Understanding Horsepower and Torque
They aren’t mutually exclusive
By Jack Kane, firstname.lastname@example.org
One of the most hotly contested but least understood controversies concerning engines in aircraft is that of horsepower versus torque. How often have we heard the phrase, “It’s torque that turns a prop”? It’s one of those statements that have been repeated so often that most people now accept it as fact. While there’s some truth to it, it’s a very misleading concept.
A Bit About the Author
Jack Kane has been actively involved in piston engine development since the early 1950s and has configured, built, and modified successful engines for a wide variety of specialized applications and winning race cars. As founder and CEO of EPI Inc., he has been responsible for the design and development of a line of liquid-cooled aircraft engines, propeller reduction drives, and accessory drive units for various applications. He writes technical articles for Race Engine Technology magazine.
Beyond engineering, Jack is an accomplished machinist, commercial pilot, and certificated flight instructor. He’s done development, modification, and overhaul work on certified (Lycoming, Continental, Orenda) aircraft engines. In his younger days, he was a winning driver in a variety of automobile racing categories, including rear-engine formula cars, sports cars, midgets, and stock cars, and has won several championships.
Jack was a dean’s list student in the electrical engineering program at the U.S. Air Force Academy. Later, he earned bachelor of mechanical engineering (cum laude) and master of science degrees (summa cum laude) from civilian universities. He holds a master of engineering degree.
Prior to starting EPI Inc., Jack spent 31 years in design and development, including research and design work on the first generation of full authority digital engine controllers (FADECs); design and development of several complex real-time computer operating systems, control systems, and data communication networking systems; and subsystems in computer-based controls including GPS, nuclear systems, data acquisition systems, missile flight control systems, and satellite communication networking systems. He was a pioneer in the use of multilevel finite state automata to build bulletproof real-time control software systems.
Jack currently holds a commercial pilot rating in multi-engine, single-engine, and glider aircraft, and a private helicopter rating. He has been a certificated flight instructor for single- and multi-engine aircraft and has worked as a Part 135 charter pilot, corporate pilot, glider-tow pilot, ferry pilot, and instructor in high-performance, tailwheel, and aerobatic aircraft.
He has pilot-in-command experience in a wide variety of aircraft as well as test-flight and demo work in various high-performance experimentals.
Power, Torque, Work, and Energy
In order to discuss powerplants in any depth, it’s essential to understand the concepts of power and torque. However, in order to understand power, we must first understand work and energy. Here’s a quick review of work, energy and torque.
Suppose the engine of your car stalls while you’re in line to exit from a flat, level parking lot. You try several times to restart it, but it just won’t start. Since you’re a considerate person, you decide to push your car out of the way of the people behind you. You get out and go round back and begin to push on the car. Suppose also that you’re a fairly strong person, so you exert a horizontal force of 100 pounds on the rear of the car. The car doesn’t move. But you’re also a persistent person; you continue to push on the car for two whole minutes, exerting the same 100 pounds of force. The car still won’t move. Although you’ll probably be quite tired, you’ll have done no work. Why? Because work is defined as a force operating through a distance. The car didn’t move, so although there was force, there was no motion.
Now you get smart and release the parking brake, and again you push with the same constant 100-pound force. The car moves, and it travels 165 feet in two minutes. In that case, you’ll have produced 16,500 foot-pounds of work (100 pounds of force x 165 feet of distance).
Later, you’ve recovered from the parking lot ordeal and are working in your shop. You need to install a 3-inch-long spring into a 2-inch space. The nature of this particular spring is that it takes 600 pounds of force to compress it 1 inch.
Using a lever-operated spring compressor, you pull on the lever with a force of 100 pounds and you move the lever 6 inches, causing the compressor to squeeze the spring and shorten it by 1 inch. The spring is now pushing on the compressor with a force of 600 pounds.You’ve stored the work you did on the compressor lever (100 pounds x 6 inches = 600 inch-pounds) in the spring in the form of energy (600 pounds x 1 inch = 600 inch-pounds).
