Part 38

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.

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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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BY THE SAME AUTHOR.