various substances a body contains. 
 
 A volumetric or quantitative analysis by volume determines 
 their respective volumes and is applicable to liquids and gases. 
 From the known volumes the weights can be calculated by the 
 laws of chemical equivalence. 
 
 In the metric system the unit of weight or mass is the gramme 
 and the unit of volume for solids and liquids is the citbic centimetre, 
 and for gases and vapours the litre or cuhic decimetre. 
 
 The British units of weight and volume are t\\QpoiLnd and grain, 
 and the cuhic foot or cid}ic inch. 
 
 The gramme is (very nearly) the weight of i c.c. of distilled 
 water at 4° C. and 760 mm. pressure. This system has the advan- 
 tage of expressing weight and measure at the same time, so that 
 I c.c. represents the corresponding weight of water in grammes, 
 and the weight of i c.c. indicates both the density and specific 
 gravity of any substance. At temperatures of iS-S"" C. or 60° F., 
 instead of at 4° C, the difference in weight is so insignificant that 
 the error may be neglected, except in very accurate estimations. 
 
 In the British system the pound and cubic foot have not the 
 convenient relation between the mass and bulk of water as in the 
 French system — now almost universal in scientific work. The 
 British units of mass and volume give different numbers expressing 
 density and specific gravity, and are not directly convertible. 
 
 In chemical calculations the unit of weight for gases and vapours 
 is the weight of a litre of dry hydrogen at 0° C. and 760 mm. 
 (32'' F. and 30 in.). This is 0.08958, or shortly 0.0896 gramme. 
 The weight of i c.c. of hydrogen is therefore y^Vo ^f this, and is 
 equal to 0.0000896 gramme, "N.T.P." 
 
 The cubic centimetre is too small a unit for estimating the 
 volume of gases, and weight per litre has the disadvantage of 
 expressing density in fractional parts of a gramme. 11.16 litres 
 are therefore used as the unit of volume for vapours and gases and 
 the densities are expressed as whole numbers. 
 
 The absolute density of a gas is the mass of 11.16 litres at 
 N.T.P. By Avogadro's law : " The molecular weight in grammes 
 of any gas at 0° and 760 mm. occupies a volume of 22.32 litres." 
 
 One litre of hydrogen under normal conditions weighs 0.0896 
 
CHEMISTRY -"> 
 
 gramme, therefore by the following equation we find the volume 
 of I gramme of hydrogen (N.T.P.) is 11.16 litres : 
 
 o.o8q6 : I : : I : a?. x=^ 7^^= 11. 16 
 
 0.0896 
 
 and that of 2 grammes of hydrogen (N.T.P.) is 22.32 litres. 
 
 0.0896 : 2 : : I : x. 
 .T = 22.32 litres. 
 
 The latter is known as the " gramme-molecular volume of hydro- 
 gen," as it expresses the volume of its molecular weight in grammes. 
 
 " Equal volumes of gases at the same temperature and pressure 
 contain the same number of molecules, and the masses or weights 
 of their molecules are in the same ratio as the densities of the 
 gases to which they belong " (Avogadro). 
 
 Therefore under normal conditions : 
 
 22.32 litres of hydrogen weigh 2 grammes. 
 
 22.32 ,, oxygen weigh 16x2 = 32 grammes. 
 
 22.32 „ carbon dioxide weigh 22 x 2 = 44 grammes. 
 Therefore, if the molecular weight of any element or compound is 
 X, 22.32 litres of it will weigh x grammes. 
 
 The density of nearly all elements in the state of gas or vapour 
 corresponds to their atomic weights. Taking hydrogen as i, the 
 
 •22 . 44 
 
 vapour-density of oxygen is or 16, and that of COois — ^ = 22. 
 
 The density of any gas, simple or compound, is half its molecular 
 weight. The weight of i litre of hydrogen under N.T.P. is 
 0.0896 gramme, and the weight of i litre of nitrogen under 
 similar conditions is 0.0896 x 14 (the atomic weight of nitrogen) 
 or 1.25 grammes. The weight of i litre of oxygen is 0.0896 
 
 . , 18 
 x 16 = 1.43 grammes, i litre of aqueous vapour weighs — 
 
 44 
 X 0.0896 = 0.806 gramme, and i litre of 000 = — x 0.0896 = 
 
 1.97 grammes. Air is 14.44 times heavier than hydrogen. 
 
 Air contains ajyproximately 21 per cent, of and 79 percent. N 
 by volume. 
 
 O. 21x16= X'i6\ 1 c • 
 
 T.T "'^^ -1442 grammes per 100 volumes 01 an\ 
 
 N. 79x14=1106) ^^ & ^ 
 
 100 X I = 100 grammes per 100 volumes of hydrogen. 
 .*. I volume of air =14.42, i volume of H=i; i.e.^ air is 
 14.4 times heavier, 
 
 and I litre of dry air under standard conditions weighs 0.0896 x 
 14.44=1.293 grammes {vide p. 4). If the weight of air is 
 
CALCULATIONS i:: HYOIENE 
 
 taken as the unit (as in Meteorology and Physics), the relative 
 weight or specific gravity of an equal volume of hydrogen is 
 
 14.44 tiiiies less than i, and is equal to =^0.0693. 
 
 If the density of two gases or vapours relative to hydrogen is 
 known, the ratio between them can be calculated at once. 
 
 Thus, the density of CO., to hydrogen is or 22, and that of 
 
 air is 14.44 5 therefore, 
 
 22 
 density of CO^ : density of air : : 22 : 14.44, or as = 1.52. 
 
 Similarly the vapour density of water relative to air is first ex- 
 pressed in " terms of hydrogen " and the ratio is then made ; this is 
 
 9 
 14.44 ^ ^ 
 
 The molecular weight of a compound is the total weight of the 
 atoms in a molecule of it. It is calculated from the chemical formula 
 by multiplying the atomic weight of each element in it by the num- 
 ber of atoms of that element and adding these numbers together. 
 
 Example. — Nitric anhydride is represented by NoO^. The 
 atomic weight of N= 14, 0= 16, and there are in the compound 
 2 and 5 atoms of each. Therefore 14X 2 + 16x5 = 28 + 80= 108, 
 which is the molecular weight. 
 
 Sulphuric acid = H^SO^. 
 
 .'. 1x2 + 32 + 16x4 = 98, is the molecular weight. 
 
 Sodium sulphate = Na2S04 I oH^,0. Its molecular weight is: 
 23X? f32 + i6x4+iox 18 = 322. 
 
 Oxjdicacid: C.,H.,0^.2H,0 : 12x2 + 1x2 + 16x4+18x2 = 1 26. 
 These parts by weight may be " grammes," " pounds " or " grains." 
 
 The percentage composition of each element is found by simple 
 proportion from the molecular weight : 
 
 Taking nitric anhydride : 
 
 The percentage amount of nitrogen and oxygen it contains is 
 
 calculated thus : 
 
 o o 2800 
 108 : 100 : : 28 : x. x = ^ =25.9 nearly. 
 
 loS : 100 : : 80 : r. .r = 74. i approximately. 
 
 And 25.9 + 74.1 = 100. 
 
 Oxalic acid : C= 19.0 per cent. 
 H= 4.8 „ 
 = 76.2 „ 
 
 00.0 (grammes or grains). 
 
CHEMISTRY 7 
 
 A chemical equation expresses the results of chemical action and 
 indicates the parts by weight of the molecule or molecules of each 
 substance employed in the reaction. 
 
 The weights may be of any system, but must be of the same 
 nature throughout the equation. 
 
 Atoms do not exist as such in the free state, but are in combina- 
 tion to form one or more molecules. Mercury is an exception, 
 its molecule consists of one atom only. 
 
 Thus : 
 
 SO, + I. + 2Hp = H^SO, + 2HI. 
 
 /i molecule = 64\ , /i moIecure=2S4\ i (^ mulecules =\ _ /i molecule =\ , /= molecules =\ 
 V parts by v/eightj T" Vparts by weight/ t- V 36 pans ) \ 98 parts ) ^ \ 256 parts / 
 
 If 10 grains of H^SO^ are required, how many grains of SO.^ 
 and I, would be needed ? 
 
 98 : 10 : : 64 : a;. a; = -g° = 6 53 grains (nearly) of SO3. 
 
 98 : 10 : : 254 : x. .t = ^^g° = 26 grains (nearly) of iodine. 
 
 2HgO = 2Hg + O3 
 
 (2 molecules = 432) = (i molecule = 400) + (i molecule = 32) 
 How much oxygen will 5 grammes of HgO yield ? 
 
 ••• 432 : 5 : • 32 : a'- a; = 0.37 gramme of oxygen. 
 
 In calculations of the volumes of gas formed by chemical com- 
 bination the relations between molecular weight and volume of 
 the respective gases must be known. One molecnle of a gas under 
 N.T.P. occupies the same volume as one molecule or two atoms of 
 hydrogen i.e., i volume or 22.32 litres, which is the gramme- 
 molecular volume expressing both weight and volume. 
 
 The volume of the gas in the last example is easily calculated 
 from the equation : the gramme-molecular volume Oo at N.T.P. 
 is 22.32 litres, and 432 grammes of HgO yield that volume as 
 per equation. 
 
 Therefore : 432 : 5 : : 22.32 : x. ^^^^2 =^-^^^ ^^^^^ ^^ oxygen. 
 
 The molecule of a gas occupies i volume at o'' C. and 760 mm. 
 and each gramme-molecule is 22.32 litres, but the volume alters 
 with change of temperature. 
 
 Changes of Volume and Temperature, Pressure remain- 
 ing the same.— By Charles' or Gay-Lussac's law the volume of gas 
 
 at constant 2)ressure expands or contracts -—( = 0.003665) of its 
 
 volume at 0° C. for each increase or decrease of 1° C. 
 
 In the Fahrenheit sc:ile gases at 32^' F. increase or dimmish 
 
8 CALCULATIONS IN HYGIENE 
 
 — - (0.002^ of their volume for every rise or fall of 1 F., the 
 
 491 
 
 pressure remaining the same. These fractions are " coefficients of 
 
 expansion " of gases in the respective scales. 
 
 In other words : 
 
 At 0° C. 273 volumes of a gas under constant pressure become : 
 „ 1° C. = 273 + I volumes, and at - i "* C. = 273 - i volumes. 
 
 „2''C.-273 + 2 „ „ -2°C. = 273-2 
 
 „T"C. = 273 + T „ „ -TC. = 273-T 
 
 Similarly at 32"" F. 491 volumes of gas under constant pressure 
 become : 
 
 at 33° F. =--491 + I volumes, and at 31° F. =491-1 volumes. 
 „ 34° F. =491 + 2 „ „ 30" F. =491-2 „ 
 
 „32 + T- = 49i+T „ „ 32-T°=49i-T . 
 
 As the volumes of all gases vary directly with absolute tempera- 
 ture (the zeros being - 273° C. and - 491° F.), 273 or 491 must be 
 added to all observed or required temperatures in making calcu- 
 lations and corrections for changes of volume so as to express 
 them as degrees on the "absolute scale." Thus, 15° C. becomes 
 2734-15, and- 150. = 273- 15. 62° F. = 491 +62, and -20° F. 
 = 401 - 20. 
 
 V 273 + T 
 
 The formula for the Centigrade scale is ^j- = -7?r, where 
 
 V, 273 -M2' 
 
 Tj and V^ represent the original temperature and volume and T^ 
 
 and v., the temperature and volume for which the calculation or 
 
 correction is made. 
 
 Example. — What will be the volume at- 10° C. of 120 c.c. 
 
 of a gas measured at 1 5 ° C. ? Here Yj = 1 20. Tj = 1 5, T^, = - i o. 
 
 v., is to be calculated. 
 
 120 273 4-15 120(273-10) ^,0^^ 
 
 —^ = --7 x. V.,= , = lOQ.c^b c.c. 
 
 ^■> 273 + (-io) - 273-M5 ^^ 
 
 (2) 15 c.c. of CO^, are measured at 12" C. What would the 
 volume be at 36^ C. ? 
 
 It; 273H-I2 -r,- ^ ^ 
 
 -^=—L^ Y.,= 16.26 C.C. 
 
 ^2 273-1-36 
 
 In Fahrenheit's scale the formula is t^ = ^^—\rJ^ I 
 
 V, 49i-F(T,-32) 
 
 or by deducting 32 from 491 at once : — J = i^^ — -' 
 ' ^'> 459 + ^> 
 
CHEMISTRY 9 
 
 . y _ (459 + T,)xY , 
 
 ■■ ' 459 + T, 
 
 Example. — What volume will 128 cub. in. of nitrogen measured 
 at 32° F. occupy at 122° F. (constant pressure)? 
 
 £28^ 459±3^^ J -J eubic inches. 
 
 To Correct the Volume of a Gas for Normal Temperature. 
 
 Let Vj = observed volume of the gas 
 
 Tj — observed temperature of the gas. T = 0° C. 
 
 .*, v., = volume at 0° C. 
 
 . V._ 273 + T, 
 
 ' ' V, 273 FO" 
 
 V(273 + T0 = 273xV, .-. V, = |g^ 
 
 Example. — Find the volume which 147.8 c.c. of a gas originally 
 measured at 50° C. will occupy when the temperature falls to 
 o°C. 
 
 ^j 273 X 147.8 
 
 Y.^ = —^ ^t^^— = 124.9 c.c. 
 
 273 + 50 
 
 Fahrenheit scale : ^ = '\^l±^L.Zll} but T., = 32 " 
 V, 49i+(l\,-32) 
 
 ._ V^_ 49i+(Ti-^32) _459 + T, 
 v., 491+0° 491 
 
 y.= 
 
 49T X V, 
 
 459 + '-l\ 
 
 Example. — Calculate the volume which 147.8 cubic inches of a 
 gas measured at 122° F. will occupy at normal temperature, 
 pressure remaining unaltered. 
 
 401 X 147.8 491 X 147.8 , . 
 
 T,= ^2° F. y,= ^^ , — ±1-1^ -^*^^ = 124.9 cb.m. 
 
 2 '^ - 49T+(i22-32) 459 + 122 
 
 The density of a gas varies inversely as the absolute tempera- 
 ture if pressure remains unaltered. 
 
 Let Dj and D, be the densities at temperatures Tj and T.,. 
 
 Dj_273 + T, 
 •'• D,-273 + T/ 
 
 Example. — At what temperature will air having a density of 
 14.44 (relative to hydrogen as i) be of the same density as oxygen 
 at 0° C. or 32° F., pressure being constant? 
 
10 CALCULATIONS ]N HYGIENE 
 
 T., is to bo found. 
 
 1444 ^ 27 3 +^T,, _ 2 7 3 + T, 
 16 273 + T^ 273 + 
 
 i6(273 + TJ= 14.44x273. 
 
 T,= -26.6 C. 
 
 In the Fahrenheit scale : 
 
 14.44 491 + CA> - 32 ) _ 459 + T o _ 459 + T, 
 16 ~49i+(l\-32) 459 + Ti 459 + 32* 
 
 .-. 16(459 + 1:0=14.44x491. 
 
 a>- 15.87° F. 
 
 Changes of Volume and Pressure, Temperature remaining 
 the same. — By Boyle's and Mariotte's law the volume of a gas 
 varies inversely as the pressure, temperature remaining constant. 
 
 The formula is ^~- = ^\ 
 
 Example. — 237 litres of a gas are measured at 755 mm. pressure. 
 Find the volume if the pressure be changed to 915 mm., the tem- 
 perature being unaltered. 
 
 Vj = 237 litres. V., is to be found. 
 P. = 9i5- Pi = 755- 
 
 , M7^9i5 y^^ 23^^15^ 1-,^.^,^ 
 
 V, 755 ' 9^5 ^^^ 
 
 To Correct the Volume of a Gas for Normal Pressiire. 
 
 Let Vj =^ observed volume. 
 
 Pj = observed pressure (height of barometer in mm.). 
 V2 = V„ = volume at normal pressure (760 mm.). 
 P,= 760 m.m. 
 
 v,_p., y, X p, y^ X p, 
 
 •*• v,~p; •*• "~ p, " 760 • 
 
 Example. — Correct for normal pressure the volume of i litre of 
 a gas measured at 740 mm. pressure, temperature being un- 
 altered. 
 
 I litre = 1000 c.c. P,, = normal pressure == 760 mm. 
 
 1000 760 1 000 X 740 , 
 
 • —r^= . y = ^-^ =07^-6 c.c. 
 
 •• \^„ 740 " 760 ^'^ 
 
CHEMISTRY 11 
 
 In the Fahrenheit scale : 
 
 Let Vj = looo cubic inches. 
 
 v., = Y„ = volume at normal pressure. 
 Pj = 29.2 inches. 
 P, ^ 30 inches. 
 
 1000 _ 30 
 
 1000 X 29.2 ... , 
 
 Y = — = 973.3 cubic inches. 
 
 Under constant Temperature Density varies directly 
 
 D P 
 as Pressure. — The ratio is expressed thus : --' = ^\ 
 
 We have already seen that density varies inversely as the 
 absolute temperature if pressure is constant, the ratio being : 
 D. 273 + T ,. 
 1^^273 + ^^ 
 Example. — The density of air at 760 mm. pressure is 14.44, at 
 what pressure will it equal that of normal oxygen ? 
 
 D^= 14 44. D^ =- 16. Pi = 760. P, is to be calculated. 
 14.44 760 
 •■• 16 ~ P/ 
 
 ^ 16 X 760 ^ 
 
 P,= = 842.1 mm. 
 
 14.44 
 
 Expressing pressure in inches of mercury instead of as milli- 
 metres, the same example becomes : 
 
 -^^ = 1?. P,= ^^'2 inches. 
 16 Po ^ ^"^ 
 
 V P 
 
 By Boyle's and Mariotte's law : y' = p^ (P- io)> ^nd as density 
 
 P, d/ ' ^1 ?2_5.> • 
 
 varies directly as pressure p^ = -jy, therefore ^ - p - jj ' ^-^-j 
 
 temperature remaining constant, volume varies inversely as 
 density, and vice versa density varies inversely as volume. 
 
 Example. — 250.4 volumes of nitrogen have a density of 14. 
 Find the volume when the pressure is diminished and the density 
 becomes 12. 
 
 Yj= 250.4 r)i=i4 
 
 Yg is to be found. D2=i2. 
 
12 CALCULATIONS IN HYGIENE 
 
 250.4 12 ^^ 2^0.4 X 14 
 •i^ = — . V, = - = 202. 1 volumes. 
 
 y, 14 - 12 
 
 Change of Volume with simultaneous Changes of Tem- 
 perature and Pressure. — The formuUe of Charles' and Boyle's 
 laws are used together thus : 
 
 (•) V-V-,^- Or: V,:V,::(273 + T,):(273 + T,). 
 
 (2) ^ = '^. Or:V, :V,::P„:P, 
 
 Combining these two formul.ie for correcting the volume for 
 temperature and pressure simultaneously : 
 
 ^=r^7^1xF- Or:V,:V,::(273 + T,)xP,:(273 + T,)xP, 
 
 *! (273+ -Ij) ^i 
 
 ■■ -•-'^'''(273+'',) i^, 
 
 , V, 49i+(T, -32) P., 
 In the Fahrenheit scale : y- ~ '7^rsrr\r^T2\ ^ P" 
 
 (459 + 'r,)xP, 
 
 •• -'^ ■ (459 + 'J',) P, 
 
 Example. — 250 litres of a gas are measured at 745 mm. pressure 
 
 and 15° C. Calculate the volume at 25° C. and 765 mm. 
 
 pressure. 
 
 2c;o 27-^ + I s 
 (i) By Charles' law: J- = ^^i^. 
 
 2^0 76c: 
 
 (2) By Boyle's law: -A- = ^^. 
 
 . 250^(273 + 15)^^765 y _ 250 X (273 + 25) ^^745 
 
 ^, (273 + 25) 745* ' (273+15) 765 
 
 = 251.9 litres. 
 
 Example. — 250 cubic inches of dry air are measured at 29.4 
 inches and 5"" F. Find what the volume would be at 30.2 inches 
 and 77 F. 
 
 250 (459 + 5)^30-2 ,. 25ox(459 + 77) 29.4 ^o,,^- 
 
 'Tf~ = /' ; \X . v., = / V - X - =20 I.I c. in. 
 
 ^2 (459 + 77) 29.4 -' (459 + 5) 30-2 
 
CHEMISTRY 13 
 
 To Correct for Normal Temperature and Pressure 
 simultaneously. — The above formulae are only modified. 
 LetV3 = Y = volume at N.T.P. 
 T, = o^' C, or 32" F. 
 
 and 273 + T3 = 273 + 0'^ C. 
 
 and 459 + T,- 459 + 32° F. = 49i. 
 
 P^=:P„ = 760 mm. (C.°), or 30 inches (F."). 
 
 . •. ~i == y-^ '[ X -^- for Centigrade. 
 
 V„ (273 + 0) Pi 
 
 273 .. P, 
 
 V„ = V 
 
 1^273 + Tj 760- 
 
 T. 1.1. 1. > ^1 (459 + T.) ..3o (459 + T,) 30 
 For Fahrenheit : xr = 7 ; \ x x>~ = x ly- 
 
 v„ (459 + 32) Pi 491 Pi 
 
 ^ 459 + J-i 30 
 Example. — Find the volume which 500 c.c. of a dry gas or 
 vapour measured at 15^ C. and 750 mm. pressure will occupy at 
 normal temperature and pressure. 
 
 500^ (273+15) ^ 760 
 V„ (273 + 0)'' 750 
 
 c:oo X 273 7SO 
 
 The same example converted into the Fahrenheit scale of tem- 
 perature and inches of pressure, and taking the original volume as 
 equal to 500 cubic inches, gives : 
 
 15^ C. = 59^ F. and 750 mm. = 29.6 inches. 
 
 . 500^(459 + 59)^ 30 
 V, (459 + 32) 29.6 
 
 500X491 29.6 ^ ^ u- • 1 
 
 y,, = ^^- X -^- =467.6 cubic inches. 
 
 518 30 
 
 Sometimes the Fahrenheit scale is used for temperature, and the 
 metric system for pressure, so that the problem may be : 
 
 Find the volume under N.T.P. of i cubic foot of dry air 
 measured at 60° F. and 730 mm. pressure. 
 
 Without converting 730 mm. into inches : 
 
 Vj_76o (459 + 60) _ _i__7<^o 519^ 
 ^7^^(45'9T^2) •• V^" 730 ^491" 
 
U CALCULATIONS IN HYGIENE 
 
 7|o ^ 49^x_i ^ ^^^^^.^ ^^^^ 
 
 " ?6o 519 ^ 
 
 If the normal pressure is to be expressed as 30 inches and not 
 as 760 mm., then 730 mm. must be converted into inches 
 thus : 760 : 730 : : 30 : a;. 
 
 x= 28.8. 
 
 •. Y„ = — ^ X ~ = 0.Q08 cubic foot (as before). 
 
 " 30 519 
 
 To Find the Weight of a given Volume of Gas at a given 
 Temperature and Pressure. 
 
 (i) Find the volume corrected for N.T.P. 
 
 Let W = required weight. 
 
 Yj := original volume, Y., = required volume. 
 
 Pj = original pressure, P^ = required pressure (normal). 
 
 Tj = original temperature, T., = required temperature = 0' C. 
 
 Centigrade Scale : 
 
 y ^ Zl , "^73 -^^i 
 ' 76o''273+T^ 
 
 (2) Multiply the " corrected " volume by the weight of i litre 
 (unit vohime) at N.T.P. 
 
 Example. — Find the weight of i litre of dry air at 15" C. and 
 730 mm. pressure. 
 
 (0 V, = x litre. V, = P°X(-^^^po.9, liu-e. 
 
 (2) I htre of dry air N.T.P. -= 1.293 grammes : 
 I : 0.91 : : 1.293 • ■^' = 0.91 x 1.293 = 1-176 grammes. 
 Fahrenheit Scale : 
 
 ^ ^ ' 30 459+-^! 
 
 (2) Multiply the result by the weight of i cubic foot (unit 
 volume) at N.T.P. 
 
 Example. — Calculate the weight of i cubic foot of dry air at 
 75"" F. and 29 inches pressure. 
 
 I cubic foot of dry air at N.T.P. = 567 grains (approximately). 
 
 29 491 X I . 
 
 ••• ^ = 3^^ ~^^ X 567 = 503-96 grams. 
 
 Example. — Find the weight of i cubic foot of COjat 62° F. and 
 28.5 inches pressure. 
 
CHEMISTRY 15 
 
 W 2 30 (459+62) ^ ' ' 
 
 (2) To express the weight of CO.^ in terms of dry air relative to 
 hydrogen. 
 
 Weight of I cubic foot, CO., 22 
 
 Weight of I cubic foot of dry air 14.44 
 
 .-. Weight of I cubic foot 00^ = weight of i cubic foot of dry 
 air X 1.52. 
 
 Weight of I cubic foot of dry air = 567 grains approximately. 
 
 .-. Weight of I cubic foot CO., = 567 x 1.52 grains. 
 
 .-. 0.89 cubic foot C0^== 0.S9 X 567 X 1.52 = 767.0376 grains. 
 
 Mixture of Gases and Vapours. — By Dalton's law, if several 
 gases or vapours are enclosed in the same space, each one exerts 
 the same pressure as it would if the others were absent. This 
 " partial pressure " is known as " vapour tension," " elastic force,'' 
 or " force of expansion." The total pressure, or weight, of a 
 mixture of gases and vapours is the sum of the partial pressures or 
 weights. There is a limit to the quantity of vapour which can be 
 formed at a given temperature, and both the vapour itself and the 
 space containing it are said to be '• saturated " when this limit is 
 reached. The vapour is then at " maximum tension " and " maxi- 
 mum density." The vessel containing a mixture of gases has a 
 uniform pressure at all points of its surface. The pressure of a 
 vapour in contact with its own liquid is the same for the same 
 temperature. 
 
 Example. — 100 volumes of pure dry air at a pressure of 760 mm. 
 contain : 
 
 Oxygen . . . 20.94 parts 
 
 Nitrogen . . . 79.00 ,, 
 
 Hydrogen . . . 0.02 ,, 
 
 Carbon dioxide . . 0.04 ,, 
 
 Find the partial pressure of each. 
 
 , . ^ 760 X 20.04 
 
 Oxygen: 100 : 760 : : 20.94 : x. x = - ^-^=159.144 mm. 
 
 Similarly Nitrogen =^ '- l1 = 600 4 
 
 100 
 
 Tx 1 760 X 0.02 
 
 Hydrogen = '- = o. 1 5 2 mm. 
 
 100 
 
 CO., = - = 0-304 11^11^- 
 
 100 
 
 100 
 mm. 
 
10 CALCULATIONS IN HYGIENE 
 
 If the gas contains moisture (i.e., water vapour) and the volume 
 is to be calculated under normal conditions : * 
 
 Let 2J = pressure of aqueous vapour 
 
 P= ,, ,, the moist gas (gas and vapour) 
 .-. 'P-p= „ » „ dry gas. 
 
 The formula is V, = ^—^^-^ ^ ^^ ^.,r -. It corrects the volume 
 
 760 X (273 + I) 
 
 of a gas for temperature, pressure, and aqueous vapour. 
 
 Example. — The observed volume of gas and water-vapour is 
 68.6 c.c, the height of the barometer is 738.5 mm. and the tem- 
 perature is iS-o"" C. Find the volume under normal conditions. 
 
 It is necessary in all cases to ascertain by Vapour-tension tables 
 {e.g., Regnault's) the maximum tension {i.e., 2)ressure) of the 
 aqueous vapour present at this temperature. 
 
 It is found that this pressure at 15° 0. = 12.7 mm. Hence wc 
 have : V = 68.6 c.c. P = 738.5 mm. 2^=i2.j mm. 
 
 The true pressure of the dry gas is 738.5 - 12.7 = 725.8 mm. 
 
 721^.8 X 273 X 68.6 
 .-. Y,= '-^ 7— ^" — . -62.1 c.c. 
 ■' 76ox(273-f 15) 
 
 For the Fahrenheit scale the formula is : 
 
 ,. (P-;>) 491XV 
 
 -'" 30 ''(459 + T)' 
 
 Example. — Fii.d the volume which i cubic foot of moist air at 
 
 60" F. and 29.5 inches pressure will occupy under N.T.P. 
 
 The maximum tension or pressure of aqueous vapour at 60"^ F. 
 
 and 29.5 inches = 0.52 inch. 
 
 (20.=; - 0.1^2) X 401 X I ,. „ , 
 
 .-. V =^-- ^ , ' — y-. =0.914 cubic foot. 
 
 30 X (459 + 60) ^ ^ 
 
 Absorption of Gases in Liquids having no Chemical 
 Action on them. — The volume of gas dissolved is constant for 
 the same temperature at (approximately) all pressures ; the weight 
 of the dissolved gas is directly proportional to the pressure. 
 (Dal ton's and Henry's law.) 
 
 The ratio of the volume of gas dissolved to the unit volume of 
 water (or any other liquid) which dissolves it is, under certain fixed 
 conditions of temperature and pressure, a fixed and definite 
 quantity, and is called tlie " coefficient of absorption " of the 
 body for that particular gas. 
 
 The coefficient of absorption is the volume of gas at N.T.P. 
 which is taken up by 1 c.c. of a liquid at the same pressure. 
 
 * r/r/rp. 13. 
 
CHEMISTRY 17 
 
 The coefficient of solubility is the ratio of the amount of gas in 
 unit volume above, and in, the liquid. 
 
 Increase of temperature diminishes the coefficient of solubility. 
 At the boiling-point it is equal to zero. In a mixtuie of two or 
 more gases in contact with water (or any liquid) each gas will be 
 dissolved to the same extent as if it were the only gas present. 
 
 Solutions. — A. normal solution contains the hydrogen-equiva- 
 lent of the substance in grammes dissolved in i litre of water at 
 15.5° C. or 60" F. 
 
 The " hydrogen- equivalent" of a reagent is its weight in grammes, 
 which is chemically equivalent to i gramme of hydrogen. 
 
 The normal solution of a monovalent body contains its molecular 
 weight in grammes per litre. 
 
 £.g., the normal solution of NaHO = 23 -t- i -I- 16 = 40 grammes 
 per litre ; of HCl = i + 35.5 = 36.5 grammes per litre. 
 
 In a divalent reagent the normal solution has A^T/ the molecular 
 weight of the substance per litre, and in a trivalent body one- 
 third of the molecular weight, e.g. : 
 
 11.- x?u • I, 1 ^ Ba2(H0) 137-f 2 x(i-f 16) 
 
 Normal solution of barium hydrate = ^^ -^ = -^^ ^^ 
 
 2 2 
 
 = 85.5 grammes per litre (divalent). 
 
 Of hydrated oxalic acid = -^ — '—^ ^- = — =63 grammes per 
 
 litre. 
 
 Normal solution of sodium phosphate (trivalent) 
 
 Na,HP0,-M2H.,0 358 ,., 
 
 ;= — 2 4_' 2__ ^ op ^ J jg^ grammes per litre. 
 
 In normal solutions i c.c. of any acid of whatever '' valency " 
 corresponds to i c.c. of any alkali. 
 
 A solution of half, y^, yj^, &c., the normal strength is a 
 
 " semi-" ( ^ ), " deci-" ( — ) or "centi-normal " ( j solution. 
 
 " Standard solutions" other than "normal" may be made of any 
 required strength without reference to valency. 
 
 If a solution is not of standard or normal strength it is titrated 
 with one of standard strength and the "factor for correction" 
 calculated. 
 
 Example — 20 c.c. "normal" alkaline solution required 
 18.6 c.c. of the acid solution instead of 20 c.c. The alkaline 
 standard being of proper strength proves the acid solution to 
 be weaker than normal, and the factor for correcting the latter 
 is calculated thus : 
 
 B 
 
]« CALCULATIONS IN HYGIENE 
 
 18.6 : I : : 20 : r. x= - ^ = 1.071; ^^ "factor." 
 18.6 '^ 
 
 .*. c.c. of acid solution x 1.075 = <^f'i'iection to normal. 
 
 By working with 100 c.c. (= 100,000 mgr.) the results can be 
 expressed at once in parts per 100,000. This is the best method, 
 and is almost universally adopted in scientific research. By 
 multiplying the results by 0.7 they are stated as "grains per 
 gallon." 
 
 By taking 70 c.c. ( = 70,000 mgr.) the returns are in "grains 
 per gallon," and are changed to " parts per 100,000 " by multiply- 
 ing by — or by 1.43, a less convenient process than the above. The 
 
 statement of results in grains per gallon is practically limited to 
 Great Britain, and is more readily understood by the Public and 
 by Sanitary Authorities. Working with 70 c.c. instead of 100 
 may shorten the time in making an analysis, and this may be a 
 desideratum. In all cases the result should be expressed also in 
 parts per 100,000. 
 
CHAPTER II. 
 
 SPECIFIC GRAVITY. 
 
 The calculation is based on the principle of Archimedes, that every 
 body immersed in water or any fluid is subjected to an upward 
 pressure equal to the weight of the liquid displaced by the body. 
 There is a loss of weight equal to that of the displaced liquid. 
 Two forces influence the body : 
 
 (a) One equal to its weight acting at the centre of gravity and 
 tending to depress it. 
 
 (/3) The other at its centre of buoyancy tending to raise it. 
 I. To Determine the Specific Gravity of a Solid. 
 
 1. Heavier than and insoluble in Water. A. — (a) Weigh it m 
 air in a chemical balance in the ordinary way. (b) Suspend it by 
 a very fine fibre (the weight of which may be neglected or 
 estimated) from the hook of the balance-arm without removing 
 the scale-pan, or from the hook beneath the pan, the substance 
 being immersed in pure distilled water at 4"" C, or more usually at 
 15.5° C, the ordinary temperature of a laboratory. Air-bubbles 
 on the surface are to be brushed oflf gently. Air in the interior 
 of a porous substance is removed by the air-pump or by boiling 
 during immersion. 
 
 Weigh the substance a second time. The difterence between 
 this and the first weight in air is the loss of weight in water. 
 Weio;ht in air 
 
 — '—. ^-rz = Sp. 2T. 
 
 Loss ot weiglit in water at 4 U. 
 (or weight of an equal volume of water). 
 
 IJ.g., Weight in air= 136 grammes. 
 ,, water =121 ,, 
 
 Loss of weight in waters 136 - 12 i = 15 grammes. 
 
 13^ = 9.06 = sp. gr., or '' density relative to water." 
 Similarly for grain-weights. 
 
'20 
 
 CALCULATIONS IN HYGIENE 
 
 B. By Nicholson's hydrometer (constant immersion). A weight 
 of loo grammes placed on the tray (Fig. i) sinks the instrument to 
 the zero mark when it is immersed in distilled water, i.e., weight of 
 instrument + I oo grammes = weight of water dis^Jaced. The 
 body of which the sp. gr. is to be estimated must be insoluble in 
 water, and its weight less than loo grammes and heavier than 
 that of an equal bulk of water, so that it will sink. 
 
 (i) Place the body on the tray and add weights till the instru- 
 ment sinks to zero. JiJ.g., 55 grammes must 
 be added .'. weight of body in air = 100 - 55 = 
 45 grammes. 
 
 (2) Place the body in the "basket," under 
 water, and add additional weights to those 
 already in the tray until the instrument sinks 
 again to the zero mark. E.g., 34 grammes 
 are added. This is equivalent to the weight 
 of water displaced by the solid. 
 
 .*. Sp. gr. of solid = — = i.^. 
 
 It can also be used for estimating the sp. gr. 
 of a liquid denser than water : Let the instru- 
 ment be placed in the liquid to be tested ; 
 add weights on the tray till the zero level is 
 reached, e.g., 65 grammes. 
 
 Let the weight of the instrument itself = 
 
 Fic. I. — Mcliolson's 
 Hydrometer. 
 
 120 grammes. 
 
 .-. Sp. gr. of liquid 
 
 120 + 65 
 
 — ^=0.84. 
 220 
 
 120 + 100 
 
 2. The Body is lighter than and insoluble 
 IN Water. — A "sinker" is used to keep the body under water. 
 The weight of the sinker in air need not be known, but its 
 weight in water must be ascertained and also that of any wire, 
 etc., connecting it with the solid when under water. 
 
 Let S„ = weight of the solid in air. 
 
 S-„ = 
 
 S.^ 
 
 sinker in water. 
 
 solid and sinker together in water 
 
 The combined weights of both in water will be less than that of 
 the sinker in water by itself, owing to the foico of buoyancy tending 
 to lift the body and sinker. This force is equal to s„. minus Ss„. 
 
 .'. the weight of the li(]uid displaced, or tlie loss of weight in 
 water - (S„ + s„) - Ss„.. 
 
SPECIFIC GRAVITY 21 
 
 Example. — Weight of the body in air = 48 grain.^. 
 
 ., „ sinker in waters 123 grains. 
 
 ,, in water of body and sinker = 94 giains. 
 
 .-. Sp. gr. = -— — =0.62^ (water = i). 
 
 (48 + 123) -94 ^' ^ 
 
 3. The Solid is heavier than Water and soluble in it. — 
 The body is weighed in air and then in a fluid which will not 
 dissolve it, and the specific gravity of lohich is knoion. 
 
 Let Wa = weight of solid in air {e.g. = 7.42 grammes). 
 W^ = „ „ liquid ( = 4.34 grammes). 
 
 y = sp. gr. of this liquid relative to water ( = 0.76). 
 
 W 
 
 .". Sp. gr. of solid relative to the liquid = — — ~ 
 
 To express this relative to water : ■ ' " ' ;; = - = 7 
 
 Sp. gr. or water i 
 
 W 
 
 .-. Sp. gr. of the solid relative to water = —- — ^ x y 
 
 W^, - W; ' 
 
 ^_7:42 — xo.76 = i.83 
 7.42-4.34 ' ^ 
 
 4. The Substance is heavier than and insoluble in Water. 
 — By the specific gravity bottle. A specific gravity bottle usually 
 has engraved on it the weight of distilled water it can contain at 
 standard temperature (15.5° C. or 60° F.) when filled up to a 
 certain mark and properly stoppered. 
 
 An attached thermometer indicates the temperature of the 
 contents. 
 
 The weight of the powdered substance being known, it is 
 poured in. 
 
 Add distilled water to the powder already in the bottle, boil, 
 and fill up to the mark with freshly boiled distilled water. 
 
 Weigh again at standard temperature. 
 
 Let W = weight of distilled water at standard temperature 
 {e.g. = 50 grammes). 
 Let P = weight of dry powder (—14 grammes). 
 
