Getting Gold

Appendix

(Selected Data for Mining Men)

To Find Lost Part of Vein

Zimmermann’s rule for finding the lost part of a vein on the other side of a vein, is as follows:—

Lay down upon paper the line of strike of lode and the line of strike of the fault (cross-course), and by construction ascertain the horizontal projection of the line of their intersection; from the point where the cross-course was struck by the lode, draw a line at right angles to the strike of the former and directed to its opposite wall. Notice on which side of the line of intersection this perpendicular falls, and after cutting through the cross-course, seek the “heaved” part of the lode on that side.

Thus let AB (Fig. 49) represent, at any depth, the line of strike of a fault or cross-course dipping east, and CD the line of strike of a lode dipping south, and we will suppose that in driving from C to D in a westerly direction, the fault has been met with at D. Knowing the dip of the lode and that of the fault, it is easy to lay down on any given scale, AₖBₖ and CₖDₖ, the lines of strike of the fault and lode respectively at a certain depth, say 10 fathoms, below AB. The point Dₗ, where AₖBₖ and CₖDₖ meet, is one point of the line of intersection. Join D and Dₗ, and prolong on both sides. The line MN represents the horizontal projection of the line of intersection of the two planes. At D erect DE at right angles and directed towards the opposite wall of the fault. As DE falls south of MN, the miner, after cutting through the fault, would drive in a southerly direction and eventually strike the lode again at F. It will be at once understood that if the miner were following the lode from b to F, the perpendicular would lie to the north of the line of intersection, and following the rule, he would drive in that direction, after cutting through the fault. When several faults in succession dislocate a lode, very great complications may arise.

The Calculation of Ore Reserves4

Having finished the survey of a metalliferous mine, the surveyor is sometimes called upon to calculate the quantity of ore reserves in that mine. Various methods are employed for this purpose.

Indeed, different surveyors will not agree within wide limits as to the amount of ore reserves in the same mine. Sometimes the amount of ore in sight will be considered to be a rectangular block, limited by the outcrop of the vein, the depth of the shaft, and the extreme points of the levels, diminished by the amount extracted. Other surveyors would avoid so excessive an amount, and take but one-third of that amount.

The following method is recommended by Mr. J. G. Murphy, an experienced American mining engineer, as the fairest and most trustworthy:—

Let it be required to calculate the ore reserves in a mine opened up on a vein with a mean cross section of 6 feet; a cubic foot of the vein matter in place weighing 150 lb. The ore stopes are generally very irregular. In this case, however, it may be supposed that the stope faces are 11 feet apart and 8 feet high. There is an inclined shaft, 10 feet by 6 feet, following the dip of the vein, and six levels, each 7 feet by 6 feet, 100 feet apart. The lengths of the levels are—

I.	200 feet west 150 feet east.
II.	160 feet west 100 feet east.
III.	120 feet west 400 feet east.
IV.	100 feet west 140 feet east.
V.	165 feet west 180 feet east.
VI.	350 feet west 150 feet east.

The longest level west is 350 feet, and the shortest 100 feet.

Assuming the bounding line of the area of available ore to be at a distance west of the shaft—

If the longest level east is 400 feet, and the shortest 100 feet, the bounding line in this direction, calculated in a similar way, will be at a distance of 250 feet from the shaft.

The inclined shaft has opened up the vein for 670 feet. Deducting, say, 15 feet for the irregularity of the surface, the quantity of ore in sight will be a rectangular block 655 feet deep, 225 + 250, or 475 feet long and 6 feet wide, that is 1,866,750 cubic feet.

From this quantity, however, must be deducted the quantity of ore extracted, namely:—

									Cubic Feet
Inclined	shaft		665	x	10	x	6	=	39,900
Level	I.		350	x	7	x	6	=	14,700
Level	II.		260	x	7	x	6	=	10,920
Level	III.		520	x	7	x	6	=	21,840
Level	IV.		240	x	7	x	6	=	10,080
Level	V.		345	x	7	x	6	=	14,490
Level	VI.		500	x	7	x	6	=	21,000
Level	I.		Stopped	east	(rough	estimate)			3,400
Level	I.		″ ″	west					6,500
Level	II.		″ ″	west					7,000
Level	III.		″ ″	east					20,000
Level	VI.		″ ″	west					12,000
									-----------

							Total	181,530
							Or in round numbers	182,000

This quantity, deducted from 1,866,750 cubic feet, leave 1,684,750 cubic feet. Divided by 13½, the number of cubic feet required for a ton, this gives 124,797 tons of ore in sight.

The quantity of ore discovered in a mine may be estimated from its specific gravity and the average size of the vein. The specific gravity of the ore, with that of water taken at 1000 for standard is equal to the number of ounces in a cubic foot. Great caution is necessary to determine the proportion of the vein which may be considered solid ore. A vein 6 feet square and 1 inch thick, contains 3 cubic feet, therefore, in order to find the number of cubic feet per square fathom of a vein, it is merely necessary to multiply the thickness in inches by three.

The following example illustrates the method of finding the weight of any ore per square fathom in a vein. What quantity of galena will be produced per square fathom from a mineral vein 6 inches in width? One quarter of the vein consists of galena, the remainder of zinc-blende. One-twentieth must be allowed for cavities in the vein. The specific gravity of galena is 7·5, and a cubic foot of water weighs 1000 ounces; therefore a cubic foot of galena weighs 7500 ounces.

The vein being 6 inches thick, there are 18 cubic feet in a square fathom. One quarter of that amount, or 4·5 cubic feet, consists of galena. The weight of galena in ounces is therefore:

7500 × 4·5 = 33,750 =	2109·375 lb.
From this one-twentieth or	105·468 lb.
	-------------
must be deducted, leaving	2003·907 lb.

or 17 cwt. 3 qr. 15 lb. as the weight of lead ore per square fathom.

Estimating Ore Values

In testing a gold mine with a view to purchase, it should be remembered that as a rule the intersections of leaders or small veins with the main ore body are usually the richest portions of the lode. This the experienced prospector knows, and generally his shafts and cuttings are made at such points. For the ordinary mining investor, when inspecting with a view to purchase, these are places to avoid if endeavouring to form a correct estimate of the value of the ore in bulk. Take samples across the lode from place to place, break down and bag personally, and mark bags. Test the rich portions separately and average, estimating quantity of both.