Energy is defined as the capacity of a body to do work, by virtue of the position or condition of the body.
Now suppose there’s a 150-pound plate of steel on your bench, resting on four blocks which are 2 inches tall (so the space between the bottom of the plate and the bench is 2 inches). You install the compressed spring into that space and locate it at exactly the center of gravity of the plate, and you release the spring compressor. The spring will lift the steel plate 3/4 of an inch, so the spring has done work on the plate, thereby releasing some of the energy stored in the spring.
There are many different forms of energy. There are a few which are of particular interest with respect to powerplants: kinetic energy (the energy contained in a body by virtue of its velocity), potential energy (the energy contained in a body by virtue of its position), chemical energy (energy which can be released by a chemical reaction such as combustion), and heat energy (energy which can be used to make machines operate).
Power and Torque
It often seems that some engine people are confused about the relationship between power and torque. For example, I’ve heard engine builders, camshaft consultants, and other technical experts ask their customers, “Do you want your engine to make horsepower, or do you want it to make torque?”
The question is posed in a tone which strongly suggests that these experts believe power and torque are somehow mutually exclusive. In fact, the exact opposite is true, and we all need to be clear on the following facts. Power is defined as the amount of work done per unit time, or the rate of doing work. Torque and rpm are the measured quantities of engine output. Power is a quantity which is calculated from torque and rpm by the following equation:
hp = torque x rpm ÷ 5,252
An engine produces power by providing a rotating shaft which can exert a certain amount of torque on a load at a certain rpm. The amount of torque the engine can exert usually varies with rpm. A dynamometer (dyno) substantiates the power an engine produces by applying a load to the engine output shaft by means of a water brake, generator, eddy-current absorber, or any other controllable device capable of absorbing power. The dyno control system causes the absorber to exactly match the amount of torque the engine is producing at any given instant. It then measures that torque as well as the instantaneous rpm of the engine shaft, and from those two measurements, it calculates observed power after factoring variables (air temperature, barometric pressure, relative humidity) in order to correct the observed power to the value it would have been if it had been measured at sea level-standard atmospheric conditions (corrected power).
Torque is defined as a force around a given point, applied at a radius from that point. Note that the unit of torque (force) is one pound-foot (often misstated), while the unit of work is one foot-pound.
Referring to Figure 1 (above), assume that the handle is attached to the crank arm so that it’s parallel to the supported shaft and is located at a radius of 12 inches from the center of the shaft. In this example, consider the shaft to be fixed to the wall. Let the arrow represent a 100-pound force, applied in a direction perpendicular to both the handle and the crank arm, as shown.
Because the shaft is fixed to the wall, the shaft doesn’t turn, but there’s a torque of 100 pounds-feet (100 pounds x1 foot) applied to the shaft. Note that if the crank arm in the sketch was twice as long (i.e. the handle was located 24 inches from the center of the shaft), the same 100-pound force applied to the handle would produce 200 pounds-feet of torque (100 pounds x 2 feet).
Power is the measure of how much work can be done in a specified time. In the explanation of work and energy at the beginning of the article, by pushing the car you produced 16,500 foot-pounds of work. If you did that work in exactly two minutes, you would have produced 8,250 foot-pounds per minute of power (165 feet x 100 pounds ÷ 2 minutes).
In the same way that one ton is a large amount of weight (by definition, 2,000 pounds), one horsepower is a large amount of power. The definition of one horsepower is 33,000 foot-pounds per minute. The power which you produced by pushing your car across the lot (8,250 foot-pounds per minute) equals 1/4 horsepower (8,250 ÷ 33,000).
Okay, all that’s fine, but how does pushing a car across a parking lot relate to rotating machinery? Consider the handle-and-crank-arm sketch from Figure 1, but with these changes:
The handle is still 12 inches from the center of the shaft, but instead of being fixed to the wall, the shaft now goes through the wall, supported by frictionless bearings, and is attached to a generator behind the wall.