 !*«(;= 5j „ powder + water ( = 56.8 grammes). 
 
 14 
 
 .-. Sp. gr. of powder = 7 ^ z'o — i-94- 
 
 ^ ° ^ (14 + 50) --56.8 ^^ 
 
i>2 CALCULATIONS IN HYCJIENE 
 
 5. The "specific gravity tube" — Spreiigel's (Fig. 2) or Per- 
 kin's. One of the capillary tubes forming an " arm " has a mark 
 etched on it indicating the point to which the liquid reaches. The 
 other arm ends in a fine aperture through which the fluid to be 
 tested is drawn by suction. Small ghiss caps are used to close the 
 tubes and prevent evaporation. By means of a fine platinum 
 wire of known weight the specific-gravity tube is suspended from 
 the arm of the balance without removing the weighing-pan. 
 
 Fi(i. 2. — Spixiigers tei)ecitic (Jravity Tube 
 
 Let the weight of the empty U-tubo (clean and dry)= 10.398 
 grammes. Fill it with distilled water a little beyond the mark 
 and warm the contents in a beaker of water at '15.5° C. Draw 
 off the superfluous water with blotting-paper till the water is level 
 with the mark. Dry, and weigh the tube and the contained water. 
 Let this = 15.486 grammes. 
 
 Empty the tube of water, dry it in the air-bath, cool, and then 
 fill it as before with the fluid to be tested. Weigh as before. 
 Jjct this weight^ 14-563 grammes. 
 
 Sp. gravity of fluid 
 
 14.563- 10.39 8 ^ 4.16s 
 10.398 5.088 
 
 0.818. 
 
 15.486 
 
 This method is preferable to the sp. gr. bottle for temperatures 
 above that of the atmosphere, and for li(puds less dense than water 
 — e.g., alcohol. 
 
SPECIFIC GRAVITY 
 
 23 
 
 6. Hare's apparatus ; air is drawn out by the central tubing 
 and the li(iuids rise in each tube. 
 
 Let II, = height of the column of liquid. 
 H„. = „ „ „ » vvater. 
 
 .-. bp.gr. = g^. 
 
 As the heights of two columns of liquid in equilibrium are 
 inversely as their specific gravities, the heavier liquid has the 
 
24 
 
 CALCULATIONS IN HYGIENE 
 
 shorter column and the greater sp. gr., the scales being maiked in 
 opposite directions to indicate this. 
 
 7. The specific gravity balance of Westphal (Fig. 4) or Saitorius 
 
 (Fig. 5)- 
 
 The plummet is immersed in the liquid at a known temperature 
 (15.5^ C. or 60' F.), and the riders are applied on the graduated 
 arm till a perfect balance is obtained. The scale can be adjusted 
 to the fourth decimal, and gives the sp. gr. relative to water as i. 
 
 8. By hydrometers of variable immersion. 
 
 Fig. 
 
 -Sartorius l>al;'.uce. 
 
 Twaddell's (Fig. 6) is usually used in Britain for liquids denser 
 than water. The divisions are at distances corresponding to eqi^al 
 differences of density and are not of the same length. The 
 temperature of the liquid is to be 15.5 C. or 60^ F. 
 
 The number of degrees read oft* multiplied by 5 and added 
 
 to 1000 gives the sp. gr. relative to water as 1000. 
 being the indicated reading : 
 
 Thus 
 
 23-5 
 
 Sp. gr. of liquid = (23.5 x 5) + 1000 
 
 = 117.5 + 1000= 1117.5. 
 
 If water is taken as i, this becomes 1.1175. 
 
SPECIFIC GRAVITY 
 
 25 
 
 Baiim^'s Hygrometers. — (a) For liquids denser than water 
 (" salimeter "). The zero is at the top of the scale and is the point 
 to which the instrument sinks in distilled water at 15.5° C. or 
 60° F. = 1.000 sp. gr. 10° = 1.075, 20^ = 1. 161, 30° = I 263, 40° = 
 r.385, 50° = 1.532, 60" = 1. 714, 70° = 1-946 sp. gr. 
 
 (/3) For liquids less dense than water (" alco- 
 holimeter "). The zero is at the bottom of the 
 scale, and indicates a mixture of 10 parts by- 
 weight of salt and 90 parts by weight of water. 
 
 10° is the level in distilled water at 15.5° C 
 (60" F.) = 1.000 sp. gr. 20° = 0.928, 26° = 0.892, 
 30° = 0.871, 36° =0.837, 40° = 0.817,50° =0.761, 
 60° = 0.706 sp. gr. 
 
 II. To Find the Specific Gravity of a 
 Liquid relative to Water. 
 
 (i) By using a solid of known weight insoluble 
 in either of the fluids. 
 
 A metal ball or plummet is taken and weighed : 
 
 (a) In air (e.g. = 11 8.7 grains). ^ 
 
 (b) In distilled water ( = 75.3 grains). V at 60"" F. 
 
 (c) In the liquid itself ( = 62.4 grains).] 
 
 118. 7-62. 4 = weight of liquid displaced by the 
 ball. 
 
 118.7-75.3= „ „ water 
 ball. 
 
 ,7-62.4 
 
 Sp. gr. of the liquid = — ; 
 
 I-3- 
 
 F1G.6.— Twiukk'ir 
 Hydrometer. 
 
 8.7-75-3 
 
 (2) By the specific gravity bottle for liquids. 
 
 Let weight of empty bottle = 25.623 grains. 
 
 „ ,, bottle + distilled water at 15.5° C. = 78.658 grains 
 ,, ,, ,, +liquid at 15.5° C. = 67.581 grains. 
 
 67.581 - 25.623 = 41.958 = weight of liquid. 
 78.658-25.623 = 53.035= „ „ water. 
 
 .-. Sp. gr. of liquid = 1^^ = 0.79;. 
 
CHAPTER 111, 
 
 METEOROLOGY. 
 
 To Calculate the Weight of Aqueous Vapour or " Mois- 
 ture " present in a Mixture of Air and Vapour at a 
 given Temperature and Pressure. 
 
 1 . Find the weight of the same voUime of dry air corrected for 
 N.T.P. under similar conditions of temperature and pressure. 
 
 2. Multiply the result by the specific gravity or relative density 
 of the vapour, air being taken as unity. This is as 0.623 to 1. 
 If the pressure is not stated it must be ascertained from a Table 
 of Pressures (or " Tensions ") of Aqueous Vapour (Regnault's). 
 Opposite each degree and tenth of a degree C. is given the maxi- 
 mum tension in mm. of mercury. This is read oft* for the given 
 temperature. 
 
 Example. — Calculate the weight of i litre of aqueous vapour at 
 15-5' C. 
 
 (i) Vi= I litre, v., is to be calculated. P, is found from the 
 Table of Vapour Tensions (p. 27). Opposite " 15.5 C." is " 13.1," 
 i.e. = maximum pressure of aqueous vapour at that temperature. 
 
 Let W = weight of i litre of dry air at 15.5° corrected for 
 N.T.P. I litre of dry air at N.T.P. weighs 1.293 grammes. 
 
 W= ZaIT-J (^»Ji .^ r- \ ^ i.293 = o.oi6 X 1.293 = 0.0207 
 
 gramme. 
 
 76ox(273+i5.5) 
 
 (2) 0.0207x0.623 = 0.0128961 gramme = weight of 1 litre of 
 aqueous vapour at 15.5° C. and 13.1 mm. pressure. 
 
 If the Table of Vapour Tensions expresses Temperature and 
 Pressure only in C." and mm. and the equivalents are required in 
 F/^ and inches, the conversion to the latter is easily made. For 
 temperature, as already indicated (pp. 2, 3) ; and for pressure as 
 
 ^^^^^^'^ • I millimetre = 0.03937 inch 
 
 .-. "Tension in millimetres of mercury " x 0.03937 = Tension in 
 inches of mercury (/.e., "pressure"). 
 
METEOROLOGY 
 
 27 
 
 F = 59.9 or 60° (iiearlv). 
 
 Example. — 15.5' C = 13.1 mm. (RegnaulL). To convert to F.' 
 and inches : 
 
 i5-5_F-32 
 
 5 9 
 
 and 13. 1 X 0.03937 =0.515747 inch. 
 
 = 0.516 inch (nearly). 
 •'• "i5-5^ ^- ^^^ ^3-^ mm." correspond to 60'^ F. and 0.516 
 inch. 
 
 Example. — Find the weight of 2 cubic feet of aqueous vapour 
 at 60.8° F. 
 
 Pressure not being given, it is found from an Aqueous Vapour- 
 Tension Table : 
 
 60.8° F. = 0.5315 inch ( = maximum pressure), 
 (i) Find the weight of an equal volume of dry air, denoted by 
 " W." I cb. ft. of dry air at N.T.P. weighs 567 grains. 
 0-53ISX (459 + 32) X 2 
 
 •. W 
 
 30 X (459 + 60.8) 
 
 X 567 grams. 
 
 o.c^^i c; X 401 X 2 ^ „ 
 
 ^ J'^ J — zz_^ X q67 = 18.9 strains. 
 
 30x519.8 ^ ' 
 
 (2) 18.9 X relative density (or weight) of aqueous vapour = 18.9 
 
 0.623= 11-77 gi'ains, the required weight of aqueous vapour. 
 
 Pressure, Tension, or Elastic Force of Aqueous Vapour 
 
 from 0° to 30° C. 
 
 In mm. of mercury. 
 
 C\ Mm. 
 
 c\ 
 
 Mm. 
 
 c\ 
 
 Mm. 
 
 c\ 
 
 Mm. 
 
 c^ 
 
 Mm. 
 
 0.0. ..4.6 
 
 6.5 . 
 
 . 7.2 
 
 13.0.. 
 
 .11.2 
 
 I9-5- 
 
 .16.9 
 
 26.0. 
 
 .25.0 
 
 0.5. ..4.8 
 
 70 . 
 
 • 7-5 
 
 I3-5-- 
 
 .11.5 
 
 20.0. 
 
 .17.4 
 
 26.5. 
 
 •25-7 
 
 I.O...4.9 
 
 7-5 • 
 
 . 7.8 
 
 14.0. 
 
 .11.9 
 
 20.5. 
 
 .17.9 
 
 27.0. 
 
 •2b.5 
 
 1.5.. .5.1 
 
 8.0. . 
 
 . 8.0 
 
 14.5. 
 
 .12 3 
 
 21.0. 
 
 .18.5 
 
 27-5- 
 
 •27-3 
 
 2.0. ..5.3 
 
 8.5 •• 
 
 • 8.3 
 
 150. 
 
 .12.7 
 
 21.5. 
 
 .19.1 
 
 28.0. 
 
 .28.1 
 
 2.5---5-5 
 
 9.0.. 
 
 . 8.6 
 
 I5-5- 
 
 .13.1 
 
 22.0. 
 
 .19.7 
 
 28.5. 
 
 .28.9 
 
 3.0. ..5.7 
 
 9-5-- 
 
 . 8.9 
 
 16.0. 
 
 •13-5 
 
 22.5. 
 
 .20.3 
 
 29 
 
 .29.8 
 
 3-5---5-9 
 
 10. 0.. 
 
 . 9.2 
 
 16.5. 
 
 .14.0 
 
 23.0 
 
 .20.9 
 
 29 5- 
 
 •30-7 
 
 4.0.. 6.1 
 
 10.5.. 
 
 • 9-5 
 
 17.0. 
 
 .14.4 
 
 23-5- 
 
 .21.5 
 
 30.0. 
 
 •31.5 
 
 4.5. ..6.3 
 
 II. 0.. 
 
 . 9.8 
 
 I7-5- 
 
 .14-9 
 
 24.0. 
 
 .22.2 
 
 
 
 5.0. ..6.5 
 
 II. 5. 
 
 .10.1 
 
 18.0. 
 
 .15.4 
 
 24.5. 
 
 .22.9 
 
 
 
 5.5. ..6.8 
 
 12.0. 
 
 .10.5 
 
 18.5. 
 
 ..15.8 
 
 25-0 
 
 23-5 
 
 
 
 i 6.0. ..7.0 
 
 12.5. 
 
 .10.8 
 
 19.0. 
 
 ..16.3 
 
 25-5- 
 
 ..24.3 
 
 
 
L^S 
 
 CALCULATIONS IN HYGIENE 
 
 Pressure, Tension, or Elastic Force of Aqueous Vapour 
 from 32 to 90 ^F. 
 
 In inches of mercury. 
 
 F\ Inches. 
 
 F°. Inches. 
 
 F°. Inches. ; 
 
 ¥'. Inches. ' 
 
 ¥\ 
 
 Inches. : 
 
 32. ..0.181 
 
 44. ..0.288 
 
 56. ..0.449 
 
 68. ..0.684 
 
 80.. 
 
 .1.023 
 
 33. ..0.188 
 
 45. ..0.299 
 
 57. ..0.465 
 
 69. ..0.708 
 
 81.. 
 
 .1.057 
 
 34. ..0.196 
 
 46.. .0.311 
 
 58.. 0.482 
 
 70. ..0733 
 
 82.. 
 
 .1.092 
 
 35. ..0.204 
 
 47. ..0.323 
 
 59. ..0.500 
 
 71. ..0.759 
 
 83- 
 
 .1.128 
 
 36. ..0.212 
 
 48. ..0.335 
 
 60. ..0.518 
 
 72. ..0.785 
 
 84. 
 
 .1.165 
 
 37. ..0.220 
 
 49. ..0.348 
 
 61. ..0.537 
 
 73. ..0.812 
 
 85- 
 
 .1.203 
 
 38. ..0.229 
 
 50. ..0.361 
 
 62. ..0.556 
 
 74. ..0.840 
 
 86 
 
 .1.242 
 
 39. ..0.238 
 
 5I---0.374 
 
 63. ..0.576 
 
 75. ..0.868 
 
 87. 
 
 .1.282 
 
 40. ..0.247 
 
 52. ..0.388 
 
 64. ..0.596 
 
 76. ..0.897 
 
 88. 
 
 •1-323 
 
 41. ..0.257 
 
 53---0-403 
 
 65. ..0.617 
 
 77. ..0.927 
 
 89. 
 
 ..1.366 
 
 42. ..0.267 
 
 54. ..0.418 
 
 66. ..0.639 
 
 78. ..0.958 
 
 90. 
 
 ..1. 410 
 
 43. ..0.277 
 
 55---0433 
 
 67. ..0.661 
 
 79. ..0.990 
 
 
 
 Tensions at intermediate temperatures are approximately calcu- 
 lated by taking the Arithmetical Mean of the tensions given in the 
 table, at temperatures immediately above and below the given 
 tempei-ature. 
 
 Example. — Find the approximate tension of aqueous vapour at 
 56.5^ F. and 79.5° F. 
 
 Vide Table : 56'' F. = 0.449 inch. 
 ' F. = 0.465 „ 
 
 57 
 
 79 
 
 80^ 
 
 2)0914 
 
 0.457 inch = approximate tension at 56.5" F. 
 
 , = 0.990 inch. 
 = 1.023 „ 
 
 2)2.013 inch. 
 
 1.006 inch = approximate tension at 79.5 F. 
 
 To Calculate the Weight of a given Volume of Air satu- 
 rated with Moisture at a given Temperature and Pressure.* 
 — Consider the total volume of saturated air to consist of a volume 
 of dry air phs a volume of acjueous vapour. 
 
 I. Calculate the weight of an equal volume of each at the same 
 temperature and at their respective pressures. 
 
 * ]'uU' fp. 12, 14. 
 
METEOROLOGY 29 
 
 2. Add these results together : their sum is the weight of satu- 
 lated air. 
 
 To work out the pressure of aqueous vapour, Tables are neces- 
 sary, as already indicated. 
 
 Let P = total pressure of the combined volumes of " dry " air 
 and aqueous vapour. 
 
 Let 2^ = pressure of aqueous vapour only. 
 
 .*. P -^ = pressure of " dry " air. 
 
 Example. — Find the weight of i litre of air saturated with 
 moisture at 15° C. and 730 mm. pressure. 
 P= 730 mm. = pressure of combined volumes, 
 ^j at 15° C. = 12.7 mm. {vide Table). 
 P-jt?=73o- 12.7 = 7 1 7.3 = pressure of " dry " air. 
 
 7I7.'?X27''X I 
 
 I . (a) Volume of dry air at i n; ^ 0. and 717.^ mm. = , , — : 
 
 ^ ^ ^ ^ ' ' ^ 760 X (273 + 15) 
 
 = 0.89 litre. Its weight = 0.89 x 1.293 = 1. 15 gramme. 
 
 I. {h) Yolume of aqueous vapour at 15'' 0. and 12.7 mm. 
 12.7 X 273 XI _ -., 
 
 = -7 — ^ — — ^ = 0-158 litre. 
 
 760 X (273 4- 15) 
 
 Its weight = 0.1 58 X 1.293 = 0.204 gramme, 
 And 0.204 X 0.623 (relative density of aqueous vapour) = 0.127 
 gramme. 
 
 .*. Weight of saturated air = 1.15 + 0.127 
 
 = 1.277 gramme. 
 
 Example. — Find the weight of i cubic foot of saturated air 
 at 60" F. and 30 inch pressure. 
 
 P = pressure of combined volumes = 30 inches 
 p= „ ,, aqueous vapour at 60° F. = 0.518 inch 
 P-;j)= ,, ,, "dry" air = 30 - 0.518 = 29.482 inches 
 
 (i) {a) Volume of i cubic foot of dry air at 60° F. and 29.482 
 inches. 
 
 29.482 X (459 + 32) X I 29.482x491x1 . 
 
 = -^-^ , ^ , . = = o.Q2Q cubic inch. 
 
 30 X (459 + 60) 30x519 ^ ^ 
 
 Its weight = 0.929 X 567 = 526.743 grains. 
 
 (6) Volume of i cubic foot of aqueous vapour at 60° and 
 0.518 inch. 
 
 0.518 X49I XI ^ , . . , 
 
 = =0.016 cubic inch. 
 
 30x519 
 
 Its weight = 0. 016x567 = 9072 grains. 
 
BO 
 
 CALCULATIONS IN HYGIENE 
 
 9.072 X relative density of aqueous vnpour = 9.072 x 0.623 = 
 5.652 grains. 
 
 .•. Weiglit of saturated air at 60 F. and 30 inches 
 
 = 5-6.743 + 5-652 = 532.395 grains. 
 The weight of i cubic foot of dry air at 60° F. and 30 inches is 
 
 30 X 491 X I 
 X ^6'] = K 7,6.7.8 grains. 
 
 30 X 519 J / JO o & 
 
 So that a volume of moist or saturated air is lighter than an 
 equal volume of dryaiv under the same temperature and pressure. 
 
 A 
 
 a d 
 
 t> c 
 
 B 
 
 Yu^. 7. 
 
 The diminution in density and weight is due to expansion of total 
 volume. 
 
 Let ah c d (Fig. 7) represent i cubic foot of dry air and A B C D 
 the increased volume of this 2av j^lus moisture, i cubic foot a bed 
 of this augmented volume will weigh less than the same volume of 
 dry air because of expansion and rarefaction, but the weight of 
 the volume A B C D will, of course, be greater. 
 
 "Dry air" denotes air containing no aqueous vapour — i.e., its 
 " humidity is zero." Humidity of the air is the weight of 
 aqueous vapour pi-esent in a given volume of air, expressed as a 
 percentage of the weight of vapour at saturation which would 
 occupy the same volume at the actual temperature. (Everett.) 
 
 A volume of air at any temperature can contain a definite 
 quantity of water vapour. When the contained water vapour is 
 the greatest amount possible at that temperature, the air is 
 " saturated." It is then at maximum humidity. This is the 
 " dew-point," which is therefore the temperature of saturation 
 {vide p. 32). 
 
 Air holding less than its maximum amount of aqueous vapour 
 will be saturated by that same quantity of moisture when its tem- 
 perature falls to dew-point. If in excess, moisture is deposited 
 on solid surfaces, forming dew. 
 
 " Absolute humidity " or " absolute moisture " is the weight of 
 water vapour (expressed as grammes or grains) actually present in 
 a known volume of air at a certain temperature. 
 
 It is estimated by ascertaining from Tables the maximum pres- 
 
METEOROLOGY 
 
 81 
 
 sure or tension of water- vapour at the temperature of the dew- 
 point. 
 
 e.g. Dew-point = 49° F. Tension = 0.348 in. = Absolute humidity. 
 ' ' Relative humidity " is expressed as : 
 
 (i) Weight of water actually present in a known volume of air 
 
 Weight of water which would saturate the same volume of air 
 Or: 
 
 (2) Tension of aqueous vapour at the temperature of the dew-point 
 Tension of aqueous vapour at the temperature of the dry bulb 
 
 Fig. 8. — Daniell's HygTomotor. 
 
 By the first method the weight of water is calculated as already 
 indicated (pp. 26, 29). 
 
 In the second formula the vapour tensions are obtained from a 
 Table : 
 
 Example. — Dew-point temperature = 52° F. = 0.388 inch. 
 Dry-bulb „ =63° F. = 0.576 „ 
 
 (FzV^e Tables.) 
 
 0.388 ^ , 
 .'. Relative humidity = — -^ = 0.674 (nearly). 
 
 Here saturated air is taken as i ; in 100 parts the relative 
 humidity is 67.4 per cent, of saturation, or 0,674 x 100. 
 
 Oil the Continent " maximum moisture " denotes the maximum 
 quantity of aqueous vapour which air can take up at a certain 
 temperature. A volume of air at any known temperature can 
 contain a definite quantity of water vapour, and Tables of maximum 
 moisture at various temperatures are constructed. 
 
 The difierence in pressure between maximum (saturation) and 
 absolute moistures is known as the "deficiency of saturation." 
 
?,2 CALCULATIONS IN HYGIENE 
 
 Example. — Relative moisture =70 per cent. Temperature = 
 
 17° (J. 
 
 Find the absolute moisture. 
 
 At ly"" C. maximum moisture = 14.4 mm. (per Table). 
 
 .•. TOO : 14.4 : : 70 : £c. a;— 10.08 absolute moisture. 
 
 14.4 — 10.08 = 4.32 = deficiency of saturation (Lehmann). 
 
 In this country complete saturation is denoted as 100 parts by 
 
 Fig. 9. — Kegnault's Hygrouietcr. 
 
 weight of aqueous vapour and relative humidity is expressed as a 
 percentage of saturation. 
 
 A relative humidity of 75 per cent, means, therefore, 75 per 
 cent, of saturation. 
 
 The drying power of the air is 100 - relative humidity, so that 
 100-75 — 25 per cent., which is the drying-power of the air with a 
 relative humidity of 75. 
 
 The dew-point is ascertained (i) directly by hygrometers or 
 " instruments of condensation " — Daniell's (Fig. 8), Regnault's 
 (Fig. 9), or Dines' (Figs. 9a, 9^), (2) By hygrometers of absorption : 
 (a) 13e Saussure's, consisting of a weighted human hair free from 
 grease, which elongates as humidity increases and contracts as it 
 diminishes and moves an index ; [h) by the chemical hygrometer, 
 consisting of U -tubes (containing a dry hygroscopic substance) 
 which are weighed before and after the aspiration througli them 
 of a known volume of moist air. (3) Indirectly by the dry- and 
 wet-bulb thermometer or psychrometer (hygrometer of evapora- 
 tion) with Glaisher's or Apjohn's formula. 
 
METEOROLOGY 
 
 33 
 
 I. Directly. 
 
 In all cases take the reading twice : (a) the moment the him ot 
 moisture appears ; (b) when it disappears. The mean of these 
 observations gives the correct dew-point temperature. If only 
 one reading is taken the first would be a little below, and the 
 second a little above the true dew-point. 
 
 Fig. 9a.— Diues' Hygrometer, latest modificatiou. 
 
 r. reservoir for ice-coUl water turned ou by tap D. . . , , ^ r 
 
 r pi iln separate tube for running- ether instead of water into d.amber C 
 coitaining tiiermometer-bulb and covered by black glass slab for observing 
 moisture. 
 
 Example.— Temperature when film forms 49-6° F. 
 
 J, ,, disappears 4 9.8 .b. 
 2)99.4" 
 
 Dew-point = 49.7 
 
 E. 
 
 c 
 
u 
 
 CALCULATIONS IN HYGIENE 
 
 The mean of a series of observations is more accurate. 
 
 II. Indirectly by the dry- and wet-bulb hygrometer or 
 psychrometer. 
 
 (a) Glaisher's Formula. — This is an empirical one, founded 
 on observations extending over several years in various latitudes. 
 
 To use this method a Table of Glaisher's Factors (for each reading 
 of the dnj-huYb thermometer) is indispensable. 
 
 Fig. gb. — Diues' Hygrometer, old pattern. 
 
 Method. — (i) Take the reading of the dry- and wet-bulb 
 thermometers at the same time. 
 
 (2) Subtract the latter reading from the former. 
 
 (3) Multiply the difference by the "factor" {vide Table) corre- 
 sponding to the dry-bulb temperature. 
 
 (4) Deduct the product from the dry-bulb reading. 
 
 Let Ta = Temperature of the dry-bulb. 
 T,„= ,, „ ,, wet-bulb. 
 
 F = Glaisher's factor opposite the dry-bulb temperature. 
 
 .-. Dew-point = T,-{(T,-T,,)xF.} 
 
 Example.— T,i = 62° F. 
 
 F(at 62° F.)=i.86. 
 
 .-. Dew-point =-62 - {(62 - 51) X 1.86} 
 = 62- {11 X 1.86} 
 = 62 — 20.46 
 -41.54^ F. 
 
 { 
 I 
 
METEOROLOGY 
 
 35 
 
 TABLE OE GLAISHER'S FACTORS. 
 
 Ilygrometrical Tables, adapted to the use of Dry- and Wet-Bulb 
 Thermometers, by James Glaisher, F.R.S., &c., 1885. 
 
 Reading 
 
 
 Reading 
 
 
 Reading 
 
 
 Reading 
 
 
 of Dry- 
 
 
 of Dry- 
 
 
 of Dry- 
 
 
 of Dry- 
 
 
 Bulb 
 
 Factor. 
 
 Bulb 
 
 Factor. 
 
 Bulb 
 
 Factor. 
 
 Bulb Fa 
 
 ctor. 
 
 Thermo- 
 
 
 Thermo- 
 
 
 Thermo- 
 
 
 Thermo- 
 
 
 meter. 
 
 
 meter. 
 
 
 meter. 
 
 
 meter. 
 
 
 Fahr. 
 
 
 Fahr. 
 
 
 Fahr. 
 
 
 Fahr. 
 
 
 10 
 
 8.78 
 
 33 
 
 3.01 
 
 56 
 
 1.94 
 
 79 I 
 
 69 
 
 II 
 
 8.78 
 
 34 
 
 2.77 
 
 57 
 
 1.92 
 
 80 I 
 
 .68 
 
 12 
 
 8.78 
 
 35 
 
 2.60 
 
 58 
 
 1.90 
 
 81 I 
 
 68 
 
 13 
 
 877 
 
 36 
 
 2.50 
 
 59 
 
 1.89 
 
 82 I 
 
 67 
 
 14 
 
 8.76 
 
 37 
 
 2.42 
 
 60 
 
 1.88 
 
 S3 
 
 67 
 
 15 
 
 8.75 
 
 38 
 
 2.36 
 
 61 
 
 1.87 
 
 84 I 
 
 66 
 
 16 
 
 8.70 
 
 39 
 
 2.32 
 
 62 
 
 1.86 
 
 85 I 
 
 65 
 
 17 
 
 8.62 
 
 40 
 
 2.29 
 
 63 
 
 1.85 
 
 86 I 
 
 65 
 
 18 
 
 8.50 
 
 41 
 
 2.26 
 
 64 
 
 1-83 
 
 87 
 
 64 
 
 19 
 
 8.34 
 
 42 
 
 2.23 
 
 65 
 
 1.82 
 
 88 I 
 
 64 
 
 20 
 
 8.14 
 
 43 
 
 2.20 
 
 66 
 
 1. 81 
 
 89 
 
 ^3 
 
 21 
 
 7.88 
 
 44 
 
 2.18 
 
 67 
 
 1.80 
 
 90 I 
 
 63 
 
 22 
 
 7.60 
 
 45 
 
 2.16 
 
 68 
 
 1.79 
 
 91 I 
 
 62 
 
 22> 
 
 7.28 
 
 46 
 
 2.14 
 
 69 
 
 1.78 
 
 92 I 
 
 62 
 
 24 
 
 6.92 
 
 47 
 
 2.12 
 
 70 
 
 1.77 
 
 93 I 
 
 61 
 
 25 
 
 ^•S2> 
 
 48 
 
 2.10 
 
 71 
 
 1.76 
 
 94 I 
 
 60 
 
 26 
 
 6.08 
 
 49 
 
 2.0S 
 
 72 
 
 1-75 
 
 95 I 
 
 60 
 
 27 
 
 5-6i 
 
 50 
 
 2.06 
 
 73 
 
 1.74 
 
 96 
 
 59 
 
 28 
 
 5-12 
 
 51 
 
 2.04 
 
 74 
 
 1-73 
 
 97 I 
 
 59 
 
 29 
 
 4-63 
 
 52 
 
 2.02 
 
 75 
 
 1.72 
 
 98 
 
 58 
 
 30 
 
 4-15 
 
 53 
 
 2.00 
 
 76 
 
 1. 71 
 
 99 I 
 
 58 
 
 31 
 
 3.60 
 
 54 
 
 1.98 
 
 77 
 
 1.70 
 
 100 I 
 
 57 
 
 32 
 
 3-32 
 
 55 
 
 1.96 
 
 78 
 
 1.69 
 
 
 
 {h) Ap John's Formula. — By this method the vapour tension, 
 or pressure, at the temperature of the dew-point is first obtained, 
 and from it the dew-point is ascertained by reference to a Table of 
 Vapour Tensions. The use of this Table is therefore indispensable 
 for working Apjohn's formula. 
 
 Method. — I. Vapour tension at the dew-point, 
 (i) Observe the readings of the dry- and wet-bulb thermometers 
 at the same time. 
 
P>6 CALCULATIONS IN HYGIENE 
 
 (2) From a table of vapour tensions obtain the pressure in 
 inches of mercury at the temperature of the wet-hu\h. 
 
 (3) Note the height of the barometer in inches. If nearly at 
 sea-level pressure (30 inches) this may be neglected. 
 
 Let T^i = Temperature of the dry-bulb {i.e. of the air). 
 T«,= „ ,, wet-bulb. 
 
 V.p.^ = vapour pressure at the temperature of the wet- 
 bulb. 
 H = height of the barometer. 
 
 Vapour 2)ressiire at the dew- point 
 
 /T -T II\ 
 — y n _ (— ^ — - X — for tem|)eratuies above 32' F. 
 
 ^"^ V 87 30/ 
 
 The height at sea-level is practically 30", so that the fraction 
 
 — = -^ = I , and may therefore be neglected, and the formula 
 
 then becomes : 
 
 T — T , 
 
 Vapour pressure at the dew-point ^ V.p.^t, ~ o^ '/ . "r\ ^'•^•5 ^« the 
 
 temperature of the air (dry-bulb) is above or below 32° F, 
 
 II. Having obtained the vapour pressure at the dew-point 
 expressed in inches, ascertain from the Table the temperature 
 corresponding to this tension. This gives the " dew-point " itself. 
 
 Example. — Taking the same temperatures as in the last example 
 of Glaisher's Formula : 
 
 I. T, = 62 F., T,= 5i^F. 
 
 Vapour tension at T^y = o.374 inch, (Vide Pressure Tables, 
 
 p. 28.) 
 
 13 = 28.9 inches. 
 
 /62-51 28.9 
 .'. Vapour tension at dew-ponit = 0.374 - ( — r, ^ 
 
 = o-374-(gyx 0.963 
 
 = 0.374-0.122 
 =■■ 0.252 inch. 
 
 II. In the Table of Aqueous Vapour Pjessures 0.252 i.'s not given. 
 
 i 
 
METEOROLOGY 
 
 but it lies between 0.247 and 0.257, which respectively correspond 
 to 40" F. and 41° F. 0.252 is found to be the mean of these 
 
 pressures : ^-^- ^ + °-^57 ^0.252, and the corresponding tempera- 
 ture may be taken as approximately the mean of 40° and 41 ° F., i.e., 
 
 40 + 41 
 
 = 40.5" F. 
 
 .04 
 
 .-. *Dew-point = 4o.5'^ F. 
 
 The result obtained by Glaisher's Formula is 41.54° F^ so that 
 the discrepancy is the comparatively slight one of 
 
 Weight of Air.— Method. 
 — A glass globe fitted with a 
 stop- cock and of known capacity 
 (preferably 12 to 13 litres) is ex- 
 hausted of air and weighed in a 
 balance (" baroscope," Fig. 10). 
 After equilibrium is established 
 dry air is admitted by opening 
 the stop-cock, and the weight 
 of the globe is taken again. 
 
 Example. — Capacity of globe 
 = 13 litres. 
 
 Difference in weight between 
 globe when full of air and 
 w-hen exhausted of air= 16.809 
 
 grammes. 
 
 Weight of air = 
 
 16.809 
 13 
 
 ItoWNSON 4 MERCER 
 
 Fig. 10. — Biroscope. 
 
 1.293 gi^ammes per litre, which 
 is the weight of dry air at N.T. P. 
 
 If the density of i litre of water is taken as 1000 (at 4° 0.) the 
 
 • . i-293_ I / \ 
 
 ratio of an equal volume of dry air to it is -^^^ ~ 773 ^ "" 0.00 1 29) 
 
 — i.e., air is 773 times lighter than water; vice versd, water is 
 773 times heavier than air. 
 
 In a mercurial barometer the usual height of the column is 30 
 inches (2.5 feet); mercury being 13.6 times denser than water, 
 in a water barometer the equivalent height is 2.5 x 13.6 = 34 feet. 
 
 Glycerine is 1.26 times denser than water (=- i), the height of 
 the column of the glycerine barometer is therefore less, varying 
 inversely as the density. 
 
 To calculate it from the above data : 1.26 : i : : 34 : .t= 27 feet. 
 
38 CALCULi^LTIONS IN HYGIENE 
 
 3.6 
 
 . /I3-C 
 Again : mercury is nearly ten times denser than glycerine ^^^^ 
 
 .-. 1.26 : 13.6 : : 30 : .^=27 feet (as before). 
 
 .-. Mercurial column 30 inches = water column 34 feet = 
 glycerine column 27 feet. 
 
 Barometric Corrections. 
 
 I. Correction for Capacity.— This is not necessary in the 
 Fortin, Kew, and Siphon barometers. Howson's and McNield's 
 long-range barometers are self-adjusting, the bore of the tube is 
 noteless than one inch, and they are unafl'ected by differences of 
 level in the cistern and need no adjustment for the neutral point. 
 In the Fortin barometer the mercury in the cistern is raised or 
 lowered to the correct level (" fiducial point ") by means of a screw 
 at the base of the instrument. 
 
 The Kew barometer has corrections all along the scale, which 
 is divided into shortened, and not true, linear inches. (The 
 
 divisions are less than true inches in the ratio of -^j^p^ {t'ide 
 
 infra). 
 
 The siphon barometer adjusts itself as the rise in one limb is 
 compensated by a fall in the other, and vice versd, and the true 
 reading is the difference of level in its two scales. 
 
 Barometers not having the above adjustments have a " neutral 
 point " marked on the scale of the instrument. At this point the 
 mercurial column gives the correct reading, and the mercury in the 
 cistern is at the proper level. 
 
 Let T = internal sectional area in square inches of the barometer 
 
 tube. 
 
 Let C = area of the cistern after deducting that occupied by the 
 
 tube and its contents. 
 
 Let D=- distance in inches of the summit of the mercurial 
 column from the neutral point— above or below it. 
 
 T 
 
 Correction = D x n* 
 
 This is to be added to the observed reading if the mercurial 
 column is above the neutral point, and to be subtracted if In-low 
 that point. The "correction" is calculated and supplied with 
 each instrument by the makers. 
 
 T I 
 
 Example.— 7i = (" correction "). 
 50 
 
METEOROLOGY 3t 
 
 Observed reading = 30.560 inches, summit of column bein< 
 I of an inch above the neutral point. 
 
 Correction 
 
 ^ X — 
 
 4 50 
 
 0.015 inch. 
 
 As the mercurial column is above the neutral point, this is to 
 be added to the observed height ; 
 
 .". 30.560 + 0.015 = 30-575 inches as the correct height. 
 
 2. Correction for Capillarity. — Unless the internal diameter 
 of the barometer tube exceeds 0.6 inch, the mercurial column 
 is slightly, and appreciably, depressed by capillary action due to 
 surface tension between the surfaces of mercury and glass. If 
 the mercury has been boiled in the tube the depression is reduced 
 to half what it would be if unboiled. The correction is always to 
 be added to the observed reading, and is calculated from the 
 height of the meniscus and from the internal diameter of the tube. 
 Tables of correction are supplied with each instrument. The 
 Kew Certificate gives the capacity and capillarity corrections and 
 index error in one figure for all readings. 
 
 Table of Corrections for Capillarity (only), to be added 
 
 to all Readings. 
 
 N.B.— To be halved for Boiled Tubes. 
 
 Diameter of Tube. 
 
 Depression in Unboiled Tube. 
 
 Inch. 
 
 Inch. 
 
 0.60 
 
 0.004 
 
 0-55 
 
 0.005 
 
 0.50 
 
 0.007 
 
 0.45 
 
 O.OIO 
 
 0.40 
 
 0.015 
 
 0-35 
 
 0.021 
 
 030 
 
 0.029 
 
 0.25 
 
 0.041 
 
 0.20 
 
 0.058 
 
 0.15 
 
 0.086 
 
 O.IO 
 
 0.140 
 
 3. Correction for Index Errors. — These include (i) errors 
 in position of the zero-point, and (2) errors of graduation along 
 the scale, and are special for each instrument. " An error of zero 
 makes all readings too high or too low by the same amount." 
 The Kgw corrections include index error, capacity and capillarity 
 
40 CALCULATIONS IN HYGIENE 
 
 errors at every half-inch of the scale. No instrument is passed 
 as a " Standard" if the errors exceed o.oi inch. 
 
 4. Correction for Temperature, or " Reduction to 32° F." — 
 Variations in temperature cause expansion or contraction of both 
 the mercury and the metallic scale of graduations. 
 