California Pump

Any handy man or rough bush carpenter can make a California Pump. The prospectors in the illustration (Plate VII.) are using a home-made contrivance, which is quite effective for raising water from shallow depths for “long tom,” or ground sluicing—a wooden frame-work and open wooden wheel with handle. Over the wheel is run a belt of canvas, say, six inches wide, with wood stops about a foot apart—a long sloping box, dipping into the water, up which the stopped belt travels—and you have a California Pump which, if not a highly scientific device, is at least very serviceable.

Hydraulics

General Data Regarding Water

An imperial gallon of water weighs, at 62° F., 10 lbs. avoirdupois. Gallons × ·1606 = cubic feet. Cubic feet × 6·288 = number of gallons.

Gallons × 277·46 = cubic inches. Cubic inches × 0·003604 = gallons. Cubic feet of water × 62·28 = number of pounds weight. Pounds of water × 0·0166 = cubic feet. Gallons of water × 0·004464 = number of tons. Tons of water × 224 = gallons of water. Cubic feet of water × 0·0278 = number of tons. Tons of water × 35·97 = cubic feet of water.

A pipe 1 yard long holds approximately (actually, 1·52 per cent. less) as many pounds of water as the square of its diameter in inches; thus a 6-inch pipe holds approximately 36 lbs. of water in each yard length.

Water Pressure

Ten feet head of water gives a pressure of 4⅓ lbs. per square inch approximately.

If H = head of water in feet, P = pounds pressure per square inch:

H = P × 2·311 P = H × ·4326

Horse-power required to Pump Water.—One h.-p. indicated will raise about 3000 gallons of water per hour 50 feet high.

To pump 1 gallon of water per minute against a pressure of 4 lb. per square inch, requires:

p × ·0007 H.-P.

Boring

Rock is bored with jumpers of 10 lb. to 18 lb., used alone, or with boring bars and hammer. The former are more effective, but can only be used perpendicularly, or nearly so, and with rock of moderate hardness; they require more skill.

18 lb.	hammers are used for			3 in. boring bars.
16 lb.	″	″	″	2½ in. ″
14 lb.	″	″	″	2 & 1¾ in. ″
5-7 lb.	″	″	″	1 in. ″

The boring bars may be made of 1¹∕₈ inch bar iron of various lengths, with steel bits up to 3 inches. A bit should bore from 18 feet to 24 feet with each steeling, and requires to be sharpened once for every foot bored.

3	men with a	3-inch	bar should bore	4	feet.
3	″	2½ ″	″	6	″
3	″	2 ″	″	8	″
3	″	1¾ ″	″	12	″
2	″	1 ″	″	8	″

per day of 10 hours in hard granite.

Power, etc., Required to Work Rock Drills

H-P.	Diameter of Air Cylinder.	Diameter of Steam Cylinder.	Stroke.	No. of 3-inch Drills driven.
	Inches.	Inches.	Inches.
6	8	8½	12	1
10	10	10½	16	2
12	12	12¼	22	3
16	13	13	24	4
20	14½	14½	28	5
30	17½	17½	36	8
40	20	20	36	11
60	22	24	60	15

Power Required to Work Double Steam and Air Cylinders

H-P.	Diameter of Air Cylinder.	Diameter of Steam Cylinder.	Stroke.	No. of 3-inch Rock Drills driven.
	Inches.	Inches.	Inches.
28	12	12	22	7
32	13	13	24	8
40	14½	14½	28	10
60	17½	17½	36	15
80	20	20	36	22
120	22	24	60	30

Durability of Ropes

The average duration of flat, wire ropes is usually taken at one year, and that of round ropes at half a year.

Diamond Drilling

This drill is applicable to sinking a borehole for prospecting for minerals or water, shafts, &c., or blasting under water.

It consists of a circular row of “carbonados,” a species of diamond, set in a circular steel ring. This is attached to a hollow steel tube which is kept rotating at about 250 revolutions per minute, pressed forward by a force varying from 400 to 800 lb. according to the nature of the rock. Water is supplied through the tube which washes out the débris and cools the diamonds.

Granite and the hardest limestones are penetrated at the rate of 2 to 3 inches per minute, sandstones 4 inches, quartz 1 inch.

The diamond drill is not effective in soft strata such as clay, sand, and alluvial deposits.

Boreholes have been made at the following rates:

In Ironstone formation:

A depth of	902	feet in	54	working days
	641	″	48	″
	434	″	54	″
	640	″	60	″

Or an average of 12 feet a day.

In Coal measures:

A depth of	1008	feet in	146	days
	802	″	168	″
	700	″	48	″
	558	″	42	″

Or an average of 7½ feet a day.

Timber

To find Solidity of Round Timber.

When all dimensions are in feet: Length × (¼ mean girth)² = cubic feet.

When length in feet, girth in inches: Length × (¼ mean girth) ÷ 144 = cubic feet.

When all dimensions are in inches: Length × (¼ mean girth)² ÷ 1728 = cubic feet.

To find Solidity of Square Timber.

When all dimensions are in feet: Length × breadth × depth ﹦ cubic feet.

When one dimension is in inches: Length × breadth × depth ÷ 12 ﹦ cubic feet.

When two dimensions are in inches: Length × breadth × depth ÷ 144 ﹦ cubic feet.

For board measure, depth always equal 1 inch.

To find surface in square feet, proceed as per rules for solidity of square timber.

Laying Out of Areas

1. In Squares.—Extract the square root of the desired content, reduced to square chains (ten square chains equal one acre). The result will be the length of the required side in chains.

Thus if we wish to find the side of a square block containing 25 acres, we first reduce the acres to square chains: 25 × 10 = 250, the square root of which is 15·81, or 15 chains 81 links; the side required.

By reference to the tables of square roots on page 189, the required sides of square block for a large number of acres can be read off at once.

One acre laid out as a square must have its side made 316¼ links, or 208-71/100 feet, or 69-57/100 yards, 70 paces being a near approximation.

