286

MOTIVE POWER EMPLOYED

X 45 = 562.5 feet in a minute, and 562.5 X 60 = 33750feet an hour = 6.4 miles, or nearly.

And the resistance or load is 384 lbs. (the friction of theengine) + (40 tons X 2240 -r 200) = 448 lbs. (the frictionof the load) = 832 lbs., moved in one revolution of thewheel 12.5 feet, = 10400, the resistance: and the poweris 9 3 x .7854 x 2=127.2 area of the pistons, whichmultiplied by 50 lbs., the pressure of the steam in the boiler,we have 127.2 X 50= 6361.7, which, in one revolutionofthe wheels, is moved 4 feet, = 25446, the power.

$ 25446 power, f , „

Whence^ . > equal to 40.8 per cent.

I 10400 resistance, \ r

If now the wheels be increased to 5 feet, then in eachrevolution the space described will be 5 X 3.1416 = 15.7X 45 revolutions = 706.5 X 60 = 42390 feet, or 8 milesan hour.

But the resistance, in this case, will be different; whilethe engine is making one complete stroke, the space movedover with 4-feet wheels is 12.5 feet; and, during the samestroke, with 5-feet wheels, 15.7 feet.

Now we have before seen that, in one com-plete stroke of the engine, the same quantityof friction took place, (except that whicharose from the action of the wheels upon therail) whether the engine moves over 12.5 or15.7 feet, the same extent of attrition occur-ring from the various working parts of theengine. The retardation of friction will thenbe inversely as the spaces passed over, viz.

As 12.5 ; 15.7, or as the diameter of the wheels.