 Each of these must be corrected in a Standard barometer. 
 
 (i) Correction for the mercury only. 
 
 (a) In the Fahrenheit scale the coefficient of expansion of 
 
 mercury is- =0.0001 per i" F., ^.e., it expands — -- of its 
 
 •^ 9990 ^ > » r gggQ 
 
 length at 32" F. for each 1° F. above that temperature. 
 
 It also contracts to the same extent per each degree below 
 32° F. 
 
 Reducing to the temperature of 32° F. : 
 Corrected height of barometer at 
 
 o ^ _ Observed height of barometer 
 ^^ ' I + {o.oooi x(F.'' -32)} 
 
 " F." = temperature of the thermometer attached to the barometer, 
 which indicates the temperature of the colur)in of r)ierciiry. 
 
 Example. — Observed height of barometer = 29.5 inches. 
 
 ,, temperature of attached thermometer = 
 58.6° F. 
 
 Corrected height at 32° F. = — ^'^ ^ ^ — 
 
 ^ I + fo.oooi X (58.6-32)} 
 __29^ 29.5 . 
 
 = — —7 -rr^ — 77 = 20.42 inches. 
 
 I + {o. 0001 X 20.6] 1.00266 -' ^ 
 
 (6) In the Centigrade scale the coefficient of expansion of 
 
 mercury per 1° C. is =0.00018, and reducing to o' C. : 
 
 5550 
 Corrected height of barometer at 0"" C. 
 
 Observed height of barometer 
 ~ I + {0.00018 X C.''} 
 
 Note. — The observed height is expressed in millimetres. 
 C = reading in Centigrade degrees of attached thermometer. 
 
 Example. — Observed height= 761.75 mm. 
 ,, temperature =^8. 6 "" C. 
 
 n . 1 u • If f on 761.75 _ 761.75 
 
 Corrected heiglit at o C. = — — 7 o — ^~7i = r> 
 
 ^ I + {0.00018 X 8.6} 1. 001548 
 
 ~ 760.72 mm. 
 (2) Corrections for both the mercuiy and tlie brass-scale. 
 
METEOROLOGY 41 
 
 (a) Faiikenheit.— The coeflScierit of expansion of brass per 
 
 T^ F at 62° F. = 0.00001. (N.B. — less than the corresponding 
 
 10 
 
 coefficient of expansion of mercury.) 
 
 Corrected height at 32° F.l= Observed height of barometer 
 (i.e. True reading) J 7ni7ms : 
 
 {Observed height of barometer x (F.° - 32) x 0.000089.} 
 (F.° = temperature of attached thermometer.) 
 ;N'ote. — "0.000089 " = coefficient of expansion of Hg, mmns 
 ditto of brass, i.e., 0.000 1 - o.ooooi. 
 
 Note.— The correction must be subtracted from the " observed 
 height," which term occurs tivice in the formula as given above. 
 Example.— Observed height of barometer = 3 1 . 5 inches. 
 
 Observed temperature of mercurial column indicated 
 by attached thermometer = 58° F. 
 
 .*. Corrected height at 
 
 32" F. = 31-5 - {31-5 X (58 - 32) X 0.000089} 
 = 3i.5-{3i.5X26x 0.000089} 
 = 31.5-0.729 
 =jo.y/i inches. 
 
 (b) Centigrade.— The coefficient of expansion of brass for each 
 degree is 0.000018, and of mercury 0.00018. 
 
 Corrected height at 
 
 0° C. = Observed height - {Observed height x C.° x 0.00016}. 
 
 C = temperature of attached thermometer in Centigrade deg-ees. 
 
 Example.— Observed height = 787.1396 mm. 
 Observed temperature = 14-5^ 0. 
 
 Corrected height at 
 
 0° C = 787.1396 - {787.1396 X 14.5 X 0.00016} 
 = 787.1396-13.1609... 
 ~ 773-97^ mm. 
 Note.—" 0.00016 " = coefficient of expansion of mercury, minus 
 coefficient of expansion of brass, i.e. = 0.00018 - 0.000018. 
 
 Schumacher's formula for the correction for mercury and the 
 bras«-scale, reduced to 32° F., is as follows : 
 
 F = temperature of attached thermometer in degrees Fahrenheit. 
 0.000 1 = coefficient of expansion of mercury at 32° F. 
 0.00001= coefficient of expansion of brass-scale at standard 
 temperature of 62 F. 
 
42 
 
 CALCULATIONS IN HYGIENE 
 
 )■ 
 
 Corrected height at 5*2° F. 
 
 = Observed height, minus or j^^us : 
 
 ( o.oooi X (F. - 32) - o.ooooi (F. - 62) 
 {Observed height x 1 + o.oooi (F. - 32) 
 
 The ''Correction'' is to be subtracted {-) from the observed 
 height at 29" F. and all higher temperatures, but is to be added 
 (+') to it for all temperatures below 29^ F. See Table of Correc- 
 tions (Scott). 
 
 Corrections to be applied to Barometers with Brass Scales 
 extending from Cistern to the top of the Mercurial Column to 
 reduce the observation to 32' F. 
 
 Temperature. 
 
 Observed Height of Barometer in Inches. 
 
 Dearrees F. 
 
 28.0 1 28.5 1 29.0 1 29.5 1 30.0 1 30.5 1 31.0 
 
 27 
 
 + 0.004 All through. 
 
 28 
 
 + O.OOT 5) 
 
 29 
 
 -0.00 I ?> 
 
 30 
 
 - 0.004 , V , 
 
 40 
 
 - 0.029 
 
 0.029 
 
 0.030 
 
 0.030 
 
 0.031 
 
 0.031 
 
 0.032 
 
 so 
 
 -0.054 
 
 0-055 
 
 0.056 
 
 0.057 
 
 0.058 
 
 0.059 
 
 0.060 
 
 60 
 
 -0.079 
 
 0.080 
 
 0.082 
 
 0.083 
 
 0.085 
 
 0.086 
 
 0.087 
 
 70 
 
 - 0.104 
 
 0.106 
 
 0.108 
 
 0.109 
 
 O.I I I 
 
 0.113 
 
 0.T15 
 
 80 
 
 - 0.129 
 
 0.I3I 
 
 0.133 
 
 0.136 
 
 0.138 
 
 0.140 
 
 0.143 
 
 90 
 
 -0-153 
 
 0.156 
 
 0.159 
 
 0.162 
 
 0.164 
 
 0.167 
 
 0.170 
 
 100 
 
 -0.178 
 
 0.I8I 
 
 0.184 
 
 0.188 
 
 0.191 
 
 0.194 
 
 0.197 
 
 5. Corrections for Altitude, or "Reduction to Sea-Level." 
 
 If the atmosphere were a homogeneous medium of uniform and 
 
 constant density and incompressible, its height could be calculated 
 from the pressure at sea-level as ob.^erved by the barometer, and 
 from the densities or weights of air and mercury relative to water. 
 Weight of atmosphere at sea-level = mercurial cohimn 30 inches 
 
 high. 
 
 I 
 
 water = i . 
 
 Weight of air = o 00 1 29 
 
 „ ,, mercury = 1 3.6, /.e., 10543 times that of air J 
 .-. 0.00129 : 13.6 : : 30 : height of atmosphere (" homogeneous "). 
 
 1-^.6 X 30 ^ . , , 
 
 .-. height of atmo.spheie = -^^;:;;77:;-- =316,000 inches. 26,333 
 
 feet, or nearly 5 miles. 
 
 0.00129 
 
METEOROLOGY 43 
 
 In a homogeneous atmosphere the fall of the Imrometer would 
 be regular, the pressure diminishing with the ascent, and depend- 
 ing on the weight of the vertical column of air between points at 
 different levels. By the simple proportion : 
 
 30 inches : i inch : : 26333 ^^^^ • ^ ^'^6*- 
 The fall would be i inch for every 877.8 feet of ascent, or in 
 round numbers about 880 feet. This does not hold good, because 
 air is not a homogeneous medium, and is affected by pressure, 
 temperature, movement (wind) and moisture. The average fall in 
 the barometer is, therefore, determined by experiment, and is 
 only an approximate estimate. In Great Britain the decrease is 
 I inch for every 900 feet of ascent from sea-level— i.e., o.ooi inch 
 per foot, omitting variations of temperature, pressure, &c. 
 
 For the British Isles the standard sea-level adopted by the 
 Ordnance Survey and Meteorological Offices is the mean sea-level 
 at Liverpool — i.e., half the average range of fluctuation between 
 high- and low- water mark. From this " datum " altitudes of all 
 localities are calculated, and are marked on the Ordnance Bench 
 Marks and Survey maps. 
 
 All barometric observations must be reduced to sea-level, and 
 corrected to standard temperature (32° F.) for comparison. The 
 " correction " for places above sea-level is to be added to the 
 observed local reading ; and to be subtracted for localities below 
 sea-level which are exceptional. 
 
 If n = altitude of place above sea-level in feet. 
 X = correction for altitude in inches. 
 900 : n \ \ I \ X. x = nx 0.00 1 inch. 
 
 Example. — Altitude = 4500 feet above sea-level. 
 
 Observed reading at this altitude = 27.3 inches (corrected to 
 
 " Correction''' i : 4500 : : o.ooi : x. x = ^.^ inches. 
 Reduction to sea-lei'el= 27.3 4-4.5 ~ 3^-^ inches. 
 For correct records the temperature of the air must be taken 
 by a dry-bulb thermometer, and not by the instrument attached 
 to the barometer, and the pressures of the barometer should be 
 taken, if possible, simidtaneously at the higher altitude and at the 
 sea-level. 
 
 Correction for Unequal Intensity of Gravity. — This is 
 usually omitted. It is calculated as follows : 
 
 Let 11 = height of barometer in inches at 32° F. 
 
 G = acceleration due to gravity in feet ( = 32 ft. per second). 
 13.6 = density of mercury relative to water at 32". 
 Absolute pressure = H x G x 13.6, 
 
44 CALCULATIONS IN HYGIENE 
 
 To convert Barometric Readings in Inches into Milli- 
 metres. 
 
 I metre at o° C. = 39.37 inches or 3.3 feet. 
 I inch at 0° C. = 25.4 mm. 
 
 Therefore, a reading of 31 4 inches ^^ 31.4 x 25.4 = 797.56 mm. 
 
 A fall of I inch for every 900 feet of ascent corresponds to a 
 fall of 25.4 mm. for 274 metres, which is approximately a fall of 
 I mm. of mercury for every 10.8 (or 11) metres of altitude. 
 
 The *' correction " is therefore as follows : 
 
 Let n = altitude of place above sea-level in metres. 
 .T = correction for altitude in mm. 
 II : n : : 1 : X. 
 
 Example. — Altitude above sea-level = 468 metres. 
 
 Observed reading at 20° C. reduced to o'' C. = 
 
 740 mm. 
 
 "Correction'' 11 : 468 : : i : x. »? = 42.5. 
 deduction to sea-level = 740 + 42.5 = 782.5 mm. 
 
 Formula for measuring Heights by the Barometer. — 
 
 Altitudes are usually estimated by aneroid barometers specially 
 graduated for the purpose and corrected by the use of " Altitude 
 Tables." 
 
 If mercurial instruments are employed simultaneous readings 
 should be taken at sea-level and at the higher altitude. The 
 method is similar to the one for reduction to sea-level. 
 
 Example. — Sea-level reading = 31.5 inches. 
 
 Reading at higher altitude = 28.4 inches. 
 Difference = 3.1 inches. 
 
 Height above sea-level : i : 3.1 : : 900 : x. x — i'ji^o feet. 
 
 Strachan's Formula. — This necessitates the use of an aneroid 
 barometer. Take the upper and lower readings in inches, tenths, 
 and hundredths of an inch {i.e., to two decimal places), subtract 
 one from the other and treat the result as if it consisted entirely 
 of whole numbers (move the decimal point two places to the right, 
 i.e., multiply by 100, converting decimals into whole numbers), 
 and multiply by 9 ; the result gives the required height in feet. 
 
 Example. — Heading at lower level = 32.53 inches. 
 „ „ upper „ =28.41 „ 
 Difference = 4. 1 2. Take 4.12 as 412 (no decimals). 
 .'. 412x9 = 3708 feet = altitude above sea-level. 
 
METEOROLOGY 
 
 45 
 
 The Vernier (Figs, n and 12). — The length and subdivisions 
 of the Vernier-scale are in relation to the fixed scale of the 
 barometer. 
 
 The scale on the barometer is divided into : 
 inches and half -inches. 
 
 tenths of an inch = — = 0.1 inch, 
 10 ' 
 
 half -tenths of an inch 
 
 0.05 inch, 
 
 Fig. II.— Veruier. 
 
 Fig. 12. — Vernier. 
 
 which form the smallest divisions on the barometer scale ; and 
 twenty-four of them are equal to 0.05 x 24= 1.2 inch. 
 
 The Vernier-scale has 5 large divisions, each of which is sub- 
 divided into 5, making 25 in all. These 25 are equal in length to 
 24 of the 0.5 inch divisions of the fixed scale, and are therefore 
 equal to 1.2 inch in length, so that each small division equals — , 
 or 0.048 inch, and one large division on the Vernier-scale = 5 x 0.048 
 = 0.24 inch. 
 
 The differe7ice between the smallest divisions on the two scales 
 is equal to 0.05 - 0.048 = 0.002 inch ; therefore each small division 
 
46 CALCULATIONS IN HYGIENE 
 
 on the Vernier is 0.002 inch smaller than that on the fixed scale, 
 and each of the larger Vernier divisions indicates a difference of 
 5x0.002 = 0.01 inch. A longer line placed at these points is 
 marked i, 2, 3, 4, and 5, corresponding to 0.0 1, 0.02, &c., or 
 hundredths of an inch. 
 
 To take a reading (after noting the temperature of the attached 
 thermometer and adjusting the cistern-level, &c.), bring the lowest 
 (zero-) line of the Vernier exactly on a level with the top of the 
 mercurial column (" tangential " to the convex surface). If it is 
 on a line with one of the divisions on the fixed scale, read ofi:' the 
 height on the latter only, the Vernier-scale is not required. 
 
 If the mercury-level lies between two of the smallest divisions 
 of t\iQ fixed scale, adjust the Vernier as before, read ofi'the height 
 on the fixed scale in inches, o.i inch, and to the nearest 0.05 inch 
 which lies immediately heloiv the top of the column. 
 
 Next follow. up the smallest divisions on the Vernier-scale until 
 one of them is found exactly level with a line on the fixed scale. 
 (i) Count how many of these small Vernier divisions intervene 
 between the zero-line (level with the mercury) and that on the 
 fixed scale, multiply them by 0.002 inch, and add the result 
 to the partial reading already obtained on the fixed scale. 
 (2) The figures on the Vernier may be utilised instead of the 
 above method: Note the Vernier figure (" i," "2," "3," or 
 "4,") immediately below the "junction" of the scale-lines, and 
 consider it as 0.0 1, 0.02, 0.03, or 0.04 inch and add it to the reading 
 of the fixed scale. Multiply any smaller divisions lying between 
 this point and the ^^ junction " by 0.002 and add the result, the total 
 of these three readings is the correct one, and will be the same 
 as that obtained by the first method (i). 
 
 Example. — Fig. 11. Zero-line of Vernier is level with top of 
 mercurial column and also with line of fixed scale. The reading 
 is taken entirely by the latter, the Vernier is unnecessary. 
 
 Reading: 29.5. 
 
 Fig. 12. I. Heading hy fixed scale: 29.6 (inch- and o.i inch- 
 divisions). 
 
 ,, „ 5, ,, o 05 (immediatelyibelow top 
 
 of column). 
 
 II. Reading by Vernier-scale: (i) 17 small divisions 
 intervene between the zero-line of the Vernier and the "junction " 
 line (level with one on the fixed scale) : 1 7 x 0.002 = 0.034 inch. 
 
 Therefore 29.6 
 0.05 
 0-034 
 Correct 7-eading = 2^.684. inches. 
 
METEOIiOLOGY 47 
 
 (2) Instead of reading as above: Note that in Fig. 12 the 
 figure "3" on the Vernier-scale is immediately below tlie "junc- 
 tion"; this represents 0.03 inch. Between this point and the 
 "junction " there are two small divisions; therefore 2 x 0.002 = 
 0.004. 
 
 We have therefore by fixed scale : 29.6 
 
 0.05 
 „ Yernier-scale : 0.03 
 0.004 
 
 Correct reading = 29.684 (as above). 
 Temperature. — British Method. F.° scale. 
 
 I. One observation daily, 9 a.m. 
 
 E.g. On Feb. 21 at 9 a.m. Maximum registered 43.5' F. 
 
 Minimum „ - 29.0'' F. 
 
 Maximum reading = maximum temperature of j^revious afternoon 
 Minimum ,, = minimum ,, „ same morning 
 
 i.e. " February 20 (afternoon) -= 43.5 ° F. 
 
 ,, 21 (morning)= - 29.0^^ F." 
 
 Daily mean temperature : 
 
 II. (i) Two observations daily, 9 a.m. = 58. 6) JQ5-i o -r, 
 
 9 P.M. = 46.5/ 2 ""52.5 J^- = 
 r _f^ y ^ daily mean. 
 
 Or shade maximum ~ °'° 
 
 207.7 1 ., 
 
 = 5 1.9 = daily mean. 
 
 (2) r^^-' 
 
 ,, minimum \ ^^ 
 1 = 43-5 
 
 III. Three observations daily, 6 am. = 42.5° F.] ^ 
 
 2 PM. = 48.0° I ^ = 40.9 daily 
 
 10 P.M. = 32.2'' J ^ mean. 
 
 Continental Method. C.° scale. Three observations at 8 a.m. 
 2 and 10 P.M. marked -f- and - if respectively above and below 
 0° C. The results are added algebraically (subtracting - signs) 
 and the 10 p.m. observation is doubled, i.e., added twice, and the 
 total result is divided by 4. E.g. : 
 
 8 A.M. = - 6° C.l 
 
 2 P.M. = -1-5 
 /rp • xflO P.M. = +3 
 
 (ivvice)-^ , "^ 
 
 ^ ^(lO P.M.= 4-3 
 
 II -6= +5. - = + 1.25° C. daily 
 
 "^ mean temperature. 
 
 Sum of the daily means 
 
 Monthly mean temperature = vf r t^t- — • — n tt 
 
 •^ ^ JN umber 01 days in the month 
 
 Sum of monthly means 
 
 Annual ,, ,, = 
 
 " 12 
 
48 
 
 CALCULATIONS IN HYGIENE 
 
 Daily range or amplitude of temperature : 
 
 Shade njaximum - Shade minimum temperature for that day. 
 
 Sum of the daily ranees 
 
 Monthly mean range : ^, - , ~^, r—r\ — - — tt 
 
 •^ Number or days m the month 
 
 or. 
 
 or, 
 
 Daily maxim a of month - Daily minima of month 
 Number of days in month 
 
 Sum of monthly mean ranges 
 
 Yearly mean range 
 
 12 
 
 Shade 
 
 Daily maxima of year - Daily minima of year 
 365 
 
 Solar radiation : Sun-maximum 
 maximum temperature. 
 
 Better : Sun maximum mimes open-air 
 temperature taken when the sun-maximum 
 thermometer is at its highest point. 
 
 Terrestrial radiation = Shade minimum - 
 " grass mhiimum " reading. 
 
 Graduation of the Measuring- Glass of 
 aRain-Gauge.^ — The usual diameters of the 
 rim of the rain-gauges (Fig, 13) used in this 
 country are 5 (Symons' " Snowdon ") and 8 
 inches (Glaisher's). The latter diameter is 
 used by the Meteorological Office and by the 
 British Association. 
 
 Let D = diameter in inches. 11 = radius. 
 D = 2 R. 
 
 C = circumference of rim. 
 
 TT = 3. 141 6 ( ratio of circumference to diameter ^ 
 
 C ^ 2 ttR. Area of circle = ttR". 
 
 If D = 5 inches. 
 
 Area of receiving-surface of rain-gauge = 3.1416 x (2.5)'- 
 
 = 3.1416x6.25 
 — 19-635 sq. inches. 
 
 One inch of rainfall on this area= 19 635 x i = 19.635 cubic inches, 
 or half an inch of rainfall = 9.8175 cubic inches. 
 
 1.73 cubic inch = i fluid ounce, .". 9 8175 cubic inches = 5.675 
 ounces of rain. If this amount of water be poured into a non- 
 graduated measure-glass and the level marked with a line, it will 
 represent the height of half (0.5) an inch of rainfall on the receiving 
 
METEOROLOGY 49 
 
 area of the gauge. This height on the glass is subdivided into 50 
 equal parts, and each denotes ^L of ^ inch, or 0.02 x 0.5 inch of rain 
 = 0.0 1, or one-hundredth of an inch. The amount collected when 
 poured into the measure-glass can at once be read off in decimals of 
 an inch. If D = 8 inches, area = 3. 1416 x (4)2 = 50.2656 square 
 inches =14.5 ounces nearly (J in. of rainfall). The subdivisions 
 are made as before. 
 
 An area of any known dimensions will do for collecting the 
 rain and graduating a glass. Conversely, if the amount of water 
 required to represent half an inch of rainfall is stated, the neces- 
 sary diameter of the rain-gauge can be calculated. 
 
 Example. — One ounce of water in the measure-glass is required 
 to denote half an inch of rainfall. What must be the diameter of 
 the rim ? 
 
 Area of receiving surface enclosed by the rim = ttE,-, and this 
 area when covered with half an inch of rainfall must equal i 
 fluid ounce, i.e., 1.73 cubic inches. 
 
 1.73 
 .'. ^.1416 X R- X 0.5 = 1.73. R- = 7 = T.ii3 
 
 ^ ^ J /o 3.1416x0.5 ^ 
 
 1.055 ^^^^ (nearly) 
 2R= 2.11 inches. 
 
 The diameter of the rim must be 2. 11 inches and the surface- 
 area will be 3.49 square inches. 
 
 Velocity of the Wind. — The anemometer named after the 
 Rev. Dr. Robinson of Armagh is in general use. The length 
 of the arms measured from the centres of opposite cups varies 
 from 1. 1 2 feet to 2 feet. 
 
 In the first case 1.12 feet is the diameter of the circle described 
 by the centre of each cup as it revolves. 
 
 .-. Circumference = 2 7rR= TT x diameter = 3-1416 x 1.12 = 3.52 
 feet. 
 
 If the cups are considered to revolve at one-third the velocity 
 of the wind, wind-velocity = 3.52 x 3 = 10.56 feet per each revolu- 
 tion, and to calculate the number of rotations per mile : 
 10.56 : 5280 : : : I : X. a;=5oo revolutions. 
 
 Symons (in Stephenson and Murphy's Ti^eatise) considers the 
 velocity of the wind to be 2.5 times greater than that of the cups, 
 
 .'. 3. 52x2. 5 = 8. 8 feet per revolution, and the number of revo- 
 lutions per mile = 600. 
 
 Using 2.5 as the factor, with a diameter of 2 feet the cir- 
 cumference of the circle is 6.28 feet, the wind-velocity 15.7 feet 
 per revolution, there being 336.3 rotations per mile. 
 
 The records are not accurate, as the velocity varies with the 
 
 D 
 
50 CALCULATIONS IN HYGIENE 
 
 length of the arms, being l)elow the mark with short, and above 
 it with long arms. 
 
 Pressure of the Wind. — Pressure varies as the square of the 
 velocity. 
 
 Col. H. James' Formula : 
 
 Let P = pressure in pounds per square foot. 
 V = velocity of wind in miles per hour. 
 
 V- -.ro I 
 
 Y = j20oP. V2=20oP. P= -^=V->: — = V=xo.ooq. 
 ^ 200 200 ^ 
 
 Example. — (i) Pressure in pounds per square foot= 15.25. 
 
 V-=200X 15.25 = 3050. 
 
 V= v/3050 = 55.22 miles per hour. 
 (2) Velocity of wind = 86.8 miles per hour. 
 P = (86.8)- X 0.005 = 37-67) or 37.7 pounds per square foot. 
 Note. — (Velocity in feet per second)- x 0.0023 = Pressure ^^ 
 pounds per square foot (approximately). 
 
 ^ tj. 2 2 X K 280 
 
 Example (i) as above : -^^7 ^^^— = 80.96 feet per second. 
 
 (80.96)- X 0.0023 = 15.075, or nearly 15.25 pounds per sq. ft. 
 
 i 
 
CHAPTER IV. 
 
 ventilation: 
 
 Respiration. — Pare air contains on an average 
 Oxygen 
 Hydrogen . 
 Nitrogen . 
 Argon, metargon, tfec. 
 Carbon dioxide . 
 
 20.94 
 0.02 
 
 78.00 
 1. 00 
 0.04^ 
 
 100.00 
 
 Expired air contains on an average 1 6 per cent, of oxygen and 
 4.4 per cent, of CO2. 
 
 Amount of CO^ exhaled at each breath in excess of that con- 
 tained in the air = 4.4 - 0.04 = 4.36 i^er cent. An adult male 
 breathes about 18 times per minute, and the tidal air averages 
 about 25 cubic inches (500 c.c). Amount of CO3 expired at each 
 breath : 
 
 100 : 25 : : 4.36 \x. x= 1.09 cubic inch (21.8 c.c.) 
 
 Per minute = 1.09 x 18 = 19.62 cubic inch (392.4 c.c.) of COg 
 „ hour= 1 1 77.2 cubic inches, or nearly 0.7 cubic foot. 
 
 Parkes and de Chaumont adopted 0.6 cubic foot per individual per 
 hour as the average amount of COg exhaled in a "mixed com- 
 munity," i.e., of adults and children. 
 
 Velocity of Inflow 
 Volume of air 
 
 and Outflow of Air. — Velocity = 
 
 Sectional area of aperture = '^'^<= " terms " must be of the same 
 
 " denomination," i.e., to calculate a velocity in linear feet the 
 
 volume must be expressed as cubic feet and the sectional area in 
 
 square feet. If the latter is stated in square inches these must be 
 
 converted into square feet. 
 
 „ Flow in cubic feet 
 
 -t,.q.. Velocity m linear teet = 5^— ^ 1— -. j- — . 
 
 •^ "^ feectional area m square feet 
 
 * According to some investigators the average amount of CO., in pure 
 air is but little over 0,3 per cent., or 3 parts in 10,000 by volume. 
 
52 CALCULATIONS IN HYGIENE 
 
 Example.— (i) An outlet of 48 square inches sectional area 
 delivers 12,000 cubic feet per hour. Estimate the velocity of out- 
 flow in feet per minute and per second. 
 
 1200 „ 4000 
 
 Velocity in linear feet = — ^- = 4000 feet =-^ = 66.6 per 
 
 M4 
 
 minute,and-T^- = i.ii feet per second. If J be deducted for 
 ' 60 
 
 friction i.ii - ^-" (=1.11 x o.75) = o.83 feet per second. 
 4 
 (2) To find the velocity required to deliver 3000 cubic feet of 
 air per hour through an inlet of 2 5 square inches. 
 
 V = ?^^ = 3ooo X ^= 17280 feet per hour 
 25 25 
 
 144 
 
 17280 o r i. 
 
 = --^ — - =4.8 leet per i . 
 60 X 60 ^ 
 
 4.8 
 Deducting \ for friction : — = 1.2 
 
 .-. 4.8 - 1.2 = 3.6 feet per i". 
 
 Friction at every right angle diminishes the velocity by \. 
 .'. With 2 right angles velocity = J. 
 
 (3) The velocity of inflow is 3 feet per second, and the aperture 
 has a sectional area of 36 square inches. Calculate the volume of 
 air entering per hour. 
 
 Velocity = 3 feet per second. Area = -— = I square foot. 
 
 Let volume of in-coming air = a;. .-. 3 = T- •'^ = 1 cubic foot 
 
 per second. 
 
 = I X 60 X 60 = 2700 cubic feet per hour. 
 
 (4) In a ward for 30 beds it is desired to supply 4000 cubic 
 feet of air per head per hour at a velocity of 2 J feet per second. 
 Estimate the size of inlet required. 
 
 As the velocity is given " per second " and the volume of air 
 supply is to be estimated " per hour," the velocity must also be 
 expressed " per hour." 
 
 V = 2.5 X 60 X 60 linear feet per hour. 
 
 Total inflow = 4000 x 30 cubic feet per hour. 
 
 Sectional ai-ea in cubic feet = .%-. 
 
VENTILATION 53 
 
 4000 X 30 
 .*. 2.1^x60x60 = — . 
 
 2.5 X 60X 60a; = 4000 X 30. x= 13.3 square feet = 1915. 2 square 
 inches. 
 
 For 30 patients = 63.84 square inches per head. 
 
 Therefore 30 inlets each of 63.84 square inches will suffice, or 
 60 inlets each of 31.92 square inches. 
 
 Supply of Fresh Air. — De Chaumont estimated that 0.02 per 
 cent., or 0.2 per 1000 volumes of CO^, should represent the maxi- 
 mum amount of respiratory impurity in excess of that existing in 
 pure (external) air. As pure air contains 0.4 volume of CO^ per 
 1000, 0.4 + 0.2 = 0.6 volume of total CO^ permissible in 1000 
 volumes of air in a room. This 0.6 per 1000 volumes must not 
 be confused with the 0.6 cubic foot of 00^ exhaled per individual 
 per hour in a " mixed community." 
 
 0.2 cubic foot of added respiratory impurity per 1000 cubic feet 
 = 0.0002 per I cubic foot of air. If 0.7 cubic foot = amount of 
 CO., exhaled per male adult per hour, the calculation for estimating 
 the" supply of fresh air needed per hour is : 
 
 0.0002 : 0.7 : : I : x. x= ^"^ =3500 cb. ft. If instead of 0.7 
 ' 0.0002 
 
 we take 0.72. x= -^^^^ = 3600 cb. ft., or i cb. foot per second 
 ' 0.0002 
 
 — a convenient standard. 
 
 Let S = supply of fresh air necessary per hour in cubic feet. 
 E = amount of 00^ exhaled per individual per hour. 
 I = excess of CO, permissible (impurity) per i cubic foot 
 per hour. 
 
 De Chaumont's Formula.— De Chaumont considered 0.6 
 cubic foot CO, exhaled per individual per hour as a fair estimate in an 
 
 assemblage of men, women and children, so that b = -^^^^ = 3000 
 cubic feet of fresh air required per person per hour. 
 The above estimates are for individuals at rest 
 
 If a man is doing light work, E = 0.9. S = ^^^ = AS^o cb. ft. 
 
 hard „ E= 1.8. 8 = ^4:; = 9000 n 
 " " " 0.0002 
 
 Large supplies of fresh air necessitate large buildings or increased 
 velocity of inflow, and augment the cost. In most cases this 
 
54 CALCULATIONS IN HYGIENE 
 
 cannot be done. In a General Hospital the increase is roughly 
 about I more than in health : for a male ward E may be taken as 
 = 0.9 (as above) and S = 4500 cubic feet. For a female ward E = 
 
 0.4 
 0.8. S = 4000 cubic feet. In schools E = 0.4. S= = 2000 
 
 cubic feet. 
 
 If I = CO., added as impurity per 1000 cubic feet of air (instead 
 of per I cubic foot). 
 
 ^. E 0.6 .... 
 
 S r= — X 1000 = — X 1000 = •^000 cubic reet. 
 I 0.2 ^ 
 
 Let E = o.6 cubic foot of CO2 expired per individual per hour, 
 o 4 = 00^ present in 1000 cubic feet of fresh (external) air. 
 I = total CO^, in 1000 cubic feet of air in room minus 0.4. 
 P = number of persons occupying the room (exhahng 0.6 
 
 cubic feet of COg per head per hour). 
 H = number of hours they occupy the room. 
 S = supply of air for P persons during H hours : 
 
 o.6xPxH 
 
 ^ I {Total CO2 in 1000 cb. ft. of air in room - 0.4) 
 Any one term of this equation can be calculated if the others 
 are known. 
 
 Example. — (i) 5 men work for 8 hours in a room containing 
 6500 cubic feet of air. The total supply of fresh air per hour is 
 12,000 cubic feet. Calculate the total amount of CO., present in 
 the air of the room at the end of that period and also the amount 
 of impurity added as COg. 
 
 Total air supply for 5 men for 8 hours : 
 Pure air originally in the room = 6500 cubic feet. 
 Additional air supplied during 8 hours = 12,000 x 8 = 96,000 
 cubic feet. 
 
 6500 + 96000= 102,500 cubic feet = S. 
 Each man expires (say) 0.9 cubic foot CO.^ per hour = E. 
 X = CO^ per 1000 cubic feet in room after 8 hours .*. I = a: - o 4. 
 
 0.9 X 5 X 8 
 
 .*. io2,t^oo=— — ' ~ X 1000. 
 
 102.5 ^ C^' ~ °-4) = 0.9 X 5 X 8. 
 
 io2.5.x-4i = 36. io2.5.f=77. 
 
 £c^o.75 ft. CO, per 1000 = 0.075 per cent. 
 
 ./I <:ZcZe<:Z impurity (I)= 0.75-0.4 = 0.35 per 1000, or 0.035 per 
 cent. It should not exceed 0.2 per 1 000 according to de Chaumont's 
 formula, therefore there is an excess of 0.35 - 0.2 = 0.15 per 1000. 
 
VENTILATION 55 
 
 Carnelley's Formula. — Carnelley, Ilaklane and Anderson 
 allowed a larger excess of COg than de Chaumont did, as 
 follows : 
 
 (i) For dwelling-houses 0.6 vol. of CO^ per 1000 in excess of 
 that in fresh air. 
 
 (2) For schools 0.9 vol. of COgper 1000 in excess of that in fresh 
 air. 
 
 As fresh (pure) air contains 0.4 vol. COg per 1000, -the total 
 CO2 allowable for (i) dwellings and (2) schools is 0.6 + 0.4 =1.0 per 
 1000 ; and 0.9 + 0.4= 1-3 per 1000, respectively. 
 
 .-. I for (i)= I. o - 0.4 = 0.6 added imjmriiij per 1000 cubic 
 feet = 0.0006 per i cubic foot. 
 
 for (2) = 1.3 - 0.4 = 0.9 added imjmrity per 1000 cubic feet 
 = 0.0009 per I cubic foot. 
 
 (i) E = o6. 
 
 0.6 
 
 .*. S = 2 = 1000 (houses). 
 
 .0006 ^ ^ 
 
 (2) E = o4. 
 
 ^ = 0.^^0^ = 444 (schools). 
 
 0.45 
 If E = 0.45. S= = 500 (schools). The latter figure is 
 
 preferable, as allowance must be made for the adult staff of the 
 school. 
 
 Example. — (2) A male ward in a General Hospital has an 
 hourly supply of 1 1 2,500 cubic feet of fresh air. The atmosphere of 
 the ward must not contain more than a total amount of 0.06 CO, 
 per 100 cubic feet. How many beds should there be? 
 
 S= 112,500 cubic feet per hour. 
 
 E = o.9 cubic foot, CO^ per head per hour (male cases). 
 I = 0.6 - 0.4 (pure air) = 0.2 per looc cubic feet of ward atmos- 
 phere. 
 
 P = is to be calculated. 
 H = I hour. 
 
 0.9 X P X I 
 
 .*. 112,500 = X 1000. 
 
 '^ 0.2 
 
 112.5=^-^. 225 = 9P. P = 25. 
 
 There should be accommodation for 25 male patients. 
 (Note. — The air originally present in the ward is omitted.) 
 
56 CALCULATIONS IN HYGIENE 
 
 Example. — (3) In a school 65 children work for 5 hours, and 
 the air at the end of that time is found to contain 0.13 per cent. 
 of CO^. Find how much air is supplied during the entire period, 
 and also per individual per hour. 
 
 Taking E as 0.45 cubic foot COg evolved per head per hour. 
 P = 65. H = 5. 0.13 per cent. = 1.3 per thousand. 
 • '. I = 1.3 - 0.4 COg^ added impurity per 1000 cubic feet of air 
 in schoolroom. 
 S is to be found : 
 
 .*. b = ^XIOOO. 
 
 1.3-0.4 
 
 146.25 
 = X 1000, 
 
 0.9 
 
 = 162,500 feet per head during 5 hours ; 32,500 cubic feet 
 per hour, or 500 cubic feet per individual per hour. 
 
 According to Carnelley's estimate this is satisfactory ; by de 
 Chaumont's standard of 2000 cubic feet per child per hour it is 
 2000 — 500= 1500 below the proper supply. 
 
 Example. — In a room 20 feet long, 14 wide, and 12 feet high, 
 5 clerks work, using 3 ordinary gas-burners. How much air is 
 required for proper ventilation ? 
 
 Gas used per burner = 4 cubic feet per hour. 
 
 = 12 cubic feet per 3 hours. 
 I cubic foot of gas = 0.5 cubic foot CO^. 
 4 V feet „ =2.0 „ feet „ 
 
 ^^ >> 5) ?5 =0.0 „ ,, ,, 
 
 Taking 0.7 cubic foot COg as exhaled per man per hour. 
 
 3.5 „ feet „ is „ „ 5 men „ 
 
 6.0 + 3.5 = 9.5 COg in the room per hour. 
 
 9.5x1000 " ,. P , 
 
 b = — = 47,c;oo cubic reet. 
 
 0.2 ' -^ 
 
 The room originally contained 20 x 14 x 12 = 3360 cubic feet 
 of fresh air (not allowing for furniture, &c.) 
 
 .*. 47,500 - 3360 = 44,140 cubic feet in addition to that origi- 
 nally in the room should be supplied during the first hour of occupa- 
 tion, and for every subsequent hour (after the 3360 cubic feet are 
 exhausted) 47,500 cubic feet of fresh air are necessary. 
 
 Example. — (4) A room 20 feet long, 12 feet broad, and 10 feet 
 high is occupied by two adults and a child under six years of age. 
 If there is practically 010 ventilation, when will the limit of per- 
 missible impurity be reached ? 
 
VENTILATION 57 
 
 S = 20 X 12 X io = 2400 cb. ft. fresh air originally in the room. 
 
 E = o.6 cb. ft. CO2 per head per hour (" mixed community.)" 
 
 Two adults and one child = 2.5 adults. 
 
 .-. P = 2.5. H is to be calculated. 
 
 Total impurity = 0.6 cubic foot 00^ per 1000 cubic feet. 
 
 I = 0.6 - 0.4 = 0.2 cubic foot per 1000. 
 
 0.6 X 2.5 X H 
 
 .•. 24.00 = ^ X 1000. 
 
