| H.P. = | P r 2π R | ([Fig. 1], B) |
| 33,000 |
where
| P | = | pressure in klgs. or lbs. |
| r | = | radius on which P is acting. |
| R | = | Revolution/min. |
When P × r = M, then
| H.P. = | M.R.2π | , |
| 4,500 |
thence,
| M = | H.P. × 4,500 | = | 716.2 H.P. | in meter kilograms, |
| R2π | R |
or in English system
| M = | H.P. 33,000 | = | 5253.1 H.P. | in foot pounds. |
| R2π | R |
Now the power on the circumference of the propeller will be reduced by its radius, so it will be M/r = p. A part of p will be used for counteracting the air and bearing friction, so that the total power on the circumference of the propeller will be (M/r) × η = p where η is the mechanical efficiency of the propeller. Now η/tan α = T, where α is taken on the tip of the propeller.