2. In Rectangles.—Divide the content by the length or breadth, according to which factor is known, and the result will be the required side.

Thus 5 acres, or 50 square chains, if 10 chains long, will require to be 5 chains wide.

If the content only is given, and the length is to be a certain number of times the breadth, the content in square chains divided by the ratio of the length to the breadth, and the square root of the quotient, will give the length of the shorter side. Thus, if we wish to lay out 72 acres as a rectangle twice as long as broad: 72 acres = 720 square chains, divided by 2, the ratio given, = 360, the square root of which is 18·97 chains, the length of the shorter side. The length of the other side is therefore 18·97 × 2 = 37·94 chains, or 3794 links.

Mensuration

To find the area of a triangle when the base and perpendicular height are given: Multiply the base by half the height, or vice versâ.

To find the area of a triangle when the three sides are given: Take half the sum of the sides, subtract each severally from this sum, then multiply this and the three remainders together, and take the square root for the area.

To find the area of a rectangular figure: Multiply the length by the breadth, the product will be the area.

To find the area of a trapezoid: Multiply half the sum of the two parallel sides by the distance between them.

To find the area of a parallelogram whose angles are not right angles: Multiply the length of any one of the sides by the perpendicular.

To find the area of a trapezium: Divide it into two triangles and find the areas of the latter by the first rule.

To find the area of an irregular polygon: Divide the polygon into triangles and find the area of the latter.

To find the area of an irregular figure: Draw the figure on fine cardboard or thin sheet metal, cut the same carefully out and weigh with an accurate balance. Then this weight, compared with the weight of a piece of the cardboard or metal of a definite size, say one square inch, gives at once the area required.

Mine Surveying Problems

It would be futile to profess, in the limits of a small work of this kind, to instruct the beginner fully in the principles and practice of mine surveying, especially as the most elaborate treatise can only be of service when some actual practical experience and knowledge of instruments have been obtained.

For an exhaustive and well-arranged work on the subject, Brough’s “Treatise on Mine Surveying” can be strongly recommended, and should be carefully studied by all wishing to learn the best methods of accomplishing the accurate results that any mine-surveying worth the name demands.

The following methods of connecting underground and surface work are therefore addressed to such as are thoroughly acquainted with a dial and the method of traversing.

1. To find where a shaft should be sunk to connect with any part of the underground workings.

Should the mine be one opened by an adit, there is no difficulty in doing this, as the dial can be set up at the mouth and a sight taken to a light; this can then, by means of the vertical arc, be prolonged up the hill, and the remaining bearings and distances are then easily laid down to the desired point. When a starting-point has been obtained, the chief difficulty in these cases has been overcome. If a single shaft is the only connection to the underground workings, the magnetic bearing of the first line at bottom must first be carefully ascertained, and the position of the first station brought to the surface by means of a plumb-line, made of copper wire preferably, the plummet being put into a dish of water to steady it. The dial is then set exactly over the end of the line at the surface, and the first bearing and distance laid off.

Should it be impossible to set the instrument over the point at surface, a spot must be found by trial outside the shaft which is in the correct course. The dial is set up in the supposed direction of the line and repeated sights taken to the first point till the instrument and it are exactly in the required line, when the length of the first line can be measured along it, and the new lines proceeded with.

As local attraction frequently affects the needle at the bottom of the shaft, and so vitiates the surface survey that depends on its swinging the same at both points, a method of dispensing with the needle must be resorted to. This is the suspension of two plumb-lines from opposite sides of the shaft, the plummets hanging exactly over as much of the first underground line as the width of the shaft will allow.

The two plummet lines at surface then give the direction, and by trial the dial must be put exactly in line with them in order to prolong it correctly.

If there are two or more shafts sunk on the workings, it will be an easy matter to ascertain if the needle can be depended on for laying out any further surface work, as the underground survey connecting the shafts can be laid down on the surface, or the direct bearing and distance calculated, when its correctness is tested by the terminating point of the survey.

2. To find depth of shaft at any point, to cut a vein whose dip is known.

Rule:—Multiply natural tangent of angle of dip C (Fig. 50.) by the distance from outcrop to proposed shaft AC. The result is the depth required, AB.

By Protractor and Scale.—Rule on paper a line AC of the required distance, then at C set off the angle of dip and draw AB at right angles to AC. Then scale off AB = depth of shafts.

3. Given Depth of Shaft and Angle of Dip to find where it outcrops.—Then AC = AB × natural tangent of angle ABC. Or by scale and protractor by inspection.

4. Given Depth of Shaft AB and Dip of Vein Angle CB to find Distance BC between Bottom of Shaft and Outcrop.—BC = AC × natural sine ACB. Or by scale and protractor by inspection.

Rainfall

One inch of rain = 22,680 gallons, or 102·35 tons of water per acre.

Notes on Belting

Co-efficient of friction between ordinary leather belting and cast-iron pulleys or drums = ·423. Ultimate strength of ordinary leather belting = 3086 lb. per square inch. Belts vary from ³∕₁₆ in. to ¼in. thick, average ⁷∕₃₂ in.

Power of Single Leather Belts

To calculate the power of single leather belts, the following formula may be used: Let HP = actual horse-power. W = width of belt. F = driving force. T = working tension from 70 to 150 lb. V = velocity of belt in feet per minute.

Then F =	W × T.	HP =	V × F.	W =	33,000 × HP.
	2		33,000		F × V

Example:—A 10-inch belt running 2500 feet per minute, what horse-power will it transmit? Assuming the working tension to be 100 lb.,

F =	10 × 100	HP =	25,000 × 500	=	378 horse-power.
	2		33,000

Nystrom gives this rule:—HP = ^{V × F}⁄₅₅₀. V = velocity of belt in feet per second. F = force in pounds transmitted by belt.

The first rule gives good practical results where there is no great inequality in the diameter of the pulleys.

Double Belts transmit 1½ times as much as single belts.

Splicing Belts

Width of Belts	1	2	3	3 to 6	6 to 8	above 8 in.
Lap in inches	2	4½	5½	6	8	10

Rules for Double Leather Belting

A = covered area of driven pulley in square inches.

