also in feet, and that product by 5.1; then of the last product equal the quantity discharged in cubic feet per second. 2. When the aperture is under a given head.---Multiply the area of the aperture in feet by the square root of the depth also in feet, and by 5.1; the product is the quantity discharged in cubic feet per second. Example 1.-Required the quantity of water in cubic feet per second discharged through an opening in the side of a dam or weir, the width or length of the opening being 6 feet, and the depth 9 inches, or 75 of a foot. Square root of 75=866. Then 6.5×75×866×51×2 14-3839 cubic feet. Example 2.-What would be the quantity discharged through the above opening if under a head of water 4 feet in height? Square root of 4=2, and 2×51=10.2 feet velocity of the water per second. And 6.5×75×2×5.1=49.725 cubic feet discharged in the same time. STATICS. PRESSURE OF EARTH AGAINST WALLS. e 8 d C Let abcd be the vertical section of a wall, behind which is a bank or terrace of earth, of which a prism whose section is represented by dag would detach itself, and fall down, were it not prevented by the wall. Let ag be the line of rupture or the natural declivity which the earth would assume, were it not for the resistance of the fh wall. n a In sandy or loose earth the angle hag seldom exceeds 300, in firmer earth it becomes 370, and in some favorable cases more than 150 It has been found, however, theoretically, and confirmed experimentally, that the angle formed with the vertical by the prism of earth that exerts the greatest horizontal stress against the wall, is half the angle which the natural slope of the earth makes with the vertical. Then the resultant, I n, of the pressure of the bank behind a vertical wall is at a distance, a n, from the bottom of the wall= of da. Of the experimental results, the best which we have seen are those of M. Mayniel, from which the following are selected, viz.: That the friction, in vegetable earth, is the pressure; in sands. The line of rupture, ag, in a bank of vegetable earth, is=618 of da. When the bank is of sand, then dg=677 of da. When the bank is of vegetable earth, mixed with small gravel, then dg=646 of d a. If it be of rubbles, then dg=414 of da. THICKNESS OF WALLS, BOTH FACES VERTICAL. Brick.-Weight of a cubic foot=108 lbs. avoirdupois, bank of vegetable earth carefully laid course by course, de='16 ad. Unhewn Stone.-135 lbs. per cubic foot, bank as before d e=15 a d. Brick. Bank clay, well rammed, de 17 ad. Hewn Freestone.-170 lbs. per cubic foot, bank vegetable earth, de=14 a d. Bricks.-Bank of sand, d e=33 ad. Unhewn Stone. Bank of sand, de=30 ad. When the earth of the bank or terrace is liable to be much saturated with water, the proportional thickness of wall must be at least double. (See Gregory's Mathematics for further particulars, page 224.) MOTION. Motion, generally, is the effect of impulsive force, or the act of changing place; in Mechanics, it is understood as the act of transmitting power, or the means by which power is distributed. Equality or inequality of motion is as the diameters of the wheels or pulleys by which it is transmitted. THE VELOCITY OF WHEELS. The relative velocity of wheels is as the number of their teeth. To find the Velocity or Number of Turns of the last Wheel to one of the first. Rule. Divide the product of the teeth of the wheels that act as drivers by the product of the driven, and the quotient is the number. Example. If a wheel of 32 teeth drive a pinion of 10, on the axis of which there is one of 30 teeth, acting on a pinion of 8, what is the number of turns of the last? To find the Proportion that the Velocities of the Wheels in a Train should bear to one another. Rule. Subtract the less velocity from the greater, and divide the remainder by one less than the number of wheels in the train; the quotient is the number, rising in arithmetical progression from the less to the greater velocity. Example. What are the velocities of three wheels to produce 18 revolutions per minute, the driver making 3 revolutions per minute? 18-3-15 3-1-2-7-5; then 3+7.5=10.5, and 10.5+75=18; thus, 3.10.5 and 18 are the velocities of the three wheels. To find the Number of Teeth required in a Train of Wheels to produce a certain Velocity. Rule. As the velocity required is to the number of teeth in the driver, so is the velocity of the driver to the number of teeth in the driven. Example. If the driver has 90 teeth, makes 2 revolutions, and the velocities required are 2.10 and 18 what are the number of teeth in each of the other two? 2d wheel, 10:90::2: 18 teeth. THE TEETH OF WHEELS. To Construct a Tooth. Divide the pitch into 10 parts. Let 3.5 of these parts be below the pitch line, and 3.0 of them above. The thickness should be 47 of the pitch. The length should be 6.5 of the pitch. The Diameter of a Wheel is measured from the Pitch line. The wood used for teeth is about the strength of cast iron; therefore they should be twice the depth to be of equal strength. To find the Diameter of a Wheel, the Pitch and Number of Teeth being given. NOTE. The pitch, as found by this rule, is the arc of a circle; the true pitch required is a straight line, and must be measured from the centres of two contiguous teeth. To find the Pitch, the Diameter and Number of Teeth being given 3-1416 -2-radius. The Chord and Versed Sine being given, to find the Diameter. Rule. Add the square of half the chord of the are to the square of the versed sine, and divide by the versed sine. Given the velocities of two Wheels or Pulleys, and the number of teeth or dameter of the one, to find the number of teeth or diameter of the other, as required for the given velocity. Rule. Multiply the number of teeth or diameter given by its velocity, and divide the product by the other's velocity; the quotient is the number of teeth or diameter required. Examples.-1. The velocity of a wheel containing 120 teeth is 30 revolutions per minute: required the number of teeth in another to make 300 revolutions in the same time. 2. The diameter of a pulley is 37 inches, and making 45 revolutions per minute: what must be the diameter of another to make 28 in the same time? To determine the diameters of a Pair of Wheels in contact with each other, their velocities and the distance of their centers apart being given. Rule. Divide the greatest velocity by the least; the quotient is the ratio of diameter the wheels must bear to each other. Hence, divide the distance between the centers by the ratio, plus 1; the quotient equal the radius of the smaller wheel; and subtract the radius thus obtained from the distance between the centers; the remainder equal the radius of the other. Example. The distance of two shafts from center to center is 50 inches, and the velocity of the one is 25 revolutions per minute; the other is to make 80 in the same time: required the diameters of the wheels at the pitch lines suitable for the purpose. 80-25-3-2, ratio of velocity, and 50 3.2+1 = 11.9, the radius of smaller wheel; then 50-119-381, radius of larger: hence, their diameters equal 23.8 and 76-2 inches. TABLE, Whereby to find the Diameter of a Wheel for a gwen Pitch of Teeth, or the Pitch of the Teeth to a given Diameter. No. Teeth. Diameter. No. Teeth. Diameter. No. Teeth. Diameter. No. Teeth. Diameter. 45 14.3355 90 28-6537 135 42-9757 180 57-2986 46 14-6536 91 28-9719 136 43-2939 181 57-6169 47 14-9717 92 29-2902 137 43-6122 182 57-9352 48 15-2897 93 29-6084 138 43-9305 183 58.2535 59 15-6078 94 29-9267 139 44-2488 184 58-5718 50 15-9259 95 30-2449 140 44-5671 185 58-8901 51 16-2440 96 30-5632 141 44-8854 186 59-2084 52 16-5621 97 30.8814 142 45-2036 187 59-5267 53 16-8803 98 31-1997 143 45-5219 188 59-8450 54 17-1984 99 31-5179 144 45-8402 189 60-1633 |