"In the above example, when the length of the stroke is neglected, the radius of the beam 12 feet, and the distance between = the vertical line and the end of the radius rod "Then by the rule we have 12 + 4 = 16, and 2 x 12 +4 = 28. the above result. If the radius rod be shorter than the parallel bar, it is only necessary to use subtraction instead of addition in the first rule. Thus, suppose that the axis of the radius rod had been 4 feet on the other side of the vertical line, then 12 and 12 x 2 - 4 = 24 - 4 = 20. Hence = 8 x8 64 33 = length of the radius rod.” 48, "PRACTICAL OBSERVATIONS.--Since we have given rules for finding the length of radius rods, it now becomes necessary to show how to put them on." "Plumb the piston rod when the piston is at the top extremity of its stroke; then one end of the radius rod being moveable about E', with the other end o' describe an arc of a circle. Now, bring the piston down to the lowest extremity of its stroke, and again plumb the piston rod, and in the same manner as before describe another arc of a circle; and the point where these arcs intersect is the centre upon which the end of the radius rod is to move." "It is a practice among engineers to set the cylinder half the vibration of the beam in towards the centre of the beam; and we shall here show how to find the vibration or versed sine of the arc described." "From the square of the length of the beam, taken from the centre of motion, subtract the square of half the length of the stroke; and the square root of the remainder, subtracted from the above length of the beam, will give the vibration required." "Ex.-Given the length of the beam from the centre of motion, 5 feet; and half the length of the stroke, 3 feet: to find the vibration. "Now, 52 = 25, and 32 = 9; then, √ (25 9) = √16 4. 541. That is, the vibration will be one foot." Therefore, in this case, the horizontal distance between the centre of motion and the centre of the cylinder must be 4 feet 6 inches." "When an engine works with a vibrating pillar, the vibration is in an opposite direction, and the centre of the vibrating pillar axle must be set half the vibration in towards the cylinder." "The air-pump bucket rod must be hung on at the point E'; for the point describes the same kind of line as the piston-rod describes." "To find the length of the connecting rod: " "Set the beam at half stroke, that is, parallel to the horizon; and the distance between the centre of the pin on which the connecting rod is to move and the centre of the shaft, is the length of the connecting rod." "Some practical men set both the beam and crank parallel to the horizon, and take the distance between the centre of the pin on which the connecting rod is to move and the centre of the crank pin, for the length of the connecting rod." "This method has also been given by some writers; but it is incorrect, and when the length is so taken, the connecting rod will be found too long." ECCENTRICS.-The length of the lever on the weigh shaft being given, to find the requisite throw of the eccentric : Multiply the length of the stroke of the valve by the length of the lever on the weigh shaft for the eccentric rod, and divide the product by the length of the lever which works the valve, and the quotient will be the throw required. Ex. The stroke of a valve is 6 inches, the lever on the weigh shaft 9 inches, and the lever for the eccentric rod 10 inches; required the throw of the eccentric. B Eccentric. The throw of an eccentric is the difference of the distances between A B and A B'. The throw of the eccentric and the journey of the valve, also one of the levers on the weigh shaft being given to find the other: Multiply the throw of the eccentric by the length of lever required to work the valve, and divide by the journey the valve is to travel, which will give the length of the other lever. Or,Multiply the journey of the valve by the length of lever on the weigh shaft for the eccentric rod, and divide by the throw of the eccentric, the quotient will be the length of lever for working the valve. Ex.-Suppose a valve be required to travel 7 inches, the throw of the eccentric being 6 inches, the length of lever on the weigh shaft being 12 inches, required the length of the other. 7 x 12 84 6 THE GOVERNOR, OR REGULATOR.-The vertical distance between the point of suspension, and the plane in which the balls revolve is taken as the height of the governor. Then to find the height suited for any proposed number of revolutions per minute:: Divide the number 375 by twice the number of revolutions per minute, and the square of the quotient will be the required height of the governor in inches. Ex-What should be the height of a governor, if the number of revolutions be 40 per minute? the 40 x 280 and 375 80 = 4.6875 = 21.9 inches. Then to find the number of revolutions per minute, the height of governor being given: Divide 375 by the square root of the governor's length, and half the quotient will be the velocity required. Ex.-What number of revolutions ought a governor to make whose height is 21.9 inches? 375 80 4.6875 2 = SAFETY VALVE.-To ascertain the diameter to be given to a safety valve: of Find the area of the fire surface, and divide by the excess pressure above the atmospheric expressed in pounds per square inch, and the quotient will be the square of the narrowest diameter of the valve in inches. Ex.-Required the diameter of the aperture of a safety valve, for a boiler having 80 square feet of fire surface; pressure 12 lbs. per square inch above the atmospheric. 180 = 15 12 ✓ 15 = 3.87, or 319 inches. It will be prudent to make the aperture rather larger than the result found by this rule. Safety Valve. The lever on the safety valve of a steam-engine is of the third order. It may be required to calculate this lever in three different ways. 1st. To find the weight necessary on the lever, the distance between the fulcrum and weight, and distance of G valve from centre of motion being given, to overcome any given pressure on the valve. Ascertain the area of the valve by multiplying its diameter by 7854, then multiply by the number of pounds per square inch required upon the valve; the product will give the total weight. Then add together the effective weight of the lever, the weight necessary to balance the lever, and the weight of the valve, which deduct from the first product. Multiply the remainder by the distance between the fulcrum. and valve; divide the product by the distance between the fulcrum and intended position of the weight; the quotient will be the weight in pounds to be placed at the required distance on the lever. Ex.-The diameter of a valve is 3 inches, the length of the lever from A to B 24 inches, the distance between a and c 3 inches, weight of valve 3 lbs., weight of lever 2 lbs., and weight necessary to balance the lever at a 4 lbs. ; what weight should be placed on the lever at B to give a pressure of 30 lbs. per square inch on the valve? 32 × 7854 = 9 × ·7854 = 7·06 square inches area of valve. 7·06 × 30 = 212 lbs. whole pressure on valve. required to give a pressure of 30 lbs. per square inch on valve. 2nd. To find where a given weight must be placed upon the lever to equal a certain pressure upon the valve, the distance of valve from the fulcrum being given : Multiply the required weight per square inch upon the valve by its area in inches, the product will give the total weight. Add together the effective weight of the lever, the weight necessary to balance the lever, and the weight of the valve and spindle, which deduct from the total weight. Multiply the remainder by the distance between the fulcrum and valve, divide the product by the weight in pounds upon the lever, and the quotient will give the distance in inches for placing the weight. Ex-Suppose, as in the last example, a valve be 3 inches diameter, weight of valve 3 lbs., weight of lever 2 lbs., weight necessary to balance the lever at ▲ 4 lbs., and the distance between A and c equal 3 inches, at what distance must the weight B, equal 25.37 lbs., be placed from a to equal 30 tbs. per square inch on the valve? 3rd. To find the pressure of steam per square inch acting upon the valve; the distance of weight, distance of valve from fulcrum, and weight upon lever being given : Multiply the weight in pounds upon the lever by the distance of the same to the fulcrum, divide the product by the distance between the fulcrum and the valve, to the quotient add the weight of the valve and the effective weight of the lever; the sum is the weight upon the valve in pounds per square inch. Ex.Suppose a weight of 24 lbs. is placed upon a lever 24 inches from A to B, the distance from A to c being 4 inches; required the weight upon the valve, the weight of the lever being 6 lbs., the weight of valve being 3 lbs., and its diameter 4 inches. |