V = speed of belt in feet per minute.

H = indicated horse-power.

W = width in inches.

H = ^AV⁄_66,000 A = ^{66,000 H}⁄_V. W = A/L, where L = length of belt on driven pulley in inches.

Another authority simply says H = {70 to 80} × ^WV⁄_33,000.

And a third says W = ^{36,000 H}⁄_6VL, where L is here in feet.

Evan Leigh’s rule is W = ^{66,000 x IHP}⁄_{L x V}.

L = length of arc of contact upon smaller pulley in inches.

V = velocity of rim in feet per minute.

A belt transmits its motion solely through frictional contact with the surfaces of the pulley. The lower side of the belt should be made the driving side when possible, as the arc of contact is thereby increased by the sagging of the following side. Increase of power will be obtained by increasing the size of pulleys, the same ratio being retained. Wide belts are less effective per unit of sectional area than narrow belts. Long belts are more effective than short ones. The proportion between the diameters of two pulleys working together should not exceed six to one. Convexity of pulleys to receive belt = ½ inch per foot wide. The width of pulley should equal 1·2 times width of belt.

Speed of Belts

Belts have been employed running over 5000 feet per minute. Nothing, however, is gained by running belts much over 4000 feet per minute. About 3500 feet per minute for main belts agrees with good practice; lathe belts from 1500 to 2000 feet per minute. The life of a belt may be prolonged and its driving powers increased by keeping it in good working order. To ensure this it should be dressed on the back with castor oil every few weeks, more or less according the dryness of the atmosphere in which it works.

Weight and Bulk of Materials

The weight of a cubic foot of any material is its specific gravity multiplied by 62·425, or the weight of a cubic foot of water in pounds. To find the specific gravity of a stone, divide its weight in air by loss of weight in water of temperature of 60° F. = specific gravity.

Thus:

Quartz crystal weighs in	air	293·7	grains
″ ″	water	180·1	″ ″
		------
	Loss in weight	113·6	″ ″

Then:

^293·7⁄_113·6 = 2·59 = Specific gravity of quartz.

One ton of quartz when solid occupies 13 cubic feet, but when broken, about 20. Rocks when solid, as compared to the same when broken, usually increase in volume in the ratio of 1 to 1·5 or 1 to 1·18, the increase depending on size and form of fragments.

A dwt. of gold in a cwt. of ore = 1 oz. of gold per ton of ore.

For approximate calculation a grain of gold = two pence, and a dwt., four shillings.

In the following table of the chemical elements the standard of sp. gr. is hydrogen for the gaseous elements (hydrogen, oxygen, &c.) and water for the others.

The Chemical Elements, Their Symbols, Equivalents, and Specific Gravities

Name.	Symbol.	Atomic Weight.	Specific Gravity.
Aluminium	Al	27·5	2·56
Antimony	Sb	122·0	6·70
Arsenic	As	75·0	5·7
Barium	Ba	137·0	4·00
Bismuth	Bi	210·0	9·7
Boron	B	11·0	2·63
Bromine	Br	80·0	5·54
Cadmium	Cd	112·0	8·60
Caesium	Cs	133·0	1·88
Calcium	Ca	40·0	1·58
Carbon	C	12·0	3·50
Cerium	Ce	92·0	6·68
Chlorine	Cl	35·5	2·45
Chromium	Cr	52·5	6·81
Cobalt	Co	58·8	7·7
Columbium	Cb	184·8	6·00
Copper	Cu	63·5	8·96
Didymium	Di	96·0	6·54
Erbium	E	112·6	—
Fluorine	F	19·0	1·32
Gallium	Ga	69·9	5·9
Glucinum	Gl	9·5	2·1
Gold (Aurum)	Au	196·7	19·3
Hydrogen	H	1·0	0·069
Indium	In	113·4	7·4
Iodine	I	127·0	4·94
Iridium	Ir	198·0	21·15
Iron (Ferrum)	Fe	56·0	7·79
Lanthanum	La	90·2	11·37
Lead (Plumbum)	Pb	207·0	11·44
Lithium	Li	7·0	0·59
Magnesium	Mg	24·0	1·75
Manganese	Mn	55·0	8·01
Mercury (Hydrargyrum)	Hg	200·0	13·59
Molybdenum	Mb	96·0	8·60
Nickel	Ni	58·8	8·60
Niobium	Nb	94·0	6·27
Nitrogen	N	14·0	0·972
Osmium	Os	199·0	21·40
Oxygen	O	16·0	1·105
Palladium	Pd	106·5	11·60
Phosphorus	P	31·0	1·83
Platinum	Pt	197·4	21·53
Potassium (Kalium)	K	39·0	0·865
Rhodium	Rh	104·3	12·1
Rubidium	Rb	85·4	1·52
Ruthenium	Ru	104·4	11·4
Selenium	Se	79·5	4·78
Silicon	Si	28·0	2·49
Silver (Argentum)	Ag	108·0	10·5
Sodium (Natrium)	Na	23·0	0·972
Strontium	Sr	87·6	2·54
Sulphur	S	32·0	2·05
Tantalium	Ta	182·0	10·78
Tellurium	Te	129·0	6·02
Thallium	Tl	204·0	11·91
Thorium	Th	115·7	7·8
Tin (Stannum)	Sn	118·0	7·28
Titanium	Ti	50·0	4·3
Tungsten (Wolfram)	W	184·0	7·5
Uranium	U	120·0	18·4
Vanadium	V	51·3	5·50
Yttrium	Y	61·7	—
Zinc	Zn	65·0	7·14
Zirconium	Zr	89·5	4·15

The figures indicating the proportions by weight in which the elements unite with one another are called the combining or atomic weights, because they represent the relative weights of the atoms of the different elements. Since hydrogen is the lightest element, it is taken as the standard, and its combining or atomic weight = 1.

To find the proportional parts by weight of the elements of any substance whose chemical formula is known:

Rule.—Multiply together the equivalent and the exponent of each element of the compound; the product will be the proportion by weight of that element in the substance.

Example:—Find the proportional weights of the elements of Alcohol C₂H₆O.

