I did a math experiment today, but wanted to verify the results.

I wanted to figure out how much horsepower is needed for an aircraft to hover in midair with no wings/lifting body (1:1 thrust/weight).

So to start I found that a unit of horsepower is the amount of work required to move 550 pounds 1 foot every second.

Then I wanted to find out how many foot/pounds of force gravity places on 1 pound of material. With a static acceleration of 9.8m/s (32f/s), it seems that 1 pound has 32 foot pounds of gravity acting on it.

Now for my equation:

Hp = 550 (f/p)
Gravity = 32 (f/p)
Hp / Gravity = 17.1875

So it seems 1 hp can hold 17 pounds in the air, but to incorporate propeller inefficiency, I just used 70% of that. This leaves me with 1 hp being able to vertically lift 12 pounds when accounting for the inefficiency of a propeller.

Is this a relatively correct assumption? I know thrust changes as the aircraft accelerates, but in a hovering state the acceleration is 0 f/s as a whole. So will these numbers apply?

Edit: I initially used a specific ultralight to prove how high the numbers were, but I used wrong data from the spec sheet. So I removed that reference from my question.

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    $\begingroup$ You seem to be confusing acceleration, power and thrust, and then running with it into increasingly more erroneous calculations. What you should be asking is how much of the engine's rated power can be converted into thrust by the propeller. Thrust is a force, which you can compare to gravity. Power, on the other hand is energy per unit of time. $\endgroup$ Sep 23, 2019 at 14:20
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    $\begingroup$ Don't you need the gross weight from the spec sheet, which is ~1000lbs? According to Wikipedia, the engine alone weighs 110lbs, so it seems your estimate that the whole aircraft weights 115lbs is too low. $\endgroup$ Sep 23, 2019 at 14:25
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    $\begingroup$ @NuclearWang oh wow... yeah that needs rectifying. Maybe the math isn't as off as I thought $\endgroup$
    – YAHsaves
    Sep 23, 2019 at 14:29
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    $\begingroup$ 1 horsepower = 1 horse worth of thrust, ergo horses can fly/hover. =) $\endgroup$ Sep 23, 2019 at 15:12
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    $\begingroup$ @EricHauenstein one Clydesdale can exert a peak of roughly 17 horsepower. So they should accelerate vertically like a rocket :) $\endgroup$
    – Criggie
    Sep 24, 2019 at 9:30

5 Answers 5


Horsepower cannot be converted into thrust without knowing the speed at which this is done. In your case this is zero, so the special equation for static thrust applies. With $T$ for thrust, $P$ for power, $\rho$ for air density and $d_P$ and $\eta_P$ the propeller diameter and efficiency, respectively, this is $$T_0 = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{P}^2\cdot\pi\cdot d_P^2\cdot\frac{\rho}{2}}$$ With this equation you cannot simply say how many HP are required to lift that many pounds; instead, you need to add the propeller geometry. If we assume a very efficient propeller with a large diameter, we can combine blade size and mass into disk loading – here 8.5 kg/m² or 41.5 lbs/ft² is a good value. Now we can write, using standard atmospheric density at sea level: $$\frac{T_0^2}{P^2\cdot\eta_{P}^2} = \frac{2\cdot1.225\,\frac{\text{kg}}{\text{m}^3}}{8.5\,\frac{\text{kg}}{\text{m}^2}\cdot g} = 0.0294\,\frac{\text{s}^2}{\text{m}^2}$$ assuming for simplicity a 1 m propeller diameter. If we now use 70% for propeller efficiency, as you did in the question, the power to mass ratio becomes 8.33 Watt power per N of weight force or 81.7 Watt per kg. Converted into funny units this is 0.05 HP per pound of mass. 1 HP can hold 20 lbs in the air, but that will not be enough to even spin up the propeller unless it can adjust blade pitch. For practical usability I recommend to double that number. If the propeller diameter shrinks and disk loading increases, efficiency will drop and the power required will rise.

