Torque and BHP

There is a fixed relationship between torque, BHP and RPM, (ft/lb torque = ( ( 5252 x BHP) / RPM) such that with our example vessels’ engine producing 45 HP @ 2000 RPM we can determine that it is producing 118 ft/lbs of torque at the crankshaft. It is very important that you do not forget to deduct power used by engine peripherals such as internal and raw water pumps and alternators, this can easily total 3 BHP even on small motors, and the significance of these losses grows with smaller motors. 3 BHP to drive alternators and water pumps is not a lot to a large 50 BHP motor, while it is a very significant loss to an 18 BHP motor.

While our example gearbox has a 1 to 1 reduction ratio, users should understand that if it had a 2:1 ratio while the shaft RPM would halve to 1000 RPM torque would double to 236 ft/lb, note that the power would be unchanged, the gearbox does not generate power. This is how a heavy lorry can climb a steep hill, it exchanges wheel RPM or speed, for torque.

We also have to understand that devices like gearboxes and bearings are not 100% efficient, they all absorb some power and then give it off as heat. We can assume that a gearbox absorbs 3% of the power it carries, and a bearing 1.5% of the power it carries.

Our example vessel with two bearings and one gearbox thus loses 6% of it’s power, since the ratio is 1:1 we can determine via the formula above that at the propeller there is 42 horsepower available, at 2000 RPM and 111 ft/lb of torque.

This is the amount of power that we have available at the propeller to drive the boat, and in the interests of safety and economy as well as efficiency we must use all of it and use it effectively.