MAIN LIBRARY TN 371 Main REESE LIBRARY OF THE UNIVERSITY OF CALIFORNIA Deceived Accession No. / Jj /^ Class No. PRACTICAL NOTES ON HYDRAULIC MINING. BY GEO. H. EVANS, M. E. SECOND E OF THB UNIVERSITY San Francis JOHN TAYLOR & Co., 63 First Street Vtf ~\ i >-* INDEX To Practical Notes on Hydraulics, by G* H. Evans. A. Air Valves . . .19 j* Angles, loss of head 27, 28 7 Area of Ditches 10, 13, 14 B. Banks of Ditches 13 I Beams, strength of. 34, 39 Bends, loss of head 27, 28 Breast Wheels 32, 33 2 Bull Wheels 29 ? Bursting Strain of Pipes, Plates, etc 42, 43 d C. Capacity of Pipes and Nozzles , 14 ^ Chains, strength of 39, 40 Construction of Nozzles 15 J Creeks, measuring supply 4 . " to measure supply 4 , Current Wheels 30, 31 J D. Dams, pressure of water 15 * Diameter of Valves 22 * Discharge of Pipes 36 ,J Discharge through Gates 20, 21 J Discharge through Sluice Valves 21 Ditch By- Washes 7 I Ditch Sidelings 7 ' Ditches, allowance for leakage 11 4 area of 10,13,14 -J " capacity of 9 * discharge of 10, 11 -J ' ' friction in same 8 * gradeof. 9,10 " height of sides. 8 ; " in porous ground 9 ^ " laying out line of 7 ' " springs on line of 7 " timber work on 7 " tunnelling through spurs, etc 7 " weak banks 13 *' wetted border 10 M Elevator, Hydraulic 44, 47 Evaporation, loss due to ., 4,11 Expansion Joints 19 F. Fall of Pine Line 15 Flumes, discharge of 11 " grade of 8 height of sides 8 " iron or steel 7 Friction Head for Pipes 16 Friction, in pipes 24, 25, 26, 27 G. Gates, careless opening and shutting 20 discharge through .20, 21 head required 21 " screws of , 19 Grades, correct 5 " nature of country 6 of ditches 9 of sluices 1 47, 48 Gravel Elevators 44, 47 H. Head, due to velocity 19 loss due to bends 27,28 required for gates and valves 21 Hydraulic Elevator 44, 47 Motors 28 I. Iron or Steel Flumes 7 Iron and Steel, strength of 40, 41 J. Joints of Pipe 20 L. Laying Pipe Line 15 Leakage, loss due to 4, 11 Loss due to Evaporation and Leakage 4 M. Materials, strength of 33, 34, 42 Methods of Measuring Water Supply 4, 5 Motors, Hydraulic 28 N. Nozzles 22, 23, 24 capacity of '. 14 construction of 15 V O. Overshot Wheels,.. ...33 Perimeter or Wetted Border 10 Pipe Air Valves 19 Pipe Line, fall of 15 " " laying of 15 Pipes, bell-mouthed entrance 16 bursting strain 42, 43 capacity of 14 discharge of 16 discharge through gates 20, 21 expansion joints 19 friction head 16 friction in 24, 25, 26, 27 joints of 20 loss due to bends 27, 28 plates, etc., bursting strain 42,43 safety valves 19 velocity of water in 16, 17, 18, 19 Plates, iron and steel, strength of 42, 43 R. Reservoirs, pressure of water 15 Ropes, strength of 40, 41, 42 S. Safety Valves 19 Screws of Gates 19 Sizes of Sluices 47, 48 Sluices, grade of 47, 48 narrow vs. deep 47,48 size of 47, 48 Sluice Valves, discharge and head required 21 Strength of Beams 34, 39 of Chains 39, 40 of Iron and Steel 40, 41 of Ropes 41, 42 Supply, average of water 4 T. Timber Work on Ditches 7 Trestle Work, foundation for 7 Tunnels, form of roof 8 ;;A,. u. Undershot Wheels 31, 32 V. Valves, diameter of 22 " sluice, discharge of 21 Velocity, head 19 " mean 5 of water through pipes 16, 17,18, 19 to find in creeks, ditches, etc , 5 W. Water, average supply of 4 Elevator 44, 47 Facilities 4 pressure of J|> Wheels 28, 29, 30 Wetted Border 10 Wheels, breast 32, 33 bull 30,31 " overshot 33 undershot ....31, 32 water .. 28,29,30 Practical Notes on Hydraulic Mining. The following article was written for the MINING AND SCIENTIFIC PRESS by GEORGE H. EVANS, C. E., M. E., General Manager Cons. G. Mines of Gal., Ld., Oroville, Cal. COPYRIGHTED. Of the various kinds of mining there are none more interesting than hydraulic mining, and in connection with it there are innumerable points on which a mine manager or superintendent should be thoroughly posted, some of the most important being as follows: First Water facilities and the different methods of both roughly and accurately measuring the average amount of water available the season through for working the claim or claims. Second The nature of country through which the ditches, flumes and pipe lines have to be constructed in order to carry water for mining purposes, and the different grades suitable for such purpose. This is most important. The writer knows instances where ditches have been constructed for long distances with too heavy a grade, and consequently the water when turned in acquired too much velocity and completely washed away the greater portion of the ditch. On the other hand, by the employment of cheap or in- competent men, ditches have been constructed with too little grade, and there are cases where men have constructed ditches with the fall in the wrong direc- tion. Third The quantity of water different size pipes will carry or that can be discharged through pipes and nozzles under various heads. Fourth The friction caused by using pipes of too small diameter, and the loss of head due to this; also loss of head due to bends of short radius, and angles of all kinds in pipe lines. Fifth A full and complete practical knowledge of the different motors used in connection with hydrau- lic mining, and all particulars relative to the efficiency of the various kinds of water wheels, etc. Sixth The strength of materials, and especially of chains, hemp, manilla arid wire ropes. Seventh The bursting and working strain of iron and steel pipes of different diameters, and the strength of iron and steel plates, single and double-riveted, with punched and drilled holes. Eighth Methods of economically treating alluvial deposits in large quantities, when there is not suffi- cient grade for ground sluices, and yet enough water for piping, or in cases where, owing to the debris law, it is necessary to impound the tailings. Water Facilities. This is one of the most important matters in connection with hydraulic mining, and great care should be taken in arriving at the aver- age quantity of water available all the year round, or during the entire season, so that sluicing opera- tions can be carried on continuously. In order to do this it is necessary to correctly measure the creeks, or streams, at different points, to fairly approximate the average supply of water that can safely be relied upon from such source, not forgetting that, according to the location, allowance must be made for loss due to evaporation and leakage, which in some cases reaches as high as 20 per cent. A very simple method of measuring the quantity of water flowing in a stream is as follows: Measure the depth of water in feet, at from six to twelve points across the stream at equal distances; do this in two or three places along a fairly straight course; add all depths together, and divide the result by the number of measurements taken; this will give the average depth of the stream, and such depth multiplied by the average width in feet will give its cross section, or area in square feet, which, multiplied by the velocity of water in feet per minute, will give the number of cubic feet flowing per minute in the stream. To find the velocity, a very simple way is to step or measure off 120 feet along the bank, and in order to allow for the surface of the water flowing faster than the bottom or sides, and thus obtain the mean velocity, call the measurement 100 feet, and at the commence- ment of this 100 feet throw into the middle of the stream several pieces of paper or wood at intervals, and note the time it takes each one of them to reach the end of the measured line; then divide the total time in min- utes taken by all the floats by the number of floats, and the result will be the average time taken for each float to make the trip ; divide the average time in min- utes by the distance in feet, viz, 100, and the result will be the velocity in feet per minute, and this multiplied by the area in square feet will give the number of cubic feet flowing per minute; or, if the answer be required in miners' inches, multiply the cubic feet per minute by 2 and divide by 3. Another simple method for small streams is to put a small dam across the stream and back up the water sufficiently deep to prevent any considerable velocity, and on top of the dam place a thin board with a notch cut out of it wide enough, by estimation, to carry the whole of the water with a moderate depth of over- flow, and the following calculation will give the num- ber of gallons discharged per minute, and this result divided by 11.25 will convert the gallons per minute to miners' inches. For example: A weir with 4 inches overflowing the length of a notch which is 6 feet, or 72 inches, wide, the number of gallons per minute would be found by the following formula: G = dXV"dXlX 2.67. Where G represents gallons per minute, d = depth of overflow in inches, and 1 = length of notch in inches. In this case G will be found by multiplying 4 by the square root of 4, and by the length of notch or 72 inches, and then by 2.67, making the quantity 5 of water in gallons per minute = 1538, and this divided by 11.25 = 136.71 miners' inches. There are many other and more correct methods of measuring the flow of water in channels and streams, but I have illustrated the two most simple, in order that any person of ordinary intelligence could easily deter- mine the quantity of water running in open streams without the aid of difficult formulae. Nature of Country for Grades, etc. The nature of the country through which it is intended to carry ditches or flumes must be carefully considered in order to establish the correct grade, upon which, of course, depends the velocity, or, more plainly speaking, the destroying force of the water, and in locating the sites for water races the following points should be care- fully considered: First Ascertain by careful aneroid readings the lowest point in the stream, creek or other source of water supply that will allow sufficient grade for con- veying the water to a point suitable for working the claim or claims, and, if the maximum supply obtain- able is less than required for advantageous working, then favorable sites must be located for the construc- tion of storage dams or reservoirs capable of storing sufficient to keep up the required supply. Second Should the lowest point in the stream, creek or other source of water supply contain more than enough to meet the requirements of the mine or mines, in driest season, then the locator should select the greatest elevation that the country through which the ditch has to be constructed, and the water supply available at driest season will allow, so that the ditch when completed, will command the largest area of mining ground with the maximum head or pressure. This is an important point, as in many instances ditches of considerable length have been constructed and much money wasted in the endeavor to com- mand large tracts of mining ground, and when such ditches have been completed it was found that they 6 tapped the source of supply at such an elevation that it was impossible, except in the wet season, to get sufficient water to wash with. Third All timber work along the line of ditch should be curtailed as much as possible, and when fluming cannot be avoided the use of iron or steel should be carefully considered for ditches of a per- manent nature, as in many instances the first cost is not very much greater, but the durability and the great saving in cost of maintaining more than compensates the owner of the ditch. When timber work is found necessary, care should be taken in securing the most durable kinds, and after flumes and supports are finished they should beithor- oughly coated with a hot mixture of asphaltum, or painted with a good mineral paint, while all founda- tions for trestles, etc., should be placed in such a man- ner that they can be easily removed and renewed at all times. Fourth The line of the ditch should be carefully laid out so that it will be as short as possible, with, of course, due regard to economy, etc., and in coming around long points, or in places where sidelings are very steep and composed of loose rock, tunneling through such spurs should be carefully considered, or when it is proved by boring that such tunnels can be constructed to stand without timbering, they should always be preferred to long ditches around such spurs or points, unless the ground for ditching is exceed- ingly good and the extra distance quite short. Fifth Along the line of ditch all springs or water courses should be connected by means of short flumes or ditches, so that the loss due to leaks and evapora- tion from the main supply will be entirely or partly made up. It is also absolutely necessary for the safety of the ditch that by-washes or water gates be con- structed for the purpose of taking care of any sudden increase of water from heavy rains or melting snow along the line of ditch. 7 These by-washes must be kept in condition to, at all times, divert any water above the usual height over the gates or through the openings in the sides of the race, and in locating the points for by-washes, or safety outlets, it is necessary to carefully consider what becomes of the surplus water, as in many cases owners of ditches have rendered themselves liable for heavy damages. Sixth At the different points along the line of race when fluming has to be resorted to, allowance should be made for an increase of grade, in order that the flume can be constructed of much smaller dimensions than the ditch and yet carry all the water required. While on this subject, it is necessary to remember that the least amount of friction in ditches and flumes is developed when the least wetted border, or perim- eter, is obtained, and to do this the width of the bot- tom must be from If to 2 times the depth of the sides. These two points, if carefully studied, will save ditch owners large sums of money in both lumber and con- struction accounts. The following is a simple rule for finding the height of the sides of a ditch or flume when area of same is known, and it is desirable to follow the rule just men- tioned above : When width is to be 2J times the height of the sides, multiply the area in square inches by 4 and divide the result by 9, then take the square root of the product and that will be the height of the sides. When the width is to be If times the height of the sides, multiply the area in square inches by 4 and divide by 7, then extract the square root of the product and the answer will be the height of the sides. Seventh It is agreed by the best authorities that, when constructing tunnels, where they will stand without timbers, the best form of roof is the Gothic arch, as it stands better than the circular or any other kind of roof and is not so liable to flake. In fact, tunnels constructed with circular roof, except in very ,^ tight ground, have been noticed to flake off. until they assume nearly the section of the Gothic arch. Grades, Capacity, etc. In connection with the grades and various shapes of water races, the follow- ing points require particular attention: First As before mentioned, the character of the ground through which the ditch is constructed will have a great bearing on the grade required, but, as a guide, it will be well to remember that practical results have demonstrated that in ordinary ground, the water should travel at the rate of from 180 to 200 feet per minute. Then the grade will be determined by the dimensions of the ditch, and its intended carry- ing capacity. Second Races in which the water flows at too high a velocity through ground of a porous nature will never be free from leakage, owing to the fact that the velocity of the water will not allow any sedi- ment to settle, and in all ditches properly constructed the sediment traveling with the water at a moderate velocity is always relied upon to entirely tighten up all portions of the ditch cut through ground of a porous nature; and, again, if the velocity is too high, it will scour holes in the bottom and sides of the ditch when constructed in sandy or clay soils. By neglect- ing these points, the cost of maintaining will, un- necessarily, be increased. Third To establish the grade of a ditch when the velocity and area of same is known, one of the sim- plest methods of calculation is as follows: Multiply the velocity in feet per minute by the wetted perim- eter, in feet, and divide the result by twice the area in square feet, and the product will be the total fall in feet required to each mile. To reduce this fall to inches, for each 12 feet in length multiply by .027. Example: Suppose we have a ditch to construct for a distance of six miles, to deliver 600 miners' inches, or 900 cubic feet, per minute, or 15 cubic feet per second. To commence with, we are told that in ordinary ground the velocity should be about 3 feet per sec- ond, or 180 feet per minute. Now, knowing the velocity and the distance, the area required is ob- tained by dividing the discharge in cubic feet per second, viz, 15 by the velocity in feet per second, viz, 3, the result shows that an area of 5 square feet will discharge the quantity of water required, and, in order to have the ditch or flume constructed with least amount of friction, the width of the bottom must be from If to 2% times the height of the sides, and in this instance the section of the ditch or flume would be 3 feet in bottom, with l-foot-8-inch sides. Having now the velocity and area, we next find the wetted border or perimeter in other words, the length of so much of the bottom and sides as is wetted by the water; for instance, if a flume or ditch is 30 inches wide and 12 inches deep, its wetted perim- eter, when full, ,is 30+12+12=54 inches, or 4.5 feet, and the same ditch or flume, if empty, has no wetted perimeter at all. Now, fully understanding the meaning of wetted perimeter, we find that the ditch or flume in our example has a wetted perimeter = to 3 + 1 ft. 8 in. + 1 ft. 8 in. = 6 feet 4 inches, which, for convenience in calculating, we reduce to decimals, and have 6.33. We now have the follow- ing results, viz: Velocity, 3 feet per second; dis- charge, 15 cubic feet per second; wetted perimeter, 6.33 feet. To find the grade, we first multiply the velocity in feet per second by itself, and in this instance the result is 3X3=9, which has to be multiplied by the wetted perimeter in feet, 6.33; therefore, 9X6.33= 56.97, this total has to be divided by twice the area in square feet, viz, 5X2=10; therefore, 56.9710= 5.69, the total fall in feet per mile. This result is practically correct for flumes and ditches of short length in good ground, but allowance must be made according to the roughness and the contour of the ditch. 