Читать книгу A Catechism of the Steam Engine - C.E. John Bourne - Страница 7

24. Q.--What do you understand by centrifugal and centripetal forces?

A.--By centrifugal force, I understand the force with which a revolving body tends to fly from the centre; and by centripetal force, I understand any force which draws it to the centre, or counteracts the centrifugal tendency. In the conical pendulum, or steam engine governor, which consists of two metal balls suspended on rods hung from the end of a vertical revolving shaft, the centrifugal force is manifested by the divergence of the balls, when the shaft is put into revolution; and the centripetal force, which in this instance is gravity, predominates so soon as the velocity is arrested; for the arms then collapse and hang by the side of the shaft.

25. Q.--What measures are there of the centrifugal force of bodies revolving in a circle?

A.--The centrifugal force of bodies revolving in a circle increases as the diameter of the circle, if the number of revolutions remain the same. If there be two fly-wheels of the same weight, and making the same number of revolutions per minute, but the diameter of one be double that of the other, the larger will have double the amount of centrifugal force. The centrifugal force of the same wheel, however, increases as the square of the velocity; so that if the velocity of a fly-wheel be doubled, it will have four times the amount of centrifugal force.

26. Q.--Can you give a rule for determining the centrifugal force of a body of a given weight moving with a given velocity in a circle of a given diameter?

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?

A.--The weight of a malleable iron rim of one square inch sectional area and 7 feet diameter is 21.991 feet x 3.4 lbs. = 74.76, one half of which is 37.4 lbs. Then by the same process as before, 8,000/37.4 = 213.9, the centrifugal force in terms of the weight: 213.9 x 7, the diameter of the wheel = 1497.3, the square root of which, 38.3 x 4.01 = 155.187 feet per second, the highest velocity of the rims of railway carriage wheels that is consistent with safety. 155.187 feet per second is equivalent to 105.8 miles an hour. As 4,000 lbs. per square inch of sectional area is the utmost strain to which iron should be exposed in machinery, railway wheels can scarcely be considered safe at speed even considerably under 100 miles an hour, unless so constructed that the centrifugal force of the rim will be counteracted, to a material extent, by the centripetal action of the arms. Hooped wheels are very unsafe, unless the hoops are, by some process or other, firmly attached to the arms. It is of no use to increase the dimensions of the rim of a wheel with the view of giving increased strength to counteract the centrifugal force, as every increase in the weight of the rim will increase the centrifugal force in the same proportion.