How much horsepower to weight is required for a 1:1 thrust ratio?

Question

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.

Answer 1

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.

Answer 2

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.

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

Answer 3

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.

Answer 4

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.

Answer 5

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.