Carbon C₂	=	equivalent	12	×	exponent	2	=	24
Hydrogen H₆	=	″	1	×	″	6	=	6
Oxygen O	=	″	16	×	″	1	=	16

Of every 46 lb. of Alcohol, 6 lb. will be H; 16, O; 24, C.

To find the proportions by volume, divide by the specific gravity.

Common Names of Chemical Substances

Common Names.	Chemical Names.
Aqua fortis	Nitric acid.
Aqua regia	Nitro-hydrochloric acid.
Blue vitriol	Sulphate of copper.
Cream of tartar	Bi-tartrate of potassium.
Calomel	Chloride of mercury.
Chalk	Carbonate of calcium.
Caustic potash	Hydrate of potassium
Chloroform	Chloride of formyl.
Common salt	Chloride of sodium.
Copperas, or green vitriol	Sulphate of iron.
Corrosive sublimate	Bi-chloride of mercury.
Dry alum	Sulphate of aluminium and potassium.
Epsom salts	Sulphate of magnesium.
Ethiops mineral	Black sulphide of mercury.
Galena	Sulphide of lead.
Glauber’s salt	Sulphate of sodium.
Glucose	Grape sugar.
Iron pyrites	Bi-sulphide of iron.
Jeweller’s putty	Oxide of tin.
King’s yellow	Sulphide of arsenic.
Laughing gas	Protoxide of nitrogen.
Lime	Oxide of calcium.
Lunar caustic	Nitrate of silver.
Mosaic gold	Bi-sulphide of tin.
Muriate of lime	Chloride of calcium.
Nitre, or saltpetre	Nitrate of potash.
Oil of vitriol	Sulphuric acid.
Potash	Oxide of potassium.
Realgar	Sulphide of arsenic.
Red lead	Oxide of lead.
Rust of iron	Oxide of iron.
Sal ammoniac	Chloride of ammonia.
Salt of tartar	Carbonate of potassium.
Slacked lime	Hydrate of calcium.
Soda	Oxide of sodium.
Spirits of hartshorn	Ammonia.
Spirits of salt	Hydrochloric acid.
Stucco, or plaster of Paris	Sulphate of lime.
Sugar of lead	Acetate of lead.
Verdigris	Basic acetate of copper.
Vermilion	Sulphide of mercury.
Vinegar	Acetic acid (diluted).
Volatile alkali	Ammonia.
Water	Oxide of hydrogen.
White precipitate	Ammoniated mercury.
White vitriol	Sulphate of zinc.

Thermometer

The following are the formulæ for the conversion of degrees of one scale to those of another:—

Centigrade° × 9 + 32 = Fahr.°		Fahr.° - 32 × 4 = Réaumur°.
5		9
Réaumur° × 9 + 32 = Fahr.°		Centigrade° × 4 = Réaumur°.
4		5
Fahr.° - 32 × 5 = Cent.°		Réaumur° × 5 = Centigrade°.
9		4

Freezing, Fusing, and Boiling Points

Substances.	Réaumur.		Centigrade.		Fahrenheit.
Bromine freezes at	-17·6°	=	-22°	=	-7·6°
Oil, Anise ″ ″	8	=	10	=	50
Oil, Olive ″ ″	8	=	10	=	50
Oil, Rose ″ ″	12	=	15	=	60
Quicksilver ″ ″	-31·5	=	-39·4	=	-39°
Water ″ ″	0	=	0	=	32°
Bismuth metal fuses at	211	=	264	=	507°
Copper ″ ″	963	=	1204	=	2200
Gold ″ ″	963	=	1204	=	2200
Iodine ″ ″	95·6	=	107°	=	224·6°
Iron ″ ″	1230	=	1538	=	2800
Lead ″ ″	260	=	325	=	617
Potassium ″ ″	50°	=	62·5°	=	144·5°
Silver ″ ″	530	=	537·70	=	1000
Sodium ″ ″	76·5°	=	95·6	=	204°
Steel melts at a lower
temperature than malleable
iron	-		-		-
Sulphur fuses at	54	=	120	=	248°
Tin ″ ″	189·6°	=	237	=	459°
Zinc ″ ″	329·6°	=	412	=	773°
Alcohol boils at	59·5°	=	74·4	=	173·1
Bromine ″ ″	46·4	=	58	=	136
Ether ″ ″	28·4	=	35·5	=	96
Iodine ″ ″	140	=	175	=	347
Quicksilver ″ ″	288	=	360	=	680
Water ″ ″	80	=	100	=	212

Heat Values of Fuels

Pounds of water evaporated by 1 lb. of fuel as follows:—

Straw	1·9
Wood	3·1
Peat	3·8
Coke or Charcoal	6·4
Coal	7·9
Petroleum	14·6

Signs and Symbols Used in Expressing Formulas

= Sign of equality, denoting that quantities so connected are equal to one another; thus, 3 feet = 1 yard.

+ Sign of addition, signifying plus or more; thus, 4 + 3 = 7.

- Sign of subtraction, signifying minus or less; thus, 4-3 = 1.

× Sign of multiplication, signifying multiplied by or into; thus, 4 × 3 = 12.

÷ Sign of division, signifying divided by; thus, 4 ÷ 2 = 2.

{} () Brackets, denoting that the quantities between them are to be treated as one quantity; thus, 5 {3(4 + 2)-6(3-2) = 5 (18-6) = 60.

Letters are used to shorten or simplify a formula. Supposing we wish to express length × breadth × depth, we may put the initial letters only, thus, l × b × d, or, as is usual when algebraical symbols are employed, leave out the sign × between the factors, and write the formula lbd.

When division is to be expressed in simple form, the divisor is written under the dividend; thus (x + y) ÷ z = (x + y) / z

° ’” are signs used to express certain angles in degrees, minutes, and seconds; thus 25 degrees 4 minutes 21 seconds would be expressed 25° 4’ 21”.

√ This sign is called the radical sign, and placed before a quantity indicates that some root of it is to be taken, and a small figure placed over the sign, called the exponent of the root, shows what root is to be extracted.

Thus ²√ a or √ a means the square root of a
∛ a ” cube”
∜ a ” fourth”

ρ This sign is used to denote the force of gravity at any given latitude.

