Part 13
───────────────────────────────────────────────────────────────────────── EQUIVALENTS OF POUNDS AVOIRDUPOIS. ─┬───────┬────────────┬────────────────┬────────────────┬──────────────── │ 10 │ 100 │ 1000 │ 10,000 │ 100,000 ─┼───────┼────────────┼────────────────┼────────────────┼──────────────── │qr. lb.│cwt. qr. lb.│ton cwt. qr. lb.│ton cwt. qr. lb.│ton cwt. qr. lb. 1│ 0 10│ 0 3 16│ 0 8 3 20│ 4 9 1 4│ 44 12 3 12 2│ 0 20│ 1 3 4│ 0 17 3 12│ 8 18 2 8│ 89 5 2 24 3│ 1 2│ 2 2 20│ 1 6 3 4│ 13 7 3 12│133 18 2 8 4│ 1 12│ 3 2 8│ 1 15 2 24│ 17 17 0 16│178 11 1 20 5│ 1 22│ 4 1 24│ 2 4 2 16│ 22 6 1 20│223 4 1 4 6│ 2 4│ 5 1 12│ 2 13 2 8│ 26 15 2 24│267 17 0 16 7│ 2 14│ 6 1 0│ 3 2 2 0│ 31 5 0 0│312 10 0 0 8│ 2 24│ 7 0 16│ 3 11 1 20│ 35 14 1 4│357 2 3 12 9│ 3 6│ 8 0 4│ 4 0 1 12│ 40 3 2 8│401 15 2 24 ─┴───────┴────────────┴────────────────┴────────────────┴────────────────
TRIGONOMETRICAL FUNCTIONS.
RIGHT-ANGLED TRIANGLES.
Sin. A = (_a_)/(_b_) Sec. A = (_b_)/(_c_) Tan. A = (_a_)/(_c_)
Cos. A = (_c_)/(_b_) Cosec. A = (_b_)/(_a_) Cotan. A = (_c_)/(_a_)
Versin. A = (_b_ − _c_)/(_b_). Coversin. A = (_b_ − _a_)/(_b_).
───────┬─────────┬───────────────────────────────────────────────────── Given. │Required.│ Formulæ. ───────┼─────────┼───────────────────────────────────────────────────── _a_,_b_│ A,C,_c_ │Sin. A = (_a_)/(_b_) Cos. C = (_a_)/(_b_) _c_ = │ │ √((_b + a_)(_b − a_)) │ │ _a_,_c_│ A,C,_b_ │Tan. A = (_a_)/(_c_) Cotan. B = (_a_)/(_c_) _b_ = │ │ √(_a_^2 + _c_^2) │ │ A,_a_ │C,_c_,_b_│ C = 90° − A _c_ = _a_ × Cotan. A _b_ = │ │ (_a_)/(Sin. A) │ │ A,_b_ │C,_a_,_c_│C = 90° − A _a_ = _b_ × Sin. A _c_ = _b_ × Cos. │ │ A │ │ A,_c_ │C,_a_,_b_│ C = 90° − A _a_ = _c_ × Tan. A _b_ = │ │ (_c_)/(Cos. A) │ │ ───────┴─────────┴─────────────────────────────────────────────────────
OBLIQUE-ANGLED TRIANGLES.
_s_ = ½(_a + b + c_)
───────────┬─────────┬───────────────────────────────────────────────── Given. │ │ Formulæ. ───────────┼─────────┼───────────────────────────────────────────────── A,B,C,_a_ │ Area= │(_a_^2 × Sin. B × Sin. C) ÷ 2 Sin. A A,_b_,_c_ │ „ │½(_c_ × _b_ × Sin. A) _a_,_b_,_c_│ „ │√(_s_(_s_ − _a_)(_s_ − _b_)(_s_ − _c_)) ───────────┼─────────┼───────────────────────────────────────────────── Given. │Required.│ Formulæ. ───────────┼─────────┼───────────────────────────────────────────────── A,C,_a_ │ _c_ │ _c_ = _a_(Sin. C)/(Sin. A) ───────────┼─────────┼───────────────────────────────────────────────── A,_a_,_c_ │ C │ Sin. C = (_c_ Sin. A)/(_a_) ───────────┼─────────┼───────────────────────────────────────────────── _a_,_c_,B │ A │ Tan. A = (_a_ Sin. B)/(_c_ − _a_ Cos. B) ───────────┼─────────┼───────────────────────────────────────────────── _a_,_b_,_c_│ A │Sin. ½A = √(((_s_ − _b_)(_s_ − _c_))/(_b_ × _c_)) „ │ „ │ Cos. ½A = √((_s_(_s_ − _a_))/(_b_ × _c_)); „ │ „ │ Tan. ½A = √(((_s_ − _b_)(_s_ − _c_))/(_s_(_s_ − │ │ _a_))) ───────────┴─────────┴─────────────────────────────────────────────────
COMPOUND ANGLES.