 ^ 0.2 
 
 T c IT 
 
 2,4 = -^^ . 4.8= 15 H. 11 = 0.32 of an hour. 
 
 = 19.2 minutes. 
 
 By Carnelley's formula : 2.4 = -^r- = H = 0.96 of an hour. 
 
 = 57.6 minutes. 
 
 Fresh Air Supply for Horses and Cattle. — These animals 
 are not affected by a rapid current of air or a low temperature if 
 well stabled. 
 
 De Chaumont's formula may be adopted. 
 
 A horse exhales i . 1 3 cubic feet COg per hour. 
 
 Taking "I" as 0.6-0.4 = 0.2 cubic foot per 1000 as ^^ added 
 
 impurity " : S = — ^ = 5650 cubic feet fresh air supply per hour. 
 
 The same ratio may be used for large cattle. 
 
 Natural Ventilation. — The velocity acquired by a body falling 
 through space is proportional to the time it takes in falling. Its 
 " acceleration " is due to gravity, which causes a velocity of fall 
 equal to 32.2 (taken as 32.0) feet per second in this latitude, and 
 the motion is uniformly accelerated, 32 being a constant "factor.' 
 
 Let V = velocity of fall in feet per second. 
 G = " acceleration " = 32 feet per second. 
 T = time of fall in seconds. 
 V = GxT. 
 = 32xT. 
 
 If a body falls (from a state of rest) for three seconds : 
 V = 32X3 = 96 feet per second at the end of three seconds. 
 
 If the " height " or distance through which a body has fallen 
 be known, its final velocity in feet per second is equal to eight 
 times the square root of the distance traversed. 
 
 Let H = height (distance) of fall. 
 
58 CALCULATIONS IN HYGIENE 
 
 .-. V- = 2 GxIL G = 32. .-. V--2X32XH. 
 V- J2 X 32 x H. 
 = V64X H. 
 
 By Montgolfier's law, air and other lluid media pass through 
 an opening in a partition with the same velocity as a body falling 
 through a height equal to the difference in level of the Huids on 
 each side of the partition. This difference in level is "head." 
 
 Air rushes into a vacuum (resistance at first is nil) at the same 
 velocity as a body falling from a height of five miles or 26,333 f®^*- 
 
 .-. V = ^2 X 32 X 26333 = 8/26333 = 1296 feet per second. 
 
 As air passes into a vacuum, the internal pressure (being zero at 
 first) gradually increases with increase of air volume, and its 
 decreasing velocity would be equal to that of a body falling through 
 diminishing heights representing differences of pressure inside and 
 outside the chamber. These varying difierences in height and pres- 
 sure {i.e., weight) cannot be calculated, and are estimated approxi- 
 mately by differences in temperature of the inside and outside air. 
 
 Difference in pressure = difference in height of inlet and outlet 
 ( = height of heated column of air) x difference of temperature x 
 coefficient of expansion of air. 
 
 Let II = height of heated column of air in the flue (or vertical 
 distance between inlet and outlet). 
 
 „ T = temperature (Fahrenheit) of inner air. 
 
 ,, « - „ „ „ o^itei- » 
 
 0.002 ( = — approximately) = coefficient of expansion of air per 
 
 i" F.). V=« velocity in linear feet per second. 
 Difference in pressure = Hx(T-i)xo.oo2. 
 .-. V = 8 VHx(T-i)xo.oo2. 
 
 Example. — The external opening of a ventilating shaft is 22 feet 
 above the internal aperture. The temperature of the room is 
 60"^ F. and that of the external air 47.2"" F. Calculate the 
 velocity of the current of air up the shaft. 
 
 11 = 22 f eet. T-^ = 60 -4 7.2 = 12.8. 
 Y = 8 ,^22x12.8x0.002. 
 
 = 870.5476. 
 
 = 8x0. 74 = 5. 9 2 feet per second. 
 
 Deducting | for loss of velocity due to friction (which is the 
 same as multiplying by 0.75) 5.92 x 0.75 = 4.44 feet per second. 
 
VENTILATION ^')i) 
 
 De Chaumont's Modification of this Formula for calcu- 
 lating the Size of Inlet or Outlet, or the Quantity of Air 
 Supply per Hour. 
 
 Let S = supply of air in cubic feet per hour. 
 A = are<a of inlet or outlet in square inches. 
 H = height of heated column of air. 
 T = temperature of room or of air-column. 
 t^ „ ,, external atmosphere. 
 
 / seconds in i hr. \ . i,^ — j^ — ~r 
 
 S- 8x . -. fT x^x \/Hx(T-i)x 0.002x0.75. 
 
 \ sq. in. in i sq. it./ 
 
 S = 20oxAx ^Hx(T-«)x 0.002x0.75. 
 
 Example. — (i) The difference in height between inlet and outlet 
 is 25 feet. The inner and outer mean temperatures are 65° F. and 
 45° F. respectively, and the air supply is 3000 cubic feet per hour. 
 Estimate the size of the inlet. 
 
 S = 3000. T - ^ = 20. 3000 = 200 X A X ,725 X 20 X 0.002 X 0.75. 
 — 200 A X 5 X 0.2 X 0.75. 
 A= 11-25 square inches. 
 
 According to Parkes and de Chaumont the dimensions of the 
 outlet may correspond with those of the inlet, the increase in 
 volume of air that has been warmed is so small ( /^ to y^) under 
 the usual conditions of ventilation and heating that it may be 
 neglected. 
 
 The size of inlets and outlets per individual in health may be 
 approximately estimated as follows : 
 
 Let 7^ = number of adults to be supplied (from i to 6 as a 
 maximum) by one aperture. 
 
 .-. Area of aperture =i2X 2n square inches, therefore : 
 
 For I adult „ „ ., x 2 = 24 square inches. 
 
 2 a 
 
 dults „ „ „ X 4= 48 
 
 2 adults and I child „ 55 x 5 
 
 60 „ f child = -J 
 
 5j 3 55 5' " ,, X 6= 72 „ „ 
 
 „ 6 „ „ „ „ X 12 = 144 » 
 
 At temperatures between 55" and 60" F., a velocity of 
 1.5 linear ft. per sec. is not perceived 
 2-2.5 " » " " " by most people. 
 
 3 5. ,5 '5 felt „ ,, 
 
 3.5 ,^ „ ,, „ by all and causes a draught. 
 
 {Parkes and de Chaumont.) 
 3000 cubic feet per hour --50 cubic feet per minute = 0.83 cubic 
 foot per second. 
 
60 CALCULATIONS IN HYGIENE 
 
 Velocity of air current (linear feet) 
 
 Inflow (cubic feet) 
 
 Sectional area of inlet (square feet)' 
 Inlet = 24 square inches. Inflow = 0.83 cub. ft. per sec. 
 
 Velocity = — jj- = 4.98 lin. ft. per sec. 
 
 TTT 
 
 Inlet = 48 square inches. „ ,, ,, ,, 
 
 Q ~ 
 
 Velocity -- —^ = 2.49 lin. ft. per sec. 
 
 Inlet = 60 square inches. „ ,, ,, ., 
 
 Velocity = 1.99 (2.0) lin. ft. per sec. 
 
 Inlet = 72 square inches. ,, ,, ,, „ 
 
 Velocity = 1.66 lin. ft. per sec. 
 
 Inlet = 144 square inches. „ ,, „ ,, 
 
 Velocity = 0.83 lin. ft. per sec. 
 
 The inlets of 48 to 72 square inches give the best velocity for a 
 supply pe7' head of 3000 cubic feet per hour. 
 
 Parkes and de Chaumont considered the minimum and maxi- 
 mum sizes should be 24 and 144 square inches, and that for larger 
 supplies of air the number of openings must be proportionally 
 increased and not their size. This method facilitates uniform 
 distribution of air per person, and diminishes liability to draughts 
 or to areas of " stagnation." 
 
 Example. — (2) A heated column of air 40 feet in height has an 
 average temperature of 60° F., the air outside being at 32"' F. 
 The area of inlet equals that of outlet, and is 72 square inches. 
 Calculate the supply of air passing in, and the velocity of outflow. 
 
 T-^ = 6o-32=^8^ 
 
 8=200x72x^40x28x0.002x0.75. -i 
 
 = 200 X 72 X ^^2.24 X 0.75. 
 
 = 200 X 72 X 1.5 X 0.75. 
 
 16,200 
 = 16,200 cubic feet per hour= -. — =270 per minute, or 
 
 270 
 
 = 4.5 cubic feet per second. 
 
 To find the velocity of outflow in linear feet per second : 
 Outflow in cubic feet per second 
 e oc y — (.<g(.|^iQjjr^] j^i-g.^ Qf inlet in square feet 
 
 = -^'2 =1=9 linear feet per second. 
 
VENTILATION Gl 
 
 Natural Ventilation by Diffusion of Air. — It occurs only 
 in gases and vapours and does not affect molecular matter. The 
 rate of diffusion is inversely proportional to the square root of 
 the density. 
 
 The density of hydrogen = I .*. — ,^ = i. Rate of diffusion of 
 oxygen , = -. =4 times greater than that of hydrogen. 
 
 Rate of diffusion of nitrogen = 
 
 -= = -!- = 3.74 times . 
 V14 3-74 
 
 Rate of diffusion of air = 
 
 ; = ^ = 3.8 times „ „ „ 
 
 x/i4-44 3-8 
 Calculation of Friction in Ventilation. — It is determined 
 by the length of the shaft, its angles, and by the dimensions and 
 shape of the aperture through which the air current passes. 
 
 Length. — If the tube or shaft is of uniform bore throughout, 
 friction increases in direct proportion to increase in length. 
 Example (i).— Shaft A= 100 feet in length 1^^.^^^,^ ^^^^^ 
 
 „ B=I20 „ „ „ j 
 
 The friction in B is i more than in A. 
 
 Angles. — Every bend of 90° diminishes the velocity of the 
 current by half — e.g., a velocity of 10 linear feet will be after the 
 first right-angle bend = 5 feet, and after the second bend = J of 
 5 feet = 2 J feet, and after a third similar angle = J of 2 J = i^ linear 
 feet. This is theoretical only, as after two such bends in a pipe 
 ventilation is practically nil for hygienic requirements. Friction 
 at other angles is calculated by Trigonometrical factors. 
 
 Dimensions and Shape of Apertures. — A. Loss by friction 
 varies inversely as the diagonal or diameter. 
 
 The openings are of similar shape : 
 
 (a) Squares. 
 
 Example. — A square aperture ABCD is subdivided into 4 
 smaller squares of equal size. Calculate the friction in each of 
 these compared with that in the larger opening {vide p. 62). 
 Diagonal of ABCD - AC or BD. 
 
 ,, „ each small square = J AC or BD. 
 
 If friction in ABCD = i . 
 
 .*. friction in each small square = 2. 
 
 i.e. it is twice that in ABCD. 
 
 {h) Circles. 
 
02 
 
 CALCULATIONS IN HYGIENE 
 
 Example. — Two circular apertures have diameters of i foot 
 and 4 inches respectively. Estimate the friction in the smaller 
 relative to that in the larger. 
 The larger diameter =12 inches. 
 „ smaller „ = 4 „ The ratio of the latter to the 
 
 former is as 4 to 12 or as i to 3. 
 
 If the friction in the smaller circle = i. 
 
 .*. „ ,. ,, larirer = i. It is t. times less than that 
 
 of the smaller aperture. 
 
 B. Loss by friction varies inversely as the square roots of the 
 areas, whether the openings are of similar 
 or dissimilar shape. 
 
 Supposing the square ABCD has its 
 sides I foot in length, its area is 144 
 square inches. 
 
 Each smaller square has an area of 36 
 square inches, J 36 = 6, ^144=12: a 
 ratio of I to 2. 
 
 The loss by friction in the smaller 
 aperture is twice that in the larger one. 
 If a circular inlet has an area of 9 square 
 Fig. 14. feet and a square aperture an area of 4 
 
 square feet, J9 = S. ^4 = ^- 
 The loss by friction in the former to the latter is a.s J to J. 
 The circumference of a circle encloses maximum ^rea in mini- 
 mum periphery. If the friction in two circular openings is to be 
 contrasted, the calculation is made from the respective diameters, 
 as above. 
 
 If the shape is non-circular, ascertain its periphery and the 
 area enclosed. Calculate the length of periphery of a circle enclos- 
 ing an exactly equal area. The friction in the circular opening 
 is to that in the non-circular one as the periphery of the one is 
 to that of the other. 
 
 Example. — A square aperture has its side 14.18 inches long, 
 compare the friction in it to that in a circular opening of equal 
 area. 
 
 Area of square aperture = (14.18)-= 201.0724 square inches. 
 Periphery,, „ „ = 14.18 x 4 = 56.72 linear „ 
 
 Area of circle - ttH'^. 7rIl- = 201.0724 square inches. 
 
 K2 = - — -^-^- = 64 inches (nearly). 11 = ^64 = 8 inches. 
 
 Circumference = 27rK = 2x3. 14 16x8 = 50.2656 inches. 
 
 Friction in circular opening is to friction in square opening as 
 
 50.2656 IS to 56.72, or as i to 1.12, i.e. 
 
 the friction in the latter 
 
VENTILATION 63 
 
 is a little more than ^th greater, and the velocity is proportion- 
 ally diminished. 
 
 Artificial Ventilation. — i. By an ordinary Fire-place. — In 
 a sitting-room with an open fire-place, the chimney is the usual 
 means of ventilation. Its sectional area may be taken as i square 
 foot, and the velocity of the current of air up the flue as 4.5 linear 
 feet per second. 
 
 4.5x1=4.5 cubic feet per second, or 4.5x60x60=16,200 
 cubic feet per hour discharged by the flue. 
 
 Each adult requires 3000 cubic feet of fresh air per hour. If 
 i6j2oo cubic feet of fresh air per hour replace that amount 
 extracted by the flue, v^^e find by simple proportion : 
 
 3000 : 16,200 : : I : a.'. ^^ = 5 adults. 
 
 So that the ventilation is suflicient for 5 adults occupying the 
 room. 
 
 2. By Circular Fans. — The circumference of the circle de- 
 scribed in each revolution by the tip of a vane is calculated from 
 the diameter of the fan ; and the velocity, from the number of 
 revolutions per second, or per minute. 
 
 The tangential velocity of the particles of air leaving the fan is 
 f the velocity of the tips of the vanes. 
 
 Velocity of the air-current x sectional area of outlet = volume of 
 discharge. 
 
 Example. — (i) Diameter of fan = 2.5 feet. 
 
 Number of revolutions = 3 per second. 
 Diameter of outlet = i foot. 
 
 Calculate the velocity of the air and the volume discharged. 
 Circumference of circle described by the vanes at one revolution 
 = 3.1416 X 2.5 = 7.85 feet. 
 
 Velocity of each vane = 7.85 x 3 = 23.55 ^®®* P®^ second. 
 „ „ air current = 23.55 x J = 17.66 feet per second. 
 
 Area of outlet = tt x (radius)^= 3. 1416 x (0.5)- = 0.7854 square 
 foot. 
 
 Discharge = 17.66x0.7854=1387 cubic feet per second 
 
 = 49932 „ „ „ hour. 
 
 Example. — (2) 750,000 cubic feet of fresh air are to be supplied 
 per hour. Calculate the diameter of the fan necessary to eflject 
 this and the number of revolutions per second, the diameter of 
 the inlet being 2 feet. 
 
 Delivery = velocity of air current x area of inlet. 
 
 Radius of inlet = 1 foot. .*. area = 7r x 1 = 3.1416. 
 
 .*. 750,000 = velocity of air current x 3.1416. 
 
64 CALCULATIONS IN HYGIENE 
 
 Velocity of air current = Z 5Q>^QQ f^^^ . ]^q^^^. 
 3.1416 ^ 
 
 750,000 
 
 = 3-1416 X 60 x6o = P'"''^''°"''- 
 
 Velocity of each vano^ ^ ^5°.°°°x4 ^ ^ 
 
 3.1416X60X60X3 ^ ^ 
 
 second, or distance traversed jier second. 
 
 Fi(i. 15. — Anemometor. -i 
 
 Let D = diameter of fan in feet, and II = number of revolutions 
 per second. 
 
 3. 141 6 X D = circumference of fan in feet. 
 
 3.1416 X D X 11 = distance traversed by a vane in one second. 
 
 .-. 3.i4i6xDxK=-.88.4 feet. 
 
 -,. T, 88.4 
 .-. J ) X K = - - ^- = 28 feet per second. 
 3.1410 
 
 If the number of revolutions is 10 [:er second, D= 2.8 feet. 
 
VENTILATION 
 
 If the diameter of tJie fan is 4 feet there must be 7 revolutions 
 per second. 
 
 To Calculate the Velocity of the Current of Air and its 
 Volume. — Ascertain: (i) the area of each inlet ; (2) the mea7i 
 velocity of the current passing through it ; (3) the actual air 
 capacity of the room. 
 
 The average of several observations at the periphery and near 
 the centre of the aperture must be taken. 
 
 If the opening is circular, note the velocity of the current at a 
 point f of the diameter from the side of the shaft. 
 
 Mean velocity x area of opening = volume of air. 
 
 For greater accuracy the 
 amount of air entering 
 should be checked by esti- 
 mating the amount of air 
 leaving the room. These 
 should be equal. 
 
 The Anemometer. — (i) 
 Casella's instrument. The 
 velocity of the rotating 
 vanes is registered by the 
 long hand, recording 50 
 feet per minute on the 
 large dial divided into 100 
 feet, and on five small dials 
 showing respectively velo- 
 cities of 1000, 10,000, 
 100,000 feet, and miles. 
 The indices are started or 
 stopped by means of a small 
 
 knob, and can be thrown out of gear without checking the 
 Velocity in feet per minute 
 
 Fig. 16. — Birain's Anemometer. 
 
 rotation of the vanes. 
 
 88 
 
 velocity in 
 
 miles per hour. 
 
 {a) If the indices are not at zero, read off any previous record 
 and deduct this from the reading after the experiment. 
 
 (6) A " correction " (determined by the instrument-maker) is 
 sent with each instrument, and represents the minimum velocity 
 which will move the vanes. It is to be added to the reading for 
 each observation, and is about 30 feet per minute, or 6 inches 
 per second. 
 
 (2) Biram's Anemometer. — In the smaller instruments the 
 vanes rotate on a central axis on which are the index-dials. The 
 larger instruments have a different arrangement. 
 
(k; calculations in hygiene 
 
 Example. — The actual air capacity of a room is 1450 cubic 
 feet. Two air inlets, each of 60 square inches, supply air at an 
 average velocity of 2 feet per second. Find the volume delivered 
 per hour and the number of times it is renewed. 
 
 Total inlet area = 2 x = J square foot. 
 
 144 
 
 Current velocity = 2 x 60 x 60 = 7200 linear feet per hour. 
 
 Amount supplied per hour = 7200 x ^ = 6oco cubic feet. 
 
 Air capacity of room = 1450 cubic feet. 
 
 .*. = 4- 14 nearly, i.e., the atmosphere of the room is com- 
 
 1450 t t ^' ' 1 
 
 pletely changed a little more than four times an hour. In a 
 
 small room, unless the current is well divided and properly 
 
 warmed, it will cause a draught. 
 
 Example. — Air-meter registers 2546 linear feet from previous 
 observations. Placed in ventilating shaft r distance from side it 
 indicates " 3746 " linear feet in 5 minutes. 
 
 Sectional area of shaft =1.6 square fotit. Correction for in- 
 strument = 30 linear feet per minute additive. 
 
 3746- 2546= 1200 linear feet in 5 minutes. 
 
 Adding correction (30x5) =1350 linear feet in 5 minutes or 
 4.5 linear feet per second. 
 
 I 6x4.5 = 7.2 cubic feet per second, or 25,920 cubic feet per 
 hour. 
 
 Estimation of Superficial and Cubic Space. 
 
 Superficial measurement of : 
 S([uare = (side)-. 
 
 Rectangle = length x width. ^ 
 
 Tri.in^le -=base x h height or height x !, base. 
 Area of rectilinear surface : divide into triangles, and take the 
 ^um of their i-espective areas. 
 
 rr 1 / T N' / ^''' 22 
 
 Circle^- TT X (radius)-. (7r=.— =- =3.1416). 
 
 07-:^ -( = 0.7854) X (diametei-). 
 
 4 
 Circumference = tt x diameter = 27r x nulius. 
 
 circumference 
 Diameter = . 
 
 o , 1 , • I . cube of height 
 
 Segment 01 cu'cle = r, x chord x lieiiiiit 4- , ? — . 
 
 ^ ^ *^ 2 X chord 
 
VENTILATION 
 
 67 
 
 TT X loHijf (iiam. x short diam. /long diam. = uiajor axib 
 ™^PS^^ 4 Vshort „ = minor „ , 
 
 Cubical measurement of : 
 
 Cube or solid rectangle = length x breadth x height. 
 Solid triangle = sectional area of triangle x height. 
 Cone or pyramid = area of base x i height. 
 
 Dome 
 Cylinder 
 
 Sphere = 
 
 4 TT (radius) - 
 
 TT 
 
 or — ^ X 
 
 Tr.qiL'ziiiui. 
 
 ^ ^ (diameter)'^ 
 
 ==0.5236 X (diameter). 
 Trapezium : 
 Lengths of parallel sides 
 
 tical distance between them. 
 
 Parallelogram : length of one 
 
 side X vertical distance between 
 
 the parallel sides. 
 To Estimate the Cubic Area from the Floor-plan of a Room 
 with Dimensions: uniform height = 12.5 feet. FT) =^2 feet. 
 The furniture, &c., occupies 118 cubic feet. 
 How many adults can occupy it as a sitting-room during the day ? 
 
 A. Z4-F^ E 
 
 l';aMllcIoL;r.ii 
 Fic. K. 
 
 /4Ft 
 
 B 16 Ft^ 6FP C 
 
 Fic. i/r/.— Area of R)om. 
 
 Superficial area of ABCE = 24 x 14 sq. ft. 
 
 Cubic ,, „ ,, = 24 X 14 X 12.5 = 4200 cubic feet. 
 
 Adding cubic space of EGD : 
 
 Superficial area = x 2 = 1 1 sq. ft. 
 
 Cubic 
 
 II X 12.5 
 
 137-5 
 4337-5 
 
 OF THE 
 
68 CALCULATIONS IN HYGIENE 
 
 J)ed acting cubic space of HCC5 : 
 Superficial area = 4 x 3 = 1 2 sq. ft. 
 Cubic ,, 12x12.5 = 150-0 cubicfeet. 
 
 4187.5 
 Deducting cubic space for furniture, &c. 1 18.0 „ 
 
 4069.5 „ 
 
 For a cubic space of 1000 cub. ft. per head : suitable for 4 persons 
 
 JJ )) )) S*^*-* )' 55 " " -^3 J' 
 
 Cubic and Superficial Spaces in Hospitals. 
 
 The following data may be taken as a fair average : 
 Maximum number of cases in a ward = 30. 
 
 Hospitals for Infectious Diseases. 
 
 Cubic air space =2000 cubic feet ( = J more than in General 
 Hospitals). 
 
 Superficial floor area =144 square feet ( = J more than in 
 General Hospitals). 
 
 Wall-space ^ 1 2 linear feet (^^- more than in General 
 Hospitals). 
 
 General Hospitals. 
 
 Each to have cubic air space = 1500 cubic feet. 
 
 Superficial floor-area = 125 square feet = jV cubic space. 
 
 Wall-space = 8 linear feet. 
 
 Open-window area= 20 s(j[uare feet= i square foot per 75 cubic 
 feet of air space, about {. floor-area. 
 
 Air supply per patient per hour = 4000 cubic feet. 
 
 Each window to be 3 feet above the floor and 6 inches short ol 
 the ceiling. 
 
 Width of ward = 25 feet. 
 
 Height „ =15 „ 
 
 Length depends on the number of beds, the cubic space and 
 floor-area. If the above data are carried out for 30 patients, the 
 
 len<(th must be : =120 feet. 
 
 25x15 
 
 With an air space of 150c cubic feet and a supply of 4000 
 
VENTILATION 
 
 69 
 
 cubic feet of fresh air per case per hour the air must be renewed 
 
 4000 
 1500 
 
 = 2.6 (" recurring decimal ") times per hour. 
 
 If the ward is to be 100 feet long — let x be the number of 
 patients it can contain, the above conditions being fulfilled : 
 
 1500 X 
 
 = 1 00. x=2^ patients. 
 
 25x15 ^^ 
 
 Beds. — Length of each, 6 feet. Width, 3 feet. Average cubic 
 space occupied by bed and bedding, 10 cubic feet; by each patient 
 (adult), 3 cubic feet. 
 
 Wall of Ward 
 
 Window 
 
 4- Ft 
 
 Ift 
 
 T^^-75/^^ 
 ^9/n 
 
 m. 
 
 Wall of Ward. 
 .....6 ft ). 
 
 Sin 
 
 \IF^ 
 
 5Ft 
 
 1-5 Ft 1-5 Fii 
 
 Bed 
 
 3Ft 
 
 Fig. 18. — Spacino- of Beds in Ward. 
 
 Each bed is to be at least 5 feet clear of the next one, and is 
 best placed between two windows. 
 
 An equal number of beds is placed on each side of the ward. 
 If there are 15 a side there must be 16 windows, or 32 in all, as 
 the first and last beds on each side have one end-window (top 
 and bottom of ward) to themselves. 
 
 Length of ward =120 feet; 15 beds down each side; length 
 of wall for each bed = 8 feet, or 4 feet from the middle line of 
 each bed. 
 
 Width of bed = 3 feet. Total free space per bed = 8 - 3 = 5 feet, 
 or 2.5 feet on each side (4- 1.5 = 2.5 feet). This leaves a clear 
 space of 5 feet between adjacent beds. 
 
 If the ward is 15 feet high and each window is 3 feet above 
 the floor and 6 inches short of the ceiling, its total length = 15 - 
 3.5 feet= 1 1.5 feet. As the upper half of the window usually 
 opens, its width to obtain an open space of 20 square feet must 
 
70 
 
 CALCULATIONS IN HYGIENE 
 
 be — — X width = 20 square feet ; width = = 3-5 feet. 
 
 2 5-75 
 
 A window being phieed centrally between two beds (I.e., in the 
 
 3-5 
 centre of the intermediate space of 5 feet) or 1.75 feet of 
 
 this wall-space is occupied by the window, leaving a distance of 
 2.5 - 1.75 = 0.75 feet or 9 inches between the side of the bed and 
 that of the adjacent window, measured along the wall. As the 
 head of the bed is placed about a foot oft' the wall this distance is 
 only relative to the surface of the wall. 
 
 Distance between centres of two windows = 8 feet. 
 
 Wall-space between adjacent sides of two windows = 4.5 feet. 
 
 Example. — A ward for general cases is to contain 8 beds. 
 Calculate the dimensions and window-space, &c. 
 
 Total cubic space = 1500 x 8 
 „ floor ,, = 125 X 8 
 = 12,000 
 
 32 
 
 Length x 25 x 15 
 
 length 
 
 Wall-space = 
 
 Space between beds = 8-3 
 
 Total window area = 
 
 75 
 
 One bed between two windows 
 80 
 5 
 
 Area of each window 
 
 Total length of window 
 
 Half the length usually opens, there 
 
 fore width of window 
 
 = 12,000 cubic feet. 
 =r^ 1000 square ,, 
 = 32 feet. 
 
 = 8 feet per bed. 
 
 f = 5 feet (2.5 feet on 
 l^the sides of each bed). 
 = 160 square feet 
 (about J floor-space). 
 80 square feet on each 
 side of the ward. 
 
 / = 5 windows on each 
 [side of the ward. 
 
 = 16 square feet. 
 
 = 11.5 feet. 
 
 16 
 
 5T5 
 
 •J 
 
 Distance between edge of bed and sides 
 of adjacent windows (measured along 
 
 wa1l)=..5-^-:|^ . . . \ 
 
 If all the 10 windows were open at the 
 upper half, and formed the only mrans 
 of ventilation, total inlet-area = 160 
 square feet, and total air-supply . 
 
 2.78 feet. 
 
 foot. 
 
 = 32,000 cubic feet 
 per hour. 
 
= 12 5 linear ft. per hr. 
 0.03 „ „ ,, sec. 
 
 VENTILATION 71 
 
 32,000 r = 200 linear ft. per hr. 
 
 Velocity ot air current =-^^ . . |^^g __ ^^ ^ ,,,, 
 
 If the ward had a residual amount of 
 fresh air equal to its cubic capacity 
 (12,000 cubic feet) before it was occu- 
 pied by the 8 patients, the amount of 
 fresh air required to be supplied throuojh 
 the 10 windows would be 32,000- 
 12,000 = 20,000 cubic feet, and the 
 velocity of the air-current during the 
 
 20,000 
 first hour of occupation would be — 7 — 
 
 Estimation of CO^ in Air— Pettenkofer's Method, 
 
 I. Baryta water. 4.5 grammes Ba2(H0) per litre.* 
 Oxalic acid. Standard solution: i c.c. = 0.25 c.c. COg. 
 
 Method.— C,H,0, + ^^.f) =126 grammes. 00^ = 44 grammes. 
 
 One molecule of acid combines with as much barium hydrate as 
 one molecule of CO,. Therefore 126 of oxalic acid correspond to 
 44 CO,. 
 
 I litre CO, at N.T.P. = — x 0.08936 =1.965 grammes. 
 
 1 „„,... . =0.491 gramme. 
 
 .-. 44 : 126 :: 0.491 : x. . . . .x'=i.4i grammes. 
 
 .-. 1. 4 1 grammes of oxalic acid dissolved in i litre of water 
 are equivalent to 0.25 c.c. CO,. 
 
 The same result may be obtained by u>ing Avogadro's law : 
 
 44 grammes CO, at N.T.P. . . = 22.32 htres. 
 
 .-. 126 „ oxalic acid at N.T.P. . =22.32 „ 
 
 And = q. 64 grammes oxalic acid = i litre cf CO,. 
 
 22.32 ^ ^^ 
 .-. 5.64 grammes oxalic acid dissolved in i litre of water =1 
 litre of CO2 ; and i c.c. of this solution = i c.c. of CO,. 
 
 A solution containing ^^ = 1.41 grammes oxalic acid per litre, 
 I c.c. = 0.25 c.c. of CO,. 
 
 II. Lime water. Clear solution (saturated). 
 Oxalic acid. Standard solution, i c.c. = i mgr. CaO. CaO = 56. 
 C,H,0,-t-2H,0= 126. One molecule of acid combines with as 
 much CaO as a molecule of CO,. 
 
 56 : I : : 126 : re-. £C=2.25 grammes. 
 * Or : Ba2(H0) 28 gms. BaClo (10 % sol" ) 5 cc. Water 4 litres. 
 
72 CALCULATIONS IN HYGIENE 
 
 2.25 grammes of oxalic acid per litre : i c.c. - i mgr. CaO. 
 
 CaO + CO, = CaC03. I mgr. CaO = ^ mgr. CO.,. 
 (56) (44) (100) 5^ 
 
 I litre of CO, (or 1000 c.c.) = — x 0.0896= 1.97 grammes = 1970 
 mgrs. 
 
 1970 : — 7 : : 1000 : a;, x* = 0.4 c.c. COg. 
 
 .*. I c.c. oxalic acid solution =1 mgr. CaO = 0.4 cc. CO, (2.25 
 grammes per litre). 
 
 Example. — (i) Total air-capacity of dry jar = 4320 c.c. 
 100 c.c. pure baryta solution added (jar shaken for half an 
 hour*). 
 
 25 c.c. pure baryta solution . = 42.8 c.c. oxalicacid. solution. 
 
 25 c.c. baryta solutions- CO, of | _ ^ 
 
 air in jar . . . . " . J " ^ ^"^^ " " 
 
 .*. 25 c.c. baryta solution + CO, of \ = 3.2 c.c. „ „ „ 
 air in jar . . . . .J representing COg absorbed. 
 
 I c.c. oxalic acid solution . . =0.25 c.c. COg (standard). 
 
 .-. 3.2x0.25 = 0.8 c.c. CO, in 25 c.c. baryta solution taken 
 from jar. 
 
 0.8 X 4 = 3.2 c.c. COg in 100 c.c. baryta solution added to jar. 
 
 (As this amount is identical with the 3.2 c.c. obtained by 
 subtraction above, the former result can be taken at once as the 
 amount of CO, absorbed by 100 c.c. of baryta solution.) 
 
 Actual volume of air tested = 4320 - 100 = 4220 c.c. 
 
 Percentage of CO, present : 4220 : 100 : : 3.2 : x. .x" = 0.076 per 
 cent, of CO^ or 7.6 volumes CO, per 10,000 volumes of air "at 
 current temperature and pressure ^^ (i cubic metre = 1000 litres). 
 
 If temperature = 58.6° F., and pressure = 29.264 inches, it is 
 unnecessary to alter these to C° and mm. : 
 
 To correct for normal temperature and pressure (32° F. and 30 
 inches) : 
 
 4220 X (459 + 33) ^ 29_£64^ ^^ „f ^i^ ^^ N.T.P. 
 
 '^» (459 + 58.6) 30 ''^^' 
 
 3904 9 : 100 : : 3.2 : .r. .r = o.o8 per cent, of CO,, or 8 volumes 
 CO, per 10,000 volumes of air. 
 
 As the " correction " for N.T.P. is trifling, it is, as a rule, 
 
 * Clowes and Coleman shake up for half an hour and titrate the whole 
 of the linie-water in the bottle itself {(^nantitatirc Anali/sh). 
 
VENTILATION 7?> 
 
 only made at altitudes above sea-level where the difference of 
 pressure is considerable. 
 
 Example. — (2) Total capacity of air-jar = 4640 c.c. 
 
 100 c.c. of lime-water added (jar shaken for three-quarters of 
 an hour). 
 
 25 c.c. pure lime-water . . =3 9. 6 c.c. oxalic acid solution 
 
 25 c.c. of lime-water from jar . =34.8 c.c. „ „ ,, 
 
 ( = 4.8 c.c. „ „ „ 
 
 .-. 25 c.c. of lime-water from jar -j corresponding to CO., ab- 
 
 [sorbed 
 Oxalic acid solution : i c.c. . = 0.4 c.c. COg (as above). 
 
 f = 1.92 c.c. CO., in 25 c.c. of 
 
 4.8 X 0.4 
 
 [lime-water from jar, 
 
 J =7.68 c.c. CO., in 100 c.c. of 
 
 ^•92^4 \lime-water added to jar. 
 
 Air in jar = 4640 - 100 . . =4540 c.c. 
 
 .-. 4540 : 100 : : 7.68 : x. x= 0.169 per cent, of CO., in air, 
 or 16.9 volumes of CO, in 10,000 volumes of air, "at current 
 temperature and pressure." 
 
 Supposing temperature and pressure are to be expressed in C."" 
 and mm., and are read originally as 46.5° F. and 30.46 inches. 
 These correspond to 7.4° C. and 771.65 mm. 
 
 To correct 4540 c.c. of air to N.T.P. (o^ C. and 760 mm.), 
 
 _ 454ox(273 + o) 77 1^5 ^ ^ ,,,, ^f ^j^ at N.T.P. 
 "" (273 + 74) 760 ^^ 
 
 4506.7 : 100 : : 7.68 : X'. 0^ = 0.17 per cent, of CO, in air, or 
 17 volumes of CO^ per 10,000 volumes of air. 
 
 Estimation of COg in Air by aspiration through Petten- 
 kofer's Absorption-tubes. 
 
 Baryta solution, strength as before (p. 71), or it may be twice 
 as strong. 
 
 First tube may contain 150 c.c. or 100 c.c. of solution, and 
 second tube 100 c.c. If the solution is sufficiently turbid after 
 I to 4 litres have passed, the aspiration may be stopped and 
 the estimation of CO, made. As a rule, in the open air 10 litres 
 are passed through 'at the rate of i litre every quarter of an 
 hour or 10 litres in two and a half hours. The mean temperature 
 and pressure must be observed during the process, and if the 
 
74 
 
 CALCULATIONS IN HYGIENP] 
 
 fluctuations are not great, may be taken at " half-time " as an 
 approximation. 
 
 Example. 
 
 Pure baryta solution: 25C.C. = 42.6 c.c. oxalic acid sokit'on. 
 10 litres air aspirated. 
 
 First tube (containing 100 c.c. pure baryta solution) : 
 
 25 c.c. = 39.7 oxalic acid solution. 
 
 42.6 - 39.7 = 2.9 c.c. equal to CO., absorbed by 100 c. c. baryta sol n. 
 
 Fic. 19. — Kstiinatioii of CO^ by I'cttenkofi'r's 'riilic>-. 
 
 Second tube (containing 100 c.c. pure baryta solution) 
 
 25 c.c. = 42.2 c.c. oxalic acid solution. 
 
 42.6 (pure baryta solution) - 42.2 ■= 0.4 c.c. : 
 
 .•. 2.9 + 0.4 = 3.3 c.c. in 10 litres of air. 
 
 Supposing 10 litres of air (10,000 c.c.) corrected Jar S JW 
 
 9850 c.c. 
 
VENTILATION 75 
 
 .-. 9850 : TO,ooo : : 3.3 : .'>•• •'' = 3-35 volumes of CO, per 10,000 
 or 0.035 CO., per cent. 
 
 (This method is more accurate than the first; but though 10 
 litres of water are run oflf— or i litre made to aspirate 10 times 
 by reversing the bottles— some of the air in the apparatus diffuses 
 out.) 
 
CHAPTER y. 
 
 WATER. 
 
 Water is taken as the standard for determining the unit of heat, 
 the specitic heat and latent heat of a body, and also as the 
 standard for ascertaining the relative density or specific gravity 
 of solids and liquids, as already indicated. 
 
 The unit of heat is the amount of heat required to raise unit 
 mass of water one degree in temperature. It varies with the 
 unit of mass and scale of temperature adopted : 
 
 Centigrade Scale. — A kilogrsiuime of water at o° C. is the 
 unit of mass, and the amount of heat required to raise its 
 temperature from o° to i° C. is the unit of heat known as the 
 " Calorie^ The small calorie is the heat that will raise one 
 gramme to the same extent. 
 
 Fahrenheit Scale. — The amount of heat necessary to raise the 
 temperature of a pound of water i° F. either from 32° F. to 
 33° F. or from 60° F. to 6t° F. — the latter temperature being 
 the usual one — is the British unit of heat. 
 
