That is, the weight put on at the end of the lever must be such as to press 116·164 lbs. on the whole valve; but the leverage of the weight is 8, therefore one-eighth part of this weight will do. Thus, 116·164 ÷ 8 = 14·52 lbs. weight which must be put on at the end to press 40 lbs. per inch upon the valve. Now, to mark the lever where we will have 35 lbs., 30 lbs., 25 lbs., 20 lbs., &c. per square inch, we must proceed thus: 3.1416 35 157080 94248 109.956 9.5 number of square inches in the area of the valve. 100.456 = weight on the valve from the action of the weight. And 100-456 14.52 6.918 the leverage → which the weight must have; and if this leverage be multiplied by the distance between the fulcrum and valve, thus: 6·918 = leverage. 2 distance between the fulcrum and valve. 13.836 = the distance along the lever from the fulcrum. 16 - 13.836 2.164 inches the distance which the weight must be moved in towards the valve. And if you want 30 lbs. per square inch, move it in 2.164 inches farther, and so on, as far as you please, making the division always 2∙164 inches. TABLES OF SAFETY VALVE LEVERS. 1. Diameter of the valve 4 inches, weight of the valve, &c. 3 lbs., length of the lever 24 inches, the distance between the fulcrum and valve 4 inches, and the weight of the lever 8 lbs.; then the weight put on at the end of the lever, to give 30 lbs. per square inch upon the valve, will be 58.332 lbs., or 58 lbs. 5 oz. nearly. For 10 lbs. pressure, dist. from fulcrum 6-765 inches. 20 lbs...... .15.382 inches. 24 inches. 30 lbs...... 2 2. Length of the lever 16 inches, distance between the fulcrum and valve 2 inches, diameter of valve 2 inches, the weight of the lever 4 lbs, and weight of valve lb.; it will require a weight of 9.7185 lbs. to be put on at the end of the lever to give 30 lbs. per square inch upon the valve; and the distances in inches which the weight must be from the fulcrum to give 10, 15, 20, 25, and 30 lbs. are respectively as follow: If the weight be taken off from the lever, then the weight on the valve from the action of the lever alone will give 5 lbs. per square inch. Note.-9.7185 lbs. is 9 lbs. 11 oz. 16 3. Weight of lever 4 lbs., and weight of valve, &c. 1 lb.; whole length of the lever 24 inches, distance between the fulcrum and valve 3 inches, diameter of valve 3 inches, weight put on at the end 42.05375 lbs. Distances from the fulcrum in inches: 3.8292 8.8719 13.9146 18.9573 10 lbs. 20 lbs. 30 lbs. 40 lbs. 50 lbs. SHRINKAGE OF TIRE BARS. 24 The general allowance for shrinkage is of an inch to the foot of diameter of wheel centre. EXAMPLE.-Suppose we wish to turn a centre of driving wheel to fit a tire which is 5 feet the inside diameter, equal to 188 inches circumference; then we must turn the centre ğ larger in circumference, which is 189 inches. Shrinkage of Castings. In making all patterns of work, we make an alinch larger per foot in cast iron; for lowance of I give the following result of experiment made by me with spring steel. The bar made use of was supported on both extremities, and the weight suspended in the middle. The following are the results:-Length between the fulcrum 2 feet; width of bar 1 inch. 2 SPRING STEEL. The table shows that the deflection was equal in both cases. At the same time it appears that the squares of the thickness are to each other as the bending weights, very near, thus conform82 3 118 ing to the general theory. For calculating the Radius of a Curve when the Angle of Deflection and Chord are given. Railroad curves are always laid off with chords of 100 feet, and we often find, when speaking of curves, the angle of deflection is merely given. Now to find the radius: 5730 feet are a common radius, which is equal tò a deflection of 1°. RULE.-Divide the number of degrees deflected into 5730; the product will be the radius of the curve. EXAMPLE. We have a curve with a deflection of 6o, and the chord 100 feet. 6)5730 955 feet radius of curve, Ans. |