  • 1
    $\begingroup$ Seeing the numbers laid out like this helps me understand where all the "wasted" energy is actually going. If I understand it right my initial estimate of "12 pounds per hp" wasn't too far off, but that's assuming perfect air pressure and a very large prop. I see why real world conditions almost never meet that, considering the number of variables accounted for. $\endgroup$
    – YAHsaves
    Sep 23, 2019 at 21:40
  • $\begingroup$ DIYDrones.com listed directly measured electric motor/prop data at around 1 kg thrust/100 watts Power, fairly linear through RPM range. Yours numbers (even with a great variety of prop/rotor design), are close. $\endgroup$ Sep 24, 2019 at 3:06
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    $\begingroup$ For practical usability you really should be running at around half or 3/4 power to generate hover thrust otherwise you would not have enough power budget to maintain altitude in case of disturbance. So yeah, I strongly agree with doubling the power. For RC planes you wouldn't even attempt to hover unless your plane has 2:1 thrust-to-weight $\endgroup$
    – slebetman
    Sep 24, 2019 at 7:24
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    $\begingroup$ did you really mean to say "assuming for simplicity a 1 m² propeller diameter"? Diameter is a distance, but m² is an area. $\endgroup$
    – Michael
    Sep 24, 2019 at 17:08
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    $\begingroup$ @Michael: Better now? $\endgroup$ Sep 24, 2019 at 22:30

Thrust from a propeller aircraft will vary with airspeed, propeller efficiency, density altitude, etc.

For a very rough guideline you can use 1 hp to equal approximately 3 lbs of thrust. So theoretically a 3,000 lb aircraft could hover if it had about 1,000 horsepower.

The Lockheed XVF weighed approximately 15,000 lbs and could hover using about 5,000 hp.

enter image description here

Here is a link to some detailed testing regarding propeller thrust: Static Thrust Measurement for Propeller-driven Light Aircraft

enter image description here enter image description here

  • $\begingroup$ Does that imply that propeller aircraft have terrible efficiencies? Total energy should be preserved, so if the propeller is only accelerating 1/17th the mass of air downward that it could be based on power input, where does the other 16/17th of the power go? Is it lost to circular momentum friction and vibrations? $\endgroup$
    – YAHsaves
    Sep 23, 2019 at 14:44
  • $\begingroup$ I have edited my answer to correct wrong formation. I have also added some interesting photos and links. $\endgroup$ Sep 23, 2019 at 20:00
  • $\begingroup$ max takeoff weight: 1,370 lb (635 kg) & -A2C flat 4 piston engine, 124 hp (93 kW) so 10 pounds per horsepower, is easily seen in production machines $\endgroup$ Sep 24, 2019 at 16:17
  • $\begingroup$ @James Jenkins its pounds of THRUST per horsepower, not aircraft weight. That ratio is commonly thrust = 25-30% of aircraft weight (the miracle of wings) But we are also looking at helicopter rotors. $\endgroup$ Sep 25, 2019 at 1:01

In order to hover in the air, you have to accelerate a mass of air downwards. Usually, this is done with a rotor, and the larger the rotor, the less the power you'd need to hover. The general expression (in theory...)for the power required, based upon momentum considerations, is:

$P_{required}=T\sqrt{\frac{T}{2\rho\cdot A}}$

where T is the thrust (should be the same as the weight, in order to hover...), A is the rotor disk area, and $\rho$ is the air density.

Using SI units, thrust/weight should be in newtons, A in square meters, and $\rho$ is, at sea level, 1,23 kg/m3.

  • $\begingroup$ Seeing this equation really helps. I guess we are losing a lot of power due to the environment we are working with. However I don't know where to plug "rho" into the question. Is "rho" the "2p" or how do I apply it? $\endgroup$
    – YAHsaves
    Sep 23, 2019 at 14:46
  • $\begingroup$ Yes, rho is the 'p'...For example, if your craft has a mass of 500 kg, that would mean a weight/thrust of 4900 newton. Assuming your rotor has 8m diameter, the disk area will be 50,24 m2. Inserting those figures in the expression, the power will be 30,9 kW. In practice, you'd need 50% more at least, since reality is usually hard... $\endgroup$
    – xxavier
    Sep 23, 2019 at 14:54
  • $\begingroup$ A real world example for comparison: Robinson R22 ha MTOWof 622 kg, rotor diameter of 7.7m and max power of 124hp. Max hover altitude is 8000 (~MTOW) and assuming 3% power loss /1000ft, I'd say it takes about 95hp to hover R22 at 622kg (when not in ground effect). Feel free to correct if I made a mistake :) It's worth noting that some of that 95 hp (or what ever if I didn't get it right) is "lost" in couteracting the main rotor torque with the tail rotor. $\endgroup$
    – Jpe61
    Sep 23, 2019 at 16:08

You've made quite a lot of mistakes in your math, some less severe and some more severe; you may want to think about these.