Sin. (A + B) = Sin. A Cos. B + Cos. A Sin. B. Sin. (A − B) = Sin. A Cos. B − Cos. A Sin. B. Cos. (A + B) = Cos. A Cos. B − Sin. A Sin. B. Cos. (A − B) = Cos. A Cos. B + Sin. A Sin. B.
Tan. (A + B) = (Tan. A + Tan. B)/(1 − Tan. A Tan. B).
Tan. (A − B) = (Tan. A − Tan. B)/(1 + Tan. A Tan. B).
SLIDE RULE DATA SLIPS, COMPILED BY C. N. PICKWORTH, WH.SC.
(_It is suggested that this page be removed by cutting through the above line, and selected portions of the Sectional Data Slips attached to the back of the Slide Rule._)
¹⁄₃₂ │0·03125 ¹⁄₁₆ │0·0625 ³⁄₃₂ │0·09375 ⅛ │0·125 ⁵⁄₃₂ │0·15625 ³⁄₁₆ │0·1875 ⁷⁄₃₂ │0·21875 ¼ │0·25 ⁹⁄₃₂ │0·28125 ⁵⁄₁₆ │0·3125 ¹¹⁄₃₂ │0·34375 ⅜ │0·375 ¹³⁄₃₂ │0·40625 ⁷⁄₁₆ │0·4375 ¹⁵⁄₃₂ │0·46875 ¹⁷⁄₃₂ │0·53125 ⁹⁄₁₆ │0·5625 ¹⁹⁄₃₂ │0·59375 ⅝ │0·625 ²¹⁄₃₂ │0·65625 ¹¹⁄₁₆ │0·6875 ²³⁄₃₂ │0·71875 ¾ │0·75 ²⁵⁄₃₂ │0·78125 ¹³⁄₁₆ │0·8125 ²⁷⁄₃₂ │0·84375 ⅞ │0·875 ²⁹⁄₃₂ │0·90625 ¹⁵⁄₁₆ │0·9375 ³¹⁄₃₂ │0·96875
Circ. of circle = 3·1416 _d_.
Area „ „ = 0·7854 _d_^2.
Sq. eq. area to cir., _s_ = 0·886 _d_.
Circle eq. to sq., _d_ = 1·128 _s_.
Sq. inscbd. in circ., _s_ = 0·707 _d_.
Circsb. circ. of sq., _d_ = 1·414 _s_.
Area of ellipse = 0.7854 _a_ × _b_.
Surface of sphere = 3·1416 _d_^2.
Volume „ „ = 0·5236 _d_^3.
„ „ cone = 0·2618 _d_^2 _h_.
Radian = (180°)/(π) = 57·29 deg.
Base of nat. or hyp. log. = e = 2·7183.
Nat. or hyp. log. = com. log. × 2·3026.
g (at London) 32·18 ft. per sec., per sec.
Abs. temp. = deg. F. + 461° = deg. C. + 274°.
C.° = (5)/(9)(F.° − 32°); F.° = (9)/(5)C.° + 32°.
Cal. pr.—Ther. units per lb.: Coal, 14,300;
petrol’m, 20,000; coal gas per cu. ft., 700.
Sp. heat:—Wt. iron, 0·1138; C.I., 0·1298;
copper, brass, 0·095; lead, 0·0314.
Inch = 25·4 mil’metres; mil’metre = 0·03937 in.
Foot = 0·3048 metres; metre = 3·2809 feet.
Yard = 0·91438 metre; metre = 1·0936 yards.
Mile = 1·6093 kilomtrs.; kilomtr. = 0·6213 mile.
Sq. in. = 6·4513 sq. cm.; sq. cm. = 0·155 sq. in.
Sq. ft. = 9·29 sq. decmtr.; sq. decmtr. = 0·1076 sq. ft.
Sq. yd. = 0·836 sq. metre; sq. metre = 1·196 sq. yds.
Sq. ml. = 258·9 hectares; hectare = 0·00386 sq. ml.
Cu. in. = 16·386 c. cm.; c. cm. = 0·06102 cu. in.
Cu. ft. = 0·0283 c. metre; c. metre = 35·316 cu. ft.
Grain = 0·0648 gramme; gram. = 15·43 grs.
Ounce = 28·35 grams.; „ = 0·03527 oz.
Pound = 0·4536 kilogm.; kilogm. = 2·204 lb.