Suppose, as illustrated in the sequence in Figure 2 (above), that a constant force of 100 pounds is somehow applied to the handle so that the force is always perpendicular to both the handle and the crank arm as the crank turns. In other words, the “arrow” rotates with the handle and remains in the same position relative to the crank and handle, as shown in the sequence above. (That is called a “tangential force.”) If that constant 100-pound tangential force being applied to the 12-inch handle (100 pounds-feet of torque) causes the shaft to rotate at 2,000 rpm, then the power the shaft is transmitting to the generator behind the wall is 38 hp, calculated as follows:
100 pounds-feet of torque (100 pounds x 1 foot) x 2,000 rpm ÷ 5,252 = 38 hp.
In order to make the concept clearer, here are five examples of real-world systems which produce 300 hp.
Example 1: How much torque is required to produce 300 hp at 2,700 rpm? Since hp = torque x rpm ÷ 5,252, then rearranging the equation to solve for torque produces: torque = hp x 5,252 ÷ rpm.
Answer: torque = 300 x 5,252 ÷ 2,700 = 584 pounds-feet.
Example 2: How much torque is required to produce 300 hp at 4,600 rpm?
Answer: torque = 300 x 5,252 ÷ 4,600 = 343 pounds-feet.
Example 3: How much torque is required to produce 300 hp at 8,000 rpm?
Answer: torque = 300 x 5,252 ÷ 8,000 = 197 pounds-feet.
Example 4: How much torque does the 41,000-rpm turbine section of a 300-hp gas turbine engine produce?
Answer: torque = 300 x 5,252 ÷ 41,000 = 38.4 pounds-feet.
Example 5: The 300-hp engine in Example 4 drives a gearbox which turns a propeller at 1,591 rpm. How much torque is applied to that shaft?
Answer: torque = 300 x 5,252 ÷ 1,591 = 991 pounds-feet (ignoring losses in the gearbox, of course).
The point to be taken from those numbers is that a given amount of horsepower can be made from an infinite number of combinations of torque and rpm.
Think of it another way: In cars of equal weight, a 2-liter twin-cam engine that makes 300 hp at 8,000 rpm and 400 hp at 10,000 rpm will get you out of a corner just as well as a 5-liter engine that makes 300 hp at 4,000 rpm and 400 hp at 5,000 rpm.
In order to design an engine for a particular application, it’s helpful to plot out the optimal power curve for that specific application, then from that design information, determine the torque curve which is required to produce the desired power curve. By evaluating the torque requirements against realistic brake mean effective pressure (BMEP) values, you can determine the reasonableness of the target power curve.
Typically, the torque peak will occur at a substantially lower rpm than the power peak. The reason is that, in general, the torque curve doesn’t drop off (percentwise) as rapidly as the rpm is increasing (percentwise).
For a race car engine, it’s often beneficial (within the boundary conditions of the application) to operate the engine well beyond the power peak in order to produce the maximum power within a required rpm band. However, for an engine which operates in a relatively narrow rpm band, such as an aircraft engine, it’s generally a requirement that the engine produce maximum power at the maximum rpm. That requires the torque peak to be fairly close to the maximum rpm.
For an aircraft engine, we typically like to see the torque curve peak at the normal cruise setting, and from there, stay essentially flat up to maximum rpm. That positioning of the torque curve would allow the engine to produce significantly more power if it could operate at a higher rpm, but the goal is to optimize the performance within the operating range.
An example of that concept is shown in Figure 3 above. The three dashed lines represent three torque curves having exactly the same shape and torque values, but with the peak torque values located at different rpm values (3,000, 4,500, and 6,000 rpm). The solid lines show the power produced by the torque curves of the same color.