10- A more difficult, but correct formula, which has been obtained from actual experiments made in con- nection with ditches constructed in ordinary ground, with the usual winding course and short bends, is as follows: Velocity in feet per second = 6 times the square root of 2XGXRXS. Where G is the accel- eration of gravity, or 32.2, R is the hydraulic radius, which is found by dividing the sectional area of the ditch in feet by the wetted perimeter or border in feet, and S, the sine of inclination, or the total fall or grade in feet, divided by the total length in feet. If the ditch is constructed through rough country and the bottom or sides of same present rough sur- faces to the water, then 5 times the square root of 2 g r s will give the mean velocity in feet per second, and the velocity multiplied by the area in square feet will give the discharge in cubic feet per second, which result multiplied by 40 will give the discharge in miners' inches. Example: To find the velocity and then the dis- charge in cubic feet per second, and miners' inches from a ditch with a fairly straight course, and con- structed through good ground, having the following dimensions and fall, viz, section of ditch, 6X3 feet; fall or gradient, 8 feet to the mile; length of ditch, 15 miles. We first proceed by working out R, which we are told is the sectional area of ditch in feet divided by the wetted perimeter in feet, and in this instance il '] 6X3 18 11 1S 6+3+3=12= ^ S, or sine inclination, will be found by dividing the fall by the length, or -Jl* = .001515. uZoO Since twice G (the acceleration of gravity) is 2X 32.2, or 64.4, we have G, R and S, and our formula stands as follows: Six times the square root of 64.4X1.5X.001515, and the easiest method of calculation in this case is by logarithms, as follows: 11 Log^s. 2g=2X32.2 or 64.4=1.8089 R=1.5=0.1761 S=.001515=7.1804 SQUARE Roox=2 |T9.1654 "9.5827 X6 0.7782 10.3609=2.29 Answer 2.29 feet per second velocity, and this multiplied by the area, 18 square feet=discharge, or 41.22 cubic feet per second, or 41.22X40=1648.80 miners' inches. The above formula is also correct for flumes with sawed boards, and battens over the joints inside the boxes, but instead of using 6 or 5 as a co-efficient the formula must read SV^grs for velocity in feet per second, and 8 \X2grs X area for discharge in cubic feet per second. I might add here, that these last formulae have been practically tested by several authorities, and especially by the Government engineers of New Zea- land, to whom I believe belongs the credit of arriving at the exact co-efficients shown above. As before mentioned, it is of great importance that a safe allowance should be made for loss due to leak-- age and evaporation, more especially when the line of ditch does not pick up any small creeks or springs on its course, and it is agreed by the best authorities that a suitable allowance may be calculated by the follow- ing formula: Sectional area of ditch in feet Mean velocity in feet per secondX5280 Equals the loss in cubic feet per second per mile, where M is a co-efficient varying from 3 to 20, ac- cording to the climatic conditions of the country through which the ditch is constructed. In New Zealand, on the west coast, and, in fact, all through the middle island, good results have been ob- 12 tained by using 3 for a multiplier, but again in the North Island, where the climate more resembles this country and the loss due to evaporation is heavy, it is necessary to often use as high as 20 for M, in order to obtain satisfactory results. Fourth When owing to weak banks it is necessary to build walls on the lower side of a ditch, the ground should be removed to obtain a solid foundation, and two walls an outer and an inner should be built up, with space enough between to allow a good pud- dle clay to be rammed in; such a wall, if properly constructed, will never give further trouble. Fifth All earth, trees, roots, etc., must be moved quite clear of the lower side of the ditch, with the exception of just sufficient to make a track. Unless all waste materials are moved to such a distance that they will not become a heavy drag on the lower side of the ditch, slides will be frequent and costly. Sixth At the entrance of tunnels, commencement of flumes, and at other points where the velocity of the water is considerably retarded, the effect of water changing its form at such places is an important point, and must never be neglected; in fact, all calculations referring to the flow of water in ditches, etc., the mean velocity must be determined as accurately as possible. Before going further, many readers will appreciate the following simple method of arriving at the areas or cross-sections of the different forms of ditches, sluices and flumes, which may be calculated as follows, viz: To find the area of a section of a flume or ditch with straight sides, multiply the width of bottom (in inches) by height of sides (in inches), the product will be the area in square inches, and this divided by 144 will give area in square feet. Example : What is the area of a flume or ditch 28 inches wide and 18 inches deep? Answer: 28X18=: 504 sq. in., which, divided by 144, = 3| sq. ft. 13 To find the area of a flume or ditch with sloping sides, add the width at top and bottom (in inches) together and divide the result by 2. This answer will be the area in square inches, which, divided by 144, will give the area in square feet. Example : What is the area of a cross section of a ditch 48 inches wide at top and 26 inches wide at the bottom, with a depth of 24 inches? Answer: 48 -f- 26 = 74, which, X 24, = 1776, and this divided by 2 = 888 square inches ~ 144 = 6.16 square feet. To find the square feet of the cross section of a ditch or flume with sides sloping to a point at the bottom, multiply the width (in inches) by half the depth (in inches) and the answer will be area in square inches, which, divided by 144, gives area in square feet. Example: What is the area of a flume or ditch 70 inches wide and sloping to a point at the bottom, with a depth of 36 inches? Answer: 70X18, or half the depth in inches, = 1260 inches, or divided by 144 = 8.75 square feet. Carrying Capacity of Pipes, Discharge of Nozzles, etc. It is hardly possible to point out any portions of a hydraulic plant that are of more importance than pipes and nozzles, and in out of the way places miners have great difficulty in finding out the cor- rect sizes of pipes, and particularly the capacity of same, especially with regard to quantities of water discharged through pipes and nozzles of different diameters, there being innumerable instances at the present day where miners do not know the pressure of water is only as the head, without any regard (neglecting friction and bends) to the size of pipes. For example: A pipe line composed of 6-inch pipe, and another line of 40-inch pipe, with same fall or head, will both give the same pressure. It was only a short time since that I was asked by a miner of some experience if he could not double his pressure by doubling the diameter of his pipe. - 14 - As an illustration, this fact is easily demonstrated by attaching the same size and kind of faucet to two tanks, one, say, of 3 feet diameter and 3 feet deep, and the other as large as convenient, say 6 feet in diameter and 3 feet deep. Fill both with water to same