Work done = Force (in lbs.) × Distance (in ft.)
= 50 × 12 = 600 ft. lbs.

Power.—The rate at which work is done is a measure of the power exerted. One horse-power is exerted when 33,000 foot-pounds of work are done in one minute. The work done per minute (in ft. lbs.) divided by 33,000 gives the horse-power expended.

Example:—To propel a motor-car along a level road at a speed of 30 miles an hour requires a tractive effort or pull of 70 lbs. if the vehicle weighs one ton. Find the horse-power required, at the road surface.

Horse-power = Work done per minute in ft. lbs./33,000
= Force (in lbs.) × Distance through which it acts per minute (in ft)./33,000 = (70 × 30 × 5280/60)/33,000 = (7 × 264)/330 = 5·6

Example:—If the car in the preceding example had to climb a gradient which rose one foot for every four feet traversed by the car, find the additional horse-power needed to keep up a speed of 30 miles an hour while climbing the gradient.

Here we have to raise a weight of 1 ton vertically upwards through a height equal to one-fourth of the road surface covered, every minute.

Additional Horse-power required

= (2240 (lbs.) × (30 × 5280/60) × ¼ ft. per min.)/33,000
= (2240 × 660)/33,000 = 44·8

Total Horse-power to climb the gradient of 1 in 4 at 30 miles an hour = 5·6 + 44·8 = 50·4

Brake Horse-Power.—The length of the circumference or boundary line of a circle is 6·28 times the length of the radius of the circle or 3·14 times the length of its diameter. Hence, if an engine exerts a pull of P lbs. at the end of a brake arm of length R feet when it is maintaining a speed of N revolutions per minute (we may imagine the brake to be fitted round the rim of the flywheel), we can calculate the brake horse-power thus:—

Brake Horse-Power or B.H.P.
= (Work done on the brake per minute in ft. lbs.)/33,000

hence B.H.P = (Pull at the end of the brake arm (in lbs.)) × (6·28 times the radius of the arm (in feet)) × (the number of revolutions made by the engine (in one minute))/33,000
= (P × 6·28 × R × N)/33,000