 The specific heat of a body is the ratio of the amouiit of heat 
 taken in by the body when its temperature increases one degree 
 to the amount of heat taken in by an equal weight of ice-cold 
 water for a similar rise of temperature. 
 
 By " Dulong and Petit's law/' the specific heat of an element 
 varies inversely as its atomic weight. 
 
 In the Centigrade system the temperature of the water is taken 
 at 4° C. — i.e., at maximum density. In the Fahrenheit scale it is 
 estimated at 60° F. 
 
 The specific heat of a substance when mixed with another can 
 be ascertained by the following formula : 
 
 Let S.H.J = Specific Heat, Wj= Weight, and t^ = temperature of 
 the one substance. 
 
 Let S.H.^, = Specific Heat, W^, = Weiglit, and t, = temperature of 
 the other substance. 
 Let T = common temperature of both substances after mixing. 
 
WATER 77 
 
 S.H.,_ W,x(T-t,) 
 
 If S.H.2 = specific heat of water, it is equal to i. 
 . .XX W,x(T-t,) 
 •• ^•^•i-Wix(ti-T)- 
 
 Example. — A kilogramme of iron at a temperature of ioo° C. 
 is placed in a kilogramme of water at 4^ C. On cooling, the 
 common temperature is found to be 13.8° C. 
 
 The specific heat of iron is found as follows : 
 
 IX (13-8-4) 9-8 ^ ^^^ 
 I x(ioo-i3.8)~86.2~°-"4- 
 
 Example.— A kilogramme of ice is mixed with two kilogrammes 
 of boiling water. Find the temperature of the water when the 
 ice has melted. The specific heat of ice is 0.505. 
 
 2(T-ioo) ^ _, 
 
 T^n nn -Q-5Q5- 2T-2oo= -0.505T. 
 ^y""-^) 2.5o5T = 2ooT-79.8° C. 
 
 Latent heat is the heat absorbed by a body in changing from a 
 solid to a liquid ("latent heat of fusion"), or from a liquid to a 
 gaseous state ('' latent heat of evaporisation "). It is a trans- 
 formation of energy in the form of heat into another form of 
 molecular energy. The latent heat of water is 80 heat-units of 
 the Centigrade scale, and 143 heat-units of the Fahrenheit 
 system. 
 
 The latent heat of steam is 536 kilogramme-units C, and 
 966 pound-units F. 
 
 Calculations of Water-Supply. 
 
 Data required : Depth of water. 
 
 Area of the receiving-surface. 
 Loss by evaporation, &c. 
 Rain Water.— The depth of rainfall is calculated by the rain- 
 gauge (vide Meteorology) and by recorded observations extending 
 over many years. 
 
 Hawksley's Formula for the Estimation of Rainfall : 
 
 1. Ascertain the average rainfall for 20 years. 
 
 2. Amount of rainfall in the wettest year = average rainfall 4- J. 
 
 3. „ „ driest ,, = „ jj - 3- 
 Or amount of rainfall in the driest year = average rainfall in 
 
 3 driest years. 
 
78 CALCULATIONS IN HYGIENE 
 
 Symons' Formula (for the British Isles) : 
 Let the average rainfall =i. 
 
 Rainfall in the wettest year =1+1 = 1,5 (half above the average). 
 „ „ driest ., = i - J =| or 0.7 (one-third belowthe 
 
 average). 
 ,, ,, ,, of 3 consecutive years = i — l = o.8 (one- 
 fifth below the average). 
 
 Example. — Supposing 37.3 inches = the mean rainfall of the 
 United Kingdom : 
 
 By Hawksley's formula: Wettest year = 37.3 + 1 2.63 = 49-93 i^^- 
 Driest „ =37.3-12.6 =24.7 „ 
 
 By Symons' „ Wettest „ =37-3 + 18.65 = 55.95 „ 
 
 Driest ,, =37.3-12.6 =24.7 „ 
 
 Driest of three consecutive years = 37.3- 7.46 = 29.84 ,, 
 
 I inch of rainfall over i square yard of surface (1296 square 
 inches) = 1296 cubic inches. 277.27 cubic inches = i gallon. 
 
 .*. I inch of rainfall over i square yard of surface (1296 square 
 
 inches) = ^- ^4-67 gallons. 
 
 '' 277.27 ^ ' ^ 
 
 I inch of rainfall on i acre = 4.67 x 4840 = 226028 gallons. 
 I gallon of water at 62° F. = 10 lbs. 
 
 2 2602.8 X 10 
 
 I inch of rainfall over surface of i acre (at 62" F.) = ^^ 
 
 ^ ' 2240 
 
 = 100.9 or loi tons nearly. 
 
 Area of Receiving-surface. — The watershed may be estimated 
 from the contour-lines of an Ordnance Map. Where contiguous 
 lines are far apart the slope is small, and where they are close 
 the land is steep and the " head " of water great. The point at 
 which one contour-line crosses another is a " col " or " saddle- 
 l)ack " — the upper level descending to the lower. All water- 
 courses, marking lines of greatest incline, cut the contour-lines 
 at right angles. 
 
 If the rain-water sinks into a i)orous stratum which is not 
 overlying an impervious layer, the efl'ective leceiving-surface 
 cannot be determined by the watershed of the porous layer, but 
 only by that bounding an underlying impervious stratum, which 
 may be ascertained from a Geological Map of the district. 
 
 Area of receiving-surface in square feet x rainfall in inches _ 
 
 12 
 cubic feet of water per annum. 
 
 0.16 cb. f t. = I gallon. .-. 1 cb. ft. = - — ^ gallon = 6.25 gallons. 
 
 The usual estimate is 6.23 gallons. 
 
WATER 79 
 
 Temperature of the air, the nature of the receiving-surface, 
 growing vegetation, &c., cause a loss of water by evaporation, 
 leakage and absorption, which is to be deducted from the above. 
 These conditions vary indefinitely. As a rough estimate | may 
 be deducted from the amount, which is equivalent to multiplying 
 the result by 0.75. A more accurate result is obtained by 
 estimating at the same time the rainfall and the actual discharge 
 of streams supplied by the area. 
 
 If the surface is a sloping roof, only the horizontal area it 
 covers is to be estimated as a "receiving-surface." 
 
 Example. — Area of receiving-surface = 2000 square feet. 
 
 Mean annual rainfall = 26.7 inches. 
 
 Volume of water available per annum, after deducting | for 
 
 2000 X 144 X 26.7 
 evaporation = ^ x o-75- 
 
 2000 X 26.7 1 • p - 
 
 = X 0.75 = 3337-5 cubic leet per annum. 
 
 =- 3337-5 X 6.25 = 20859.375 gallons, or 57 gallons per 
 diem. 
 
 Hawksley's Formula for Storage in the impounding 
 reservoir. 
 
 Number of days' storage = D. 
 Mean annual rainfall in inches for three consecutive dry years 
 
 Example. — Mean annual rainfall = 49.61 inches. 
 
 Taking Symons' formula and deducting i {i.e., multiplying the 
 mean annual rainfall by 0.8) : 
 
 Mean annual rainfall of the driest of three consecutive years = 
 49.61 X 0.8 = 39.69 inches. 
 
 T^ , , 1000 1000 ,, , 
 
 Days' storage := ■ = -^— = 158.73 days. 
 
 Dr. Pole's Formula for the Area of the Collecting-surface : 
 
 Mean annual rainfall in inches ^ R. 
 
 Daily supply in gallons = G. 
 
 Area of collecting-surface in acres = A. 
 
 Lo-sby evaporation in inches ■-=^. 
 
 G = 62 A(R-E). 
 In a rainy place : i5oG = 62A(R- E). 
 
 „ dry „ : 200 G= ,, ,, ,, 
 
 Example. — 10,000 gallons are required as a daily supply. 
 Estimating a loss by evaporation of i R in all cases. 
 
80 CALCULATIONS IN HYGIENE 
 
 (i) Where 11 = 25 inches. 
 
 r K I \ ^ K 10,000 . - 
 
 10,000 = 62 A (2=5 - O. 62 A = . A = 8 acres. 
 
 ^ ^ ^' 20 
 
 (2) In a wet place with a rainfall of 50 inches : 
 
 10,000 X 150 = 62 A (50 - 10). 
 
 10,000 X 1 150 
 
 A = — = 60 c: acres. 
 
 62 X 40 -^ 
 
 (3) In a dry locality where the rainfall is 15 inches : 
 10,000 X 200 = 62 A (15 - 3). 
 
 10,000x200 
 
 A = — 7 = 2688 acres. 
 
 62 X 12 
 
 Water-Supply of a Stream. — Current- njeters are used to 
 indicate the rate of flow or the supply in gallons. 
 
 The stream is also dammed up and the water conveyed along a 
 channel of known dimensions, which may be a trough — z.e., length, 
 width and depth and the velocity of the current are estimated along 
 the length of the channel by using a float as indicator. 
 
 Outflow = velocity x width x depth. 
 
 (i) Discharge through a sluice: Area of the sluice = width 
 X height. 
 
 " Head of Water " : 
 
 (a) sluice above the lower level, = height of the upper level of 
 the water above the centre of the opening of the sluice. 
 
 (6) Sluice entirely below the lower level, = difference of level of 
 the water above and below the dam. 
 
 Discharge in cubic feet = area x 5 ^ head of water. (Poncelet 
 and Lesbro's formula.) 
 
 Example. — The sluice has a length of 10 feet and a height of 
 2 feet, the height of the upper level of water above the centre is 
 6.76 feet. 
 
 Discharge = 10 x 2 x 5^6 76. 
 
 = 100 X 2.6 = 260 cubic feet. 
 
 (2) Di -charge over weir : theoretically it is equal to that of a 
 body falling freely through the distance of surface-level above and 
 below the weir : v- = 2gh^ or v = 8 ^/ H (as in Montgolfier's formula). 
 This is the velocity of the lowest stratum of water : the average 
 velocity is | of 8^/H. 
 
 Blackwell's Formula for Discharge over a Weir : 
 
 Q = cubic feet per minute passing over weir. 
 
 IV = width of weir in feet. 
 
 d = depth in inches over weir. 
 
WATER 81 
 
 4.^ = "factor" (varying from 3.5, depth = i inch ; to 4.4, depth = 
 9 inches). 
 
 Q = 4.5 X 1(7 JcP. 
 
 Example. — Wid:h of weir = 9 feet. 
 
 Depth ,, 5, = 4 inches. 
 
 Q = 4.5 X 9 X ^/4"^ = 4.5x9x8 = 324 cubic feet per minute. 
 
 Yield of a Stream. 
 
 A^icertain the mean depth by repeated soundings along as 
 uniform a channel as possible, noting the length and breadth, 
 and the surface velocity of the current by a float. 
 
 The mean velocity is estimated as t or 0.8 (by some as f) 
 the surface velocity. 
 
 Example.— Width of stream = 16.5 feet ; mean depth = 4 feet. 
 
 Velocity of surface current = 45 feet in 70 seconds -= 38.6 feet 
 per minute. 
 
 Mean velocity = 38.6 x 0.8. 
 
 Sectional area= 16.5 x 4 square feet. 
 
 Outflow = 16.5 X 4 X 38.6 X 0.8 = 2038.08 cubic feet per minute. 
 
 2038.08 X 6.25 = 12,738 gallons per minute. 
 
 The Hydraulic Ram is employed to raise water from a stream 
 to a maximum height of about 150 feet. For any height above 
 this the proportion of water wasted exceeds the amount supplied. 
 Height to which water can be raised = 25 x height of fall (" head "). 
 A fall below 12 inches is ineffective, and one above 6 feet causes 
 too great a strain on the apparatus. 
 
 If height of fall = 6 feet, 25 x 6 = i 50 feet or maximum " lift." 
 From 50 to 80 per cent, of the available power can be utilised. 
 The waste water is troublesome to dispose of. 
 
 Problem — " Assuming that in a chalk formation the quantity 
 of retained water is 16 per cent, of the volume of the chalk, and 
 that the curve of saturation is a straight line, find the number 
 of gallons of water contained in a hill ij miles wide (i.e., 
 measured at right angles to the line of greatest slope) and in 
 which the highest point of the curve of saturation is 51 feet 
 vertically above the lowest and a mile from it horizontally." 
 (Cambridge University.) 
 
 Area of triangular surface ABC 
 
 = base X J height. 
 
 = S280 X — feet. 
 ^ 2 
 
82 CALCULATIONS IN HYGIENE 
 
 Area of solid triangle ABC £ 
 
 DEA 
 = ai-ea of A B C x A E 
 
 = 528ox^-x(ii; X 5280) ''^'^^''^ 
 
 cubic feet. 
 
 = 1,066,348,800 cubic feet 
 
 in which water =16 per 
 
 cent. 
 100 : 16 : : 1,066,348,800 : x. 
 x = 666,468 cubic feet x 6.25 
 = 4,165,425 gallons. 
 (i cubic foot of water = 6. 2 5 
 gallons.) 
 
 The Suction-pump. — Atmospheric pressure on the external 
 surface of water causes it to rise into the pump-barrel, when the 
 ascent of the piston produces a vacuum. 
 
 Maximum height of mercurial column supported by atmos- 
 pheric pressure = 30 inches or 2.5 feet. 
 
 Mercury is 13.6 times heavier than water. 
 
 .-. Maximum height of water-column supported by atmospheric 
 pressure = 2.5x1 3.6 = 34 feet. 
 
 The piston when raised to its highest point must, theoretically, 
 not exceed a height of 34 feet from the surface of water outside. 
 
 Example. — Length of piston-stroke = 6 inches. 
 
 Height of suction-valve (at bottom of pump-barrel) above 
 water = 20 feet. 
 
 " Untraversed space " between lowest point of descent of i iston 
 and the suction-valve = 1.5 inches. 
 
 Air-pressure within barrel when piston at highest point = 
 
 - — '-^ of atmospheric pressure. 
 
 Atmospheric pressure = pressure of 34 feet of water. 
 
 .-. Column of water supported by this pressure = 34 x = 
 
 6 8 feet. 
 
 .'. Maximum height to which water can be raised by pump = 
 34- 6.8 = 27.2 feet. 
 
 Practically, owing to loss of energy from friction and the 
 presence of "untraversed space" at the bottom of the barrel, the 
 height cannot be more than 25 feet above the surface of water. 
 
 Velocity of Efflux of Liquids. — " Torricelli's Theorem." 
 The velocity is equal to that acquired by a body falling freely in 
 
WATER 
 
 R.n 
 
 air from a state of rest at the upper surface of the fluid to the 
 centre of the orifice. The velocity is greater as the surface of 
 fluid is above the centre of the orifice, i.e., it increases with the 
 " head " of the liquid, or its height above the o pening. 
 
 .'. v'^2gh {vide \). ^'&). v = j2gh. (^ = 32.) 
 Example. — Let " head " at A (Fig. 2 1) =^^ i inch, and velocity = 
 3 inches per second. 
 
 ,, „ B = 4 inches, 
 velocity of efllux at B = velocity at A x v 4 
 
 = 3x2 = 6 inches per second. 
 
 Fig. 21.— Efflux of Liquids. 
 
 Fig. 22. — Syphon. 
 
 This is the velocity at the " vena contracta " of the jet, i.e., the 
 point in the section of the jet where the flow is in paiallel lines. 
 The Syphon. — Let A = level of liquid in upper vessel. 
 B= „ „ lower „ 
 
 X = highest point of syphon. 
 Vertical height of x above A = 7 inches. 
 
 A „ ,, = 25 -7 = 18 inches. 
 Atmospheric pressure on both surfaces = 30 inches of mercury. 
 Pressure on the A-side of .t= 30 - 7 = 23 inches. 
 „ B-side „ =30-25^ 5 ^' 
 
 Difference of pressure at 
 
 23-5 = 1 
 
 8 =' due to diflference 
 
 of level of surfaces A and B, and not to length of " legs." 
 
 Theoretically, the highest point {x Fig. 22) of a syphcn 
 intended for water must not be more than 34 feet above the 
 upper surface A, at sea-level, as the air-pressure then is equal to 
 
H 
 
 CALCULATIONS IN HYGIENE 
 
 34 feet of water. Practically— owing to loss of energy from 
 friction— it should not exceed about 33 feet. 
 
 The Hydraulic or Bramah Press is a practical application of 
 "Pascal's law": "Pressure exerted anywhere upon a mass of 
 liquid is transmitted equally in all directions in a closed vessel, 
 
 o//3r 
 
 Fig. 23. — Bramah Press. 
 
 and acts with the same force on all equal surfaces and at right 
 angles to them." 
 
 a = force-pump (Fig. 23). 
 
 b = water reservoir. 
 
 OL = lever. OM = junction of lever and pump " plunger." 
 
 C = connecting-pipe. 
 
 D = cistern of press. E = ram's "plunger" with "cupped 
 leather collar " to prevent escape of water. 
 
 If pressure at L = 40 lbs. 
 
 „ sectional area of E=: 100 times that of the piston in a. 
 „ OL = 3xOM. 
 
 Upward pressure of plunger = 40 x 3 x 100 - 12,000 lbs. 
 
WATER 85 
 
 CHEMICAL CALCULATIONS. 
 Total Solids. 
 
 {a) loo c.c. of sample water evaporated. 
 Weight of platinum capsule + i-olids after 
 
 drying and desiccation . . . . = 43.3 1 7 grammes. 
 
 Weight of platinum capsule previously in- 
 cinerated, cooled and " desiccated " . = 43.285 ,, 
 Total solids in 100 c.c. of sample 0.032 ,, 
 = 0.032 gramme in 100 grammes of sample (taking i c.c. = i 
 gramme 
 
 = 32 parts per 100,000 (32 x 0.7 = 22.4 grains per gallon). 
 (6) 70 c.c. of vi^aterare evaporated. 
 
 Total weight of capsule and solids = 34 42 grammes. 
 „ „ „ alone = 34-33 
 
 Weight of total solids in 70 c c. of water = 0.09 ,, 
 
 or 9 milligrammes .•. = 9 grains per gallon of total solids. 
 After ignition, cooling, &c. 
 
 Weight of capsule and residue = 34.346 gramaies. 
 34.346-34.33 = 0.016 grammes or 1.6 milligrannnes volatile 
 solids in 70 c.c. of water, = 1.6 grains volatile solids per gallon. 
 
 Estimation of Chlorine, 
 
 A colour-reaction with the formation of permanent orange-red 
 chromate of silver indicating the amount of chlorine used up. 
 Standard solution of silver nitrate i c.c. = i mgr. of CI. 
 
 AgN03-l-NaCl = AgCl-fNaN03. 01 = 35.5. 
 
 (170) (58.5) (143.5) (85) 
 
 35.5 mgr. CI : i mgr. 01 : : 170 mgrs. AgNOg : x mgis. AgN03 
 
 170 
 (monovalent). .« = =4.788 mgrs. AgNOg. i.e., 4.788 mgr. 
 
 AgN03 in I c.c. of water = i mgr. 01. Therefore 4.788 grammes 
 of AgN03 dissolved in i litre of water form a solution of which 
 I c.c. = I mgr. of chlorine. The bottle containing it may be 
 labelled "i c.c. = i mgr. 01." or "4.788 grammes AgNOg per 
 litre." If labelled 14.384 grammes per litre and a solution of the 
 above strength is required : By the equation it is found that 
 4.788grainsper litre= I mgr. 01. in I c.c. .-.4.788: 16.758 :: i : .t. 
 a:;=3.5 or i c.c. contains 3.5 mgr. 01 — i.e., the solution is throe 
 and a half times too strong. It is diluted down to the required 
 standard by taking 10 c.c. and adding 25 c.c. of distilled water, thus 
 making it up to 35 c.c. i c.c. of this will contain i mgr. of chlorine. 
 
8G CALCULATIONS IN HYGIENE^ 
 
 If using a deci-normal solution of nitrate of silver : 
 
 N 
 170 grammes AgNOg per litre = N. .'. ~ = ^7 gi'arames per 
 
 litre = 3.55 grammes CI. i c.c. of solution = 0.00355 grammes 
 CI or 3.55 mgrs. Use from 100 {Lehmann) to 250 c.c. of water 
 for testing, otherwise the colour develops too quickly. 
 
 Examples. 
 
 (rt) TOO c.c. of water took 3.5 c.c. standard solution of silver 
 nitrate (strength i c.c. = i mgr; CI.) to give permanent orange, 
 
 = 3.5 mgr. CI. in too c.c. of water, or 3.5 parts per 100,000. 
 (3.5 X 0.7 = 2.45 grains per gallon.) 
 
 (6) 70 c.c. water taken. Required 11.7 c.c. standard (i c.c. = 
 I mgr. CI.) = IT. 7 mgr. CI. or 11.7 grains per gallon, or 16.7 
 parts per 100,000 of chlorine. 
 
 N 
 
 (c) 250 c.c. of water requiied 2.8 c.c. — AgNOg solution. 
 
 (i c.c. = 3.55 mgrs. CI.) 
 2.8 X 3-55 = 9-94 mgrs. CI. in 250 c.c. water. 
 9.94 X 4 = 39.76 mgr3. per litre. 
 
 =r: 3.976 centigrammes per litre or parts per 100,000, 
 ( = 2.7 S grains per gallon). 
 As 9.94 mgrs. in 250 c.c. = parts per 250,000, to obtain parts 
 
 per 100,000 divide at once by 2.5 : ^^=^3-976 parts of chlorine 
 
 per T 00,000. 
 
 (N.B. The water, silver solution, and chromate of potassium 
 solution (" Indicator ") must not be acid, otherwise the orange 
 reaction will not be permanent — the red chromate of silver gets 
 dissolved.) 
 
 Chlorine in Terms of Sodium Chloride. 
 
 CI = 35.5, NaCl== 58.5. 
 
 .^5-5 : I • : 58-5: •'■• .^'^i^s. ^ 
 
 E.g., 3.976 parts per 100,000 of chlorine = 3.976 x 1.65 = 6.38 
 parts p'. r 100,000 of NaCl. 
 
 Hardness. 
 
 Standard solutions calculated. 
 
 (A) Standard soip solution. Strength t c.c. == i mgr. CaCO.,. 
 Standaid solution of pui'c calcium chloride for stamlardising 
 the soap-solution : 
 
 C:iCl,= iii. CaC03=ioo. roo : 1 :: iii : .f. 
 .f = i.TT grammes CaCl^,. 
 
WATER 87 
 
 I.I I grammes CaCl, per litre = i.oo gramme CaCO., per litre. 
 
 I c.c. = I mgr. CaCO^. 
 Standard soap solution : 
 
 lo grammes Castile soap. 
 
 350 (or 700) c.c. methylated spirit. 
 
 650 (or 3C0) c.c. distilled water. 
 
 Strength : i c.c. = i mgr. CaCOg. 
 
 (B) Standard soap solution. Strength i c.c. = 2.5 mgT. CaCOa. 
 
 Standard solution of barium nitrate for standardising the soap- 
 solution : 
 
 Ba2N0.5=26i. CaCOg^ioo. .-. 0.261 gramme Ba2N0, per 
 litre = 0.1 mgr. CaCO^ per litre, i c.c- o.i mgr. CaC03. 50 c.c. 
 Ba2N03 solution = 5 mgr. CaCO^. 
 
 Standard soap solution : 2 c.c. = 50 c.c. Ba2N0., solution. 
 
 .-. 2 c.c. = 5 mgr. CaCO.j. i c.c. = 2.5 mgr. CaCOj. 
 
 Total Hardness. — (A) Standard soap solution: Strength- 
 I c.c. = I mgr. CaCOg. 
 
 100 c.c. water =13.5 c.c. standard soap solution to form per- 
 manent lather. 
 
 100 c.c. distilled water = i c.c. standard. 
 
 ... 13,^ _ I = 12.5 c.c. — i.e., 12.5 mgr. CaCO.^ in 100 c.c. of water 
 -12.5 parts per ^100,000 of total hardness. (12.5x0.7 = 8.75 
 grains OaOOg) per gallon, corresponding to 8.75^ of Clark's scale. 
 
 To express CaCOg in equivalent terms as CaO : 
 
 CaC03= 100. CaO =56. 100 : I : : 56 : x. x = o.s6. 
 
 .'. 12.5 parts per 100,000 of CaCOa^ 12.5 x 0.56 = 7 parts per 
 100,000 of CaO (8.75 X 0.56 = 4.9 grains CaO per gallon). 
 
 Permanent Hardness —250 c.c. of water boiled down to 
 about 150 c.c, cooled, filtered, and made up to 250 c.c with 
 boiled distilled water, 100 c.c of clear liquid took 5.5 cc 
 soap solution, 5.5- i = 4.5 cc - 4.5 mgr. CaCO., in loocc 
 or 4.5 parts CaC03 per 100,000 (3.15 grains per gallon). 
 
 Temporary Hardness.— 12.5 - 4.5 = 8 parts per 100,000; or 
 4.9 - 3.15^0.75 grains per gallon. 
 
 Total Hardness.— 70 cc water took about 29 cc. soap solution 
 (.same strength). 
 
 To obtain a more accurate result: To 70 cc fresh sample- 
 water 70 cc distilled water were added, and needed 29.5 cc. 
 standard soap solution to give a permanent lather. 
 
 If 70 cc distilled water take i c.c soap solution for the extra 
 70 c.c of water added, 2 c.c. must be deducted. 
 
 .'. 29. 5-2 = 27. 5 mgr. CaC03 in 70 cc of sample = 27.5 grams 
 
88 CALCULATIONS IN HYGIENE 
 
 CaCO., per gall, n, of total liaidnesp, conesponding to 27.5" of 
 Clark's scale (39.3 parts per 100.000). 
 
 Permanent Hardness. — 200 c.c. of water evaporated down 
 to 100 c.c. filtered, etc., and made up to 200 c.c. as before. 
 
 70 c.c. = 16.8 c.c. soap solution. 16.8 - i = 15.8 mgr. CaCO^ in 
 70 c.c. = 15.8 grains per gallon. 
 
 Temporary Hardness.— 27 5 - 158= 11.7 grains CaC03 per 
 gallon. 
 
 (B) Standard soap solution. Strength: i c.c. = 2.5 m-r. 
 CaCOg (Parkes and de Chaumont). 
 
 Total Hardness. - 50 c.c. of water aie taken. 
 
 0.2 c.c. standard solution produces a permanent lather in 50 c.c. 
 of distilled water. 
 
 e.g., 50 c.c. ?ample-water = 3.8 c.c. s'^ap solution. 
 
 3.8 - 0.2 = 3.6. 3.6 X 2.5 = 9 mgr. CaC03 in 50 c.c. of sample = 
 18 mgr. CaCO-j in 100 c.c. or 18 parts CaCO, per 100,000. (12.6 
 grains per gallon = degrees in Claik's scale.) 
 
 (Permanent and Temporary Hardness are estimated in a 
 similar way after the usual preliminaries.) 
 
 To reduce a soap solution of strength i c.c. = 2.5 mgr. CaCOj 
 to a strength of i c.c. = i mgr. CaC03, take 10 c.c. of the former 
 and " make up " with rec-ntly boiled distilled water to 25 c.c. 
 
 Estimation of Magnesia in Water after eliminating 
 Calcium salts. E.g., 70 c.c. of water free from calcium required 
 3.4 standard soap solution (strength i c.c. = i mgr. CaC03). 
 
 3.4 - I = 2.4 grains CaC03 per gallon (vide sujyra). 
 
 .'. Hardness from magnesium salts is equivalent to 2.4 grains 
 CaC03 per gallon. 
 
 .-. 2.4 X 0.56 ^1.9 grains per gallon of MgCO.,. 
 
 Estimation of Free and Albuminoid Ammonia. 
 
 Wanklyn's Process. — Standard solution of ammonium 
 chloride. Strength i c.c. = o.oi mgr. NH3-17. NH,C1 = 53.5. 
 17 • I • ■ 53-5 ■ •*'• ^=3-^5 grammes NH3CI or i part NH3 = 
 3.15 parts N up. 
 
 3.15 grammes NH^Cl per litre = i gramme NII3 per litre. 
 
 I c.c. of this solution = i mgr. NH3. 
 
 10 c.c. of this S'jlutioii are made up to i litre with distilletl 
 water. 
 
 c.c. = 0.0: mijr. Nil 3. 
 
 Example, (i) Free Ammonia. 50 c.c. (f water tested provision- 
 ally with 2 c.c. " Ne.ssler" give a light tint, therefore 500 c.c. of 
 the sample are taken for distillation. 
 

 WATER 
 
 1st 50 c.c. distillate = 2.6 c.c. standard NH^Cl solutio: 
 
 2nd „ 
 3rd „ 
 
 J) ~^'S 5J 5J n J3 
 
 4th „ 
 
 ,, =0-0 55 „ J, ;, 
 
 .-. 500 
 
 c.c. water = 4 3 ., „ „ „ 
 o.oi (i c.c. = 0.01 mgr. NH3) 
 
 )> 
 
 „ „ = 0.043 "'gr. NH3 
 
 2 
 
 1000 c.c. 
 
 (i litre),, =0.086 „ „ 
 
 80 
 
 = 0.0086 centigramme NH3 per litre or part p^r 100,000. 
 (0.0086 X 0.7 = 0.0602 grain per gallon.) 
 
 Albuminoid Ammonia. — After adding 50 c.c. alkaline perman- 
 ganate to the remainder in the flask. 
 
 ist 50 c.c. distillate = 4.2 c.c. standard NH^Cl solution. 
 
 2nd 
 
 ?> 
 
 
 ?J 
 
 = 
 
 2.7 
 
 j> 
 
 
 33 
 
 33 
 
 33 
 
 3rd 
 
 ',■) 
 
 
 5) 
 
 = 
 
 1-3 
 
 5> 
 
 
 33 
 
 33 
 
 31 
 
 4th 
 
 7? 
 
 
 >? 
 
 = 
 
 0.2 
 
 3) 
 
 
 35 
 
 33 
 
 33 
 
 
 500 
 
 c.c. 
 
 sample 
 
 = 
 
 84 
 
 15 
 
 
 35 
 
 33 
 
 33 
 
 
 ?) 
 
 
 )) 
 
 _ 
 
 0.01 
 
 mgr 
 
 . NH3 
 
 
 
 
 = o.oi 
 
 841 
 
 
 
 
 
 
 
 
 2 
 
 
 
 
 
 1000 C.C. ( I lit.) ,, =0.168 ,, 5, 
 
 = 0.0168 centigramme NH3 per litre or part per 100,000. 
 (0.01 176 grain per gallon). 
 
 (2) 50 c.c. of water treated with 2 c.c. " Nessler " gave a 
 darker colour. 
 
 250 c.c. of the sample were put in the distillation -flask. 
 
 Free Ammonia — 
 
 ist 50 c.c. distillate = 3.6 c.q. standard NH^Cl solution. 
 
 2nd „ ,, =28,, „ ,, ,, 
 
 3rci 5) 53 — 1-4 35 33 1) 53 
 
 4th ,, ,, = o^ ,, ,, ,, ,, 
 
 250 c.c. sample ^ 8.1 „ „ ,, „ 
 
 2 
 
 500 33 33 = 16.2 ,, ,, ,, ,, 
 
 0.01 (i c.c. = 0.01 mgr. NH3) 
 
 33 33 33 = 0.162 mgr. NH3 
 
 2 
 
 1000 c.c. ( I lit.) ,, = 0.324 „ ,, or 0.0324 centigramme 
 
 NH3 per litre, or parts per 100,000. 
 0.0324 X o 7 = 0.02268 grain per gallon. 
 
90 CALCULATIONS IN HYGIENE 
 
 Albiiniinoid Ammonia. — After adding 25 cc. alkaline perman- 
 ganate to the water remaining in the distillation-flask : 
 ist 50 cc. distillate = 2.1 cc standard solution 
 2n(l „ ,, =1.5 ,, „ „ 
 
 3'J 
 4tli 
 
 ?> 55 
 
 
 -0.8 
 = 0.0 
 
 5? 
 
 
 5' 
 
 
 250 CC sam 
 
 pie 
 
 = 4.4 
 
 2 
 
 55 
 
 
 55 
 
 
 500 „ „ 
 
 
 = 878 
 
 55 
 
 
 55 
 
 
 
 
 0.0 1 
 
 (i 
 
 CC 
 
 
 55 55 55 
 
 
 = 0.088 
 
 mgr 
 
 . NH 
 
 
 
 
 
 2 
 
 
 
 = 0.0 1 mgr. Nil,) 
 
 iooocc(ilit.),, =0.176 mgr. 
 
 NH3 per litre 
 
 = 0.0176 part per 100,000 
 
 (o 0176 X 0.7 = 0.01232 grain per gallon.) 
 
 Wanklyn Nesslerises the first 50 cc of distillate in estimating 
 free ammonia, and adds t to estimate all the free ammonia in 
 500 cc of water. The albuminoid ammonia is estimated per 50 
 cc as before. 
 
 J^.g., 500 cc. are taken : 
 
 ist 50 cc distillate= 1.5 cc standard solution (i cc = o.oi 
 mgr.) ; 
 
 J of 1.5 = 0.5 cc 1.5 + 0.5 = 2.0 cc standard for 500 cc of 
 water. 
 
 2.0x0.01=0.02 mgr. NH3 in 500 cc or 0.04 mgr. NH3 in 
 1000 cc. (i litre) ^o.oo| centigramme NH3 per litre or part 
 per 100,000. 
 
 (In waters containing much organic matter the addition of ^ to 
 the free ammonia in the first 50 cc does not represent all the 
 ammonia.) 
 
 Albuminoid ammonia is Nesslerised in the usual way — each 
 50 cc separately. 
 
 Parkes and de Chaumont distilled off 130 to 150 cc. of the 
 water for estimating free or albuminoid ammonia, and took 
 100 cc of the distillate for Nesslerising (or 50 cc. diluted up to 
 100 cc with ammonia-free distilled water if much ammonia, 
 ascertained by a preliminary test of the distillate, were present). 
 
 J'J.g., 250 cc of water were taken and 140 cc. distilled over, 
 the last part being free from annnonia. 
 
 100 cc. of the distillate^ 3.5 cc standard amnion, chloride 
 (i cc. = 0.01 mgr. NII3). 
 
 100 : 140 :: 3.5 x. x^^.i) cc solution for 250 cc. of water. 
 
 4 9 X 0.0 1 = 0.049 mgr. NH3 in 250 cc of water. 
 
WATER 91 
 
 0.049x4 = 0.196 mgr. NH3 per litre, or 00196 centigramme 
 per litre = 0.0196 put NII3 per 100,000 (0.0196x0.7 = 0.01372 
 graia per gallon). 
 
 Of the three methods the best is to distil and test each 50 c.c. 
 separately both for free and albuminoid ammonia. 
 
 If in the last case i c.c. standard solution = 0.017 mgr. NH,, 
 
 100 : 140 : : 3.5 : .X'. £c = 4.9C.c. 4.9 x 0.017 x 4 = 0.3332 mgr. 
 per litre = 0.03332 centigramme per litre, 
 
 or part per 100,000 of NH^. 
 
 If there is little free ammonia, which may come off in the first 
 50 c.c. of distillate, it is a waste of time to distil over 130 c.c. 
 en hloc. 
 
 Dissolved Oxygen in Water. 
 
 I. Amount expressed as milligrammes per litre. 
 
 Thresh's Method. — The iodine liberated from an acid solu- 
 tion of sodium nitrite and potassium iodide, in the presence of 
 oxygen, is estimated by titrating with a standard solution of 
 sodium thiosulphate giving a colour-reaction with starch. 
 
 Standard solution of sodium thiosulphate. Strength : i c.c. = 
 0.25 mgr. of oxygen. 
 
 2N'a3S.,03.5H,0 + 1., = 2NaI -H Na^Sp^ 
 
 (496) 
 I„ (monovalent) corresponds to (divalent). 
 
 (16) 
 16 grammes : 0.25 gramme \ : 496 grammes thiosulphate : x. 
 
 X = —7- = 7.75 grammes thiosulphate. 
 
 .-. 7.75 grammes per litre = 0.25 mgr. per i c.c. 
 
 Let e = c.c. of standard thiosulphate used in estimating the 
 amount of oxygen in the sample of water. 
 
 /= capacity of stoppered tube in c.c. - 2 c.c. 
 
 ("-2 c.c." = deduction for i c.c. sodium-nitiite-and-potassium- 
 iodide solution -1- 1 c.c. dilute sulphuric acid.) 
 
 6 = " correction " in c.c. for oxygen contained in these two 
 solutions. 
 
 d^ „ „ „ „ V thiosulphate 
 
 solution itself. 
 
 (At ordinary " laboratory temperatures " = 0.31.) 
 
 1000 
 Amount of dissolved oxygen in mgr. per litre = ,v x (e - /> - ed). 
 
 If the capacity of the " separatory tube" be 252 c.c, /= 
 (252 — 2) -1- — J- = I, and the formula becomes {e-h - ed). 
 
92 CALCULATIONS IN HYGIENE 
 
 ExAMrLE. — 252 c.c. of shallow-well water in " separatoiy tube " 
 used 10.22 c.c. thiosulphate ; 6-2.1 c.c. f^ = o.3i. 
 
 X {10. 22-2. T -(10.22 X 0.31)} = 1 X {10.22-2.41682} 
 
 4(252-2) 
 
 = 7.8 milligrammes per litre. 
 
 II. Expressed volumetrically as c.c. per litre. 
 
 Winckler's Method. — A solution of manganous chloride is 
 oxidised to manganic chloride, and the iodine, liberated by the 
 action of potassium iodide, is estimated by titration with a standard 
 solution of sodium thiosulphate with starch as the indicator. 
 
 Standard solution of sodium thiosulphate. Strength : i c.c. = 
 0.1 c.c. oxygen. 
 
 2Na,S,03. 5H3O + 1, = 2NaI + Na.Sp^. I^ = = 1 1 . 1 6 litres. 
 
 (496) 
 1 1. 1 6 : I : : 496 : x. a: = 44.4 grammes thiosulphate :== i litre 
 
 oxygen. 
 
 44.4 grms. thiosulphate per litre : i c.c. = i c.c. oxygen. 
 
 4.44 ,, ,; )j ?) = o I c.c. ,, 
 
 MnCl., solution = 40%. KI to grms. in 100 c.c. of SXNaHO. 
 
 Example. — After adding to the water 2 c.c. manganous chloride, 
 and 2 c.c. potassium iodide-and-caustic soda solutions, and finally 
 3 c.c. strong HCl : 
 
 250 c.c. water =20.5 c.c. standard thiosulphate solution. 
 