depth, then, after placing buckets of equal capacity under each faucet, turn both on at the same time. To the surprise of any person not acquainted with hydraulics it will be seen that, although the larger tank contains four times more water than the smaller one, both buckets will be filled at the same time. It is well to thoroughly understand the principle of hydrostatics in building storage dams and reservoirs, remembering that there is the same pressure on the bank of a reservoir with water 3 feet deep and ex- tending back for a distance of 10 feet, as there would be if the water dammed back 10 miles, so long as the depth remained the same. In reference to nozzles, they require great care in construction so as to be of correct form, in order that the water leaving them will be in a solid stream, instead of scattering and thus losing its power, which is the case with nozzles of improper construction. In order to get the best effect it is absolutely neces- sary that the head of the pipe conveying the water should be at least 3 or 4 feet under water in order to prevent any air getting into the pipe, which also causes the water to scatter when leaving the nozzle. All pipe lines should be laid as straight as possible, or with curves having as large a radius as can be obtained, and it must be remembered that new pipes well coated will carry more water than old pipes that have been rusted inside; therefore, allowance must be made accordingly. It is good practice to allow one-sixth the diameter of pipes under 6 inches, and 1 inch on all diameters over 6 inches. The first thing necessary for the miner to do is to ascertain the fall available for the pipe line and its length in feet. Knowing this and the quantity of water he is going to use it is easy to determine the diameter of the pipe, always bearing in mind that in order to secure efficiency and economy in con- struction water should flow in the pipe at a velocity of not more than 3 feet per second. A few of the simple methods of determining the discharge of pipes are as follows: First To obtain the velocity, first multiply the diameter in feet by the effective head in feet and divide the result by the length of the line in feet, then take the square root of the product and multiply by 50; this will give the velocity in feet per second, and the velocity multiplied by the area of the pipe in square feet will give the quantity of water discharged in cubic feet per second, which, multiplied by 40, will give the number of miners' inches. Second To find the velocity in feet per minute, multiply the number of cubic feet of water discharged per minute by 144 and divide the product by the area of the pipe in inches. For example: An 11-inch pipe discharging 150 cubic feet of water per minute, the velocity would be 150X144-4-95.03 (the area of pipe in inches) or 227.6 feet per minute, Or 227.6 Xf =151.7 miners' inches. Third In all cases the pipe will require a funnel or bell-shaped entrance, and an additional head to put the water in train in addition to correct dimensions for overcoming friction, assuming, of course, the total head available be required. To obtain the additional head commonly termed the velocity head, a simple method is to square the velocity in feet per second and divide by 64.4 and then divide that product by 0.70. The answer will be the extra head in feet re- quired. We are told by some authorities that in cases where the length of the pipe exceeds 1000 diameters the head due to velocity and even bends may be neglected, but in practice I find it better to err on the right side, and in no case neglect working out the heads due to those losses and including them in all estimates. -16- Another simple and approximate method of find- ing the velocity is by multiplying the number of .miners' inches discharged by 11 and divide the product by three times the square of the diameter of the pipe. For example: A 10-inch pipe discharging 400 miners' inches; the velocity will be 400X11=4400, divided by three times the square of the diameter, or 3X100=300. Answer, 14.6 feet per second. By means of the same formula the number of miners' inches discharged through a pipe of a known diameter and velocity will be found as follows: Multiply the velocity in feet per second by 3 and the product by the square of the diameter, then divide by 11 and the result will be the discharge in miners' inches. For example: A pipe with a diameter of 20 inches, discharging water at a velocity of 3 feet per second, the number of miners' inches discharged will be as follows: 3X3X400 (the square of the diameter) or 3600^-lli=327 miners' inches. Fourth A more complicated but accurate formula for determining the velocity per second of water in pipes is: 140 times the square root of RXS minus 11 times the cube root of RXS where R is the hydraulic radius, which is found by dividing the diameter of pipe in feet by 4, and S the sine of inclination, which is found by dividing the total fall in feet by length of pipe line in feet. This formula has been tested in a thoroughly prac- tical manner by Mr. Gordon, the Government En- gineer in New Zealand, and he found it could not be relied upon when calculating high velocities, as it gave too great a discharge, but with low velocities and small diameters of pipe, it was deemed fairly accurate. Fifth Another set of formulae given in "Practical Hydraulics," by Thomas Box, are as follows: Where d = diameter of pipe in inches. L = length in yards. H = head in feet G = gallons discharged per minute. 17 - ,~ x. ^ l/5 . T (3d)*XH d = I ?? I -*- V k=r H ^ G 2 yfMrMO V H^i r^'4 V T G=| L Example i. Find the diameter of pipe required to discharge 300 gallons per minute with 80 feet head; length of pipe, 200 yards: 80 Log's. 300 = 2.4771 2 X 200 = 2.3010 7.2552 80 == 1.9031 5 | 5.3521 Fifth root = 1.0704 ~-3= 0.4771 .5933 = 3.92 inches diameter. Example 2. Find the number of gallons discharged by a pipe 10 inches in diameter, 900 yards in length, with a head of 50 feet: X.10) 5 X 5 I 8.025 j Where G = gallons per minute, H = head or depth of water from surface to center of sluice opening, A = area of opening. Example: How many gallons per minute will be discharged from a reservoir through a sluice gate with side walls, when the depth of water above the center of the opening is 7 feet and the opening is & feet wide and 1 foot high? Answer: G = 8.025 X ^HX .6 X A X 6.23 X 60 = E. A 8.025 X V? X .6 X 3 ft. X 6.23 X 60 = 14281.785 gal- -20- Ions per minute. Which -f- 11.25 = 1269.49 miners' inches. Again, if it is required to find the head or height of water above center of opening in a sluice gate necessary to discharge 14,300 gallons per minute through an opening 3 feet by 1 foot, we proceed as follows: f- 1 1 2 H -= { V6.23 X 60 A^-^- .6 > and in this instance 8.025 ) ( f 14300 H = < V6.2,_ oi = 2.64 2 or 6.96, or nearly 7-inch head. In many instances a sluice valve is used instead of a gate, and when the pipe attached to the valves is comparatively short, say of a length not exceeding more than three diameters, the following formula may be used: Where G = gallons per minute, H = head in feet or height of water above the center of the valve open- ing, d = diameter of valve opening in inches. Example I. How many gallons per minute will a sluice valve 10 inches in diameter discharge when the height of water is 3 feet above the center of valve opening? Answer : G = T/H X d 2 X 10 = 1/Tx 10 2 = 10 = 1732 gallons per minute, or 1732 H- 11.25 = 153 miners' inches. Example 2. With same measurements find the head required to discharge 1732 gallons per minute. Answer: G vL_ 1732 21 Example j. To find diameter of valve necessary to discharge same quantity of water with same head as in examples 1 and 2. Answer: inches diameter. In connection with the last formula it must be borne in mind that if the pipe leading to the sluice valve to reservior is much longer than three diameters, allow- ance must be made for friction, etc. Nozzles. Nozzles require great care in construc- tion so as to be of correct form, and perfectly smooth in bore, in order that the water leaving them will be in a solid stream, instead of scattering and thereby losing power, as is the case with nozzles of improper construction. In order to get the best effect from any kind of nozzle, it is absolutely necessary that the head of the supply pipe should be 3 or 4 feet under water, to avoid air getting into the pipe and causing the water to scatter when leaving the nozzle. To determine the velocity and discharge in cubic feet per second for well-made nozzles, either of the following simple methods may be followed: First Multiply the square root of the hydrostatic or effective