Help!!! I need some answers to a design problem

So? You have equipped your toy plane wih the sort of power that would make a spitfire pilot drool.

That doesn't alter the fact that to stay aloft and cruise requires very similar power to weight figures on all aircaft irresepective of size.

Reply to
The Natural Philosopher
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Horsepower per pound.

The power per unit weight required for level flight is essentially

(drag to lift ratio) x (cruising speed)

The power to weight to attain a given climb rate in excess of the above is essentially the climbrate itself.

So total power to weight is

[climbrate] + [(drag to lift ratio) x (cruising speed)]

This leads directly to an understanding of the design parameters of all aircraft.

High wing loadng means high cruising speed, means more power to weight for level flight.

Low drag slippery airframes - like a sailplane or airliner - means lower power to weight in cruise.

Excessive power can be used to generate extremely high top speeds, or rates of climb.

Reply to
The Natural Philosopher

Like if you pick up any motoring magazine the dominant vehicle type will be some turbocharged coupe?

:-)

Mind you Paul, noy ion te mags I buy: The dominant type of plane in QEFI is usually something like a speed 600 sized plane.

Reply to
The Natural Philosopher

That was pounds per HP. A .40 turns out about 3/4 HP. Even with the enormous inefficiencies of tiny propellers, that airplane still outperforms the real thing.

Reply to
Dan_Thomas_nospam

| Doug McLaren wrote: ... | > | My son's .40-powered, 7-lb model has a power-to-weight ratio of | > | about 9 to one, and will outclimb a 172 in real numbers | > | > Power to weight? The units don't work. Did you mean static thrust to | > weight ratio? (9 is way too high for that.) Horsepower per lb? Wing | > loading in lb/sq ft? | > | Horsepower per pound.

Actually, pounds per horsepower makes more sense if it's 9 lbs/hp. So he really meant `weight to power' ratio (and he gave the units later in another post after being asked.)

The reason I asked is that the poser I was responding to went from static thrust/weight of a full scale plane to power/weight of a model like it was some sort of Apples:Apples comparison. It is not.

But seriously guys, units! If you're going to make some sort of scientific or engineering assertion, state your units. Imperial or metric, it doesn't matter -- but state your units!

You can't have a dimensionless ratio like `9:1' unless the units on the top and bottom are the same. A power to weight ratio of simply `9' is gibberish, because the units of power and weight are different. A lift/drag ratio of 9, that makes more sense, because the units are the same.

And you can't assume that people know which units you're using either.

1 hp/lb is 0.6 watts/gram -- the difference is rather small. And even if you stick to imperial units, wing loading is often given in lbs/sq ft or oz/sq in, and the numbers differ by a factor of 9.

| The power per unit weight required for level flight is essentially | | (drag to lift ratio) x (cruising speed)

At least here the units work! :)

| So total power to weight is | | [climbrate] + [(drag to lift ratio) x (cruising speed)]

And the units work there as well.

Reply to
Doug McLaren

| That doesn't alter the fact that to stay aloft and cruise requires very | similar power to weight figures on all aircaft irresepective of size.

Actually, no.

Using your own formula (I haven't verified it's correctness beyond working out the units, but for now I'll assume it's correc) --

The power per unit weight required for level flight is essentially (drag to lift ratio) x (cruising speed)

simply doubling the cruising speed will double your power needed, assuming you can keep the drag/lift ratio constant.

Very small planes (models) tend to create more drag because the differences in Reynold's numbers, but I'm not sure how large this effect is. I'll ignore it for now.

Perhaps more important than size is the shape of a plane. A powered glider will fly with a lot less power than a more traditional plane of the same wight. Mostly because the glider is designed to fly slower (large aspect ratio) and they're more careful about removing sources of drag.

As an extreme example, compare the power needed per pound for Helios (the NASA solar powered plane) to a fighter plane. Helios weighed

1322 to 2048 lbs, depending on payload. Cruising speed = 19 to 25 mph (at low altitude.) Helios could also cruise on 1/3rd of the power generated by it's solar cells. Wing area = 1976 sq ft.