 .'. 1000 c.c. water 82 c.c. ,, ,, ,, 
 
 I c.c. standard = 0.1 c.c. oxygen. 
 ,, (i litre) water = 8. 2 c.c. dissolved oxygen. 
 
 Oxidisable Organic Matter. 
 
 I. Tidy's Process. — Oxygen absorbed from star\,dard potassium 
 permanganate is estimated by titration with sodium thiosulphate. 
 
 Standard solution of potassium permanganate. Strength : 
 10 c.c. = I mgrm. of oxygen. 
 
 4KMnO, + 6U,^0, = 2K,S0, + 4MnS0, + 5O., + 6H,0. 
 (632) (160) 
 
 160 : I : : 632 : x. 0^ = 3.95 grms. KMnO^= i grm. oxygen. 
 
 3.95 grms. per litre = i grm. per litre : i c.c. = i mgr. oxygen. 
 
 0.395 grm. „ =0.1 „ „ : 10 c.c. = I „ 
 
 Sodium thiosulphate solution : usually i grm. Nn2S.,0.,.5H20 
 per litre. 
 
 If the required strength is " 40 c.c. = i mgr. O "' : ] 6 : i : : 496 : x. 
 ;-c-=3i grms. = i grm. O. 3.1 grm. per litre = i mgr. O per litre, 
 3.1 grms. in 4 litres : 40 c.c. = i mgr. 0. 
 
WATER 9?) 
 
 Example (i). — 250 c.c. water at 26.7'' C. (80" F.)+ioc.c. 
 standard permanganate + i o c.c. dilute H2SO^=-2 7.5 c.c. thio- 
 sulphate solution. 
 
 250 c.c distilled water similarly treated = 39.3 c.c. thiosulphate 
 solution. 
 
 = 10 c.c. standard permanganate. 
 = 1 mgr. oxygen. 
 39.3 - 27.5 = 1 1.8 c.c. thiosulphate solution equivalent to oxygen 
 taken up by organic matter. 
 
 39.3 : 11.8 : : I mgr. : x. x^o.t, mgr. taken up by 250 c.c. 
 of water, = 1.2 mgr. per litre = 0.1 2 centigramme of oxygen per 
 litre, or parts per 100,000. 
 
 (2) 200 c.c. distilled water treated as above ^38.0 c.c. thio- 
 sulphate solution. 
 
 = 10 c.c. standard permanganate. 
 = I mgr. oxygen. 
 200 c.c. sample water treated as above = 22.7 c.c. thiosulphate 
 solution. 
 
 38.0-22.7 = 15.3 c.c. thiosulphate representing oxygen ab- 
 sorbed by organic matter. 
 
 38.0 : 15.3 : : I : x. 3^ = 0.4 mgr. absorbed in 200 c.c. of water. 
 = 0.2 „ „ 100 c.c. „ 
 
 or 0.2 part of oxygen per 100,000. 
 
 Nitrites. 
 
 Griess' Test. — A colour reaction by nitrites acting on meta- 
 phenylene-diamine and dilute sulphuiic acid. Matched by standai d 
 solution of potassium- or sodium-nitrite. Strength i c.c. = 0.01 
 mgrm. ^.fl^. 
 
 Example. — 100 c.c. water 4- 1 c.c. metaphenylene-diamine solu- 
 tion-!- I c.c. dilute HgSO^ were matched by 8.5 c.c. standard nitrite 
 in 100 c.c. distilled water similarly treated. 
 
 T c.c. standard = 0.01 mgr. N2O3. 
 
 .*. 8.5 c.c. = 0.085 ™&^- " ^^ 100 c.c. of water. 
 
 = 0.085 P^^"^ " P®^ 100,000. 
 
 o 
 
 N, = 28. N,03=76. N,:N,03::28:76. K, = -^ of ^^03 = 0.37 
 
 X ^,03. 
 8.5 c.c. = 0.085 X 0-37 = ° °3^ P^^"* P^^' 100,000 of nitrogen. 
 
 Nitrates and Nitrites. 
 
 By the "zinc-copper couple," or aluminium and caustic soda 
 method, all oxidised nitrogen = ammonia ; this is estimated by 
 Wanklyn's process. 
 
94 CALCULATIONS IN HYGIENE 
 
 Example. — 500 c.c. water originally contained 0.006 part per 
 T 00,000 of free ammonia. 
 
 250 c.c. from the same sample after the zinc-copper process by 
 Nesslerisation = 0.28 part per 100,000 of ammonia. 0.28-0.006 
 = 0.274 part of ammonia per 100,000 from nitrates and nitrites. 
 
 To express as " nitrogen in nitrates and nitrites " : N = 14. 
 
 NH3=i7. N: NH,:: 14:17. N = HnH3. 
 
 14 
 .'. 0.274 X — = 0.23 nitrogen per ioo,coo of water. 
 
 Nitrates. 
 
 Phenol-Sulphonic Method.^Colour-ieaction obtained by 
 acting on nitrates with phenol-sulphonic acid and ammonia, and 
 matching the colour with a standard solution of potassium nitrate 
 similarly treated. 
 
 Standard potassium nitrate solution. Strength : i c.c. = 0.1 mgr. 
 nitrogen. 
 
 KNO,=N. -°^- = 7.22. 
 (ioi)'(i4) ^4 
 
 .-. 0.722 gm. KNO3 P®^ litre, i c.c. = 0.1 mgr. N. 
 
 Example. — (A) 25 c.c. water-sample evaporated^ Made up to 50 
 
 to dryness I c.c. with dis- 
 
 + 2 c c. phenol-sulphonic acid tilled water in 
 
 4- 20 c c. (or q.s.) strong NH3 solution. J Nessler-glass. 
 
 (B) 10 c.c. standard potassium-nitrate solution^ 
 
 -f 2 c.c. phenol-sulphonic acid V -» ,, 
 
 -1-20 c.c. (or q.s.) strong NII3 solution. J 
 
 The latter (B) gave the darker colour, and was carefully 
 "poured ofi " from the Nessler-glass till 20 c.c. remaining had 
 the same colour as the contents in the other glass (A). 
 
 20 
 
 of 10 c.c. = 4 c.c. standard nitrate solution matched 
 
 colour due to nitrates in 25 c.c. of sample-water. 
 
 I c.c. = o. I mgr. N. .".4 c.c. = 0.4 mgr. N in 25 c.c. of sample- 
 water. 
 
 = 1.6 mgr. N in 100 c.c. of sample-water, 
 or 1.6 parts „ 100,000 ,, ,, 
 
WATER 95 
 
 Quantitative Estimation of Lead, Copper, Iron 
 and Zinc. 
 
 Standard Solution of Lead Acetate. 
 
 Pb!?(0,H30,).3H,0 = 379.Pb = 207. ^ = 1.83 lead acetate. 
 
 1.83 grms. acetate per litre = i grm. lead per litre. 
 
 I c.c. = I mgr. of lead. (0.183 gi'^^a- per litre : i c.c. = 0.1 grm. Pb.) 
 
 Standard Solution of Copper Sulphate. 
 
 CuS0^.5H20 = 249.2. Cu = 63.2. ->-^:^ 3.94 copper sulphate. 
 
 3.94 grms. per litre = i grm. copper per litre. 
 
 I c.c. = I mgr. of copper. (0.394 grm. per litre: i c.c. = 0.1 grm. Cii.) 
 
 Standard Solution of Ferrous-Ammonium-Sulphate. 
 
 Fe2(NH,).2(SOJ.6H,0 = 392. Fe = 56. ^ = 7 ferrous 
 
 ammonium sulphate. 
 
 7 grms. per litre = i grm. iron per litre, i c.c. = i mgr. iron 
 
 (0.7 grm. per litre, i c.c. = 0.1 mgr. Fe.) 
 
 Standard Solution of Zinc Sulphate. 
 
 ZnSO,.7H20 = 287. Zn = 65. ^ = 4415. 
 4.415 grms. per litre = i grm. zinc per litre, i c.c. = i mgr. zinc. 
 (0-4415 ;. 5j » I c.c. = 0.1 mgr. Zn.) 
 
 Example. — 200 c.c. sample water evcq^oQ-ated down 
 
 to 100 c.c. (containing lead) In 100 c.c. 
 
 + 2 drops acetic acid \ Nessler- 
 
 + 2 drops concentrated ammonium-sulphide glass, 
 solution. 
 Is matched in tint by 100 c.c. distilled water 
 
 + 1.5 c.c. standard lead-acetate solution 
 
 (i c.c. = I mgr. Pb.) 
 + 2 drops acetic acid 
 + 2 drops ammonium-sulphide solution. 
 .'. 1.5 X 0.1 = 0.15 mgr. of lead in 200 c.c. of sample-water. 
 
 = 0-075 5J J> 'J » ^°° " " " " 
 
 = „ part per 100,000 (0.075x0.7 = 0.525 grain 
 per gallon.) 
 
CHAPTER VI. 
 SOIL. 
 
 Percentage of Air in Soil. 
 
 Loose Soil. — c.c. of dry soil : loo : : c c. of water used : per- 
 centage of air. 
 
 ^ , p . c.c. of water used 
 
 Percentage ot air= --; -— x loo. 
 
 c.c. or dry sou 
 
 Example. 
 
 Soil dried at ioo° C. and powdered, measured in burette = 
 25 c.c. 
 
 Water from second burette rising to upper level of soil ~ 
 7.5 C.C. 
 
 25 : 100 : : 7.5 : X. x = -^ x too = 30 % of air. 
 
 Porous rock. 
 
 Let W^ = weight of dry rock in air. 
 
 W,,= „ „ „ „ „ water. 
 
 Wj = ,, „ saturated rock in air. 
 
 ^a - W7y=-loss of weight in water (weight of an equal volume 
 
 of water). 
 
 W^ - W^ = weight of water absorbed. 
 
 Loss of weight in water: 100 : : weight of water absorbed : 
 
 percentage of air, 
 
 Weight of water absorbed „ . 
 
 or -f ^ . , ^ ■ 7 — X 1 00 = peicentage or an*. 
 
 Loss 01 weight in water ^ * 
 
 Example. 
 
 Weight of dry rock in air = 1 1 2 gi ms. 
 
 ,, ,, rock „ waters 72 „ 
 
 ,, ,, saturated rock in air = 125 grms. 
 
 1 1 2 - 72 = 40 grms. loss of weight in water. 
 
 125 - 112 = 13 ,, weight of water absorbed. 
 
 40 : 100 : : 13 : percentage of air. 
 
 II 0/ 
 
 ^-^x 100=^32.5 %. 
 40 
 
SOIL 97 
 
 Percentage of Moisture in Soil, 
 
 " Moisture " = air + water. 
 
 Weight of soil before drying ; loo : : loss of weight after 
 drying ; percentage of moisture, 
 
 T) 4. c ' . loss of weight after dryine: 
 
 Percentage of moisture = v—, — r—n -. — r^ — - x loo. 
 
 weight before drying 
 Example. — Moist soil = lo grms. 
 
 After drying at ioo° = 6.5 grms. Loss of weight = lo.o - 6.5 
 = 3.5 grms. 
 10 : 100 : : 3.5 : Percentage of moisture = 35. 
 
 Specific Gravity of Soil. 
 
 1. "Apparent Specific Gravity.'' (Soil containing air.) 
 
 _ Weight of known volume of soil 
 ~ Weight of an equal volume of water ^^^ ^^"^ ^^' 
 Example. — Weight of dry cylinder + dry soil= 1206 grammes. 
 „ „ cylinder + distilled water = 943 „ 
 „ „ dry cylinder (empty) = 387 „ 
 
 1206 -387 = 819 grammes = weight of soil. 
 943-387 = 556 „ = „ „ water. 
 556 : I : : 819 : apparent sp. gr. = 1.473 (water = i). 
 
 m CI -^ /-* -^ Weight of soil. 
 
 2. True Specific Gravity = -,^^ • .^ . — ^ — ; 7^ — 7- 
 
 W eight or an equal volume or water 
 
 ( = weight of water displaced by soil). 
 Example. — Weight of soil = 5 grammes. 
 
 Weight when sp. gr. bottle is full of distilled water =25 grammes. 
 >> of „ „ empty =12.25 „ 
 
 » J) „ „ + soil + distilled 
 
 water (both boiled 
 and freed of air) =40.2 „ 
 40.2 - 12.25 = 27.95 grms. (soil and water in bottle) 
 (5 + 25)-27.95 = weight of water displaced = 2.05 grammes. 
 
 True sp. gr. = = 2.4 (water = i). 
 
 3. Pore- volume, or Volume of soil occupied by air : 
 
 Apparentsp.gr. 1.473 
 
 — m = = 0.0 of total bulk occupied by soil. 
 
 I - 0.6 = 0.4 or 40 % = "pore-volume." 
 
 G 
 
98 CALCULATIONS IN HYGIENE 
 
 4. Water Capacity of Soil (Percentage of pore-volume that 
 can be filled with water by capillarity). 
 
 Weight of dry cylinder (with perforated base)= 377 grammes. 
 „ „ „ + dry soil =1102 „ 
 
 n n M +Wet „ ^ = 1277 „ 
 
 II02- 377 = 725 grammes dry sand in cylinder. 
 
 1277-1102 = 175 ,, water absorbed by 725 ,, 
 
 dry sand. 
 
 725 : 100 : : 175 ;x=28 per cent, weight of water absorbed. 
 
 "Pore-volume = 40 per cent." .-. 40 : 28 : : 100 : x='jo per 
 cent, of pore- volume = water capacity of soil. 
 
CHAPTER VIT. 
 SEWERAGE. 
 
 Circular Pipes. 
 
 " Sectional area " : that of a transverse section of the fluid, or of 
 the interior of a pipe = irr- (the area of a circle). 
 
 " Wetted perimeter " : length of arc, in a transverse section, 
 wetted by the contained fluid. 
 
 _ , _ , , Sectional area of fluid 
 
 Hydraulic mean depth = Wetted perimeter • 
 
 In a circular pipe the " H M.D." is always \ the diameter, 
 whether the sewer is running full or half-full. 
 
 If running full : Sectional area of fluid = internal circular area 
 
 of pipe = Trr^. Wetted perimeter = circumference of circle = 27rr. 
 
 Diameter 
 r = radius = . 
 
 2 
 
 ^^ , ^ _ irr- r Diameter 
 H.M.D. = — = - = . 
 
 2'KT 2 4 
 
 , « ,, -.^ ,, -r^ 2 TT?'"^ Diameter 
 If running \ full : H. M. D. = — = — = • 
 
 " ^ 27rr 27rr 4 
 
 Velocity of Flow : V = 55 v/2 D x F. 
 
 Eytelwein's Formula founded on that of De Chezy. 
 
 Y = velocity in feel per minute. 
 
 D = Hydraulic mean depth in feet. 
 
 F = fall in feet per mile. 
 
 y X Sectional area in feet = Discharge per minute in cubic feet. 
 
 Example. — (i) A 9-inch drain is to have a velocity of 3 feet per 
 second. The required fall or ^' gradient " is to be calculated. 
 H.M.D. = J of 9 inches =j?^ foot. 
 Velocity = 3 feet per second =180 feet per minute. 
 
100 CALCULATIONS IN HYGIENE 
 
 By the above formula : 
 
 --^ i8o ,^ /i8o\- 
 
 F = L^ X - = ^^ = 28.56 feet per mile. 
 121 3 121 ^ 
 
 (i mile =5280 feet). .*. fall=iini85. 
 
 2. A 6 -inch drain has a gradient of i in 60. What is the 
 
 velocity of flow and discharge per second if the pipe is half full ? 
 
 60 : 5280 : : I :ic = 88 feet per mile. 
 
 H. M. D. = iof J = Jfoot. 
 
 V = 55v/7xpr88 
 
 V = 55v/22. 
 
 = 55 X 4.69 = 258.0 feet per minute. 
 .'. velocity = 4.3 „ ,, second. 
 
 If the drain is running full the secfcional area^Vr^; 6-inch 
 diameter = ^ foot radius ; 
 
 sectional area = 3. I4i6x(i)- sq.ft. = 3. 141 6x^15- sq.ft. = 0.1 96sq.ft. 
 When half full the sectional area = 0.068 square foot. 
 Discharge = sectional area x velocity = 0.098 x 4.3. 
 
 = 0.42 14 cubic foot per second. 
 (25.28 per minute; 1516.8 per hour) 
 = 9404 gallons per hour (i cubic foot = 6.2 gallons). 
 
 (3) A circular sewer having a fall of i in 200 has a current 
 velocity of 2.5 feet per second. Calculate the necessary diameter. 
 Fall= I in 200; per mile : 200 : i : : 5280 : x= 26.4 feet. 
 Velocity 2.5 feet per second = 150 feet per minute, 
 diameter 
 
 H.M.D. = 
 
 4 
 
 / diameter , , 
 
 150 = 55 Ji2i'2 X diameter. 
 
 I -^ I = 13.2 X diameter in feet. 
 
 30V 
 
 -j =13-2 X „ „ „ 
 
 900x12 ^ • • 1 
 
 = diameter in inches = 7.3. 
 
 121 X 13.2 ' ^ 
 
 Maguire's Formula: To calculate the fall or gradient in 
 
 circular drains and sewers (diameter 3" -10'). Diameter in 
 
 inches X 10 : 
 
RuiJiiiiig full 
 
 SEWERAGE 101 
 
 4" = I ill 40 ; 6" = I in 60 ; 9" = i in 90, &c. 
 When drain-pipes are running half- or more than half-full the 
 incline may be less ; usually they are not more than J or J full, 
 and the inclines given should be carried out (W. C. Tyndale.) 
 
 Bailey Denton and Baldwin Latham recommended smaller 
 gradients : 
 
 Velocity 
 per second. 
 Diameter. 2 feet, 3 feet. 4 feet 
 
 4 Fall = I in 194 92 53 
 
 6 „ 292 137 80 
 
 9 „ 437 206 119 or J full. 
 
 12 ^ ^ „ ^ ^ 5^3 275 159 J 
 
 Velocity in feet x inclination = length of sewer. 
 U.g., velocity = " 4 feet per second." Fall = " i in 119." 
 Required length of sewer = 4x119 = 476 feet. 
 
 Civil engineers in planning drainage systems are guided by the 
 following practical rules : 
 
 As regards main sewers : 
 
 I. On the Separate System — admitting sewage proper and 
 rainfall from back roofs and yards only : The allowance is 40 
 gaWons per head, or 200 gallons per house, made up as follows: 
 
 25 gallons per head for water-supply (which practically all goes 
 into the sewer) ; 15 gallons per head for rainfall. 
 
 The usual allowance per head for backyards and roofs (in the 
 sepanxte system) is 100 square feet, and | inch of rain upon an 
 impervious surface of this extent yields 13 gallons. Deducting J 
 for loss by evaporation and absorption ( = 3.25 gallons) the actual 
 fiow-oft*is 9.75 or 9I gallons. The allowance for rainfall is pur- 
 posely taken high to allow for thunderstorms. 15 gallons = | inch 
 of rainfall daily on a surface of 100 square feet. 
 
 As the flow of sewage is variable, half the total daily volume is 
 allowed to be discharged in 6 hours, and sewers are constructed of 
 such a size that will discharge this amount when running half full. 
 
 The Local Government Board requires that all sewers on the 
 separate system shall be capable of discharging six times the dry- 
 weather flow when running full ; after which storm-water over- 
 flows may discharge the excess into the nearest water-course. 
 The dry-weather flow of sewage is practically the volume of the 
 drinking-water supply — usually calculated at 25 gallons per head. 
 
 25x6 = 150 gallons. This is what a sewer must be capable of 
 discharging at a minimum velocity of 2 feet per second. 
 
 2. On the Combined System : i. The areas to be estimated 
 (from Ordnance maps or actual measurements) are : 
 
 (a) the area built over, and (b) the area of the roads. 
 
102 CALCULATIONS IN HYGIENE 
 
 2. Allowance must be made for the maximum known rainfall 
 of the district for 24 hours, taking it at the highest recorded rate 
 jyer hour. 
 
 3. Add to this the flow of sewage proper — i.e. 25 gallons per head. 
 The result is the total volume to be discharged per hour, and 
 
 from this the size of the sewer is determined. 
 
 Allowing for a rainfall of i J inches per hour, a 4-inch drain 
 laid at a uniform inclination of i in 60 has a discharging capacity 
 sufficient for the drainage of a building containing 20 inhabitants, 
 and an area of 11,000 square feet ; a 5-inch drain for 50 inhabi- 
 tants, and an area of 19,000 square feet ; and a 6-inch drain for 
 100 inhabitants, and an area of 30,000 square feet (Lawford). 
 
 I am indebted to Mr. G. Maxwell Lawford, M. Inst. C.E., of 
 London, for the following data and for the Scale-Drawing : 
 
 Lawford's Formulae for the Velocity and Discharge of Sewers 
 and Water-Mains : 
 
 1. For velocity: Y = ?^ x R"-^ x S"-^ 
 (R°-7 = R/a = ;^e7 and S° ^ = SJ = Vs) 
 
 2(7 
 
 2. For discharge : Q = — x R^-^x S"SX4695.7. 
 
 Y = velocity in feet per second. 
 ^ = acceleration of gravity — 32.2 feet per second, 
 m = coefficient for roughness of surface. 
 
 = 0.5 for new or clean asphalted cast-iron pipes, glazed 
 
 stoneware pipes and glazed or vitrified brickwork, tkc. 
 
 = 0.55 for cast-iron pipes after some use, well laid concrete 
 
 tubes, &c. 
 = 0.6 for encrusted iron pipes, rough brick-work, <fcc. 
 R = hydraulic mean radius (or " depth ") in feet. 
 
 diameter ^ . , . ' 
 
 = for circular sewers, pipes, &c. 
 
 4 _ 
 sectional area , . . 
 
 = — ——5 : — 7 — for egg-shaped sewers. 
 
 wetted perimeter ^'^ ^ 
 
 S = slope of water surface. 
 
 head of water in feet 1 , i- •,• ^ 
 
 = ; — - — : : — J — 7 = hydraulic mean gradient. 
 
 length ot pipe or sewer in reet '' ^ 
 
 Q = quantity of water or sewage discharged in gallons per 
 
 minute when running /a//. 
 
 Example. — What diameter is required for a glazed stoneware 
 pipe sewer to provide for a population of 10,000, the available 
 fall being i in 500 ? 
 

 SEWERAGE 
 
 103 
 
 Total daily discharge = 10,000 x 25 x 6. 
 
 = 1,500,000 gallons per 24 hours. 
 
 Maximum flow = half in 6 hours. 
 
 = 750,000 gallons in 6 hours. 
 = 2083 gallons per minute = Q. 
 
 — = -^=128.8. S = -^ = o.oo2. S°-5= ^0.002 = 0.0447. 
 m 0.5 500 ^ 
 
 k 1\00 - >j 
 
 o 
 
 A 
 
 Fig. 24. — Eg-g- Shaped Sewers. Diagram showing pro- 
 portious of dimensions and radii. The horizontal diameter 
 being i, the diameter of the invert is 0.5, the total depth is 
 1.50 and the radii of the sides is 1.50. 
 
 OA = OD = OAi = i.oo. 
 
 Curve FK = arc of circle having centre at A, and radius 
 AF = i.5o. 
 
 Similarly, curve BC = arc of circle having centre at Ai. 
 
 R 
 
 ~V 128.^ 
 
 2083 
 
 = 0.383. 
 
 .8 X 0.0447 X 4695-7 
 
 and 0.383 X 4= 1.53 feet = 18 inches (nearly). 
 .'. diameter required = 18 inches. 
 Example. — What size of brick sewer is required for a popula- 
 tion of 80,000, the available fall being i in 1000? 
 
 Maximum flow (calculated as above) = 16,666 gallons per 
 minute. 
 
 0.55- 
 
 = 117.09. S = o.ooi. S°-'^ = n/o.oo I =0.0316. 
 
104 CALCULATIONS IN HYGIENE 
 
 / 16,666 
 
 ^^V 117.09 xo.o3i6x4695.7 = °-^^^4''4^^-^'^'^ 
 
 or 4 feet diameter. 
 
 An egg-shaped sewer of the same sectional area 
 
 = 5 feet X 3 feet 4 inches. 
 
 Deducting J from diameter of circular sewer (vide sujyra) 
 
 = 4 feet less 8 inches = 3 feet 4 inches. 
 
CHAPTER VIII. 
 
 DIET AND ENERGY. 
 
 Data necessary (i) a standard diet, "ordinary," "rest," or 
 " hard work " for the energy required. 
 
 (2) the percentage composition of the food. 
 
 Water-free ordinary diet for an adult ( = 300 foot-tons of 
 energy * daily). 
 
 Ounces. 
 
 Proteids . 
 Fats 
 Carbo-hydrates . 
 
 Salts 
 
 4-5 
 
 3-0 
 
 14.0 
 
 + 2 ounces for " hard-work.' 
 
 
 *' rest." 
 
 " hard- work. 
 
 "rest." 
 
 
 22.5 ounces. 
 
 
 
 
 In terms of nitrogen and carbor 
 
 L : 
 
 
 
 
 
 
 
 Grains. 
 
 Grains. 
 
 Diet for rest 
 
 , 
 
 , 
 
 200 
 
 4000 
 
 M 
 
 ordinary work 
 
 . 
 
 300 
 
 5000 
 
 )> 
 
 hard work . 
 
 • 
 
 400 
 
 6000 
 
 Percentage composition 
 
 of foods 
 
 , &c. 
 
 
 
 
 Bread. 
 
 Milk. 
 
 Meat. 
 
 Butter. 
 
 Cheese. Oatme£ 
 
 
 
 (uncooked) 
 
 
 Water 
 
 40.0 
 
 88.0 
 
 75-0 
 
 12.0 
 
 37-0 
 
 Proteids . 
 
 8.0 
 
 4.0 
 
 15-0 
 
 I.O 
 
 33-5 16.5 
 
 Fats . 
 
 1-5 
 
 3-0 
 
 8-5 
 
 84-5 
 
 24.0 6.5 
 
 Carbo-hydrates . 
 
 49.0 
 
 4-3 
 
 
 1.0 
 
 — 63.0 
 
 Salts 
 
 1-5 
 
 0.7 
 
 1-5 
 
 1-5 
 
 5-5 2.5 
 
 lOO.O 100. 100. 100. lOO.O 
 
 Food for a child : 
 Age 2 years : y\ (0.3) of the adult's standard diet for ordinary 
 work. 
 * " Foot-ton "=amoimt of energy that will raise i ton i foot in height. 
 
106 CALCULATIONS IN HYGIENE 
 
 Age 3-5 years : -/^ (0.4) of the adult's standard diet for ordinary 
 
 work. 
 Age 6-9 years : j\ (|) of the adult's standard diet for ordinary 
 work. 
 Example. — To Calculate the Quantity of Bread, Butter 
 and Cheese necessary for an ordinary Day's Work of 300 
 Foot-tons of Energy. 
 
 Proteida. Fats. Carbo-hydrate*. 
 
 Standard ordinary diet 4.5 3.0 14.0^67' ce«^.( = total diet). 
 
 Composition of bread . 8.0 1.5 49.0 „ 
 
 „ „ butter i.o 85.0 — „ 
 
 „ „ cheese 33.5 24.0 — „ 
 
 oz. oz. prda. 
 
 Let a: = oz. bread — by Simple Proportion:! 00: ic:: 8 :proteidsinbread. 
 
 8a; 
 
 Let y = oz. butter 
 
 
 100 " 
 
 )) 
 
 :5 
 
 ioo:2/::i: » 
 
 butter. 
 
 
 y 
 
 100 " 
 
 >» 
 
 55 
 
 loo'.z'.'.ss.s: „ 
 
 cheese. 
 
 
 33-52 
 100 " 
 
 )) 
 
 33-5« 
 
 5 (proteids). 
 
 
 Let ;:; = oz. cheese 
 
 and + + 
 
 100 100 100 =4 
 
 Similarly '-:5i?^ 85?/ J45 
 
 and -^ =14 (carbo-hydrates). 
 
 Reducing these fractions (multiplying by 100) : 
 
 (i) 8a: + f/ + 33.5« = 45o- ^ 
 
 (2) i.5a; + 85y-f-24x; = 3oo. 
 
 (3) 49^= 1400. x= 28.6 ounces of bread. 
 Reply cing in (i) the value of a: : 
 
 (8x28.6) + 2/ + 33.5^ = 45o. 
 
 (A) 2/ + 33-5- = 22i. 
 Replacing in (2) the value of x : 
 
 (1.5 X 28.6) + 85// +24;^; = 300. 
 
 (B) 85.v+24^ = 257. 
 Multiply (A) by 85 : 85^ + 2847.5:^-18785. 
 
 Subtract (B): 852/+ 24.0;:;= 257. 
 
 2823.52= 18528 
 
 - = 6.5 ounces cheese. 
 
DIET AND ENERGY 107 
 
 Replace in (2) the respective values of x (28.6) and ;:; (6.5) : 
 .-. 42.9 + 852/4-156 = 300. 
 85?/+ 198.9 = 300. 
 
 y=i.2 ounces butter. 
 The required quantities are: 28.6 ounces bread; 1.3 ounces 
 butter; 6.5 ounces cheese, 
 
 To Calculate if Ij Pounds of Bread and 60 Ounces of 
 Milk will be a suflacient Diet for a man doing an ordinary- 
 Day's "Work. 
 
 Carbo- 
 Proteids. Fats. hydrates. 
 
 100 oz. bread = 8 oz. 1.5 oz. 49 oz. 
 
 8 X 24 1.5 X 24 ^ 49 X 24 
 
 24 „ = =I.Q20Z. = 0.3607. =II.7 0Z. 
 
 ^ " 100 ^ 100 "^ 100 ' 
 
 100 oz. milk = 4.0 „ 3.0 „ 4-3 J) 
 4x60 3x60 Q 4.3x60 
 
 60 „ = = 2.4 „ =1.8 ,, = 2.5 „ 
 
 " 100 ^ " 100 " 100 ^ 
 
 Total(iilb.and6ooz.) = 4.32 „ 2.16,, i4'2 ?> 
 
 Ordinary diet =4.50,, 3-oo ?> 140 » 
 
 The above diet is therefore somewhat deficient in proteids and 
 fats, but is more than enough in carbo-hydrates. 
 
 To Calculate the Amount of Meat and Bread necessary 
 for a Daily Diet in Terms of Nitrogen and Carbon. 
 
 Nitrogen. Carbon. 
 
 Ordinary day's diet (moderate work) = 300 grains. 5000 grains. 
 I pound of meat =190 ,, 1900 „ 
 
 I „ bread = 90 „ 2000 ,, 
 
 Let X = pounds of meat required (i) 190a; + 90^ = 300 grains N. 
 Let 2/ = pouii(is of bread required (2) 1900.^ + 2000?/ = 5000 
 grains C. 
 
 Multiply (i) by 10 : 1900.^ + 900?/ = 3000 
 Subtract the result from (2) : i iooy= 2000. 
 
 1/= 1.8 pounds bread. 
 Substituting the value of 2/ in (i) 190X + 162 = 300. 
 
 x = o.'j pound = 11.2 ounces meat. 
 
 Energy evolved. 
 
 Internal work of the body (circulation, respiration, &c.) = 2800 
 foot-tons. 
 
 External ,, „ „ (average) Light 150 foot-tons. 
 
 Moderate 300 ,, 
 Hard 450 „ 
 
 Laborious 600 „ 
 
108 CALCULATIONS IN liYOIENE 
 
 De Chaumont's Formula for calculating external Work : 
 
 Ordinary = (300 x 5) = 1500 foot-tons. 
 
 Above „ =(300 X 5) + (300 X - j = 1500+ 1050) = 2550 ft. -tons. 
 
 Hard = (300 X 5) + f3oox -j + (300 x -j + 1350 = 3600 „ 
 
 Laborious = (300 x 5) + (300 x - j -|- (300 x -j -f (300 x -g-j = 
 
 3600 + 412.5 = 4012.5 foot-tong. 
 
 In round numbers 300 xf5-f--+- + -y-+ . . .) = 300 x 14 = 4200. 
 
 .*. Total external work = 4200 foot-tons. 
 ,, internal „ = 2800 ,, 
 
 7000 „ 
 The food necessary for 450 foot-tons of productive work must 
 provide 2550 foot-tons of potential energy for external work + 
 2800 foot-tons for internal work = 5350 foot-tons. 
 
 Therefore , or - of potential energy for external work, is 
 
 available for productive labour. 
 
 De Chaumont's Formula for productive Work : 
 
 Ordinary : 300 x i = 300 foot-tons. 
 Above,, „ „ -h (300 x J) = 450 foot-tons. 
 
 Hard: „ „ „ -f (300 x i) = 525 foot-tons. 
 
 Laborious: „ „ „ „ +{3^oxl) = 
 
 562.5 foot-tons. 
 Example. — A man is doing light work (e.r/. 250 foot-pounds); 
 how much cooked meat would he require to provide the necessary 
 amount of energy ? 
 
 100 ounces of cooked meat contain 28 ounces of proteids and 
 15 ounces of fats. 
 
 I ounce proteid = 173 foot-tons potential energy. 
 I ounce fat= 378 ,, ,, ,, 
 
 28 X 173= 4844 ,, n n \ 
 
 15x378-- = 5670 » n M 
 
 100 ounces cooked meat= 105 14 „ ,, ,, 
 
 Deducting 2800 for internal work 2800 
 
 7714 »» »> »» 
 
 available for external work, and of this oidy ^ is represented by 
 actual work, or about 1543 foot-tons of actual work. 
 
DIET AND ENERGY 109 
 
 Let a; = ounces of meat necessary for 250 foot-pounds of actual 
 work : 
 
 1543 : 250 : : 100 ; cc= 16.2 ounces cooked meat. 
 
 (By a previous example 28.6 ounces bread, 1.3 ounces butter 
 and 6,5 ounces cheese were found to be sufficient for 300 foot- 
 tons of work, therefore 23.6 ounces bread, 1.8 ounces butter, and 
 5.4 ounces cheese will provide 250 foot-tons of work.) 
 
 Energy expressed in terms of heat- value, or "calorific capacity." 
 
 I calorie = amount of heat necessary to raise i gramme of 
 water 1° C. 
 
 ^ or |- total potential heat = actual heat. 
 ^ or J „ „ = „ work. 
 
 Potential energy available from diet : 
 
 Proteids. Fats. 
 
 I ounce. I gramme. i ounce. i gramme. 
 
 Foot- tons 173 calories 4.1 | Foot-tons 378 calories 9.3 
 
 Carbo-hydrates. 
 1 ounce. I gramme. 
 
 Foot-tons 138 calories 4.1 
 
 To Calculate the Heat-value of a Food in calories. 
 
 Example. — Oatmeal contains per 100 parts (taken as grammes) 
 Proteids 1 2.5(1 gramme = 4. 1 calories) = 12.5 x 4.1 = 5 1.25 calories. 
 Fats 6.5 „ =9.3 „ = 6.5x9.3= 60.45 „ 
 
 C.-H. 63.0 „ =4.1 „ = 6.3x4.1 = 258.30 „ 
 
 370.00 „ 
 
 Calculation of Mechanical Work. — Height x weight = foot- 
 pounds or foot-tons of work. 
 Let W = weight in pounds. 
 
 H = vertical height in feet. 
 
 W X H = foot-pounds of work. 
 If H = height in miles, 5280 feet= i mile : 
 
 W X 5280 H=/oo^pounds of work. 
 
 To express foot-pounds as foot-tons : 2240 lbs. = i ton. 
 
 /TTT o ^ W X 5280 - 
 
 2240 : ( W X 5280) : I : x= foot-tons. 
 
 ^ ^ J / 2240 
 
 Allowance made for ''traction " or resistance : 
 
 Moving along a level at 3 miles per hour = lifting the entire 
 weight vertically -j-^ of the distance traversed, or lifting 2^^^ weight 
 the whole distance At 4 miles per hour this " co-efficient of 
 traction " becomes yy ; at 5 miles, J^. 
 
110 CALCULATIONS IN HYGIENE 
 
 W X 5280 
 Energy at 3 miles per hour alonglevel road = x ^V^t.-tons. 
 
 WX5280 ^ 
 " ^ '' " " " " 2240 ^ ^"^ " 
 
 WX5280 ^ 
 " ^ " " " " ~ 2240 "^ ^^ " 
 
 " W " denotes the evtire iveight carried and includes the weight 
 of the individual and of all impedimenta — e.g.^ clothes, &c., which 
 weights are to be added to the body-weight. 
 
 If there is an ascent, the "rise" must be known, and the 
 additional energy is to be calculated and added to the work done 
 on level-ground. 
 
 E.g,^ supposing the *' rise " is i in 400 feet, and the entire 
 distance walked = 5280 D feet : 
 
 400 : 5280 D : : I : x = ft. oi vertical distance = 13.2 D/^ 
 
 W 
 
 The energy for this additional distance = i ^.2 D x foot-tons. 
 
 ^•^ ^ 2240 
 
 The total energy is the sum of both : 
 
 WX5280D ^ . , . / W \ 
 
 X coemcient 01 traction + i ^. 2 D x 
 
 2240 \ »^ 2240/ " 
 
 Example. — A soldier 10 stone in weight carries a kit, &c., of 
 60 lbs. and marches, at the rate of 3 miles per hour, a distance of 
 7 miles, the ascent being i in 500 feet. Calculate the amount of 
 work done in foot-tons. 
 
 W - 10 X 14 -f 60 = 200 lbs. 
 
 200x5280x7 I . c ^. 1 , , 
 
 X — = 1 6 15 root-tons along a level. 
 
 2240 20 -^ ^ 
 
 500 : (5280 X 7) : : 1 '.x = = 73.92 ft. of vertical ascent 
 
 200 
 equal to : 73.92 x- — — = 6.6 foot-tons of energy. 
 
 .-. Total energy = 165 -I- 6.6 = 171.6 foot-tons. 
 
CHAPTER IX. 
 FOODS. 
 
 Milk. 
 
 Average specific gravity at 60° F. = 1031 (water = 1000). 
 
 Correction for Temperature. 
 
 The sp. gr. of milk falls 1° for each rise of 10° F. above 60° F., 
 and, vice versd, rises 1° for every fall of lo"^ F. below 60° F. — i.e., 
 inversely as the temperature. 
 
 To correct: add or subtract 1° for every difference of 10° F. 
 above or below 60° F. 
 
