| Length of rubber | 36 yds. |
| Weight " | 2 7/16 oz . |
| Number of turns | = 200. |
12 oz. were raised 19 ft. in 5 seconds.
i.e. ¾ lb. was raised 19 × 12 ft. in 1 minute.
i.e. 1 lb. was raised 19 × 3 × 3 ft. in 1 minute.
= 171 ft. in 1 minute.
i.e. 171 ft.-lb. of energy per minute. But actual time was 5 seconds.
∴ Actual energy developed by 2-7/16 oz. of rubber of 36 yards, i.e. 36 strands 1 yard each at 200 turns is
= 171/12 ft.-lb.
= 14¼ ft.-lb.
This allows nothing for friction or turning the axle on which the cord was wound. Ball bearings were used; but the rubber was not new and twenty turns were still unwound at the end of the experiment. Now allowing for friction, etc. being the same as on an actual model, we can take ¾ of a ft.-lb. for the unwound amount and estimate the total energy as 15 ft.-lb. as a minimum. The energy actually developed being at the rate of 0·0055 h.p., or 1/200 of a h.p. if supposed uniform.
§ 24. The actual energy derivable from 1 lb. weight of rubber is stated to be 300 ft.-lb. On this basis 2-7/16 oz. should be capable of giving 45·7 ft.-lb. of energy, i.e. three times the amount given above. Now the motor-rubber not lubricated was only given 200 turns—lubricated 400 could have been given it, 600 probably before rupture—and the energy then derivable would certainly have been approximating to 45 ft.-lb., i.e. 36·25. Now on the basis of 300 ft.-lb. per lb. a weight of ½ oz. (the amount of rubber carried in "one-ouncers") gives 9 ft.-lb. of energy. Now assuming the gliding angle (including weight of propellers) to be 1 in 8; a perfectly efficient model should be capable of flying eight times as great a distance in a horizontal direction as the energy in the rubber motor would lift it vertically. Now 9 ft.-lb. of energy will lift 1 oz. 154 ft. Therefore theoretically it will drive it a distance (in yards) of
8 × 154/3 = 410·6 yards.