They're not really clear on exactly how much power the solar cells develop (though some assumptions and some math leads to 183 kilowatts, which seems way too high.) Let's just assume that it cruises at full power out from the motors. The 14 motors are rated at 1.5 kW each, so we'll assume that it cruises at 21 kW, or 28 hp -- though that's really peak power, not cruising power.

(Helios specifications came from :

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Well, I was hoping to compare that to a jet plane, something like the SR-71, but I can't find the specifications I was hoping for.

In any event, Helios weighs about as much as a small general aviation plane. A Cessna 172 comes stock with a 160 hp engine -- much larger than the Helios's max 28 hp. (Alas, I don't know much of each's power is needed for cruising at a fixed altitude.) And a jet fighter would use much more power/weight than the Cessna. (A jet airliner like a

747 would probably use more, but it wouldn't be as extreme as a jet fighter.)
Reply to
Doug McLaren

Doubling cruise will more than quadruple the power needed. Drag increases by the square of the increase in speed, so to go twice as fast you need four times the thrust. An engine four times more powerful is apt to be nearly four times as heavy, too, so the added weight increases induced drag, the larger cowl increases form drag, and more fuel capacity increases both as well. The structure has to be stronger and therefore heavier to travel at twice the speed and to carry the extra weight, and that adds more to the drag. All these can be easily seen at work in the higher-HP versions of an airplane like the Cessna

172, where adding power doesn't do much for cruise, but improves takeoff and climb performance. After all, Newton tells us that the acceleration of a mass is dependent on the accelerating force, so a shorter takeoff roll can be expected, and a better climb from increased thrust results, since the drag at climb speeds is much lower than that at cruise. Airliners get their cruise speeds through every effort at drag reduction, from the good aspect ratio of the wing, and from engines that don't suffer thrust losses at speed like propellers do, and they cruise at altitudes where the thin air offers little drag. Even a turbocharged piston single can get big cruise numbers if it can reach those altitudes.

Older 172s had 150 HP, and before 1968 they had 145. Newest ones have an option of 180 hp, and there were versions in the early 80's that had

190. A 150 hp 172 can cruise on as little as 45% power, but the speed will be low. A typical cruising speed of 115 MPH at 5000' takes 75%.

Dan

Reply to
Dan_Thomas_nospam

Doubling the power doesn't double the speed. Not even close. A standard P-51 stripped and clean does about 425mph with the standard 1700HP Merlin. Upping the power to 3300 (Dwight Thorn engine) will only get you another 100 mph at best.

Reply to
Paul McIntosh

In article , wrote: | >simply doubling the cruising speed will double your >power needed, | >assuming you can keep the drag/lift ratio constant | | Doubling cruise will more than quadruple the power needed. Drag | increases by the square of the increase in speed, so to go twice as | fast you need four times the thrust.

You missed my big assumption -- `assuming you can keep the drag/lift ratio constant'. (And the other assumption is that TNP's formula is in fact accurate.)

And it's not quite as simple as you make it sound anyways. If you take a _specific plane_, and double it's speed and yet maintain a constant altitude, parasitic drag will quadruple as you've suggested. But induced drag will generally go down. The total drag is parasitic drag + induced drag, so what the total drag does will depend on several factors, but in most cases it'll be less than quadruple rather than more.

Here's some references for you --

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| An engine four times more powerful is apt to be nearly four times as | heavy ...

We were talking about power required/weight. TNP made the assertion that all planes require a similar amount of power per weight to stay aloft, and I said this was not true. I think that TNP was thinking specifically of small planes (models) vs. large planes, and in that case the figures may be similar for similar planes, but his assertion falls apart when comparing planes like Helios to a jet fighter.

| >A Cessna 172 comes stock with a 160 hp engine -- >much larger | >than the Helios's max 28 hp. (Alas, I don't know >much of each's | power | >is needed for cruising at a fixed altitude.) | | Older 172s had 150 HP, and before 1968 they had 145. Newest ones have | an option of 180 hp, and there were versions in the early 80's that had | 190. | A 150 hp 172 can cruise on as little as 45% power, but the speed | will be low. A typical cruising speed of 115 MPH at 5000' takes 75%.