π The Greek letter pi is invariably used to denote 3·14159, that is, the ratio borne by the diameter of a circle to its circumference.

When the figure 2 is affixed to any number, as diameter² or 12², the number is to be squared, as 12 × 12 = 144, the square; and with ³ affixed, the number is to be cubed—i.e., multiplied twice by itself, as 6³ = 6 × 6 × 6 = 316, the cube of 6.

English Weights and Measures

Measures of Length

12	lines		=	1 inch.
12	inches		=	1 foot.
3	feet		=	1 yard.
6	feet		=	1 fathom.
16½	feet		=	1 pole.
220	yards		=	1 furlong.
		}
8	furlongs
1760	yards		=	1 statute mile.
5280	feet
6086	feet			=	1 naut. mile.
7·92	inches		=	1 link.
		}
100	links
66	feet		=	1 chain.
22	yards

Square Measure

144	sq. inches		=	1 sq. foot.
9	sq. feet		=	1 sq. yard.
		}
30¼	sq. yards		=	1 sq. rod or pole.
272¼	feet
40	rods			=	1 sq. rood.
			}
4	roods
160	rods
4840	yards	=		1 acre.
43560	feet
10	sq. chains			=	1 acre.
1	hectare		=	2·471 acres.
640	acres		=	1 sq. mile.
30	sq. acres		=	1 yard of land.
Sq. ins.	× 0·007		=	square foot nearly.
Sq. yds.	× 0·00021		=	acres nearly.
113·0977	sq. ins.		=	1 circular foot.
183·46	circular ins.		=	1 square foot.

Solid or Cubic Measure

1728	cubic inches	=		1 cubic foot.
27	cubic feet	=		1 cubic yard.
			}
40	cub. ft. of rough, or	=		1 ton or load.
50	cub. ft. of hwn. tmbr.
				}
128	cub. ft. of timber	=			1 cord of wood.

Avoirdupois Weight

16	drachms	=	1 ounce.
16	ounces	=	1 lb.
14	lb.	=	1 stone.
28	lb.	=	1 qr. cwt.
112	lb.	=	1 cwt.
20	cwt	=	1 ton.
lbs.	× 0·009	=	cwt. nearly.
lbs.	× 0·00045	=	tons.
7000	grains	=	1 lb. avdp.
437½	grains	=	1 oz.

Troy Weight

24	grains	=	1	dwt.
20	dwt.	=	1	ounce.
12	oz.	=	1	lb.
5760	grains	=	1	lb. troy.
480	grains	=	1	oz. ″

Apothecaries’ Fluid Measure

Gallon (C) = 8 pints (O); 1 pint = 20 fluid ounces (oz. weight of water). Ounce (f ℥) = 8 drachms (f ʒ) = 480 minims (♏) = 720 drops (gtt.).

One wine glass = 4 tablespoonfuls = 16 tablespoonfuls = 2 ounces.

Symbols.—f. or fl. fluid; s.s. one half; a.a. for each. Thus f℥ss. ½ a fluid ounce.

Apothecaries’ weight, formerly used for dispensing medicines, superseded in 1864. 20 grains = 1 scruple; 3 scruples = 1 drachm; 8 drachms = 1 ounce; 12 ounces = 1 lb. (troy).

Liquid Measure

			Cubic in. nearly.
4 gills	= 1 pint	=	34¾
2 pints	= 1 quart	=	69⅓
4 quarts	= 1 gallon	=	277·123

French Weights and Measures

Weights

Gramme	15·432349 grams troy.
Décagramme (= 10 grammes)	5·6438 drachms av.
Hectogramme (= 100 grammes)	3·527 oz. av.
Kilogramme (= 1000 grammes)	2·204621 lbs. av., or
	2·679227 lbs. troy.
Quintal (= 100 kilogrammes)	220·462 lbs. av.
Tonne (= 1000 kilogrammes)	2204·621 lbs. av.
Decigramme (= ¹∕₁₀th of a gramme)	1·5432 grain.
Centigramme (= ¹∕₁₀₀th of a gramme)	0·15432 grain.
Milligramme (= ¹∕₁₀₀₀th of a gramme)	0·015432 grain.

Lineal Measure

Mètre	3·2808992 feet.
Décamètre (= 10 mètres)	32·808992 feet.
Hectomètre (= 100 mètres)	328·08992 feet.
Kilomètre (= 1000 mètres)	1093·633 yards.
Myriamètre (= 10,000 mètres)	6·2138 miles.
Decimètre (= ¹∕₁₀th of a mètre)	3·937079 inches.
Centimètre (= ¹∕₁₀₀th of a mètre)	0·39371 inch.
Millimètre (= ¹∕₁₀₀₀th of a mètre)	0·03937 inch.

Superficial Measure

Centiare (= 1 square mètre)	1·196033 square yard.
Are(= 100 square mètres)	0·098845 rood.
Hectare (= 10,000 square mètres)	2·471143 acres.

Measures of Capacity

Litre	(= 1 décimétre cube)	1·760773	pint (61·027 cubic inches).
Décalitre	(= 10 litres)	2·2009668	gallons.
Hectolitre	(= 100 litres)	22·009668	″
Kilolitre	(= 1000 litres)	220·09668	″
Décilitre	(= ¹∕₁₀th of a litre)	·17607	pint.
Centilitre	(= ¹∕₁₀₀th of a litre)	·017607	pint.

Solid Measure

Stère	(= 1 cubic mètre)	1·31 cubic yard.
Décastère	(= 10 stères)	13 cubic yards, 2 feet, 21 inches.
Décistère	(= ¹∕₁₀th of a stère)	3 cubic feet, 918·7 cubic inches.

Fresh and Salt Water Compared

		Fresh.	Salt.
1	cubic foot at 40° weighs	62·425 lbs.	64 lbs.
1	cubic inch at 40° weighs	·036,126 lbs.	·037,037 lbs.
1	cubic foot at 40° equals	6·242 gallons	6·2 gallons.
1	ton equals	35·943 cubic ft.	35 cubic ft.
1	ton contains at 62°	224 gallons	217 gallons.