 IJ.g., Sp. gr. at 40° F. = 1029. 1029 - 2| ^ ^^ ^^ ^^o ^ 
 „ „ 70 F. = io26. 1026 + 1 j ' 
 
 For differences of temperature less than 10° F. the same pro- 
 portion may be taken as approximately correct : 
 
 ^.^., Sp.gr. at 46° F. = 1030. 60-46 = 14° F. 
 
 10 : 14 : : I : a;. = 0.4. 1030 - 0.4 = 1029.6 = sp. gr. at 60° F. 
 
 Total Solids. 
 
 Weight of 10 c.c. milk + capsule =44-58 grammes. 
 
 „ only = 34.32 „ 
 
 = 10.26 
 
 Weight of total solids + capsule after evaporation. 
 
 = 35.67 grammes. 
 „ „ capsule =34-32 » 
 
 I-3S 
 10.26 : 100 : : 1.35 : a: = 13.15 per cent, of total solids. 
 Adam's Process. — Method. — 10 grms. of milk are absorbed 
 by fat-free bibulous paper, dried and extracted (12 syphonings) 
 with ether in Soxhlet's apparatus. The ether is evaporated off 
 and the residual fat is dried and weighed in a tared tiask. The 
 percentage of fat is calculated. 
 
 Example. — 10 grms, milk are treated as above. 
 
112 
 
 CALCULATIONS IN HYGIENE 
 
 Weight of tared flask + fat = 28.867 grammes. 
 „ „ alone =28.594 „ 
 
 „ fatin logms. milk= 0.273 » 
 
 „ „ 100 „ = 2.73 
 
 = 2.73 per cent, of fat. 
 3.0 - 2.73 = 0.27 per cent, below the standard. 
 
 Werner-Schmidt Process. — A known quantity (10 gms.) 
 of milk is placed in a specially graduated tube 
 IP^ (Stokes', Fig. 25, or Schmidt's) and boiled 
 
 with 10 c.c. of strong hydrochloric acid. The 
 casein is destroyed. The contained fat is 
 extracted with ether (added up to the 50 c.c. 
 mark) and estimated after evaporating the 
 ether, and correcting for the residue left in 
 the tube. 
 
 Example, — 10 c.c. milk+ 10 c.c. strong HCl. 
 Boil and cool. Add ether up to the 50 c.c. 
 mark; shake. Sp. gr. of milk= 1 031, 20 c.c. of 
 ether were placed in a weighed dish, evaporated 
 off and the dish dried. 
 
 Weight of fatty-residue + dish = 39.016 gms. 
 dish =38.752 „ 
 
 Fio. 25.— Stokes' Tube. 
 20 : 28.6 : : 0.264 
 
 Sp.gr. = 1031. . 
 
 Fat in 20 c.c. of ether = 0.264 ,, 
 
 = 28.6 c.c. total 
 amount of ether 
 containing all the 
 fat. 
 
 a; = 0.377 gms. fat in 10 c.c. of milk. 
 = 3-77 M n 100 „ 
 
 Ether left in tube = 8.6 c.c. 
 „ pipetted from tube 
 = 20.0 c.c. 
 
 . , , o 1031 
 
 weight ot 100 c.c. = — ^^= 1 03. 1 grammes. 
 
 77 : .-« = 3.65 per cent, of fat, or 0.65 per cent. 
 
 volume and 
 
 .*. 103. 1 : 100 : : 3 
 above the standard. 
 
 If, instead of calculating its weight from its 
 specific gravity, 10 gms. of milk are weighed out: 
 
 0.365 gm. fat in 10 gms. of milk. 
 
 3.65 per cent, of fat. 
 
 Hoppe-Seyler's Process. — Fat, casein and earthy 
 phates are precipitated from a solution of milk by addi 
 
 phos- 
 ^ a little 
 acetic acid and passing a cuirent of CO2. The precipitate is 
 filtered, and the retained fat extracted with ether in Soxhlet's 
 
FOODS 113 
 
 apparatus, dried and weighed. The filtrate contains serum- 
 albumen and lactose. The former is separated by boiling aud 
 filtering and weighed. The second filtrate contains sugar and 
 salts. The former is estimated by titration with a standard 
 solution of Fehling (lo c.c. = 0.067 g"^- lactose). 
 
 Ritthausen's Method. 
 
 (i) 10 c.c. of milk are diluted with 200 c.c. of distilled water 
 and neutralised with a standard solution of copper sulphate and 
 caustic potash (62.82 gms. CuSO^ per litre ; i c.c. = 0.1 gm. CuO). 
 
 The precipitate, consisting of fat and albuminate of copper, is 
 collected on a filter-paper of known weight by suction. 
 
 The fat is extracted by Soxhlet's method and weighed. 
 
 (2) The albumen is calculated by the difference in weight of 
 the filter-paper before and after the extraction of the fat, 
 deducting the weight of copper oxide in the precipitate. 
 
 (3) The filtrate contains lactose, which is calculated by titrating 
 with a standard solution of Fehling (10 c.c. = 0.067 g™- lactose). 
 
 Example : 
 
 1. Weight of fat-flask + extracted fat= 19.957 gms. 
 
 „ „ „ alone =- 19-635 » 
 
 difference = 0.322 gm. fat in 100 c.c. 
 
 milk. 
 = 3.22 per cent, of fat. 
 
 2. Weight of filter-paper (in test-tube) 1 
 
 „ „ albumen - = 17.251 gms. 
 
 ,, „ copper oxide J 
 
 „ „ filter-paper (in test-tube) only= 16.730 „ 
 
 difference = 0.521 gm. albu- 
 men and copper oxide in 10 c.c. milk. 
 0.521 — 0.1 = 0.421 gm. albumen. 
 
 -= 4.21 per cent, of albumen. 
 
 3. Filtrate from (i) made up to 300 c.c. 
 
 44.8 c.c. were required in titrating 10 c.c. of Fehling's- solution 
 ( = 0.067 S^- lactose). 
 
 .'. 44.8 : 300 : : 0.067 • ^^ = 0.44 gm. lactose in 10 c.c. of milk. 
 = 4.4 per cent, of lactose. 
 
 Richmond's Formula.— To calculate the percentage of fat, 
 total solids and sp. gr. being known. 
 
 Percentage of fat = (Total solids x 0.859) - (" G " x 0.2186). 
 
 '' G " = last two units of the specific gravity and any decimal ; 
 or, = specific gravity - 1000. 
 
 H 
 
114 
 
 CALCULATIONS IN HYGIENE 
 
 E.g., total solids = 10.8. Sp. gr. at 60" F. = 
 1031.5. "G" = 3i.5. 
 
 Percent, of fat=^ (10.8x0.859) -(31.5 x 0.2186). 
 = 9.2772-6.8859. 
 = 2.39 per cent, of fat. 
 
 By the above formula the third term can be 
 calculated if the other two are known : 
 
 E.g., percentage of fat = 3. Sp. gr. = 1032. 
 
 To find total solids : 
 
 3 = 0.859.x-. - (32 X 0.2 186). 
 
 3 + 6.9952 = o.859ic. a; =11.63 P®^ cent, of 
 total solids. 
 
 A more recent formula for calculating total 
 solids is : 
 
 Total solids = 1.2 X percentage of fat + o.i4 + 
 025 G. 
 
 The above example would give : 
 
 1.2X3 + 0.14 + 0.25x32 = 11.74 per cent, 
 of total solids. 
 
 Bichmond's Slide-Scale (Fig. 26). — If two 
 terms are known the third can be found. The 
 sliding-scale in the middle indicates specific 
 gravity ; the upper one, fat ; and the lower, 
 total sohds. 
 
 If specific gravity and fat per cent, are 
 known, place the arrow-head (of the sliding- 
 scale) under the figui-e denoting the per cent, 
 of fat, and the specific gravity figure will coincide 
 w^ith that for total solids. 
 
 If specific gravity and total solids per cent, 
 are known, let these figures coincide on the 
 scale, and the arrow-head will indicate the 
 percentage of fat. The scale is used in con- 
 junction with other methods {e.g., Leftman 
 Beam's process for the estimation of fat) as an 
 approximate check on the results. 
 
 Milk Standards. — Fat = 3 per cent. 
 
 Solids not fat = 8.5 per cent. 
 
 To Estimate the Amount of Fat ab- 
 stracted, the percentage of fat present being 
 known : 
 
 E.g., sample contains 2.18 per cent. fat. 3.0 
 - 2. 18 = 0.82 per cent, removed. 
 
FOODS 115 
 
 Or: 
 
 Q 
 
 3 : 2.18 : : 100 : x = = 72.7 per cent, of the original fat 
 
 remains, and 100 -72. 7 = 27. 3 per cent, has been abstracted. 
 
 To Estimate the Quantity of Water added. ( i) From non- 
 fatty solids. — This is dene by calculating the amount of <' solids 
 not fat" in the sample, as they are less variable in quantity than 
 the fat in a genuine sample of milk. 
 
 E.g.^ sample contains 7.25 per cent, of total solids. 
 
 7 2 1\ 
 8.5 : 7.25 : : 100 : .^ = -0-^^85.3 per cent, of pure milk. 
 
 100 - 85.3 = 14.7 per cent, of water added. 
 
 (2) From the ash (which also varies little in a genuine sample). 
 
 E.g., ash after ignition of sample = 0.6 per cent. 
 
 600 ^ 
 7 : 6 : : 1 00 : ic = — — == o5 • 7 
 
 100 - 85.7 = 14.3 per cent, of water added. 
 
 Butter. 
 
 Moisture. — Should not exceed 16 per cent. 
 
 Example. — Weight of dried capsule = 21.53 grammes. 
 ,, butter taken = i.oo „ 
 
 22.530 
 Weight after evaporating, drying and coolings 22.417 „ 
 
 = 11.3 per cent, moisture. 
 
 Soluble and Insoluble (Volatile and Fixed) Acids. 
 
 Method. — The melted fat is saponified in methylated spirit 
 with caustic potash, and the volatile acids, set free by dilute 
 
 sulphuric acid, are distilled over into — NaHO, and estimated by 
 
 N 
 titration with — oxalic acid. 
 10 
 
 The fixed acids, after the addition of sulphuric acid, are 
 
 evaporated, dried and weighed (not distilled over). 
 
 Example. — Volatile Acids. 
 
 2.5 grammes butter-fat 4- 5 grammes caustic potash 4- 50 c.c. 
 methylated spirit (Saponified). After evaporating ofi" the spirit 
 
IIG CALCULATiOKS IN HYGIENE 
 
 the residue is dissolved in distilled water, mixed with dilute 
 
 sulphuric acid and distilled. 
 
 N 
 it^o c.c. distilled oflfinto 20 c.c. — NaHO. 
 ^ 10 
 
 N 
 Titrated with — oxalic acid (i c.c. = 8.8 nigr. butyric acid). 
 
 N N . . 
 
 On trial 20 c.c. — NaHO= 18 c.c. — oxalic acid. 
 10 10 
 
 N N 
 
 Distillate + 20 c.c. — NaII0-=2.o c.c. — ,, 
 10 10 " 
 
 N 
 .*. 18-2 = 16 c.c. — oxalic acid not used up. 
 10 ^ 
 
 16 X 8.8= 140 mgrs. butyric acid in 2.5 {-^jj of 100) guis. butter. 
 
 140x40 = 5600 mgrs. in 100 grammes butter = 5.6 per cent. 
 
 of butyric acid. 
 
 Fixed Acids : 5 grammes butter-fat saponified similarly as 
 
 above and treated Vvith dilute sulphuric acid, evaporated, dried 
 
 and weigned in capsule of known weight : 
 
 Weight of capsule + fatty acids = 57.774 grammes. 
 
 „ „ alone = 53-i54 » 
 
 Difference = 4.620 ,, fixed acids in 
 
 5 grammes butter-fat. 
 
 = 92.4 per cent. 
 
 Specific Gravity of Butter-fat. 
 
 (i) By a specific gravity bottle at 35"^ C. or at 100^ F. 
 (2) By Westphal's balance (p. 24). 
 / V Weight of melted butter-fat at loo^ F. 
 ^'^ Weight of distilled water at 100" F. (''''^^^1^= ioo°)- 
 Example. — Weight of empty specific gravity bottle =11.85 
 grammes. 
 
 Weight of bottle + distilled water at 100° F. = 35.6 grammes. 
 „ ,, -f melted fat at 100° F. = 33.415 ,, 
 
 „ melted fat at 100° F. = 33.415 - 11.85 = 21.565 
 
 grammes. 
 ,, distilled water at 100° F. = 35.6 - 11,85 = 23.75 
 
 grammes. 
 
 Specific gravity of fat at 100° F. = - ' =0.908 
 
 = 908 (water = 1 000) 
 Lowest specific gravity of pure butter-fat =910 
 Highest ,, ,, foreign fat = 904 
 
 Difference = 6 
 = 100 per cent, of adulteration. 
 
FOODS 117- 
 
 In the example, difference = 910- 908 = 2. 
 
 6 : 2 : : 100 : .q:^. =33-3 per cent, cf adulteration with other fat. 
 
 Estimation of Albumenoids in Meat, Cereals, &c. 
 
 Kjeldahl's Method. — Organic matter is powdered, and boiled 
 with concentrated sulphuric acid till colourless. Potassium per- 
 manganate is added to oxidise it into ammonium sulphate. On 
 cooling, distilled water and caustic soda solution are added, and 
 
 N 
 the ammonia is distilled into — hydrochloric or oxalic acid 
 
 N 
 and titrated with — alkaline solution, and the nitrogen deter- 
 
 mined. 
 
 Example. — 0.2 gramme oatmeal + 10 c.c. strong sulphuric acid 
 digested till straw-coloured + KMnO^ and boiled till colourless; + 
 250 c.c. 10 per cent. NaHO solution +100 c.c. distilled water, 
 
 N . . 
 
 iqo c.c. distilled into 10 c.c. — oxalic acid, and titrated with 
 ^ 10 
 
 N 
 
 — NaHO. 
 
 10 
 
 N N . . 
 
 On titrating, 10.2 c.c. — NaHO = 10 c.c. — oxalic acid. 
 ^' 10 10 
 
 N N 
 
 10 c.c. of — oxalic acid + distillate from flask = 6.7 c.c. — NaHO. 
 10 10 
 
 10.2 : 6.7 : 10 : x. = 6.6 c.c. — oxalic acid. 
 ' 10 
 
 N 
 10 - 6.6=^3.4 c.c. — oxalic acid neutralised by NH3 distilled 
 
 over. 
 
 N 
 I c.c. — oxalic acid= 1.7 mgr. NH,. . 
 10 / o ^ 
 
 3.4x1.7 = 5.8 
 
 14 
 5.8 X — = 4.8 mgr. nitrogen. 
 
 4.8 X 6.25 = 30 mgr. albumen in 0.2 gm. of oatmeal. 
 = 15 mgr. in o.i gramme. 
 = 15 per cent, albumen. 
 
 Alcohol. 
 
 Absolute alcohol. Specific gravity at 60° F. = 0.79. 
 
 ,, „ +16 per cent, water = rectified spirit. 
 
 „ „ +42.95 per cent, water = proof „ 
 
118 
 
 CALCULATIONS IN HYGIENE 
 
 Proof- spirit is taken as the standard. Its specific gravity at 
 60° F. = 0.92. A spirit containing less alcohol than proof-spirit 
 is *' underproof " ; if containing more, ^' overproof." 
 
 Proof-spirit = 57.05 per cent, absolute alcohol, volume in volume, 
 
 in distilled water. 
 „ =49.25 per cent, absolute alcohol, weight in weight, 
 
 in distilled water. 
 „ = 42.46 per cent, absolute alcohol, weight in volume, 
 
 in distilled water. 
 
 To calculate the ratio of alcohol to proof-spirit as : 
 
 (i) volume in volume : 57.05 : i : : 100 : x. = 1.753. 
 
 (2) weight in w^eight = 49.25 : i : : 100 : x. =2.03. 
 
 (3) „ volume = 42.46 : I : : 100 : x. =2.35. 
 
 To calculate degrees " over-" and *' under-proof " : 
 (i) A sample of whisky is 25° under- proof : 
 1.753 : I : : (100 - 25) : x. =42.8 per cent, absolute alcohol, 
 volume in volume. 
 
 Fi<;. 27. — Sikc's Hydrometer. 
 
 (2) Brandy 15° over-proof: 
 
 2.03 : I : : (100 -t- 15) : x. =56.6 per cent, absolute alcohol 
 as weight in iveiyht. 
 
 (3) A sample of gin 35' under-proof : 
 
 2.35 : I : : (100- 35) : x. =27.6 per cent, absolute alcohol 
 as vmght in volume^ 
 
FOODS 119 
 
 (4) A spirit contains 28 per cent, of alcohol, volume in volume : 
 
 57.05 : 28 :: 100 : .'c. =49.08. 100 - 49.08 = 50.92 under-proof. 
 
 In Great Britain Sike's Hydrometer (Fig. 27) is used in dis- 
 tilleries and breweries. It is supplied with Tables giving the 
 percentage of alcohol corresponding to the readings. The 
 instrument floats at zero in strong spirit, specific gravity = 0.825, 
 and the heaviest disc will make it float at zero in distilled water 
 — giving a range of 500° between these. 
 
 The amount of alcohol in beer is determined as follows : 
 
 Mulder's Method. — Determine the specific gravity of the 
 beer at 15.5 ' 0. or 60 F. Take 300 c.c. and distil ofi" 200 c.c. 
 
 Make up the distillate to 300 with distilled water, and take 
 the specific gravity of the mixture, referring to the Tables for the 
 percentage of alcohol indicated by the reading. 
 
 To verify this result : (i) make up the residue in the distilla- 
 tion flask (100 c.c.) to 300 c.c. and take the specific gravity; 
 (2) subtract from this the specific gravity of the original beer; 
 finally subtract this result from 1000. This figure ought to 
 correspond with the specific gravity of the distillate (200 c.c.) 
 when made up to 300 c.c. 
 
 Example. — Specific gravity of original sample of beer = " 1015." 
 
 300 c.c. are placed in a distillation-flask ; 200 c.c. are distilled 
 off and are made up to 300 c.c. with distilled water. 
 
 Specific gravity of distillate (made up to 300 c.c.) = " 995-" 
 
 By Tables the reading " 995 " = 3.35 per cent, alcohol. 
 
 Specific gravity of residue in flask (100 c.c.) made up to 300 c.c. 
 = 1020. 
 
 .-. 1020-1015 = 5. 1000-5 = 995. 
 
 Acidity of Beer. Method. — 10 c.c. are diluted with distilled 
 
 N 
 water and titrated with — NaHO. The result is expressed 
 
 N 
 in terms of lactic acid. -— oxalic acid is used for testino- the 
 10 '^^ 
 
 N 
 
 — NaPIO. Phenol-phthalein is the " indicator." 
 
 N 
 Lactic acid = C^fi^ = 90. — = 9 grammes per litre. 
 
 I c.c. = 9 mgr. 
 
 Example. — 10 c.c. beer are treated as above. 
 
 On titration = 2.1 c.c. — NaHO. 
 
 10 
 
 N N . . 
 
 On testing — NaHO. 10.^ c.c. = 10 c.c. — oxalic acid. 
 ° 10 10 
 
120 CALCULATIONS IN HYGIENE 
 
 .*. 10.3 : 2.1 : : 10 : rr. = 2.04 c.c. — oxalic acid. 
 
 but 2.04 X 9= 18.36 mgr. lactic acid in 10 c.c. beer. 
 = 0.1836 per cent, lactic acid. 
 The acidity of wine is expressed as tartaric acid. The method 
 of calculation is similar. 
 
CHAPTER X. 
 
 LOGARITHMS AND LOGARITHMIC TABLES. 
 
 The logarithm "^ of a number is the "index " of the power to which 
 a constant number, called the base, must be raised to equal the 
 number of which the "index" is the logarithm. 
 
 If a" = cc, n is the logarithm of the number x to the base a. 
 
 The logarithm of 64 to the base 4 is 3, because 4^ = 64; the 
 logarithm of 64 to the base 8 is 2 : 8^ = 64. 
 
 It is expressed thus : Logj, x = n ; Log^ 64 = 3 ; Logg 64 = 2. 
 
 The "base " is placed between the letters " Log " or " L." and 
 the number of which the logarithm is given. 
 
 In the common system of Logarithms the base employed is 10, 
 and the power to which 10 is raised to produce any number is 
 the logarithm of that number. As this base is in general use for 
 all calculations, it is not written down, so that if no base is 
 indicated it is understood to be 10. In the Napierian system the 
 base is 12, indicated by e, but this method is not in use. 
 
 Log 269 = 2.4297523, means therefore that the base 10 raised 
 to the power 2.4297523 is equal to 269. 
 
 10 being taken as the base, the logarithm of 10 {i.e. of the base 
 itself) = I as 10^ = 10, and log 10=1; therefore 10^=100, and 
 log 100 = 2 ; and log io^= 10,000, and log 10,000 = 4, &c. 
 
 If, instead of multiplying 10 by itself, it is divided by itself : 
 
 — = 1, indicated thus : 10° = i ; therefore loe^ 1=0. 
 10 ° 
 
 ■ = — ^ = 0.1 = 10 ' log 0.1= - I . 
 
 10 X 10 10 • o 
 
 Similarly: =o.oooi = io~4 log 0.0001= -4 
 
 •^ 10,000 • ° 
 
 The logarithm of 10 being i, that of all numbers less than i 
 consists entirely of decimals, there being no whole number. 
 
 The integral part of a logarithm (the whole number or 
 numbers to the left of the decimal point) is known as the 
 
 * The number corresponding to a given logarithm is termed its "anti- 
 logarithm." 
 
122 CALCULATIONS IN HYGIENE 
 
 Characteristic, and the decimal part (figures to the right of the 
 decimal point) as the Mantissa. 
 
 E.g. log 4176 = 3.6207605; the Characteristic is 3 and the 
 Mantissa .6207605. The latter usually contains seven figures. 
 
 Numbers consisting of one whole number and any decimals 
 have zero as a Characteristic. 
 
 E.g. log 2.83 = 0.4517864. 
 log 1.386 = 0.1417632 
 
 The Characteristic in all cases is omitted from Logarithmic 
 Tables and must be prefixed by the calculator himself, as can 
 easily be done on inspection by the following Kules : 
 
 1. If the logarithm to he found is that of a number containing 
 
 ONE OR MORE INTEGERS : 
 
 The Characteristic is one less than the number of integral 
 figures in the number. 
 
 E.g. log 864 = 2.9365137. 
 log 86.4=1.9365137. 
 log 8.64 = 0.9365137. 
 It is to be observed that so long as the figures in the number 
 remain the same the Mantissa also remains the same ; the Cha- 
 racteristic alone changes, according to the position of the decimal 
 point, i.e. as the number of integers. 
 
 2. If the logarithm to be found is that of a number containing 
 decimals only and no integers, the Characteristic is the same as 
 the place to the right of the decimal point which the first signi- 
 ficant figure (not a zero) of the number occupies. The Cha- 
 racteristic is negative, and to distinguish it from a positive 
 Characteristic has a negative sign or " bar " placed above it — not 
 in front. 
 
 E.g. log 0.854=1.9314579, 8 being the^Vs^ significant figure 
 after the decimal, i is the Characteristic with the negative 
 "■ bar " over it. (It follows that a positive Characteristic indicates 
 a whole number.) 
 
 Log 0.0854 = 2.9314579, where 8 is in the second place — zero 
 not counting as a significant digit. Similarly log 0.00023 = 
 4.3617278. 
 
 The Characterislic of a logarithm may therefore be positive ( + ) 
 or negative ( - ), but the Mantissa is never a negative q^iantity, it 
 iii always positive. {Vide ^. 127.) 
 
 Tables of Logarithms. 
 
 In these only the Mantissa of the numbers (indicated in the 
 first column on the left of each page under the heading " No.") is 
 
LOGARITHMS AND LOGARITHMIC TABLES 123 
 
 given. As already explained, the Characteristic is prefixed on 
 inspection. 
 
 The descriptions here furnished are applicable to the Mathe- 
 matical Tables published by Messrs. W. & R. Chimbers, Ltd., 
 which are in universal use ; but the methods are, of course, applic- 
 able to all similarly-constructed Tables. 
 
 In these the Mantissa of each Number is in a line with it, and 
 vice versa. The first three or four figures following the decimal 
 point are in numbers after 999 in the first column to the right 
 of the number and are to be prefixed to all the gioups of four 
 figures (under the columns headed "o," "i," "2," " 3," "4." . . . 
 " 9 "), whether these groups are on the same line with them or 
 on at lovjer level but above the next group of initial figures. 
 
 The numbers i to 999 are each given separately. 
 
 Thus 
 
 
 
 
 No. 
 
 Log. 
 
 No. Log. 
 
 751 
 
 8756399 
 
 951 9781805 
 
 752 
 
 8762178 
 
 996 9982593 
 
 
 So that log 752 - 
 
 = 2.8762178. 
 
 
 log 0.951 
 
 = 1. 9781805. 
 
 Example.— 
 
 -In the Tables we find 
 
 : 
 
 No. 
 
 
 012 
 
 3 ^^^ 
 
 1067 
 
 028 
 
 1644 2051 2458 
 
 2865 &c. 
 
 * 
 
 # 
 
 * * # 
 
 
 69 
 
 
 9777 0183 0590 
 
 0996 &c. 
 
 70 
 
 029 
 
 3838 4244 4649 
 
 5055 ^<^- 
 
 
 That is^ log 1067 = 
 
 3.0281644 
 
 
 „ 10670 = 
 
 = 4. „ 
 
 
 „ 10671 = 
 
 = 4.0282051 
 
 
 „ 10673 = 
 
 = 4.0282865 
 
 
 
 „ 1069 = 
 
 = 3.0289777 
 
 but log 10691 =4.0290183 
 
 log 1070 = 3 0293838, and similarly the others. 
 
 Note that the first three figures, e.g. " 028," are carried on for 
 all the columns till another set of three, e.g. " 029," is met with, 
 the exception being where there is a line drawn over the last four 
 figures of the Mantissa, e.g. 0183, 0966, &c., as above. In these 
 cases the three first figures of the next Mantissa (below) must be 
 prefixed, it being a matter of convenience to denote the alteration 
 in this way rather than to have a broken line of figures in the 
 Tables, and also to economise space. 
 
 After the number 99999, the first four figures of the Mantissa 
 are supplied, but the method of working is the same. 
 
124 CALCULATIONS IN HYGIENE 
 
 Note the logarithm of i, lo, loo, looo, etc., is represented by 
 o ; that is, there is no Mantissa, but only a Characteristic, which, 
 as already explained, is to be o, i, 2, 3, &c., and is put down on 
 inspection, as before. 
 
 To Find the Logarithm of a given Number. — i . For numbers 
 containing less than four figures : the Mantissa is read off at 
 once, and lies by the side of the number as already indicated. 
 
 2. For numbers of five figures : the Mantissa is found under 
 *' o " — the first four decimals are a little to the left of the zero 
 in the first column of figures, and the rest directly under the 
 figures o .... 9 at the top of the page. 
 
 E.g. log 9062^ = .957 (traced upwards opposite '' 9058 "), 
 
 and 2528 (the next four decimals under " 6 "), 
 
 entire Mantissa = .9572528 
 ,, logarithm = 4.g^'j 2 ^28 
 but log 0.00906 is found under " 906," and is equal to 3.9571282. 
 
 For numbers of six figures. The Mantissa is found for the first 
 five figures of the number in the same way as before. To obtain 
 the sixth figure subtract the Mantissa of the first five figures from 
 that of the next higher number (i.e. from the next higher 
 Mantissa). The difference will coincide with the figures at the 
 top of the adjacent "column of proportional parts " under the 
 heading "Diflf" (at the extreme right of every page). Find in 
 this column (numbered i to 9) the sixth figure of the given 
 number, and opposite to it will be found the figures which must 
 be added to the last digits of the Mantissa first found {the 
 Mantissa of lower value). 
 
 Example. — To find log 268354. 
 
 The logarithm of the first five figures of this number is easily 
 found in the usual way opposite to the figures 2683, and under 5, 
 and as the sixth figure 4 cannot be read ofl[', the Mantissa corre- 
 sponding to the first five figures must be subtracted from the 
 Mantissa of 26836 — i.e., the first figures to the right of the last 
 
 log 2683d = 4.4287i'7c? 
 log 268^ = 4.428/016 
 
 Difference = 162 
 
 This difference corresponds to the figure " 162 " at the top of 
 the column under " DifiV Opposite "4" in this column, which is 
 the required sixth figure of the given number, is found " 65,'" 
 
LOGARITHMS AND LOGARITHMIC TABLES 125 
 
 wliicli must be added to the last digits of the lower Mantissa first 
 found, thus: log 26835 ^4-4287016 
 65 
 
 ,, 268354 = 5.4287081 which is the required log. 
 
 (Note. — The sixth figure of the numbers from looooi to 10800 
 inclusive can be ascertained directly from the Tables.) 
 
 To Find the Logarithm of a Number containing Seven 
 Figures. 
 
 Example. — Find log 5067958. 
 
 log 50680 = 4.7048366 
 log 50679 = 4.7048280 
 
 Difference = 86 coinciding with 86 in the " Dili?' column. 
 
 Opposite " 5 " in this column (which is the sixth figure in the 
 given Number) is " 43," and opposite 8 (the seventh figure of 
 the Number) is -'69." 
 
 Therefore: log 50679 = 4.7048280 (the lower Mantissa) 
 
 43 
 
 log 506795 =5.7048323 
 69 
 
 log 5067958 = 6.70483290 
 
 Note 69 is placed, for the seventh figure, one decimal place 
 farther to the right. 69 may be taken as 7.0^ and adding 7 to the 
 last figure, 3, of the Mantissa we have log 5067958 = 6.7048330. 
 
 Example. — Find log 317.1626 
 
 log 317.17 = 2.5012921 
 log 317.16 = 2.5012784 
 
 Difference = 137 
 
 Diff. 
 
 log 317-16 =2.5012784 
 
 137 
 
 2 27 
 6 82 
 
 27 
 
 
 log 317.162 =2.5012811 
 82 
 
 
 log 317.1626 = 2.5012819^ 
 
126 CALCULATIONS IN HYGIENE 
 
 To Find the Logarithm of a Number of Eight Figures. 
 
 Example. — Find log 23453487 
 
 log 23454 = 4-3702169 
 log 23453 = 4.3701984 
 
 Difi: 
 
 erence = 
 
 185 
 
 J)iff. 
 185 
 4 
 8 
 
 7 
 
 74 
 148 
 130 
 
 log 
 
 23453 = 
 
 = 4.3701984 
 
 
 
 4 
 8 
 
 74 
 148 
 
 
 
 7 
 
 130 
 
 
 log 
 
 23453487- 
 
 = 7.370207410 
 
 
 To Work with negative Characteristics (denoting that the 
 numbers of v hich they are the logarithms are decimals) the 
 ordinary Algebraical methods of addition and subtraction are used, 
 as in all logarithmic calculations. 
 
 Examples. — 
 Addition: 5.2657845 3.0624316 2.3461573 
 
 2.4983106 7.2713769 1. 8692317 
 
 3.7640951 4-3338085 0.2153890 
 
 (i carried over to i 
 
 Subtraction : 
 
 = 2 + 2 = 0) 
 
 [The sign of the negative Characteristic which is to be subtracted 
 i-; changed to + and the two Characteristics are added as in 
 Algebra ; the Mantissa is subtracted in the ordinary way, being 
 positive. 
 
 E.g., 3« + 2^ - (« - 6) = 3a + 26 - a + 6 (the two negatives before 
 h changing to + ) = 2a + 36. 
 
 Examples. — 
 Subtraction. 4.6290016 5.0986437 4.5641925 
 
 (wimws5-5) 5.3751147 7.4352071 6.6580496 
 
 9.2538869 1.6634366 1 1. 9061429 
 
 (i carried over is subtracted 
 
 .•.5-1 = 6, and 7 + 6=1.) 
 
LOGARITHMS AND LOGARITHMIC TABLES 127 
 
 Multiplication. 2.7460423 
 
 5 
 
 7.7302115(2 X 5 = 10 + 3 carried over =7). 
 
 Division. As the Mantissa must remain positive, the Character- 
 istic must be completely divisible by the divisor, and nothing is 
 to be carried into the Mantissa. If the Characteristic is divisible 
 as it stands, the quotient is written down in the usual way with 
 the negative bar ; if it is not divisible, a negative number is to be 
 added to it to make it so, and to the Mantissa is prefixed a 
 positive integer of equal value, so that the - and + correct each 
 other and leave the value of the logarithm unaffected, and the 
 division is carried out as usual. 
 
 Example. — log 8.1626540-^4 = 2.o4o_6635 
 
 -^ 7 = (8 + 6) + 6. 1626540 H- 7 
 = log 2.8803791. 
 
 To Find the Number from the given Logarithm. — The 
 method is the reverse of the one for finding the logarithm of a 
 number. Look up the Mantissa under the appropriate columns 
 in the Tables — i.e., under the cyphers o to 9 (at the top of the 
 page). If the decimal part is found exactly, the corresponding 
 number is to be read off in the first column (under No.) and the 
 decimal point placed as indicated by the Characteristic of the 
 given logarithm. These integers will be numerically one more 
 than the Characteristic. 
 
 E.g., ".7291648" found under "o" corresponds with the 
 figures 5360 (under No.), but the position of the decimal point 
 and the value of the figures of the number can only be ascer- 
 tained from the Characteristic of the logarithm. 
 
 .-. 0.7291648 = 5.360 (5.36) 
 1.7291648 = 53.6 
 5.7291648 = 536000. 
 
 If the given Mantissa is not found in the Tables, take out the 
 next lower Mantissa and subtract it from the Mantissa of the 
 given logarithm. The difference will be found exactly or 
 approximately in the right-hand column of figures of the Table of 
 Proportional Parts (under " Diff."), and the figure opposite to it 
 is the sixth figure of the required number. 
 
 " If the difference is not exactly found among the proportional 
 parts, take the next lower part, and the figure opposite to it is the 
 sixth figure of the number. 
 
 " Subtract this part from the given difference, annex a cypher to 
 the remainder, consider it as a new proportional part, and find 
 
128 CALCULATIONS IN HYGIENE 
 
 the corresponding figure as before. It will be the seventh figure 
 of the number." 
 
 Example. — To find the number corresponding to the given 
 logarithm 5.9173597- . 
 
 The given Mantissa is not exactly stated in the Tables, therefore 
 taking the next lower and subtracting 
 
 IP' ! 
 
 5-9173597 
 
 5-91735^4 corresponding to the number 826720 
 
 013". = 13 2 \i » ,, » 02 (6th figure) 
 
 10 6 31 
 
 006 (7th figure) 
 
 30 (cypher annexed, the near- 8267226, whichisthe 
 est in the Table of Proper- required number, 
 tional Parts = 31) 
 
 Logarithms are valueless for the Addition and Subtraction of 
 numbers. They are serviceable only for performing multiplica- 
 tion, division, raising to any power, and for extracting any root. 
 The results in most cases are a close approximation, and not 
 absolutely correct unless the numbers are represented by a 
 perfect value in the logarithm. 
 
 In all cases the logarithm of the number must be known. 
 Multiplication of numbers = addition of their logarithms. 
 Division ,, = subtraction ,, „ 
 
 Raising to any power = multiplication of the logarithm of the 
 number by the figure denoting the power to which it is to be 
 raised. 
 
 Extraction of any root = division of the logarithm of the given 
 number by the figure denoting the desired root. 
 
 Thus : X X Y = log X 4- log Y = addition of logarithms. 
 
 Y' 
 
 = logX- 
 
 -log 
 
 Y = 
 
 = subtraction 
 
 M 
 
 
 X" = 
 
 = n times log 
 
 X = 
 
 = multiplication 
 
 of the logi 
 
 iritlim. 
 
 ^x. 
 
 logX 
 
 
 _ 
 
 = division 
 
 »> 
 
 5» 
 
 Processes are thus shortened considerably by the aid of Tables : 
 the logarithms of the numbers are easily found, and vice versd 
 the logarithms being known, the numbers are ascertained. 
 
 It is to be noted that logarithms themselves are not multiplied 
 or divided by each other. 
 
LOGARITHMS AND LOGARITHMIC TABLES 120 
 
 Examples. — Multiplication of numbers 102718X 91627. 
 
 log 102718 = 5.01164655 
 log 91627 = 4.96202350 
 
 9.9736700^ 
 The Characteristic shows there are 10 integers in the number. 
 As the above Mantissa is not found in the Tables, the next loiver 
 is taken out and subtracted from it. 
 9.9736700 
 9.9736681 =log of 941 1 700000. 
 
 Diif. 19 
 
 46 18 .'. 6th figure = 4. 
 
 4 lo 
 
 29 10 7bh ,, =2. 
 
 .•. 9411742000 = Antilogarithm. 
 Division of Numbers. 
 67564 
 
 83619 
 4.8297154 
 4.9223050 
 
 = log 67564 -log 83619. 
 
 1.9074104 = log 0.S08, which is the required decimal. 
 Raising to a Power (Involution). 
 To find the value of (12. 6^ 
 
 = 6 X log 12.6 
 
 = 6 X 1. 1 003 705 = 6.6022230 
 which is the logarithm of the number 4001500. 
 Extraction of the Root (Evolution). 
 To find the value of ^58726 
 
 log C8726 4.7688^04 ^^ 
 
 = _*-5_J ^ILJ. ^— 1^0.11922076 
 
 10x4 10 X 4 
 
 corresponding to the number 1.3159- 
 
CHAPTER XI. 
 
 POPULATION. 
 
 Estimation of Population by Logarithms. — The increase is in 
 (ieometriccil Progression. 
 
 Let P = Population in any given year. 
 r = factor of annual increase. 
 P X r = increase in one year. 
 P X r*'= „ „ two years. 
 Pxr"= „ „ n „ 
 
 The rate of increase or decrease is calculated from the data of 
 the two previous Censuses. 
 
 Tlie assui-ivption is that either has continued at the same rate 
 since the last Census as between the last and the previous Census. 
 
 Registrar- General's Method of Estimating a Population. 
 
 Log Census Population + log Quarterly increase + n times log. 
 Annual increase = log Population at the middle of the n^^ year 
 since the last Census {i.e., the ?^'''' post-censal year). 
 Formula. — By Geometrical Progression : 
 If 2/ = population by the last Census. 
 
 x= „ „ „ previous,, (lo years before). 
 
 Rate of Decennial increase = '- = log ^ - log x. 
 