head in feet by 8.03. This will give the velocity in feet per second, and that multiplied by the area of the discharge end of the nozzle in square feet will give the discharge in cubic feet per second, which, multiplied by 40, will give the answer in miners' inches. Example: Nozzle 4 inches in diameter, dis- charging water under an effective head of 400 feet, find velocity and discharge. The square root of the head or 400 feet is 20, and 20X8.03 equals the velocity in feet per second, viz, 160.60. Area of 4-inch nozzle in square feet = .087266, and this multi- plied by 160.60=14 cubic feet per second, or 14X40= 560 miners' inches. The result obtained by this rule is nearly the theoretical discharge, while for ordinary 22 practical results the actual discharge will be from 75 to 85 per cent of the answer obtained by this rule. Second To find the discharge in gallons per min- ute use the following formulae: G = V h XX .24. Where G = gallons discharged per minute, h = hydrostatic or effective head on nozzle in feet, and d = the diameter of nozzle in -Jths of an inch. Example: Find the discharge from a nozzle 3 inches in diameter with a head of 205 feet. The square root of the head, viz, 205 is 14.317 and the diameter in ths of an inch is 3X8=24, which squared is 24X24, or 576. Therefore, G=14.317X576X.24 equals 1979 gallons per minute, and this divided by 11.25=176 miners' inches. For accurate results the following two rules may be followed : First Discharge in cubic feet per second V 2Xgh XaX0.96, where g is the acceleration of gravity in feet per second, commonly accepted as 32.2 h = the hydrostatic or effective head in feet, and a = the area of nozzle discharge in square feet. Second Discharge in cubic feet per second = VhXd 2 Xc where h = the effective head in feet, d = diameter of nozzle in -Jths of an inch, and c = a vari- able co-efficient from .00064 to .00066. It is a well known fact that water issuing from a nozzle should, theoretically, attain the height of the head. For instance, a nozzle with 300 feet effective head should throw a stream a height of 300 feet, but we all know this efficiency cannot be reached in prac- tice, owing to the resistance of the air, and other causes, but the difference has been found by experi- ment to vary nearly in inverse ratio to the diameter of the jet, and in "Practical Hydraulics," by Thos. Box, we have a formula for approximately calculat- ing the loss of head for each case, which is as fol- lows: h' =^- X .0125. Where H = the effective d head on the nozzle in feet, h' = the difference between -23- the head and the height of discharge column from nozzle, d = the diameter of the jet in ths of an inch. Mr. Box goes on to remark that as a result of this rule each size of nozzle attains a maximum height with a certain head, and, when the head is increased beyond that point, the nozzle does not throw the stream so far, but, on the contrary, the efficiency of the nozzle greatly diminishes; a good deal owing to the fact that an excessive head, or, more plainly speak- ing, a head out of proportion to the diameter of the nozzle tends to scatter the issuing stream and cause it to meet with more resistance from the air than a jet of solid water issuing with a moderate head. Adopting Mr. Box's formula we will work out the following, and see the result: First At what distance will a well-formed nozzle of 2-inch diameter throw a stream of water having an effective head of 200 feet? Answer: V= X -0125or 4 -^ x .0125 = 31.25. ID lb ; ; . , . Therefore, the height the nozzle will throw is 200 31.25, or 168.75 feet. Second Take the same nozzle, with a head of 450 feet, and the loss will be X .0125 or - X .0125 = 158.20 feet. That is to say, the water will be discharged to a height of 450158.20, or 291.80, feet, instead of 450 feet, the theoretical height minus all friction due to the head. Friction in Pipes. This is a most important matter to those who are connected in any way with mining or other enterprises in which water is used under pressure, and very few miners are conversant with the principles relating to friction of water in pipes, etc. Most people connected with water supply have a knowledge of the fact that when large quantities of water are discharged from pipes of small diameter the pressure is greatly reduced, but few know how to arrive at a correct method of finding out the exact loss due to friction. Were it otherwise there would not be in evidence so many palpable blunders in the construction of pipe lines used for hydraulic mining and other purposes, and in many instances success would be the rule in place of failures, many of which are due entirely to errors made in bringing the water supply to the claim, and laying down pipes of too small diameter, thus reducing the effective head or pressure (in instances I have known) to less than one- half that available with pipe lines properly propor- tioned. In a previous paragraph I stated that water flow- ing through pipes should not exceed 3 feet per second, or 180 feet per minute, and if all users of water for hydraulic mining or power purposes had their pipes of the correct diameter to insure a velocity not ex- ceeding that named above there would be very little trouble, as both efficiency of water and economy in construction of pipe line would be attained. It must be understood that the friction of water in pipes increases as the square of the velocity, and also depends upon the condition of the pipes whether they are foul and rusty, or are new, or in good condi- tion. Even the rivet heads in a pipe line of consider- able length cause a good deal of friction, and, con- sequently, loss of head. There are several formulse for determining the friction in pipes, but most all of them are difficult and too complex for ordinary miners. But Mr. William Cox has simplified Weisbach's formula, and yet gives identical results. Besides it is easy to work out. It is as follows : H - X (4 X V\+ 5 V - 2) when H = friction head in feet, d = diameter of pipe in inches, L = length of pipe in feet, and v = velocity of water in feet per second. Example: What is the loss in head of a pipe line discharging 400 miners' inches or 600 cubic feet per minute, diameter of pipe being 12 inches, and length of line 5,000 feet We must first find the velocity in feet per second, and to do this we use a simple formula, given in remarks on velocity of water through pipes, in a previous paragraph, viz, multiply number of cubic feet of water discharged per minute by 144 and divide the product by the diameter of the pipe in inches. Therefore, in this case, velocity = 600 X144113.10 the area in inches, or 763.9 feet per minute or 12.73 feet per second. Now knowing the velocity, diameter and length, we will find H, or friction head, as follows: 4V2 + 5V - 2 > 500 X-3472 1200d 12 X 1200 4 X 12.73 2 -f 5 X 12.73 2 = 709.86 and this X .3472 = 246.44 feet or friction head. That is to say, if we had in this example a fall of 500 feet, and constructed a pipe line with pipes 12 inches in diameter, having a total length of 5,000 feet, our actual head of 500 feet would be reduced to 500246.44 or 253.56 feet, or, putting it more plainly, we would have a pressure of 110 pounds to the square inch instead of 217 pounds, and this loss is due entirely to using pipes of too small a diameter. Another formula I use, and which gives clear re- sults, is as follows: H =2 2ga H = loss of head by friction in each 100 feet of pipe. p = the perimeter, or circumference of the pipe in feet. 1 = 100 feet. c a variable co-efficient from .00406 to .01338, according to the nature of the pipe, and velocity of water. v = velocity of water in feet per second. g = the acceleration of gravity, or 32.2 feet. a = the sectional area of pipe in feet. 26 Example : A pipe line 5000 feet in length, of newly- riveted pipe, 20 inches in diameter, with a head of 650 feet between the supply and discharge ends, and de- livering 400 miners' inches of water, what is the loss of head? First determine the velocity, which is 4.58 feet per second, then p 1 c v 2 _ 5.235 X 100 X .00506 X 20.9 _ 55.372 2 X 32.2 X 2.181 ~ 140.456 ~ .394 feet, or .394 feet loss for each 100 feet of line, and there being 5000 feet of pipe, the loss will be 5000-^100X.394, 19.70 feet, and the actual pressure head would in this case be 650 feet, less 19.70 feet frictional head, or 630.30 feet. It is well to remember that it makes no difference whether the water is flowing up hill or down, or whether the pressure is great or small, the total fric- tion will be materially the same, and that in wooden pipes the friction is nearly double that of iron or steel. Loss of Head Due to Bends