The Chord and Versed Sine being given, to find the Diameter. Rule.-Add the square of half the chord of the arc 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 diameter 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=32, ratio of velocity, and 50 3.2+1 11.9, the radius of smaller wheel; then 50-119-38 1, radius of larger hence, their diameters equal 23·8 and 76-2 inches. TABLE, Whereby to find the Diameter of a Wheel for a given 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. 50 15.9259 95 51 16.2440 96 52 30-2449 140 30.5632 141 16.5621 97 130.8814 142 45.2036 187 59-5267 53 16.8803 98 31.1997 143 45.5219 188 59-8450 54 17.1984 99 31.5179 45.8402 189 60-1633 44.5671 185 58.8901 59.2084 144 Rules.-1. Multiply the diameter in the table by the given pitch the product is the diameter of the wheel at the pitch circle. 2. Divide the given diameter by the diameter in the table, and the quotient is the pitch of the teeth. Examples.-1. Required the diameter of a wheel at the pitch circle to contain 48 teeth of 2 inches pitch. 15.2897 × 2.5=38.224 inches. 2. The diameter of a wheel at the pitch line to be 86 inches, and is required to contain 90 teeth what will be the pitch of the teeth? By which to determine the Number of Teeth or Pitch of small Wheels on the Manchester Principle. NOTE. The pitch is reckoned on the diameter of the wheel in place of the circumference, and distinguished as wheels of 8 pitch, 12 pitch, &c. Example. To find the number of teeth that a wheel of 16 inches diameter will contain of a 10 pitch. 16×10=160 teeth, and the circular pitch='314 inches. The diameter of a wheel for a 9 pitch of 126=126=14 inches diameter, the circular pitch of which is 349 inches. TABLE Of Dimensions of the Teeth of Wheels, that will safely transmit a given Number of Horses' Power at a given Velocity. The Diameter of the great Wheel of a Clock or other Wheel-work being 3-2 Inches and 96 Teeth, and the Pinion which it is to drive has 8 Leaves: Required the Distance their Centers should have so as they may pitch properly together. Rule. As 98.25 is to 3200, the diameter of the wheel; so is half the sum of the wheel and pinion added together to a number which shall be the distance required. Example.-96+8=104, and 104÷2=52. Then 98.25: 3200 :: 52: 1693 6, the distance which the centers ought to have, which is 16 inches, and a very small fraction (93%) besides. Having the Distance of the Centers, it is required to find the Diameter of a Wheel of 96 Teeth and the Diameter of a Pinion of 8, which it is to drive, so that they may pitch properly together. The half sum of the wheel and pinion is 52, the distance of the centers is 1693 6, and the wheel 96+2:25; then say as 52 : 1693 6 :: 98 25:31999, which number equals the distance of the wheel, 96. To find the Diameter of the Pinion, = Say as 52 1693 6: 9.5: 309-4 the diameter of the pinion. NOTE. The above rules are equally correct for clock-work, mill-work, or any kind of wheel-work, and their extreme accuracy is rarely to be met with. To divide the Circumference of a Circle into any given Number of Parts, whether Even or Odd. Example. As there are odd numbers of teeth in some wheels, it is proper to show how to divide the circumference of a circle into any given odd or even number of parts, so as that number may be laid upon the dividing-plate of a cutting engine. There is no odd number but from which, if a certain number be subtracted, there will remain an even number easy to be subdivided. Thus, supposing the given number of equal divisions of a circle on the dividing-plate to be 69, subtracting 9, there will remain 60. Every circle is supposed to contain 360 degrees; therefore say, as the given number of parts in the circle which is 69 is to 360 degrees, so is 9 parts to the corresponding arc of the circle that will contain them; which arc, by the rule of Proportion, will be found to be Therefore, by the line of chords on a common scale, or rather on a sector, set off 46 (or 46%) degrees with the compasses in the periphery of the circle, and divide that arc or portion of the circle into 9 equal parts, and the rest of the circle into 60, and the whole of the circle will be divided into 69 equal parts, as was required. Again, suppose it were required to divide the circumference of a circle into 83 equal parts, subtract 3 and 80 will remain. Thus, as 83 parts are to 360, so, by the rule of proportion, are 3 parts to 13 degrees and one-hundredth part of a degree, which small fraction may be neglected. Therefore, by the line of chords and compasses, set off 13 degrees in the periphery of the circle, and divide that portion or arc into 3 equal parts and the rest into 80, and the work will be done. Having the Diameter of a Wheel which is proposed to be cut into a given Number of Teeth, to find the Thickness of a Cutter suitable, so as to give the Teeth and Space in due Proportion. Example-A wheel of 48 inches in diameter is wanted to have 144 teeth cut on it: suppose the depth of the teeth to be one-tenth of an inch, then the diameter of the wheel taken at the bottom of the teeth and space will be 46 inches; and it is intended to have the teeth and spaces alike at the bottom, which will allow the teeth to be broader from the bottom upwards. To find the circumference of a diameter for 46 inches, say as 113 is to 355 so is 4-6 to its circumference, which in this case is 14:4513 inches. Reduce this to thousand parts, and we shall have 14451 to divide by 288, the number of teeth and spaces taken together, the quotient will be 50-2 nearly; call it 50 hundredths of an inch or half a tenth, and this is what the thickness of the cutter ought to be. TABLE, By which to determine the Number of Teeth of small Wheels, at a given Number Inch of the Diameter. per Rule.-Multiply the given diameter by the number of teeth to the inch and subtract 2 for the pitch line the remainder will be the number of teeth. To find the Breadth of the Teeth Rule.--Divide the horses' power by the velocity of the wheel at the pitch circle, per second, in feet, and twice the quotient is the breadth in inches. Example.--Required the breadth of the teeth for a first mover from a 15-horse engine, the velocity of the wheel being 6 feet per second |