Velocity of Falling Fluids

Falling fluids are governed by the same laws as falling bodies.

Fluid falls 1 foot in a ¼ of a second, 4 feet in ½ of a second, 9 feet in ¾ of a second, and so on.

The velocity of a fluid, flowing through an aperture in the side of a reservoir, is the same that a heavy body would acquire by falling from a height equal to that between the surface of the fluid and the middle of the opening.

The velocity of a fluid flowing out of an aperture is as the square root of the height of the head of the fluid. The theoretical velocity, therefore, in feet per second is as the square root of the product of the space fallen through in feet and 64·333; consequently, for 1 foot it is √64·333 = 8·02 feet. The mean velocity is about 5·4 feet, or 0·673.

Pressure of Water at Different Heads

H. = head in feet. P. = pressure in lbs. per sq. foot. p. = pressure in lbs. per sq. inch.

H.	P.	p.	H.	P.	p.
1	62·4	·4333	9	561·6	3·9
1·25	78	·5416	10	624	4·3333
1·5	93·6	·65	20	1248	8·6666
1·75	109·2	·7538	30	1872	13
2	124·8	·8666	40	2496	17·3333
3	187·2	1·3	50	3120	21·6666
4	249·6	1·7333	60	3744	26
5	312	2·1666	70	4368	30·3333
6	374·4	2·6	80	4992	34·6666
7	436·8	3·0333	90	5616	39
8	499·2	3·4666

To Find the Contents of a Tank

To find the number of gallons contained in a tank, multiply the length, width, and depth together, in feet. This gives the contents in cubic feet; multiply by 6·24, and the result is the number of gallons contained. If the dimensions are in inches, use ·003607 in place of 6·24.

Sizes and Weight of Corrugated Galvanised Iron Sheets

Thickness B.W.G.	Size of Sheets	Weight Per Square Foot.	Square Foot per Ton.
	Feet.	lb. oz.
16	6 × 2 to 8 × 3	2 1	800
17 × 18	6 × 2 ″ 8 × 3	2 4	1050
19 × 20	6 × 2 ″ 8 × 3	1 12	1300
21 × 22	6 × 2 ″ 7 × 2½	1 7	1600
23 × 24	6 × 2 ″ 7 × 2½	1 3	1900
25 × 26	6 × 2 ″ 7 × 2½	1 0	2250

Thickness and Weight of Galvanised Sheet Iron

Size of sheet, 2 feet in width by from 6 to 9 feet in length.

Wire Gauge.	Weight Per Square Foot.	Wire Gauge.	Weight Per Square Foot.
No.	Oz.	No.	Oz.
30	10	22	21
29	11	21	24
28	12	20	28
27	14	19	33
26	15	18	37
25	16	17	43
24	17	16	48
23	19	14	60

Qualities of Different Ropes Expressed Relatively

	Strength.	Rigidity.	Weight Dry.	Stretching.
Italian Hemp	—	1	1	}
Baltic ″	0·7 ·9	0·8 to 0·9	1	} ¹∕₇ to ¹∕₁₂
Manilla ″	0·9 to 1	0·75	0·88	}
Flax	0·9	low	—	¹∕₇₅
Iron Wire	3	high	4	—
Steel	6	high	4	—

Steel wire rope stretches about ¹∕₃₆₀, ¹∕₂₅₀, and ¹∕₁₃₀, of ¼, ⅓, and ½ the breaking weight.

The Atmosphere

The composition of the atmosphere is by volume, oxygen 20·8, nitrogen 79·2; by weight, oxygen 23, nitrogen 77. There are also minute quantities of carbon dioxide, aqueous vapour, and ammonia.

The barometer falls about ½” for each 500’ increase of altitude the mean temperature being 50° Fahr.

Weight or Pressure

= 14·706 lbs. per square inch.
= 29·92 inches of mercury.
= 33·7 feet of water.

Squares, Cubes, Square Roots, and Cube Roots

No.	Square.	Cube.	Square Root √	Cube Root ∛
1	1	1	1·0	1·0
2	4	8	1·414	1·259
3	9	27	1·732	1·442
4	16	64	2·0	1·587
5	25	125	2·236	1·709
6	36	216	2·449	1·817
7	49	343	2·645	1·912
8	64	512	2·828	2·0
9	81	729	3·0	2·080
10	100	1000	3·162	2·154
11	121	1331	3·316	2·223
12	144	1728	3·464	2·289
13	169	2197	3·60	2·35
14	196	2744	3·74	2·41
15	225	3375	3·87	2·46
16	256	4096	4·0	2·51
17	289	4913	4·12	2·58
18	324	5832	4·24	2·62
19	361	6859	4·35	2·66
20	400	8000	4·47	2·71
25	625	15625	5·0	2·92
30	900	27000	5·47	3·10
35	1225	42875	5·91	3·27
40	1600	64000	6·32	3·41
45	2025	91125	6·70	3·55
50	2500	125000	7·07	3·68
55	3025	166375	7·41	3·80
60	3600	216000	7·24	3·91
65	4225	274625	8·06	4·02
70	4900	343000	8·36	4·12
75	5625	421875	8·66	4·41
80	6400	512000	8·94	4·30
85	7225	614125	9·21	4·39
90	8100	729000	9·48	4·48
95	9025	857375	9·74	4·56
100	10000	1000000	10·0	4·64
110	12100	1331000	10·48	4·79
120	14400	1728000	10·95	4·93
130	16900	2197000	11·40	5·06
140	19600	2744000	11·83	5·19
150	22500	3375000	12·24	5·31
160	25600	4096000	12·64	5·42
170	28900	4913000	13·03	5·53
180	32400	5832000	13·41	5·64
190	36100	6859000	13·78	5·74
200	40000	8000000	14·14	5·84
210	44100	9261000	14·49	5·94
220	48400	10648000	14·83	6·03
230	52900	12167000	15·16	6·12
240	57600	13824000	15·49	6·21
250	62500	15625000	15·81	6·29
260	67600	17576000	16·12	6·38
270	72900	19683000	16·43	6·46
280	78400	21952000	16·73	6·54
290	84100	24389000	17·02	6·61
300	90000	27000000	17·32	6·69
310	96100	29791000	17·60	6·76
320	102400	32768000	17·88	6·83
330	108900	35937000	18·16	6·91
340	115600	39304000	18·43	6·97
350	122500	42875000	18·17	7·04
360	129600	46656000	18·97	7·11
370	136900	50653000	19·23	7·17
380	144400	54872000	19·49	7·24
390	152100	59319000	19·74	7·30
400	160000	64000000	20·0	7·36
410	168100	68921000	20·24	7·42
420	176400	74088000	20·49	7·48
430	184900	79507000	20·73	7·54
440	193600	85184000	20·97	7·60
450	202500	91125000	21·21	7·66
460	211600	97336000	21·44	7·71
470	220900	103823000	21·67	7·77
480	230400	110592000	21·90	7·82
490	240100	117649000	22·13	7·88
500	250000	125000000	22·36	7·93
510	260100	132651000	22·58	7·98
520	270400	140608000	22·80	8·04
530	280900	148877000	23·02	8·09
540	291600	157464000	23·23	8·14
550	302500	166375000	23·45	8·19
560	313600	175616000	23·66	8·24
570	324900	185193000	23·87	8·29
580	336400	195192000	24·08	8·33
590	348100	205379000	24·29	8·38
600	360000	216000000	24·49	8·43
610	372100	226981000	24·69	8·48
620	384400	238328000	24·90	8·52