and Angles. This loss is also an important one, and in many instances is great, owing to the number of sharp bends or changes in the direction of a line of pipe, carrying water for min- ing, or other purposes. In a pipe line there should be no bends having a radius of less than five diameters. To calculate the loss of head due to the resistance of a right angle bend, the simplest rule is to obtain the velocity of water flowing in feet per second due to the head, and multiply the square of such velocity by .0152. For example: What is the loss of head due to the resistance of a 90-degree elbow, with water flowing at a velocity of 15 feet per second? Answer: 15 2 X.0152, or 3.42 feet. Where the radius of the bend is greater, or more than 5 diameters, the head required to overcome the resistance can be found by multiplying the square of the velocity in feet per second by the number of de- grees in the angle, and dividing the product by 88489. For example: Velocity 10 feet per second, what is the resistance of a bend having an angle of 120? When the radius is less than 5 diameters, the resist- ance would be as per following rule: Mean velocity squared, divided by 64.4, multiplied by the square of half the angle of deflection, multiplied by 2.06 times the 4th power of the same angle. For fairly accurate results, this formula may be simplified by multiplying the square of the velocity in feet per second by C, C being equal to the following co-efficients for the various angles, viz : C = .000109 for angle of 10 degrees. C = . 000466 " " " 20 " C = .001134 " " " 30 " C = . 002158 " " " 40 " C = .003634 " " " 50 " C = . 005652 " " " 60 " C = . 008276 " " " 70 " C .011491 " " " 80 " C = . 015248 " " " 90 " Hydraulic Motors, Water Wheels, etc. There is no power easier handled or less complicated than water power, and it would take too much space to give in detail a history of all the various forms of motors, but I -will endeavor to describe the particulars and effi- ciency of the most popular methods. The power of a fall of water is easily calculated, and is found as follows: Multiply the number of cubic feet per minute by the weight per cubic foot, or 62-J pounds, and the product by the fall in feet, then divide by 33,000. For example: What is the horse-power in a body of water equal to 60 cubic feet per minute, or 40 miners' inches, having a fall of 200 feet? A 60 X 62.5 X 200 Answer: 33000 - which is the total power in the water, and from this result allowance must be made for friction, etc. Another easy method of calculating the power is to remember that one cubic foot of water flowing per minute and falling 1 foot is equal to .0016098 horse- power, and that one miners' inch falling 1 foot is equal to .0024147 horse-power. These multipliers will give result equal to about 85 per cent of the theoretical power. Example i. What horse-power can be obtained from 40 cubic feet of water per minute falling 300 feet, using a motor giving about 85 per cent efficiency? Answer:^ .0016098X40X300=19.31 H. P. Example 2. What horse-power can be obtained from 100 miners' inches of water falling 60 feet, using a motor giving about 85 per cent efficiency? Answer: .0024147X100X60=14.48 H. P. The most common forms of wheels used by many miners are current or bull wheels, undershot wheels, breast wheels and overshot wheels. The efficiency of these various wheels vary about as follows: Current, or bull wheels, 20 to 50 per cent; undershot wheels, 27 to 35 per cent; breast wheels, 45 to 60 per cent; overshot wheels, 60 to 75 per cent. The next class of motors includes the various forms of turbines, of which there are numerous varieties, many of them giving as high as 85 to 87 per cent. Approximate and simple rules for finding quantity of water, height of fall and horse-power developed at an efficiency of 75 per cent are as follows : Quantity of water in cubic feet per minute is found by multiplying the horse-power by 706 and dividing the product by the fall in feet. Example : How much water is required with 200 feet of a fall to develop 10 horse-power, using a motor giving 75 per cent efficiency? Answer: 10X706-f-200=35.3 cubic feet per minute, or about 24 miners' inches. To find how much fall is required to generate a required horse-power, with a known quantity of water, multiply the horse-power by 706 and divide by the quantity of cubic feet per minute. Example: Having a supply of 60 cubic feet per minute, and requiring 20 horse-power from motor giving 75 per cent efficiency, what fall is necessary? Answer: 20X706^60=235.3 feet. To find the horse-power in a fall of water when the fall and quantity are known, multiply the number of cubic feet per minute by the height of fall and divide by 706. Example: Having a fall of 100 cubic feet per min- ute, and a fall of 75 feet, what horse-power can be obtained from a motor giving 75 per cent efficiency? Answer: 100X75-^-706=10.6 H. P. Should the motor be of a class that would give only 65 per cent efficiency, such as most overshot wheels, it is necessary to use 815 as a multiplier instead of 706. Current or Bull Wheels. To calculate the horse- power of a current or bull wheel, it should be under- stood that in all such motors the velocity of the periph- ery of the wheel, or, more plainly speaking, the num- ber of feet per second the rim of the wheel is traveling, should never vary much from half the velocity of the stream, or half the velocity due to the head of water. In this class of wheels the diameter is seldom less than 6 feet, or greater than 16 feet, and the number of floats 7 to 13. The inclination of floats from radial lines should be between 20 and 30 degrees, depth of floats from 10 to 16 inches, and they should be immersed for about one-half their depth. The horse-power of this class of motors is found by the following formula: H = .0028XVXMXAX (V M). M = VX.55. x = cosine of angle between the floats, multiplied by the radius minus the radius, or the distance below a horizontal line produced from under the extremity of the vertical float. Where H = horse-power. V = velocity of current in feet per second. M = the mean velocity of the periphery of the wheel in feet per second. A = the immersed area of the floats in ;'*** square feet. *P. - so- To obtain the angle between the floats, divide 360 (the number of degrees in a circle) by the number of floats on the wheel. Example: A current or bull wheel, 16 feet in diam- eter, having 12 floats, each of which are 8 feet long, with a maximum immersion of 15 inches, what is the horse-power of the wheel, when the stream has a ve- locity of 7 feet per second? Answer: Angle between floats = 360 .-4- 12 or 30 degrees. x = the cosine of the angle of 30 degrees, or . 86603 X by the radius or 8 feet = 6.92824 the radius, 8 feet = 1.0717 or 12.86 inches. This is the distance that the second float will be above the hori- zontal line produced from under the extreme edge of the vertical float, thus showing that although the maximum immersion of any of the floats is 15 inches the adjoining floats would be 12.86 inches higher, and to get at the area of the immersed floats, the depth of the second float will be 15 12.86=2.14 inches, and the area of the three immersed floats=15+2.14-j-2.14 inches, or 1.606 feet multiplied by length of floats, or 8 feet=12.84 feet area, and the velocity being 7 feet per second, M or the mean velocity of the periphery of the wheel=7X.55 or 3.85 feet per second. V M A VM Now, H = .0028X8X3.85X12.84X (7 3.85) = 3.49 H. P. Undershot Wheels. To determine the horse-power of an undershot wheel, with a rim .velocity equal to about one-half, or .57 times the velocity due to the head of water, or .57XV2gh: Where g is the accel- eration of gravity, commonly taken as 32.2, and h is the head of water in feet above the bottom of the wheel, H =. 00066 Qh or ^ X 0.35. 1511 X H - 31 - Where h = head of water. Q = quantity of water in cubic feet per minute. W = weight of water in pounds. H ^effective horse-power. Example: what horse-power is obtainable from an undershot wheel, with about 35 per cent efficiency, using 1500 cubic feet per minute, with a head of 2 feet? Answer: H=.00066X1500X2=1.98 H. P., or Q = s ~ =1496 cubic feet per minute. Breast Wheels. The following calculations will ex- plain how to arrive at the effective horse-power of breast wheels: Low breast wheels : H =. 