# How/why does engine torque and horsepower affect aircraft speed? [duplicate]

I'm curious as to why torque and horsepower affect an aircraft's speed. AFAIK, it's only the engines RPM that affect speed, right?

Does this mean an engine's torque/horsepower only affects the speed it can effectively operate at or am I wrong?

Horsepower is what propels an airplane through the air; more horsepower means more speed. Power is neither torque nor RPM by itself: it is their product. (Torque)x(rpm)x(conversion factor) = horsepower.

• how does horsepower propel the airplane through the air faster? Jan 31 at 13:33
• @alekslacroix, horsepower is a measure of how fast a machine can perform work. in a flying airplane, the work is used to overcome air friction, which is created by moving through the air. Jan 31 at 18:02
• so if the horsepower of an engine was too low the propellers would slow down due to air friction? Jan 31 at 22:02
• @alekslacroix You are correct. This tiny motor and this F1 engine are both rated for 15,000RPM - and yet one outperforms the other. Clearly, RPM isn't the only factor affecting performance; torque is equally important. You cannot replace the V6 with the tiny motor and expect the same performance from it - all that RPM is of no use if it doesn't have enough torque to turn the wheels. Feb 1 at 10:37
• Similarly, to turn a propeller through the air, you need torque - thousands of foot-pounds of torque. Since both RPM and torque determine the performance of an engine, we use the product of the two, known as horsepower, to compare the performance of engines. Feb 1 at 10:41

Horsepower and torque affects RPM.

For a given torque, an aircraft needs greater RPM to achieve greater speed. And that means more horsepower.

For a given RPM, an aircraft with a higher blade pitch travels faster, but this requires more torque, due to the more sideways direction of the force in the blades. More torque means more horsepower.

For the same reason a car's engine and its transmission affect its speed. You can have an engine in a car producing constant horsepower that never changes, but given a flat road, its speed will depend on what gear has been selected in the transmission. More or less horsepower doesn't always translate to more or less speed.

Horsepower is a measure of the total power output, this is proportional to RPM multiplied by torque. How this relates to static thrust depends on the diameter and pitch of the propeller; a bigger diameter will move more air and a steep pitch will move that air faster. An aircraft designer will try to optimise these so that the full power of the engine can be used at an appropriate engine speed. This is complicated by many factors. The combination of pitch and RPM gives the speed at which the propeller will try to move the air, as the aircraft nears this speed the thrust will decrease and so a faster aircraft will call for a steeper pitch.

Perhaps, in context, this example may be helpful. We have, say, a piston engine turning a propeller sufficiently to produce enough thrust from that propeller to maintain an aircraft in flight. Consequently, the two things we have are a) torque horsepower from the piston engine, and b) thrust horsepower from the propeller. The propeller is not perfectly efficient, but nearly so. Therefore, we can state with confidence the following relationship

$$Efficiency = \frac{Thrust Horsepower}{Torque Horsepower}$$ When we fly the plane, we set the throttle to a specific fuel flow at a specific manifold pressure to maintain a specific rpm. The propeller being turned by that engine does its thing and produces the thrust required for flight. So how much thrust is that, you may ask. Well, the thrust is equal to the drag produced by the aircraft. We can measure the drag or specifically estimate the drag for a given aircraft velocity, and thereby, determine the power produced by that propeller. The relationship is simply $$\frac{D \cdot\ V}{550} = Thrust Horspower$$ where D is the drag in pounds and V is the velocity in feet per second. If our propeller is 89 percent efficient, then the torque horspower required to develop that thrust would be 1/0.89 times the thrust horse power, or alternatively, since we know the power setting of the piston engine turning that propeller, and we know the propeller efficiency, we can estimate very closely the thrust horsepower as 89 percent of the torque horesepower. Also notice that this is all balanced on the velocity of the aircraft. Hence, the following must be true -

$$V = \frac{Thrust Horsepower \cdot\ 550}{D}$$ or, specifically,

$$V = \frac{Torque Horsepower \cdot\ 550 \cdot\ 0.89}{D} .$$