A.—­Yes.  If the velocity in feet per second be divided by 4.01, the square of the quotient will be four times the height in feet from which a body must have fallen to have acquired that velocity.  Divide this quadruple height by the diameter of the circle, and the quotient is the centrifugal force in terms of the weight of the body, so that, multiplying the quotient by the actual weight of the body, we have the centrifugal force in pounds or tons.  Another rule is to multiply the square of the number of revolutions per minute by the diameter of the circle in feet, and to divide the product by 5,870.  The quotient is the centrifugal force in terms of the weight of the body.

27. Q.—­How do you find the velocity of the body when its centrifugal force and the diameter of the circle in which it moves are given?

A.—­Multiply the centrifugal force in terms of the weight of the body by the diameter of the circle in feet, and multiply the square root of the product by 4.01; the result will be the velocity of the body in feet per second.

28. Q.—­Will you illustrate this by finding the velocity at which the cast iron rim of a fly-wheel 10 feet in diameter would burst asunder by its centrifugal force?

A.—­If we take the tensile strength of cast iron at 15,000 lbs. per square inch, a fly-wheel rim of one square inch of sectional area would sustain 30,000 lbs.  If we suppose one half of the rim to be so fixed to the shaft as to be incapable of detachment, then the centrifugal force of the other half of the rim at the moment of rupture must be equal to 30,000 lbs.  Now 30,000 lbs. divided by 49.48 (the weight of the half rim) is equal to 606.3, which is the centrifugal force in terms of the weight.  Then by the rule given in the last answer 606.3 x 10 = 6063, the square root of which is 78 nearly, and 78 x 4.01 = 312.78, the velocity of the rim in feet per second at the moment of rupture.

29. Q.—­What is the greatest velocity at which it is safe to drive a cast iron fly-wheel?

A.—­If we take 2,000 lbs. as the utmost strain per square inch to which cast iron can be permanently subjected with safety; then, by a similar process to that just explained, we have 4,000 lbs./49.48 = 80.8 which multiplied by 10 = 808, the square root of which is 28.4, and 28.4 x 4.01 = 113.884, the velocity of the rim in feet per second, which may be considered as the highest consistent with safety.  Indeed, this limit should not be approached in practice on account of the risks of fracture from weakness or imperfections in the metal.

30. Q.—­What is the velocity at which the wheels of railway trains may run if we take 4,000 lbs. per square inch as the greatest strain to which malleable iron should be subjected?