00104 Q X h. 961 X H H _WXh ^ " ~"h~ r ' * ' "33000" High breast wheels: H = .00108 Q X h. ^ 928 X H WXh -_ n -hT- r ' H = : ^3000 Where Q=quantity of water in cubic feet per minute. h = head of water in feet. H = effective horse-power. W = weight of water in pounds. Example I. A breast wheel 16 feet in diameter, using 1500 cubic feet of water per minute, under an 8-foot head, what will be the horse-power of the Q h wheel? Answer: H=.00104X1500X8=12.48 H. P. Example 2. How much water will be required under a 10-foot head to generate 25 horse-power with a breast wheel 21 feet in diameter? Answer: Q = 961 XH = 961-21 minute. In either of the above cases if the wheel takes the water above one-half its diameter, of course the power would be increased, and calculations should be made by using the formulae given for high breast wheels. For example : A wheel 22 feet in diameter, using 500 cubic feet of water per minute, under a head of 16 feet (that is to say, the water goes into the buckets at a point 16 feet from the bottom of the wheel), what horse-power will the wheel give? Answer: H = .00108 X 500 X 16 8.64 H. P. ; and Q = - - = 500 cubic feet per minute. Overshot Wheels. I think I am safe in stating that this class of wheel is more popular than any other class of rough-and-ready water motors. To find the horse-power multiply .00123 by the quantity of water in cubic feet per minute used on the wheel. Then multiply the result by the head of water in feet. To find the quantity of water required in cubic feet per minute to generate a given horse-power with a known fall, multiply the horse-power by 815 and divide by the fall, or head, of water in feet. Example I. What horse-power can be obtained from an overshot wheel 12 feet in diameter, using 200 cubic feet per minute? Answer: .00123X200X12= 2.95 H. P. Example 2. How many cubic feet of water per min- ute is required to generate 20 horse-power, using an overshot wheel 22 feet in diameter? Answer: 20X 815^-22=741 cubic feet Strength of Materials. All those engaged in mining should thoroughly understand this important subject, and especially that branch relating to the breaking and working strain of ropes, bolts, chains, etc. The number of accidents and fatalities arising from ignorance on this subject should be sufficient to com- pel all owners of mining property to insist upon super- 33 intendents and foremen being able at all times to pro- vide ropes, bolts, chains, etc., of proper dimensions for the work required, and in this manner more work would be accomplished, and with far less risk and ex- pense, than is generally the case. I will first deal with wooden beams, and as a guide to further calculations the following table compiled from various authorities will be found useful. The columns marked S, N, C, E have the following inter- pretation : 6* Breaking load at center of beam when sup- ported at both ends. N Breaking load when placed at one end and the other end fixed. C Safe load in center when beam is supported at both ends. E- Safety load at end of beam when the other end is fixed. The following table is based on a factor of safety of 7; that is to say, the safe load shown in table is only one-seventh of the calculated breaking load, and even this high factor of safety should be increased when using beams not free from knots and shake. In addition to the factor of safety freedom from knots and shake, it must also be remembered that seasoned tim- ber resists crushing much better than green timber, in many cases twice as well, and the figures given in this table are for good samples of timber; therefore, the factor of safety (7) should be adhered to. The following figures and results are obtained from experiments with small pieces of timber, and, there- fore, considerable allowance must be made for beams that are not of a uniform texture. I have only men- tioned the various kinds of American woods that are in general use. Their various breaking strength and safe loads are given in round numbers, so that in working out different problems calculations may be made as simple as possible: -34 - Breaking Safe load Weight in Ibs. load in los. in Ibs. Name of Wood. per. cu. ft. S. N. C. E. Ash 45 590 147.50 84.29 21.07 Beech, white 43 440 110 62.86 15.71 Beech,red 44 570 142.50 81.43 20.36 Birch, black 45 680 170 97.14 24.29 Birch, yellow 44 440 110 62.96 15.74 Cedar, white 35 250 62.50 35.71 8.99 Fir, black 42 340 85 48.57 12.14 Hickory 50 700 175 100 25 Hickory, bt. nut 40 480 120 68.57 17.14 .Larch 35 300 75 42.96 10.71 Oak, live 54 621 155.25 88.71 22.18 Oak.red 53 562 140.50 80.28 20.17 Oak, white 49 581 145.25 83 20.75 Pine, red 40 509 127.25 72.71 18.18 Pine,pitch 41 576 144 82.28 20.87 Pine, yellow 33 395 98.70 56.43 14.71 Pine, white 34 410 102.50 58.57 14.64 Pine, Virginian 38 485 121.25 69.28 17.32 Teak 56 673 168.25 96.14 26.03 The above figures represent the number of pounds required to fracture the various kinds of wood having an area or cross section of 1 square inch by 1 foot in length. In finding the strength of beams the follow-; ing proportions of strength must be observed: VALUES OF X. (1.) With a beam fixed at one end and loaded at the other=l. (2.) With a beam fixed at one end and the load distributed uniformly=2. (3.) With a beam supported at both ends and loaded at the center=4. (4.) With a beam firmly fixed at both ends and loaded at the center=6. (5.) With a beam supported at both ends and uni- formly loaded=8. (6.) With a beam firmly fixed at both ends and uniformly loaded=12. In calculating the strength of beams, the whole weight of the material must be included when the beam has a uniform load and only half the weight of material when the load is placed at the center. Let S=tabular number of breaking load in pounds on a beam supported at both ends and loaded at the center. - 35- N~tabular number for breaking load in pounds on a beam loaded at one end, and the other end firmly fixed. C=Safe load in pounds on center of beam sup- ported at both ends. E=safe load in pounds on a beam fixed at one end and loaded at the other. b = breadth of beam in inches. d = depth of beam in inches. 1 = length of beam in feet. w = breaking load in pounds. R = w divided by 7=safe load in pounds. x = proportion of strength due to position of load, and method of fixing the ends of the beam. Square Beams. The formula for finding breaking load is as follows: bd 2 N w = = X x Example I. What is the breaking load on a beam of American yellow pine, 12 inches deep, 12 inches broad and 20 feet long, one end of same being firmly fixed and the load at the other end? Answer: b d 2 N 12 X 12 2 X 98.7 _ 12 X 144 X 98.7 _ w== " ~20~ ~20~ 1 85027.68 Ibs., or 37.95 tons. And the safe load would 37.95 , , ^ be = or 5.42 tons. Example 2. With same dimensions, but beam sup- ported at both ends and loaded in the center, the breaking load will be: b d 2 N 12 X 1 ?n X 98 ' 7 X * = 37.95 tons X 4 = 151.80 tons. 20 , no l ff U 151.80 And the safe load would be ^ or 21.68 tons. Example j. If the beam in example 1 was uni- formly loaded, the safe load would be the same, viz, 5.42 tons multiplied by x, and by referring to the different proportions of strength tabulated we find in this instance where the beam is fixed at one end and uniformly loaded, x=2. Therefore the safe load would be 5.42X2=10.84 tons, thus showing that a beam will safely stand double the load when uniformly loaded than it will with load in the center. Example 4. A beam of American pitch pine, 6 inches wide and 10 inches deep by 15 inches long, supported at both ends and uniformly loaded, the breaking and safe loads will be as follows: w = - X x Safe load = -= b d 2 N fi V 10 2 V 144- x w = g_ A * = 576 lbs - X 8 = 4608 lbs -> or \ 20.57 tons. 20.57 And the safe load would be ^ or 2.94 tons. Example 5. Take the same beam as used in ex- ample 4 and lay it flatwise, or, that is, call it 10 inches wide and 6 inches deep, then the breaking strain would be: b d 2 N 10 V ft 2 V 144- x IkS. L^ = 3456 X 8 = 27648, or 12.34 tons. 1 tons. And the safe load would be '= = 1.76 tons, or a little more than one-half the strength of the same beam laid edgewise. Round Beams In order to find the strength of a circular beam it is necessary to first work out the breaking load of a square beam of which each side is equal to the diameter of the circular beam, and multi- - 37 ply this load by .589, so the formula for the break- ing load will read: W = J*1J? XXX 0.589. Example I. Having a round beam of American cedar, 40 feet long between supports and uniformly loaded, with a diameter of 10 inches, the breaking load would be: 1 7362 3.29 tons, and the safe load would be -=- = 1052 Ibs. Oval Beams In this case first find the load for a rectangular or square beam, with sides equal to the two diameters of the oval beam, and multiply the result by 0.6. Example. Having an oval Learn of American white pine, firmly fixed at both ends and loaded in the center, having 15 feet between supports, the smallest diameter 10 