Table to Calculate Wages and Other Payments

Year.	Per Month.			Per Week.			Per Day.
£	£	s.	d.	£	s.	d.	s.	d.
1	0	1	8	0	0	4½	0	0¾
2	0	3	4	0	0	9¼	0	1¼
3	0	5	0	0	1	1¾	0	2
4	0	6	8	0	1	6½	0	2¾
5	0	8	4	0	1	11	0	3¼
6	0	10	0	0	2	1¾	0	4
7	0	11	8	0	2	8¼	0	4½
8	0	13	4	0	3	1	0	5¼
9	0	15	0	0	3	5½	0	6
10	0	16	8	0	3	10¼	0	6½
11	0	18	4	0	4	3¾	0	7¼
12	1	0	0	0	4	7½	0	8
13	1	1	8	0	5	0	0	8½
14	1	3	4	0	5	4½	0	9¼
15	1	5	0	0	5	9¼	0	9¾
16	1	6	8	0	6	1¾	0	10½
17	1	8	4	0	6	6½	0	11¼
18	1	10	0	0	6	11	0	11¾
19	1	11	8	0	7	3½	1	0½
20	1	13	4	0	7	8¼	1	1¼
30	2	10	0	0	11	6½	1	7¾
40	3	6	8	0	15	4½	2	2¼
50	4	3	4	0	19	2¾	2	9
60	5	0	0	1	3	1	3	3½
70	5	16	8	1	6	11	3	10
80	6	13	4	1	10	9¼	4	4½
90	7	10	0	1	14	7½	4	11½
100	8	6	8	1	18	5½	5	5¾

If the Wages be Guineas instead of Pounds, for each Guinea add 1d. to each month, or ¼d. to each week.

Handy Universal Interest Table

(By Alfred Fryer.)

Showing the Interest, to the 100th part of a farthing, on any Sum from £1 to £1,000,000, for any number of days, at any rate per cent.

	£	s.	d.	f.	100ths.		£	s.	d.	f.	100ths.
1	0	0	1	2	3	50,000	136	19	8	2	85
2	0	0	1	1	26	60,000	164	7	8	0	22
3	0	0	1	3	89	70,000	191	15	7	1	59
4	0	0	2	2	52	80,000	219	3	6	0	96
5	0	0	3	1	15	90,000	246	11	6	0	32
6	0	0	3	3	78	100,000	273	19	5	1	70
7	0	0	4	2	41	200,000	547	18	10	3	40
8	0	0	5	1	4	300,000	821	18	4	1	9
9	0	0	5	3	67	400,000	1095	17	9	2	79
10	0	0	6	2	30	500,000	1369	17	3	0	49
20	0	1	1	0	60	600,000	1643	16	8	2	19
30	0	1	7	2	90	700,000	1917	16	1	3	89
40	0	2	2	1	21	800,000	2191	15	7	1	59
50	0	2	8	3	51	900,000	2465	15	0	3	29
60	0	3	3	1	81	1,000,000	2739	14	6	0	99
70	0	3	10	0	11	The Rule.—Multiply the number of £ upon which the interest is to be calculated by the number of days, and then multiply this product by the rate of interest, strike out the last two figures to the right hand, and find from the table the interest on the remaining figures.
80	0	4	4	2	41
90	0	4	11	0	11
100	0	5	5	3	1
200	0	10	11	2	3
300	0	16	5	1	4
400	1	1	11	0	5	Example.—What is the interest of £271 for 90 days at 7 per cent. per ann.?
500	1	7	4	3	7
600	1	12	10	2	8	Multiply		£271
700	1	18	4	1	10	by		90	number of days
800	2	3	10	0	11	then multiply		24390
900	2	9	3	3	12	by		7	rate of interest
1,000	2	14	9	2	14			171730
2,000	5	9	7	0	27
3,000	8	4	4	2	41
4,000	10	19	2	0	55
5,000	13	13	11	2	68
6,000	16	8	9	0	82
7,000	19	3	6	2	96	strike out last 2 figures,
8,000	21	18	4	1	10	viz.: 30, and in the table	£	s.	d.	f.	100ths.
9,000	24	13	1	3	23	against 1000 you find	2	14	9	2	14
10,000	27	7	11	1	37	″ 700 ″	1	18	4	1	10
20,000	54	15	10	2	74	″ 7 ″	0	0	4	2	41
30,000	82	3	10	0	11		£4	13	6	1	65
40,000	109	11	9	1	48

If the product of the number of £ on which the interest be calculated × number of days × rate of interest is, after elision of the last two figures, a multiple of a million, multiply the figures in table against 1,000,000 by the number of that multiple.

Notes

4 Bennett H. Brough’s “Treatise on Mine Surveying,” sixth edition, p. 165.