Notation mistakes

These mistakes are not so severe; we can still tell what you're talking about.

The unit of energy you're talking about is called the "foot pound" or the "foot-pound," not the "foot/pound." A foot pound is a foot multiplied by a pound. A slash in a unit indicates division, so a "foot/pound" is a foot per pound, or, in other words, a foot divided by a pound.

You can abbreviate "foot pound" as "ft lbf," but certainly not as "f/p."

Note that in the phrase "food pound," the word "pound" means "pound-force."

Unit mistakes

These mistakes are more severe; as a result of these mistakes, your calculations are completely wrong.

Let's go paragraph by paragraph.

So to start I found that a unit of horsepower is the amount of work required to move 550 pounds 1 foot every second.

That's correct.

Then I wanted to find out how many foot/pounds of force gravity places on 1 pound of material.

There's no such thing as "foot pounds of force." Foot pounds are a unit of energy, not force. You can ask how many foot pounds of energy, or how many pounds-force of force (or just "pounds of force" for short), but not how many foot pounds of force.

With a static acceleration of 9.8m/s (32f/s), it seems that 1 pound has 32 foot pounds of gravity acting on it.

There are two mistakes here. The acceleration due to gravity is $9.8\ \mathrm{m}/\mathrm{s}^2$ (meters per second per second) or $32\ \mathrm{ft}/\mathrm{s}^2$ (32 feet per second per second), not $9.8\ \mathrm{m}/\mathrm{s}$ (meters per second) or $32\ \mathrm{ft}/\mathrm{s}$ (32 feet per second).

The second mistake is that you've changed feet per second into foot pounds. A foot per second is not the same thing as a foot pound.

There are two ways of saying this correctly:

  • 1 pound-mass has 1 pound-force of gravity acting on it.
  • 1 pound-mass has 32 pound-mass feet per second per second of gravity acting on it.

Now for my equation:

Hp = 550 (f/p)
Gravity = 32 (f/p)
Hp / Gravity = 17.1875

The numbers are correct, but since the units on the first two quantities are wrong, the units on the last quantity are wrong, too.

With the correct units, the calculation is:

$$\text{Power} = 550\ \mathrm{ft\ lbf}/\mathrm{s}\\ \text{Gravity} = 32\ \mathrm{ft}/\mathrm{s}^2\\ \begin{align}\text{Power} / \text{Gravity} &= 17.19\ \mathrm{lbf\ s}\\ &= 17.19\ \mathrm{slug\ ft}/s\\ &= 550\ \mathrm{lb_m\ ft}/s\end{align}$$

(No need for more than four significant digits of precision.)

See the definition of a "slug" on Wikipedia.

So the correct interpretation of this result is that one horsepower is enough to raise 17.19 slugs (equivalently, 550 pounds) one foot per second; or 1 slug 17.19 feet per second; or 1 pound-mass 550 feet per second. This calculation is good for raising the mass using pulleys or something; it doesn't work for aircraft.

So it seems 1 hp can hold 17 pounds in the air

Here you've made yet another mistake: the result of your last calculation was the unitless number 17 (and it was supposed to be 17 slug feet per second), but you changed that to 1 pound.

I don't think you made any other mistakes besides these.


Work is force × distance, and power is work per time. If distance travelled is zero (object is in motionless hover), work performed and power applied also are zero.

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    $\begingroup$ If that were the entire story, you'd be able to shut down a helicopter engine while hovering and stay in the air. So this is an oversimplification. $\endgroup$
    – Hobbes
    Oct 18, 2019 at 7:58

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