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Plane: Weighs 200 pounds. 16 FT wingspan, 36in. by 5 in. airfoil

Engine: 3,000 RPM, 2 HP

Propeller: 4ft with 4 blades and weighs 4 lbs.

Is there perhaps an equation that could determine this?

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There are methods of calculating the power required to fly a particular plane. In your case, assuming that you'll also need to carry ~18 kg of fuel and a pilot (roundly 75 kg), the answer is NO. A 2 hp (1.5 kw) power plant will be insufficient to give safe takeoff and climb performance -- with the possible exception of a high performance sailplane airframe (some self-launching sailplanes use ridiculously small engines).

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2hp will give you about 10 lbs of thrust, tops (it's generally around 4-5 lbs/hp). If you have an ultralight with an all up weight of 400lbs, say, that maybe has a 10:1 L/D, you will need 40 lbs of thrust just for level flight at best L/D speed, requiring about 8-10 hp. That's with no surplus power for, you know, taking off and climbing and such.

So a weed wacker won't get you off the ground in the first place in any kind of craft that can carry a human, unless of course the wings are like those on a human powered aircraft with a span that goes off into the next county.

And your ultralight has a small wing with only 48 sqf of area that will have a wing loading of 9ish psf and will need to go 50 mph just to fly, which it ain't gonna do on 10lbs of thrust. Faggeddaboudit...

For an ultralight, you need wings big enough to get the wing loading down around 3lbs/sqf, to get a stall under 30 mph, and a thrust level equal to about 2 to 3 times the drag at max L/D. If your ultralight has 40lbs of drag at max L/D, you will want to have at least 120lbs of thrust, or about 25hp minimum. You can get by on less power if your ultralight is very clean, with long wings, like a Lazair or Mitchellwing, but not that much less.

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"Is it possible to power an ultralight airplane with a single-cylinder weed-whacker engine?" -- I've seen someone actually try it. The engine was still attached to the weed-whacker. The string was replaced by a propeller. The unit was strapped to the pilot's waist, projecting backwards. The aircraft was a paraglider. The pilot foot-launched off a mountain and then started the contraption in the air. Unfortunately I never heard what sort of climb rate he was able to achieve once he flew away from the ridge lift.

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  • $\begingroup$ It would give him a flatter glide. A bit flatter. Paramotors get by on a rock bottom minimum of about 100 lbs of thrust and most are 140-160. 2hp would give you about 10lbs tops and would require a reduction drive. $\endgroup$ – John K Nov 26 '19 at 22:00
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Here is how to determine the answer (which, as pointed out by others here, is no):

We assume you know the weight of your plane, its gliding airspeed, and the glide ratio. This means you know the sink rate in feet/second which multiplied by the mass of the plane, yields the power dissipation rate during a glide which can then be converted into horsepower.

To hold altitude constant then requires that the power application rate equal the dissipation rate. To climb requires that the power application rate exceed the dissipation rate by some convenient margin (your choice).

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  • $\begingroup$ So power required = speed * weight / lift-to-drag-ratio (multiplied by some horrible constant for non-metric units, presumably) $\endgroup$ – Robin Bennett Nov 27 '19 at 10:54
  • $\begingroup$ Not sure. the answer is yes if the lift to drag ratio is exactly equal to the ratio of the forward speed to the vertical speed but that is an assumption I do not feel qualified to justify. Peter Kaempf on the aviation stack exchange would certainly know the answer. $\endgroup$ – niels nielsen Nov 27 '19 at 17:25

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