Ton = 1·016 tonnes; tonne = 0·9842 ton.
Mile per hr. = 1·466 ft., or 44·7 cm., per sec.
Lb. per cu. in. = 0·0276 kilogram per cu. cm.
Kilogram per cu. cm. = 36·125 lb. per cu. in.
Lb. per cu. ft. = 16·019 kilogm. per cu. mtre.
Grain per gall. = 0·01426 gramme per litre.
Gramme per litre = 70·116 grains per gall.
Ultimate Strength│Lb. per Sq. in. „ │Tens’n.│Comp’n. ─────────────────┼───────┼─────── Wt. iron │ 50,000│ 50,000 Cast „ │ 16,000│ 95,000 Steel │ 80,000│ 70,000 Copper │ 21,000│ 50,000 Brass │ 18,000│ 10,500 Lead │ 2,500│ 7,000 Pine │ 11,000│ 6,000 Oak │ 15,000│ 10,000
Weight of Metals.│ Cub. In. │ Cub. Ft. │12 Cu. In. ─────────────────┼──────────┼──────────┼────────── Wt. iron │ 0·277│ 480│ 3·33 Cast „ │ 0·260│ 450│ 3·12 Steel │ 0·283│ 490│ 3·40 Copper │ 0·318│ 550│ 3·82 Brass │ 0·300│ 520│ 3·61 Zinc │ 0·248│ 430│ 2·98 Alumin’m │ 0.096│ 168│ 1·16 Lead │ 0.411│ 710│ 4·93
Lb. per sq. in. = 2·31 ft. water = 2·04 in. mercury = 0·0703 kilo. per sq. cm. Atmosphere = 14·7 lb. per sq. in. = 33·94 ft. water = 1·0335 „ „ Ft. hd. water = 0·433 lb. per sq. in. = 62·35 lb. per sq. ft. = 0·0304 „ „ Cub. ft. of water = 62·35 lb. = 0·0278 ton = 28·315 litres = 7·48 U.S. galls. Gall. (Imp.) = 277·27 cu. in. = 0·1604 cu. ft. = 10 lb. water = 4·544 litres. Litre = 1·76 pints = 0·22 gall. = 61 cu. in. = 0·0353 cu. ft. = 0·264 U.S. gall. Horse-power = 33,000 ft.-lb. per min. = 0·746 kilowatt = 42·4 heat units per min. Heat unit = 778 ft.-lb. = 1055 watt-sec. = 107·5 kilogrammetres = 0·252 calorie. Foot-pound = 0·00129 heat unit = 1·36 joules = 0·1383 kilogrammetres. Kilowatt = 1·34 H.P. = 44,240 ft.-lb. per min. = 3412 heat units per hour.
Footnote 1:
It will be recognised that n is the characteristic of the logarithm of the original number.
Footnote 2:
The special case in which the numerator is 1, 10, or any power of 10 must be treated by the rule for reciprocals (page 27).
Footnote 3:
The possible need for traversing the slide, to change the indices, when using the C and D scales, is not considered as a setting.
Footnote 4:
The reader may be reminded that cross-multiplication of the factors in any such slide rule setting will give a constant product, _e.g._, 20 × 94·5 = 27 × 70.
Footnote 5:
In this case cross _dividing_ gives a constant quotient, _e.g._, 8 ÷ 3 = 4 ÷ 1·5. Since the upper scale is now a scale of reciprocals, the ratio is really
O ⅛ ¼ ─────────── D 1·5 3
Footnote 6:
These lines should not be brought to the working edge of the scale but should terminate in the horizontal line which forms the border of the finer graduations, their value being read into the calculation by means of the cursor (see page 55).
Footnote 7:
The same principle may be applied to the cursor.
Footnote 8:
Philosophical Transactions of the Royal Society, 1815.
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TRANSCRIBER’S NOTES
Page Changed from Changed to
24 the right, so the number of the right, so the number of digits in the answer = 3 − 2 × 1 digits in the answer = 3 − 2 + 1 = 2 = 2
116 grammes, we have the equation, grammes, we have the equation, _x_ × (Cl.)/(Ag.Cl.) × _x_ = (Cl.)/(Ag.Cl.) × (_a_)/(_s_). Hence, the mark (_a_)/(_s_). Hence, the mark
● Typos fixed; non-standard spelling and dialect retained. ● Used numbers for footnotes, placing them all at the end of the last chapter. ● Enclosed italics font in _underscores_. ● Enclosed bold font in =equals=. ● The caret (^) serves as a superscript indicator, applicable to individual characters (like 2^d) and even entire phrases (like 1^{st}). ● Subscripts are shown using an underscore (_) with curly braces { }, as in H_{2}O.