 Annual 
 
 Vi 
 
 log y - lo, 
 
 lO 
 
 , ^ log y - log X 
 „ Quarterly „ - 1 of ^ \^ ^ ' 
 
 log y - log X 
 ~ 4x10 
 
 The Census population for the middle of the Census-year is, 
 therefore, the actual population on March 31 of that year 2^^}^^ 
 or minus the hypothetical increase or decrease calculated (with 
 
POPULATION 131 
 
 logarithms) by Geometrical Progression from April i to June 30 
 inclusive. It is not the real mid-year population, but only an 
 approximate one. 
 
 Increasing Population. 
 
 Example. — 1881. Population = 462303. 
 1891. „ =505368. 
 
 Estimate the Population in 1898 (mid-year). 
 
 log 462310 = 5.6649333 1 I^iff- 
 
 log 462300 = 5.6649239^ J 94 
 
 3 28/ ' 
 
 log 462303 = 5.6649267 = log Population 1881. 
 
 log 505370 = 5-7036095 1 ^^"^• 
 
 log 505360 = 5.7036009I j ^^ 
 
 8 69/ ^ ^9 
 
 log 505368 = 5.7036078 = log Population 1891. 
 5.6649267= „ „ 1881. 
 
 DifFerence = 0.038681 1 = ,, Decennial increase. 
 YQ = 0.0038681 = ,, Annual ,, 
 
 J = 0.0009670= ,, Quarterly ,, 
 
 log 0.0038681 X 7 = 0.0270767 = ., 1892-8 „ 
 
 5.7036078= ,. Population 1 89 1. 
 0.0009670= ,, Quarterly increase. 
 
 5.7316515= „ Population 1898. 
 539070 = 5. 7316452 
 
 539077 63 BifiF. 
 
 57 
 
 7 57 
 
 539077.7 60 
 
 or 539078 = population for 1898. 
 
 Example. — Population 1891 = 531247 
 „ 1901=985476 
 
 To find the mid-year Population for 1907. 
 
 log 531250 = 5.7252989 1 ^^f 
 
 log 531240 = 5.72529081) ^ ^^ 
 
 ii 
 
 7 57}' 
 
 log 531247 = 5.7252965 =log Population 1891. 
 
 I 
 
132 CALCULATIONS IN HYGIENE 
 
 10-985480 = 5.9936478 \ Diff. 
 
 log 985470 = :> 9936434 \ ) 6 '^'^-6 
 
 6 26) 
 
 log 985476 = 5.9936460= log Population 1901. 
 
 5.72 52965= „ „ 1891. 
 
 Difierence 0.2683495 = ,, Decennial increase. 
 
 -1^ = 0.0268349= ,, Annual ,, 
 
 1 = 0.0067087= ,, Quarterly ,, 
 
 log 0.0268349 X 6 = 0.1610094= ,, 1902-7 ,, 
 
 5.9936460= ,, Population 1891. 
 
 0.0067087= ,, Quarterly incrense. 
 
 6.1613641 = ,, Population 1907. 
 1449900 = 6.1613380 
 
 Q 7^. Diff. 
 
 O 201 o 
 
 8 240 
 I 240 I 30 
 
 „ I Population 
 1449981= ^f^/ 07. 
 
 21 
 
 907. 
 
 Decreasing Population. 
 
 Example. — A population of 552508 in 1891 was fouid to have 
 decreased in 1901 to 517980. 
 
 To calculate the population in 1906 on the hypothesis that it 
 will decrease at the same rate : 
 
 log 552510 = 5-7423401 I ^^f' 
 
 log 552500 = 5.7423323) j 7« 
 
 8 62_) ^ ^^ 
 
 log 552508 = 5.7423385 = log Population 1891. 
 log 517980 = 5.7143130 = „ „ 1901. 
 
 Difierence 0.0280255 = „ Decennial decrease. 
 Yy = 0.0028025 = ,, Annual ,, 
 
 ;^ = 0.0007006^ = „ Quarterly „ 
 
 0.0147 1 31 l^g l^otal decrease 
 to middle of 1906. 
 log 0,0028025 X 5 -=^ 0.0140125/ =log Decrease from 1901-6. 
 
 5-7143130 
 >S ubtracting : o . o 1 4 7 1 3 1 
 
 5.6995999 = log Population for 1906. 
 5.6995 949= ., 500720 
 
 Diir. 
 
 50 5 5 44 
 
 44 7 7 61 
 
 60 500725-7 
 
 or 500726 = Population for 1906. 
 
POPULATION ]?,?, 
 
 Estimation of a Population by Arithmetical Progression. 
 Example. — 1901. Pojulation = 50742 
 1891. ^ „ =47256 
 
 To estimate the Popul. iii 1 908 : 
 
 3486 = Decennial increase. 
 tV= 348.6 = Annual 
 1= 87.15 = Quarterly ,, 
 348.6x7 =2440.20=1902-08 „ 
 
 T901 Population + Quarterly Increase + 7 times Annual Increase 
 
 = 53269.35 = Mid-y ear Population for 1908. 
 Working the above hy Geometrical Progression and by Log- 
 arithms : 
 
 log 50742 = 4.7053676 log Population of 1901. 
 log 47256 = 4.6744570 „ „ „ 1891. 
 
 0.0309106= ,, Decennial increase. 
 Yo =0-0030910= ,, Annual ,, 
 
 1 = 0.0007727= ,, Quarterly ., 
 log Annual Increase X 7 = 0.0216370= ,, 1902-8 ,, 
 
 47053676= „ Population of 1 90 1. 
 
 Adding the last three : 4-7277773= ,, ,, ,, 1908. 
 
 Corresponding Number = 53429.00 = Population of 1908 
 
 53269.35 „ as above by A.P. 
 
 Disparity = 159.65 
 
 (This is a small difference, because the increase is a slow one in 
 a comparatively small Population. In such a case A.P. is 
 applicable with fairly accurate results. It is inadmissible for a 
 large Population.) 
 
 Estimation of Population from the Birth-rate (per 1000 
 living). — The method suggested by Dr. Newsholme is useful for 
 checking the estimate of a '• present " population. It is assumed 
 that the birth-rate remains for some years the same as it was 
 when the last Census was taken. 
 
 Example. — Birth-rate (per 1000 living) 1 892-1 901 inclusive 
 
 = 30.2 Births during 1902 = 4678 
 
 30.2 : 4678 : : 1000 : .r= 154900. 
 
 . Mean Annual Population 
 
 Weekly population = 
 
 ^ ^ ^ 52.177 
 
 ^ ., Mean Annual Population 
 
 Daily „ = 7 — 
 
 ^ " 365-24 
 
134 CALCULATIONS IN HYGIENE 
 
 Marriage-rate. — It is calculated by Simple Proportion on the 
 actual population, and is expressed per looo living at all ages. 
 
 J'Lg. Population in 1899 = 18426. 
 Marriages ,, „ = 284. 
 18426 : 1000 :: 284 : .r= 15.4 per 1000 living. 
 
 Birth-rate.* — General Formula : 
 
 Mean Annual Population: 1000 : : Annual Births : x. 
 Crude Birth-rate : per 1000 of estimated (mid year) population 
 at all ayes. 
 
 Annual Birth-rate : Mid-year population = 29542. 
 
 Births registered during the year = 865 . 
 29542 : 1000 : : 865 : .'« = 29.2 per 1000 living. 
 Quarterly Birth-rate : Taking the same population. 
 
 Births registered during quarter in 
 question = 186. 
 (i) 29542 : 1000 : •. 186 : .« = 6.3 
 
 and (2) 6.3 X 4 = 25.2 per 1000 = Quarterly Birth-rate. 
 [Note. — The result denotes what the Quarterly rate would be 
 per annum if it went on at the same rate for one whole year per 
 1000 living.] 
 
 Weekly Birth-rate : A\^. Population = 28530. 
 
 Births during week in question =19. 
 Weeks in the year = 52.177. 
 (i) 28530 : 1000 : : 19 : .'« = o.67. 
 (2) 0.67 X 52.177 = 34.96 per 1000 living. 
 
 The Birth-rate is preferably calculated on the female population 
 at the child-hearing age, per 1000 married and per 1000 unmarried. 
 Death-rates. — The crude (general or gross) death-rate is that 
 of the mid-year population (at all ages) per 1000 living. 
 
 E.g. Population = 18500. Deaths during the year = 206. 
 Annual Death-rate : 18500 : 1000 : 206 : x= ii.i per 1000. 
 For the same population, deaths during a particular week = 7. 
 Annual Death-rate /or that week : 
 
 (i) 18500 : 1000 : : 7 : .^ = 0.38. 
 (2) 0.38 X 52.177 = 19.8 per 1000. 
 A Weekly Death-rate estimates the number who would die per 
 aiviium per 1000 living if the death-rate of that week continued 
 at the same rate throughout one year. A Quarterly Death-rate 
 is estimated in a similar way. 
 
 * Birth-rate, Deatli-rate at all ages, and the Net Deatli-rate (columns 4, 
 8, and 13 of the M.O.H.'s Vital Statistics Tables) are calculated per 1000 
 of estimated population. 
 
POPULATION 135 
 
 " Corrected Death-rate " of the Registrar- General. 
 
 The correction is made by multiplying the local Recorded Death- 
 rate of the Town (or crude death-rate) by the factor supplied to 
 it annually by the Registrar-Gonerrvl. It neutralises errors in 
 death-rates caused by the disparity of age- and sex-distribution, 
 and raises or lowers the local crude death-rate to what it would 
 be if the age- and sex- distribution of the town were the same as 
 for England and Wales generally. The same method is carried 
 out throughout Great Britain. 
 
 To obtain the Registrar- General's "Factor." 
 
 1. A local stcmdard death-rate is calculated for each town. 
 The local distribution of ages and sexes is obtained from the last 
 Census. To this local 2iopulation is applied the annual Death- 
 rate of England and Wales for the previous lo years {i.e., as if the 
 people had died at the same anjiual rate as for England and 
 Wales during the last lo-year interval, and not at the local death- 
 rate). 
 
 2. The annual recorded death-rate (at all ages) for England 
 and Wales for the previous Decennium divided by this local 
 standard death-rate gives the " factor '' for that town : — 
 
 "I Recorded Death-rate of England and WaVs 
 Registrar-General's | _ during previous Decennium 
 
 factor for the year j Standard Death-rate of the 
 
 j town for the year 
 
 Thus: Annual Death-rate for England and Wales from 1891-90 
 (inclusive) = 19.15 per 1000. 
 
 Standard Death-rate \ ^ , ^-^l^-^ 1.0656 Factor for 1899. 
 
 of London (1899) J '^' 17.97 ^ ^^ 
 
 Standard Death-rate 1 _ _ . 1915 _ o 
 
 of Liverpool (1899)/ ~ '^''^^ • • iy.44- i-°9&o „ „ „ 
 
 Standard Death-rate ) _ 19^5 _ 
 
 of Plymouth (1899)/- '9-7 "ig^ -°-972o o » .. 
 
 3. The " recorded death-rate for the town " {i.e., the crude death- 
 rate) X the " factor " = " corrected death-rate.'^ 
 
 Taking the above-named cities : 
 
 Recorded (crude) Death-rate of London (1899) 
 
 = 19.78 X 1.0656 = 2 1.077 = "Corrected Death-rate." 
 „ Death-rate of Liverpool (1899) 
 
 = 26.38 X 1.098 = 28.965 = " Corrected „ „ 
 
 „ Death-rate of Plymouth (1899) 
 
 = 21.72 X 0.972 = 21.111 = " Corrected „ „ 
 
 Correction for Non-residents. 
 
 Deaths of Residents of the District dying in Public Institutions 
 
J 
 
 vy 
 
 13G CALCULATIONS IN HYGIENE 
 
 are added to the Returns, ( + ), and those of Non-residents are 
 subtracted ( - ). In private cases this is not done, as it is 
 impracticable. 
 
 4. The Comparative Mortality Figure, 
 llecorded Death-rate of England and Wales for the i/ear in 
 question : 1000 : : corrected local death-rate : x. 
 
 Corrected local death-rate _ Comparative Mor- 
 
 ' Death-rate for the whole country ~ tality Figure. 
 
 Taking the same cities as before : 
 
 Recorded Death-rate of England and Wales during 1899 = 18.33. 
 
 Corrected Death-rates as already calculated : 
 London: 18.33 : 1000 : : 21.077 • a; =1150. CM. Figure, 1899. 
 Liverpool: ,, : ,, : : 28.965 : a;=i58o. ,, ,, 
 
 Plymouth: ,, : „ : : 21.1 :«;=ii52. „ „ 
 
 Infantile Mortality is estimated on the annual number of 
 registered deaths of children under one year of age per 1000 
 registered births during t)ie same year : not on the total 2)02)ulation, 
 nor on the total number of deaths at all ages. (Still-born births 
 are not registered.) 
 
 Formula. — Births during the year : 1000 : : Deaths of children 
 under i year : x. 
 
 E.y. Births registered during the year = 2372. 
 
 Deaths ,, ,, ,, ,, under i year of age = 284. 
 
 2372 : 1000 : : 284 : «;. 119 deaths per 1000 births = infant 
 mortality for the year in question. 
 
 Zymotic Death-rate. — It may be for the entire group of 
 infectious diseases or for each one in particular, and states the 
 proportion of notified cases per 1000 of population. 
 
 E.g. Population = 2 1685. Deaths from diphtheria =12. 
 
 21685 : 1000 : : 12 : x. =0.55 per 1000 living. 
 
 Proportion of Deaths from Special Diseases to Total 
 Deaths from all Causes. 
 Total deaths = 6018. 
 Deaths from Principal Zymotic diseases = 41 1. 
 
 ,, ,, Small-pox = 72. 
 6018 : 1000 : : 411 : X. 68.3 per 1000 deaths (Zymotic). 
 „ : : 72 : ,r. 11.9 „ „ „ (Small-pox). 
 
 f Incidence of Disease. — Proportion of cases per 1000 of 
 population. 
 
 E.y. Population = 765720. Scarlet fever cases = 2408. 
 765720 : Tooo : : 2408 :./•-= 3.14 per 1000 ( = Zymotic case-rate). 
 
POPULATION 
 
 C as 3 -Mortality. — Proportion of deaths per loo, or per 
 cases. 
 
 E.g. Of 2408 cases of scarlet fever 32 were fatal. 
 2408 : 100 : : 32 : x. Case-mortality = 1.3 per cent. 
 
 Population. Number of cases. Incidence of Disease. 
 
 187 
 
 [OOO. 
 
 v/ 
 
 1898. 52630 
 
 1899. 54748 
 
 1900. 58676 
 
 1901. 64526 
 
 1902. 67325 
 
 241 
 
 258 
 
 234 
 297 
 340 
 
 4.6 per 1000 of estimated populati 
 
 4.7 11 3? " " 
 
 = 3-9 '> 
 = 4.6 ,, 
 
 = 5-o '5 
 
 297905 1370 
 
 Above the Mean 
 
 Below 
 
 Fig. 28, 
 
 Dividing by 5 : 59581 = average pop. per ann. for 5 yrs. (by A.P.) 
 
 274= ,, no. of cases „ „ 
 
 Average incidence = 4.6 per 1000 of average pop. ,, 
 
 During 1898 and 1901 the Incidence of Disease coincided with 
 the average, although in 190 1 there were 56 more cases than in 
 1898 ; these were counterbalanced by an increase of population. 
 
 In 1899 and 1902 it was in excess of the mean : m the first 
 instanceby 4.7— 4.6 = 0.1 per 1000, or o.oi per cent. ; and in the 
 second by* 5.0— 4-6 = 0.4 per 1000, or 0.04 per cent. 
 
 In 1900 there were only 7 fewer ca^es than in 1898, and the 
 
138 CALCULATIONS IN HYGIENE 
 
 proportion of attacks was 4.6 — 3.9 or 0.07 per cent, below the 
 average. 
 
 Combined Death-rates. — If the populations are numerically 
 equal, the respective rates of mortality per 1000 are added 
 together and divided by the number of towns in question. 
 
 Three towns each containing 25000 inhabitants have death-rates 
 of 22, 35, and 31.5 per 1000 respectively. 
 
 Mean or Combined Death-rate = — '- = 29.5 per 1000. 
 
 If the populations are diflferent their proportion to each other 
 must be estimated : 
 
 Population 19000 at 22 deaths per 1000 = 418 deaths. 
 
 „ 22000 „ 35 „ „ „ =770 „ 
 „ 28000 „ 31.5 „ „ „ =882 „ 
 
 69000 : 1000 : : 2070 : x 
 
 Combined Death-rate = 30.0 per 1000. 
 Density of Population is important in connexion with the 
 Death-rate of the same area, which it influences. 
 
 ( 1 ) Unit of area = i square mile. 
 
 Total number of square miles : i : : total population : x. 
 
 1 ^. M Population 
 Mean population per sq. mile = ^q ; -^ — 
 
 25000 
 Pop. = 25000. Area = 94 sq. m. Density = = 266 persons 
 
 ^4 per sq. m. 
 
 (2) Unit of area= i acre. 
 
 Population : i : : total number of acres : x. 
 
 Total acres 
 
 Mean area per person = ^5 ^r^. — 
 
 ^ ^ Population 
 
 Taking the above example : 94 sq. ms. = 94 x 64b = 60 1 60 acres. 
 
 60160 
 Density = = -4 acres per person. 
 
CHAPTER Xir. 
 
 LIFE-TABLES. 
 
 Data required : i. Census Returns to ascertain («) the mean 
 population, (b) numbers living at each age-period. 
 
 2. Death-returns showing : the mean annual number of deaths 
 for the corresponding age-periods. 
 
 A. Mortality per unit for each year or each age-period 
 
 Deaths (per annum or per age-period) 
 '~ Mean population (during same period) 
 (In a Life-Table = " Annual Mortality '' per unit at age x. " D," 
 or " M.;'.) 
 
 B. Mortality per looo living for each year or each period : 
 Mean Popul. at age-period : looo : : Deaths (at same period) : x. 
 
 _ Deaths (per annum or per age-period) 
 Mean population (during same period) 
 Let this--D. Assuming D to be equally distributed through- 
 out the year or age-period : Rate during ist half-period = — 
 
 1 -5 
 
 11 11 ^nd ,, „ — 
 
 .*. lOOO living (survivors) in the middle of the year or age-period 
 numbered I ooo-l- — beginning „ ist half ,, ,, ,, 
 
 and looo ending „ 2nd ., ,, ,, ,, 
 
 .-. the " Ratio of final to initial population " 
 
 D ^ 
 
 1000 -— 2000 — D 
 _ Su rvivors at the end ^ £ ^ 
 
 ^Survivors at the beginning j^^^^i> 2000 + D 
 
 2 
 (In Life-Tables this calculation give, the " Probability of Living 
 one year from each age" denoted by the sign ';?.v'') 
 
UO CALCULATIONS IN HYGIENE 
 
 Example. — During ist year infant mortality (per looo births) 
 
 = 130. 
 
 1000 : I million : : 130 : .'■= 130,000 deaths. 
 
 .'. I million children after i year (at beginning of next year) 
 
 = I million - 130,000 = 870,000 survivors at end of ist year. 
 
 During 2nd year infant mortality = 100 (per 1000 births). 
 
 ( = /> of the Formula) 
 
 .*. By Formula : the probability of each survivor living through 
 
 . 2000 - 100 1900 
 
 one year is , = ( = o.noi^) 
 
 '^ 2000+100 2100'' ^ •^' 
 
 .'. 870,000 X- =787-^So survivors at the end of the 2nd 
 
 ' ' 2100 ' '^-^ 
 
 year. 
 
 (In Life-Tables found under the heading "Number born and 
 living at each age," and denoted by /^..) 
 
 Infant mortality = 50 per 1000 during 3rd year : 
 
 2000- Ko 1950 , , 
 
 ^=-^^ ( = 0.9512) 
 
 2000 + 50 2050 ^ ^^ ^ 
 
 •*• 787350 ^ — = 748943 survivors at the end of the 3rd year. 
 
 Similarly till none survive. 
 / If instead of annual periods 5-year ones are taken (quin- 
 ,/ quennia) the method of working is similar, but the Formula is 
 raised to the 5th power. 
 
 Let Pq = survivors at commencement of quinquennium. 
 
 ,2000 - D 
 P, = P. X 
 
 1 "0'' \2000 + Dy 
 
 Example. — Of the above survivors, supposing 675,000 were 
 
 living at the end of the 5th year : 
 
 And Death-rate for the 5-10 quinquennium = 6 per 1000. 
 
 2000-6 
 Formula for one year = -— 
 
 ^ 2000 + 6 
 
 5 years 
 
 / 2000-6 Y ^ / £9^94 y' 
 
 \2000 + 6/ \2oo6 / 
 
 = 5x(log 1 994 -log 2006) 
 
 = 5x(3 2997252-330^3309) 
 
 = 5 X 1-9973943= i-98^^97i5==o.9704 
 675000 X 0.9 7 04 = 655020 survivors. 
 
 The calculation is similarly repeated for each quinquennium 
 until there are no surviv^ors. 
 
LIFE-TABLES 
 
 For a decennial age-period the death-rate per looo is 
 
 141 
 
 2000 - 1) 
 2000-f D 
 
 Similarly : P, = R x P°°^ " ^ Y" (substituting decennium for 
 •^ ^ " \2O0o + DJ quinquennium) 
 
 Expectation of Life. — The average number of years a person 
 of a given age is likely to live as calculated in a Life-Tahle. 
 
 Q " 
 
 Expressed symbolically as " E^ =~ 
 
 " Q^ = Sum of years of life lived at age x and upwards 
 1^ = Number of survivors at each age." 
 
 Or: 
 
 Expectation of Life at Age x 
 
 Sum of total survivors after age x x ^ 
 Survivors at age x 
 
 (2.5 = Half-quinquennium). 
 
 -H2.5. 
 
 Example. — To Calculate the Expectation of Life for 
 Males at 40 Years of Age. 
 
 ^t Age 
 
 Male Survivors 
 
 40 
 
 604923* 
 
 45 
 
 564437^ 
 
 
 50 
 
 517639 
 
 
 55 
 
 462981 
 
 
 60 
 
 398400 
 
 
 65 
 
 322482 
 
 
 70 
 75 
 
 238632 
 153890 
 
 ^=2775918 
 
 5 
 
 80 
 
 80023 
 
 85 
 
 29866 
 
 13879590 
 
 90 
 
 6786 
 
 
 95 
 
 752 
 
 
 100 
 
 3o> 
 
 
 13879590 , ^ ^ 
 
 — -yr, p,4 4-'7C = 7C/11 VPni'S 
 
 y 
 
 604923 
 
 Tatham's English Life-Tables (1881-90). 
 
 ■<:%. 
 
142 
 
 CALCULATIONS IN HYGIENE 
 
 For 
 
 a Female-life at the same age : 
 
 At Age 
 40 
 
 45 
 50 
 55 
 60 
 
 65 
 70 
 
 75 
 80 
 
 85 
 90 
 
 95 
 
 TOO 
 
 I 6048 I 60 
 638912 
 
 Female Survivors 
 638912 
 604007' 
 564299 
 
 516375 
 457682 
 
 385503 
 
 299220 
 
 204208 
 
 II4536 
 
 48133 
 
 13418 
 
 2124 
 
 157 
 
 + 2.5 = 25.11 + 2.5 = 27.61 years. 
 
 3209632 X5 
 = 1 6048 1 60 
 
 111 the above Examples 22.94 and 25.11 are known as the 
 " Curtate Expectation of Life ; " and 2.5 as the " Duration of Life 
 in the Quinquennium of Death " ( = half a quinquennium or 2 J 
 years). These added together give the " Complete Expectation 
 of Life," as stated in Life-Tables under the heading " Mean after 
 Lifetime at each age x = 'E^." In Tables giving Annual, and not 
 Quinquennial age-periods, the number of survivors is added as 
 above (the result is not multiplied by 5) to obtain the Curtate 
 Expectation, and 0.5 is added (in place of 2.5) to get the Complete 
 Expectation of Life. 
 
 " Mean Duration of Life " = The Expectation of Life at Birth, 
 or at "Age zero." 
 
 By Farr's Formula, the expectation of life at birth per 1 000 
 livins: : 
 
 Example. 
 
 I 1000 2 1000 
 
 3 Birth-rate 3 Death-rate 
 
 —Birth-rate per 1000 living 
 Death 
 
 1000 
 
 55 
 2000 
 
 2000 
 
 30-5 
 17-3 
 
 3x30.5 3x17, 
 
 91-5 51-9 '^ ^ 
 
 = 49.46 years, 
 the expectation of life at any later 
 
 " Mean after lifetime 
 age than at birth. 
 
 By Willich's Formula : the expectation of life at cmi/ aye 
 between 25 and 75 years = - (80 
 
 Age 
 
LIFE-TABLES 148 
 
 Example. — Expectation of life at 45 = - (80-45 )• 
 
 = 23.3 years. 
 " Probable Duration of Life " : the age at which half a given 
 number of children (born hypothetically at the same time) will 
 have died. 
 
 ^ , „ Sum of the Ages at Death 
 '•Mean Age at Death : Number of Deaths 
 
 Poisson's Formula for estimating the liability to error. 
 
 ist series of observations = m^ 7n + n = fj.^ Total number of 
 
 2nd „ ,, „ =nj observations. 
 
 m 
 Probability of ?/i series being constant = — 
 
 71 
 
 r 
 
 True proportion of )n to \i lies between : — ± 2^ — rr- 
 
 i.e., within a possible range of 4^/ — rr- 
 
 ■J 
 
 ^271171 
 
 Similarly for the ?i series the true proportion of n to /^i lies 
 
 , 71 / -127)171 
 
 between - ±\/ — r~- 
 
 fl V f.i 
 
 Example. — Of 500 cases, 425 recovered and 75 died. 
 
 425 
 Probability of recovery = — = 85 per cent. 
 
 „ „ death =— - = 15 „ „ 
 
 13 -ui f 732x4 25x85 
 Possible 7'a7iye or error = w 7 \3 • 
 
 ^ IOC 
 
 or 9.62 per cent. 
 
 / 32 X 17 X 17 v/38.08 
 ~ ^ 100 X 100 X 100 1000 
 9.62 
 
 100 
 
 9*62 
 
 .•. Probability of recovery varies between 85 ± — 
 
 or ,, 89.81 and 80.19 percent. 
 9.62 
 „ „ death „ „ 15^-7" 
 
 „ ,, 19.8 and 10.2 ,, 
 
J 
 
 APPENDIX 
 
 Humidity (p. 32). 
 
 Example. — Relative humidity at 60" F. = 70 per cent. 
 By Tables : 60° F. = 5.8 grains per cubic ft. of aqueous vapour. 
 Absolute humidity : 100 : 70 : : 5.8 : ;>j = 4.o6 grains per cb. ft. 
 Drying-power: 5.8-4.06=1.74 „ „ 
 
 Case-Mortality (p. 137). 
 
 Example. — A Hospital contains 500 beds, \ of which are con- 
 stantly occupied. 
 
 Average period in Hospital per patient = 3 weeks. 
 „ number of deaths per annum = 142. 
 
 Supposing the cases to be uniformly distributed over the whole- 
 period, what is : 
 
 (rt) the death-rate per bed, 
 
 (6) ,, ,, ,, 1 00 cases admitted ? 
 
 Beds constantly occupied = | of 500 = 375, 
 Death-rate per bed „ = -^ = ^.38. 
 
 As each case remains 3 weeks, in one year there are : 
 
 52 
 
 — =17.3 patients per bed. 
 
 375 >^ 17-3 = 6487 cases admitted, of which 142 die. 
 
 .•. 6487 : 100 : : 142 : a; = 2.2 deaths per cent, of admissions. 
 
INDEX 
 
 Absorption of Gases, i6 
 
 Adam's process, in 
 
 Air : Velocity of Inflow and Outflow, 51, 60 ; Percentage Composition 
 
 (of Pure and Expired), 51 ; Supply of fresh, 53; for Horses and 
 
 Cattle, 57 ; COo in, 71-75 
 Alcohol: 117; Proof-, Over- and Under-proof Spirit, 118; Mulder's 
 
 method, 119 
 Ammonia : Free and Albuminoid, 88-91 
 Anemometers : Robinson s, 49 ; Casella's, 65 
 Apjohn's Formula, 3$, 36 
 Avogadro's law, 4, 5 
 
 Barometers : 37 ; Corrections for, 38-43 
 Baroscope, 37 
 Beer: Acidity of, 119 
 Birth-rates, 134 
 Blackwell's Formula, 80 
 Boyle's and Mariotte's law, 10 
 
 Butter: Moisture; Volatile and Fixed Acids, 115 ; Specific Gravity 
 of Fat in, 116 
 
 Calories, 76 
 
 Case-Mortality, 137 
 
 Charles' and Gay-Lussac's law, 7 
 
 Chlorine, 85, 86 
 
 Coefficient of : Absorption, 16 ; Expansion, 8 ; Solubility of Gases, 17 
 
 Combined Death-rates, 138 
 
 Comparative Mortality Figure, 136 
 
 Correction of Volume for : Pressure, 10 ; Temperature 7 ; Temperature 
 
 and Pressure. 10 
 Curtate Expectation of Life, 142 
 
 Dalton's and Henry's law, 16 
 Death-rates, 134. 135 
 
UQ INDEX 
 
 Density : Absolute :ind Relative, 3 : under Pres.sure, ii ; of Population, 
 
 138 
 
 Dew-point, 30, 32, 33, 34 
 Diets : 105. Calculations of, 106, 107 
 Division by Logarithms, 129 
 Dulong's and Petit's law, 76 
 
 Efflux of Liciuids, 82, 83 
 
 Energy: 107, 108 ; de Chaumont's Forniuhe, 108; in Calories, 109 ; in 
 
 Foot-tons, log, no 
 Equilibrium of Balance, i 
 Evolution by Logarithms, 129 
 Expectation of Life, 141 
 Eytelwein's Formula, 99 
 
 Fare's Formula, 142 
 
 Foods : Composition and Value of, 105-109 ; Examination of, 111-120 
 
 GlaIvSHEr's Formula, 34 ; Factors, 35 
 Gramme-Molecule, 5 
 
 Hardness, 86-88 
 
 Hare's apparatus, 22 
 
 Hawksley's Formula, 77, 79 
 
 Heat : Latent, 77 ; Specific, 76 ; Unit of, 76 
 
 Height by Barometer. 44 
 
 Koppe-Seyler's process, 112 
 
 Humidity : Absolute, 30 ; Relative, 31 
 
 Hydraulic : Mean Depth, 99 ; Press, 84 ; Uani, 81 
 
 Hydrogen : Weight of, 4, 5 ; Eciuivaleut, 17 
 
 Hydrometers : 20, 24, 25 
 
 Hygrometers : 31-33 
 
 Incidence of Disease, 136 
 
 Infantile Mortality, 136 
 
 Inlets and Outlets : de Chaunionfs Formula, 59 ; Size and Shape 
 
 of, 61 
 Involution by Logarithms, 129 
 
 James' Formula, 50 
 
 Kjeldahl's process, 117 
 
 Lawford's Formuko, 102 
 
INDEX 147 
 
 Life Tables, 139 
 Lojiarithmic Tables, 122 
 Logarithms, 1 21-129 
 
 Magnesia (in water) , 88 
 
 Maguire's Formula, 99 
 
 Marriage-rate, 134 
 
 Mass, 3 
 
 Mean Duration of Life, 142 
 
 Meteorology, 26 
 
 Milk: Chemical calculations, 111-115; Fat abstracted and Water 
 
 added, 115 
 Mixtures of Gases and Vapours, 1 5 
 Montgolfier's law, 58 
 Mulder's method, 119 
 Multiplication by Logarithms, 129 
 
 Nitrites and Nitrates, 93, 94 
 
 Normal : Pressure and Temperature, 4 ; Solutions, 17 
 
 OxiDISABLE Organic Matter, 92 
 Oxygen, dissolved, 91, 92 
 
 Pascal's law, 84 
 
 Phenol-sulphonic method, 94 
 
 Poisson's Formula, I43 
 
 Pole's Formula, 79 
 
 Population : Estimation of, 130-134 
 
 Pressure : Correction for, 10 
 
 Kadiation : Solar and Terrestrial, 48 
 Ptain-gauges : 48 ; Graduation of, 48 
 Registrar-General's Method, 130; Factor, 135 
 Respiration, 51 
 
 Richmond's Formula, 113; Slide-Scale, 114 
 Ritthausen's Method, 113 
 
 Solutions : Normal, 17 ; Factor for, 17 ; Standard, 95 
 Space : Superficial and Cubic, 66-70 ; in Hospitals, 68 
 Specific Gravity of Solids and Liquids, 19-25 
 Sprengel's tube, 22 
 Strachan's Formula, 44 
 Symon's Formula, 78 
 Syphon, the, 83 
 
148 INDEX 
 
 Temperature : Scales, 2 : Correction for. 7, 10 ; Record of. 47 
 
 Tension of A(iueous Vapour, 26-29 
 
 Thresh's method. 91 
 
 Tidy's process, 92 
 
 Torricelli's Theorem. 82 
 
 Total Solids in : Milk, iii ; Water, 85 
 
 Units of : Heat, 76 : Volume and Weight, 4 
 
 Ventilation : Artificial, 63 ; Carnelly's and de Chaumont'; 
 
 Formulne, 55, 57 ; Friction in, 63 ; Natural, 57, 61 
 Vernier, the 45, 47 
 
 Water : Area of receiving surface, 78 ; Chemical calculations, 85- 
 95 ; Head of. 80 ; Rain-, 77 : Supply of, by Stream. &c., 77-81 
 
 Weight : 3 ; of Air, 5 ; of Aqueous Vapour, 5. 26, 28 ; of Gases, 14 
 Gramme-molecular, 5 ; Molecular, 6 
 
 Werner-Schmidt process, 112 
 
 Westphal's Balance, 23, 24 
 
 Wetted Perimeter, 99 
 
 Willich's Formula, 142 
 
 Winckler's method, 92 
 
 Wind : Pressure, 50 ; Velocity. 49 
 
 Zymotic Death-rate, 136 
 
 Printed by IJai.i.antynr, Hanson &^ Co. 
 London &^ Edinburgh 
 
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 the ti^e-a-tdveezstt oif 
 
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 General Characteristics of Urine. — Inorganic Constituents. — Organic 
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 Matter. — Bile Pigments. — Bile Acids. — Adventitious Pigmentary and 
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 MAMMALIAN DESCENT: Man's Place in Nature, Being 
 
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 of the Respiratory System — of the Renal System — of the Digestive System — 
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 Diseases — Constitutional Diseases — General Information and Tables Useful 
 to the Practitioner— Rules for Prescribing — Prescriptions — Appendix. 
 
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 107 
 
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102 aSARLES ORIFFIN S 00/8 PUBLICATIONS 
 
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 Fourth Edition, Thoroughly Revised and largely Re-written. In Extra 
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 Being a Manual for the Physiological Laboratory, including- 
 
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 PART I.— CHEMICAL PHYSIOLOGY. 
 
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 " The volume proceeds from a master in his craft. . . . We can confidently re- 
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 — Central Nervous System — Reproduction — Chemistry of the Body. 
 
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 A ZOOLOGICAL POCKET-BOOK; Or, 
 
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 AN INTRODUCTION TO THE STUDY OF 
 
 Ijy^ I ID "V7" I IT- E I^ IT. 
 
 For the Use of Young Practitioners, Students, 
 and Midwives. 
 
 BY 
 
 ARCHIBALD DONALD, M.A., M.D., C.M.Edin., 
 
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 DISEASES OF WOMEN. 
 
 A CONCISE HANDBOOK FOR STUDENTS. 
 
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 "Dr. Phillips' Manual is written in a succinct style. He rightly lays stress on 
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 throughout in his views on therapeutics. He supplies an excellent series^of simple but 
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 youmal. 
 
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 \* Formerly published under the title of " PHARMACY AND 
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 AM» JLM 
 
 Appendix on the Management of the Sick room, and many H!nU tor tb0 
 Diet and Comfort of Invalldfl. 
 
 In its new Form, Db. Spknckr Thomson's "Diotionart ow Dommtio Midioini 
 fully augtaina its reputation as the 'Representative Book of the Medloal Knovl«d«r« 
 and Practice of the Day " applied to Domestic Eequirementa. 
 
 The most recent Improvements in the Treatment of the Sick — in Appliances 
 for the Relief of Pain — and in all matters connected with Sanitation, Hygiene, anc 
 the Maiwtenancb of the General Hbalth — will be fonnd in the Present Issue in oloai 
 and full detail ; the experience of the Editors in the Spheres of Private Practice, of 
 Hospital Treatment, of Sanitary Supervision, and of Life in the Tropics respectively, 
 eombining to render the Dictionary perhaps the most thoroughtly practical work of the 
 kind in the English Language. Many new Engravings have been included — improved 
 Diagrams of different parts of the Human Body, and Illustr tions of the newest M edical, 
 Surjjical, and Sanitary A pparatun. 
 
 *^* All Directions given in such a form as to he readUy and safely followed. 
 
 FROM THE AUTHOR'S PREFATORY ADDRESS. 
 
 " Without entering upon tliat difficul ground which correct professional knowledge and educated 
 Judgment can alone permit to be safely trodden, there is a wide and extensive field for exertion, and for 
 usefulness, open to the unprofessional, in the kindly oflices of a true DOMESTIC MEDICINE, the timely 
 help and solace of a simple HOUSEHOLD SURGERY, or, better still, in the watchful care more gener- 
 ally known as ' SANITAR,Y PRECAUTION,' which tends rather to preserve health than to cure disease. 
 'The touch of a gentle hand' will not be less gentle because guided by knowledge, nor will the safe 
 domestic remedies be less anxiously or carefully administered. Life may be saved, suffering may alwayj 
 be alleviated. Even to the resident in the midst of civilisation, the ' KNOWLEDGE IS POWER,' to do 
 (X>od ; to the settler and emigrant it is INVALUABLE." 
 
 "Dr. Thomsoa has fully succeeded 'u conveying to the public a vast amount of useful protesiional 
 knowledge." — Dublin Journal of Medica>, Sci«nce. 
 
 "The amount of useful kuowledge conveyed in this Work is surprising. " — Medical Timet and GazetU. 
 ' Worth its weight in qold to families and the clergy."— 0«/brd Herald. 
 
 LONDON: CHARLES GRIFFIN & CO., LIMITLD, EXETER STREEI, STRAND. 
 
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