Note that, with a torque peak of 587 pounds-feet at 3,000 rpm, the pink power line peaks at about 375 hp between 3,500 and 3,750 rpm. With the same torque curve moved to the right by 1,500 rpm (black, 587 pounds-feet torque peak at 4,500 rpm), the peak power jumps to about 535 hp at 5,000 rpm. Again, moving the same torque curve to the right another 1,500 rpm (blue, 587 pounds-feet torque peak at 6,000 rpm) causes the power to peak at about 696 hp at 6,500 rpm.
Using the black curves as an example, note that the engine produces 500 hp at both 4,500 and 5,400 rpm, which means that the engine can do the same amount of work per unit time (power) at 4,500 as it can at 5,400. However, it will last much longer at 4,500, and it will burn less fuel to produce 500 hp at 4,500 rpm than at 5,400 rpm, because the parasitic power losses (power consumed to turn the crankshaft, reciprocating components, valve train) increases by the square of the crankshaft speed (rpm).
The rpm band within which the engine produces its peak torque is limited. You can tailor a high peak with a very narrow band, or a lower peak value with a wider band. Those characteristics are usually dictated by the parameters of the application for which the engine is intended. An example of that is shown in Figure 4 below.
It’s the same as the previous graph,except that the blue torque curve has been altered (as shown by the green line) so that it doesn’t drop off as quickly. Note how that causes the green power line to increase well beyond the torque peak. Such a change to the torque curve can be achieved by altering various components – lobe profiles and separation, runner length and/or cross section to name a few. Alterations intended to broaden the torque peak will inevitably reduce the peak torque value, but the desirability is based on the application.
Because the power which a piston engine can produce is directly related to the mass flow of air through the engine, then it stands to reason that any changes which cause the engine to pump more air (“breathe” better) will enable it to make more power. Those changes obviously include the simple approach of operating the engine at a higher rpm. However, that approach has its limitations, as you might have already deduced from the preceding discussion of torque curve location.
The target operating range of a high-performance engine is usually determined by deciding the maximum operating speed the engine can survive for the required number of hours, or seconds, as is the case with many drag-racing engines. If you’re an engine designer, you make that determination based on sophisticated and extensive finite element analysis (FEA), stress and thermal analysis, followed by extensive dyno-cell testing. If you’re less sophisticated, perhaps you rely on “rules of thumb,” experience, or wishful thinking.
As a point of reference, in 2006, Formula One engines routinely reached 20,000 rpm (That’s not a typo: 20,000!) with a mean piston speed in excess of 5,200 feet per minute (fpm), producing over 750 hp at 19,250 rpm from 146 normally aspirated cubic inches (2.4 liters) of displacement. By comparison, the Continental A-65 produces 65 hp at 2,700 rpm with 170 cubic inches of displacement. ~Pat
These engines would survive for several hours of racing without substantial degradation in power. The cost of each was more than the cost of the average home.
One of the common rules of thumb for aircraft engines (proven by countless examples) is to limit mean piston speed to less than 3,000 fpm. Mean piston speed (MPS) is a fundamentally meaningless empiricism, calculated by the following equation: MPS = stroke (inches) x rpm ÷6. So, for example, an engine with a 4-inch stroke at 4,000 rpm has an MPS of 2,667 fpm and 3,000 fpmat 4,500 rpm. However, with the quality of the componentry available today, bottom-end concerns at 3,000 fpm aren’t generally a worry. Of much greater importance is the longevity of the valve train. As engine speed increases, the stresses and deflections on the valve train components go up exponentially, and generally those components are the first to fail.
Having selected an appropriate operating speed, one then designs the air passages (intake manifold plenum and runner size, path and taper, cylinder head port sizes, shapes, tapers, valve sizes, combustion chamber shape, exhaust tube size, length, configuration, etc.) and the breathing apparatus (cam lobe lift, duration, velocity, acceleration; valve train configuration, spring rates, loads, stresses and resonant frequencies; etc.) to achieve the torque curve which produces the desired power curve.
