A.--Experiments upon the force requisite to twist off cast iron necks show that if the cube of the diameter of neck in inches be multiplied by 880, the product will be the force of torsion which will twist them off when acting at 6 inches radius; on this fact the following rule is founded: To find the diameter of a cast iron fly wheel shaft:--multiply the square of the diameter of the cylinder in inches, by the length of the crank in inches, and extract the cube root of the product, which multiply by 0.3025, and the result will be the proper diameter of the shaft in inches at the smallest part, when of cast iron.
354. Q.--What was Mr. Watt's rule for the necks of his crank shafts?
A.--Taking the pressure on the piston at 12 lbs. pressure on the square inch, and supposing this force to be applied at one foot radius, divide the total pressure of the piston reduced to 1 foot of radius by 31.4, and extract the cube root of the quotient, which is the diameter of the shaft: or extract the cube root of 13.7 times the number of cubic feet of steam required to make one revolution, which is also the diameter of the shaft.
355. Q.--Can you give any rule for the strength of the teeth of wheels?
A.--To find the proper dimensions for the teeth of a cast iron wheel:--multiply the diameter of the pitch circle in feet by the number of revolutions to be made per minute, and reserve the product for a divisor; multiply the number of actual horses power to be transmitted by 240, and divide the product by the above divisor, which will give the strength. If the pitch be given to find the breadth, divide the above strength by the square of the pitch in inches; or if the breadth be given, then to find the pitch divide the strength by the breadth in inches, and extract the square root of the quotient, which is the proper pitch in inches. The length of the teeth is usually about 5/8ths of the pitch. Pinions to work satisfactorily should not have less than 30 or 40 teeth, and where the speed exceeds 220 feet in the minute, the teeth of the larger wheel should be of wood, made a little thicker, to keep the strength unimpaired.
356. Q.--What was Mr. Watt's rule for the pitch of wheels?
A.--Multiply five times the diameter of the larger wheel by the diameter of the smaller, and extract the fourth root of the product, which is the pitch.
STRENGTH OF MARINE AND LOCOMOTIVE ENGINES.
357. Q.--Cannot you give some rules of strength which will be applicable whatever pressure may be employed?
A.--In the rules already given, the effective pressure may be reckoned at from 18 to 20 lbs. upon every square inch of the piston, as is usual in land engines; and if the pressure upon every square inch of the piston be made twice greater, the dimensions must just be those proper for an engine of twice the area of piston. It will not be difficult, however, to introduce the pressure into the rules as an element of the computation, whereby the result will be applicable both to high and low pressure engines.