inches, the largest diameter 12 inches, and placed so that it would be 10 inches wide and 12 inches deep, the breaking strain or load would be: bd 2 N ' '* ' --- j- X x X 0.6. Here, on referring to the multiplier given in pro- portion of strength, we find x=6, and, therefore, w i b oxiWi(L j x 6 35424 or 15 1 15 8 15.8 tons, and the safe load would be ^ = 2.26 tons. Triangular Beams To find the breaking or safe load for this class cf beams, first find the strength of a square beam with equal sides and divide the result by 3. According to experiments made by Barlow and others, it was found that triangular beams were J- stronger where the base of the triangle was up, and 38 it was found necessary to provide in the support a triangular notch in which to place the sharp edge of the beam. Strength of Chains. The strength of chains varies, owing to the natur- of the iron from which they are made and their mechanical construction. The strength also varies as the square of the diameter of the iron from which the links are made. Experiments show that a single-link chain from good iron carefully welded, made from 1 inch diameter round bars, has a safe working strain of six tons. Great care must be exercised in using chains on loads that would cause disaster in case of breaking, and each link should be carefully examined, always bearing in mind that the strength of the chain is only equal to the strength of the weakest link. Many serious accidents have been caused by not paying proper attention to this fact. There are many formulae for determining the safe load for and breaking strength of chains, all of which are only approximate and depend upon careful exam- ination prior to attaching the load. For crane chains the breaking load can be found by multiplying the square of the circumference of the link in inches by 32 A, and for the safe load divide the result by 6. Example: What is the breaking load of a chain made with links from a round bar of iron f of an inch in diameter? Answer: 75 2 X32.4=18.225 tons, and , t , 18.225 _.. safe load = - = 3.04 tons, nearly, b Another simple and approximate result is to divide the square of the diameter of the iron from which the links are made by 9, and the result is the safe work- ing strain or load. Example : A chain made from round iron J inch in diameter, the safe load would be 4 (the number of eighths of an inch in -J inch) squared, or 16 divided by 9=1.8 tons, the safe load. The same result, nearly, is obtained by squaring the diameter of the link iron in inches and multiplying by 7.111. Taking the above example the answer would be: .5 2 or .25X7.111=1.77 tons. To find the diameter of the iron in eighths of an inch that the links should be made from, to safely support a given load, proceed as follows: Multiply the weight to be hoisted or hauled in tons by 9 and extract the square root of the product, and the answer will be the number of eighths of an inch there should be in the diameter of the links. Example: What sized iron should the links of a chain be made from to safely support a load of two tons? Answer: V2X9=V 18=4.24 eighths of an inch, or a little over inch in diameter. Some readers will no doubt recall instances to their minds where they have lifted much heavier loads with such chains as shown in the examples given. Al- though this is often done without any bad results, nevertheless these rules should be followed whenever possible, and especially when chains are used for long pulls and subject to heavy strains, as it is much better to err on the safe side. While dealing with the strength of chains, it may be of interest and information to readers to have the following table which I have in my pocketbook and often find of great value in making calculations as to strength of iron, bolts, bars, rivets, etc.: Tensile Strength and Shearing Strain of Iron and Steel. Tensile Strength Shearing Strength per Sectional In. per Sectional In. Cast Iron 7 9 Wrought Iron rolled bars 25 20 Best Lowmoor rivets 29 23 Cast Steel, best quality for tools 52 39 Double Shear Steel 40 30 Cast Steel Boiler Plates 48 36 Puddled Steel Boiler Plates 42 31^ Bessemer Steel Boiler Plates 32 24 Steel Bars 45 31Vi Cast Steel Rivets 49 37 Wrought Iron Plates , length way 22& 18 Wrought Iron Plates, crosswise 20V 1654 It is as well to explain, for the benefit of some readers that tensile strength of any material is the weight attached to the end of a bar that will tear it asunder, and the shearing strength is the weight or pressure that will cut the material through. 40- In calculating the strength of screw bolts, of course, a proper allowance must be made for the thread, and an approximate allowance is to deduct A from the diameter of small bolts, and from ^ to i for large bolts. For instance, a bolt % an inch in diameter, deduct -J of an inch for thread and calculate the strength of said bolt as if it were a f bolt, instead of a ^ inch, and with a bolt 2 inches in diameter, deduct and call it If, when calculating its working or safe load. In order to fully explain the use of the above table of tensile strengths and shearing strains, I give an example as follows: What is the tensile strength of a bar of iron 2 inches in diameter? Here 2 inches in diameter has an area or cross- sectionto 3.141 sectional inches, and 3.141X25= 78 52 78J tons, and the safe load would be ^=11.22 tons. Strength of Hemp, Manilla, Iron and Steel Rope. This subject is a most important one, and every super- intendent and mine foreman should be thoroughly conversant with the mode of calculating the breaking strain, and, more particularly, the safe load that ropes of different material will stand, as the lives of the men employed, especially in deep mines, are dependent entirely upon the safety of the ropes used in hoisting, etc.; in fact, men working in mines have a right to demand that the employers have a thorough knowl- edge of this subject. A simple test for the purity of manilla or sisal ropes is as follows: Take some of the loose fiber and roll it into balls and burn them completely to ashes, and, if the rope is pure manilla,. the ash will be a dull gray- ish black. If the rope be made from sisal, the ash will be a whitish gray, and, if the rope is made from a combination of manilla and sisal, the ash will be of a mixed color. For calculating the breaking strain of round ropes of different materials, the following table is one of several. Where B = breaking strain in tons, and C = circumference of rope in inches. 41 B = C 2 XO.m for hemp rope. B = C 2 X0.2 ordinary fiber rope. B = C 2 X1.5 iron wire rope ordinary. B = C 2 X2.5 steel wire rope. B = C 2 X2.09 flexible galvanized wire rope. B = C 2 X2.60 extra flexible galvanized wire rope. B = C 2 X4.18 plough steel rope. The working or safe load should be taken as about J ~D or y of the breaking load B, or - 6. The weight of ropes can also be approximately cal- culated from the circumference, as follows: Where W weight of each 100 feet in pounds, and C=cir- cumf erence of rope in inches, as follows : W=C 2 X4.16 for each 100 feet of hemp or fiber rope. W=C 2 X14.54 for each 100 feet of iron or steel rope. A splice weakens a rope about one-eighth, and it is well to remember that a three-strand rope is about one-fifth stronger than a four-strand one of the same dimensions. The Bursting and Working Strain of Iron and Steel Pipes, Plates, etc. Under this head I will deal only with wrought-iron and steel pipes as are generally used in connection with mining work, and in arriving at the safe working strain or pressure it must be under- stood that the following rules depend upon good workmanship, correct diameters and distance apart in riveting, etc. In making wrought-iron pipe care must be taken to have the plates rolled lengthwise, as it generally affects the strength of the longitudinal seams that is to say, the plates should be rolled across the grain, and not with it. The simplest method of calculating the pressure that wrought-iron and steel pipes will stand is as follows : Where P~safe working pressure. T = tensile strength of plates, taking iron at 48,000 pounds to the square inch and steel at 75,000 pounds. t = thickness of plates in inches or decimals of an inch. R = radius of pipe in inches. f = proportional strength of plates, as follows: when double-riveted=0.7 and single-riveted=.5. c = a co-efficient or factor of safety usually taken at 3. p = pressure in pounds per square inch due to head of water. Example: What is the safe working pressure for a 36-inch pipe, double-riveted along the longitudinal seams, and made from wrought-iron plates rolled across the grain, and $ of an inch thick? Answer : P = ^^ ^ I X I A 0223