∴ Work done is 90 ft.-lb.
= 90/33000 = 0·002727 h.p.
The weight of the spring was 6¾ oz. (this is taken from an actual experiment), i.e. this motor develops power at the rate of 0·002727 h.p. for 3½ seconds only.
§ 22. To Ascertain the H.P. of a Rubber Motor. Supposing a propeller wound up to 250 turns to run down in 15 seconds, i.e. at a mean speed of 1200 revolutions per minute or 20 per second. Suppose the mean thrust to be 2 oz., and let the pitch of the propeller be 1 foot. Then the number of foot-pounds of energy developed
= 2 oz. × 1200 revols. × 1 ft. (pitch) / 16 oz.
= 150 ft.-lb. per minute.
But the rubber motor runs down in 15 seconds.
∴ Energy really developed is
= 150 × 15 / 60 = 37·5 ft.-lb.
The motor develops power at rate of 150/33000 = 0·004545 h.p., but for 15 seconds only.
§ 23. Foot-pounds of Energy in a Given Weight of Rubber (experimental determination of).