For a real example, consider a 5.0-liter Ford Windsor small-block, built with high-quality bottom-end components. Initially, the engine was equipped with Edelbrock Victor Jr. aluminum heads, a Performer-rpm dual-plane intake manifold, a 750-CFM four-barrel carburetor and a fairly mild camshaft (216-0.544 / 224-0.555 / 112). On the dyno, that combination produced a peak torque of 328 pounds-feet at 4,800 rpm. It’s not a very impressive BMEP, 164 pounds per square inch (about the same as an angle-valve Lycoming 360 or 540), and peak power of 331 hp at 6,000 rpm – 31 more ponies than an angle-valve Lycoming IO-540 turning 2,700 rpm.
In order to move the torque curve higher and thereby produce more power, the heads, intake manifold, and cam were replaced with the following components: AFR 205 heads, an Edelbrock Victor Jr. single-plane intake manifold, and a more aggressive camshaft (248-0.614 / 254-0.621 / 110).
That combination produced a torque curve with essentially the same shape as the first combination, but shifted to a higher rpm band. Peak torque of 335 pounds-feet at 5,600 rpm, peak power of 388 hp, at a mean piston speed of 3,400 fpm, but at a valve train-killing speed of 6,800 rpm. Both configurations curves are shown in Figure 5 below.
Derivation of the Power/Torque Equation
Occasionally, after hearing power and torque explained, someone will ask: “Okay, if hp = rpm x torque ÷ 5,252, where does the 5,252 come from?” Here’s the answer, but don’t be put off by the arithmetic; at worst, it requires only 8th grade algebra.
By definition, power = force x distance ÷ time.
Using the example in Figure 2 above, where a constant tangential force of 100 pounds was applied to the 12-inch handle rotating at 2,000 rpm, we know the force involved, so to calculate power, we need the distance the handle travels per unit time, expressed as:
power = 100 pounds x distance per minute.
Okay, how far does the crank handle move in one minute? First, determine the distance it moves in one revolution, which in very simple terms is the circumference of a 24-inch-diameter (12-inch-radius) circle.
Distance per revolution (circumference) = pi x diameter.
Distance per revolution = 3.1416 x 2 = 6.283 feet.
Now we know how far the crank moves in one revolution. How far does the crank move in one minute? Distance per minute = 6.283 feet per revolution x 2,000 rpm = 12,566 fpm.
Now we know enough to calculate the power, defined as:
power = force x distance ÷ time
power = 100 pounds x 12,566 fpm = 1,256,600 pounds-feet per minute. Swell. But how about horsepower? Remember that one horsepower is defined as 33,000 foot-pounds of work per minute. Therefore hp = power (foot-pounds per minute) ÷ 33,000. We’ve already calculated that the power being applied to the crank wheel above is 1,256,600 foot-pounds per minute. How many hp is that?
hp = (1,256,600 ÷ 33,000) = 38.1 hp.
Now we combine some stuff we already know to produce the magic 5,252. We already know that:
torque = force x radius. If we divide both sides of that equation by the radius, we get:
(a) force = torque ÷ radius.
Now, if distance per revolution = radius x 2 x pi, then
(b) distance per minute = radius x 2 x pi x rpm.
And we already know that:
(c) power = force x distance per minute.
So if we plug the equivalent for force from equation (a) and distance per minute from equation (b) into equation (c), we get:
power = (torque ÷ radius) x (rpm x radius x 2 x pi).
Dividing both sides by 33,000 to find hp:
torque ÷ radius x rpm x radius x 2 x pi
By reducing, we get:
torque x rpm x 6.28
Since 33,000 ÷ 6.2832 = 5,252, therefore:
hp = torque x rpm ÷ 5,252.
Note that at 5,252 rpm, torque and hp are equal. At any rpm below 5,252, the value of torque is greater than the value of hp; above 5,252 rpm, the value of torque is less than the value of hp.
And with all of that, I certainly hope this helps clear up the controversy.