I don't think TNP was talking about cruising speed so much as the speed where a the minimum amount of power is needed to maintain altitude. At least I was. I think this would be the `airspeed of minimum sink' and it sounds like your 45% figure would be at this speed.

I suspect that Helios could maintain altitude with it's motors running at 45% capacity too (13 hp) but it's hard to get hard figures on that.

In any event, a Cessna 172 is a relatively efficient plane, but a good deal less efficient than Helios was. A supersonic jet would be much less efficient -- requiring much more power/weight to maintain altitude (at subsonic speeds, at whatever speed requires the least power) than the Cessna, because it's wings are designed to work well at supersonic speeds, not at low subsonic speeds.

Reply to
Doug McLaren

| Doubling the power doesn't double the speed. Not even close. A standard | P-51 stripped and clean does about 425mph with the standard 1700HP Merlin. | Upping the power to 3300 (Dwight Thorn engine) will only get you another 100 | mph at best.

You missed my stated assumptions as well.

| > Using your own formula (I haven't verified it's correctness beyond | > working out the units, but for now I'll assume it's correct) -- | >

| > The power per unit weight required for level flight is essentially | >

| > (drag to lift ratio) x (cruising speed) | >

| > simply doubling the cruising speed will double your power needed, | > assuming you can keep the drag/lift ratio constant.

In order to make the drag/lift ratio the same at twice the speed would generally require changing the wings -- it wouldn't be the same plane any more.

This discussion has really veered off on a tangent. At least it's still somewhat related to model airplanes, however :)

Reply to
Doug McLaren

Yes, but the Helios flies at slower speed, which is as you pointed out why it needs less power.

What I have been trying to say all along is that there isn't some magic huge increase OR decrease needed in power to weight as models come down in size.

To an extent the lower flying speed of our models is offset by the greater drag at low reynolds numbers.

My electrics fly at an output power of betwqeen 15W/lb for a very slow parkflyer, and about 150W/lb for a grossly overpowered machine of similar performance to a tuned glo type plane.

That range of power I would almost gurantee you also covers Helios to Airbus: Only the very fast military jets exceed that by any great margin. I can't remember what it comes out at, but the English Electric Lightning with no weaponry and half full tanks would go through the sound barrier in a vertical climb. It was something like 1000W/lb.

The only point is that power to weight is far more a function of actual performance required, than the size of the aircraft.

Lets face it, imagine a 1/10th scale airbus, needing 600 yards of runway to get airborne...about 20ft wingspan, and 15mph take off speed :-)

Doesn't sound like a huge amount of power to weight does it?

Reply to
The Natural Philosopher

It does at a constant lift to drag ratio.

However as speeds go up, of course profile drag starts to dominate, and the lift to drag ratio falls...

...but up to cruising speed, its more induced drag, and the L/D ratio is

*fairly* constant.
Reply to
The Natural Philosopher

Its not 'my formula' -its basic bloody PHYSICS.

Power = force times velocity AND/OR power = rate of gain of potential energy.

Its really that simple. Power is drag times speed plus weight times climbrate.

With the prop inefficiences factored in

working around C/L max where teh L/D ratio is lowest and gfairl;y constant gives you waht O quoted.

Thank you.

Yes. And in fact I stated it this I think.

And 45% throttle is what in terms of POWER. Certainly on a typical electric model its about 22% of POWER and about 50% of RPM.

Hence swing wings.

Reply to
The Natural Philosopher

I didn't miss the assumptions, I just ignored it for the purposes of being realistic. In the real world, it doesn't work like that so no reason to even use it as an assumption. ;^)

Reply to
Paul McIntosh

Up to "crusing speed"? What, pray tell, is "crusing speed"? Using the popular term for most light planes, that is the speed achieved at approx 65% throttle. The speed varies wildly depending on the airframe and engine.

Using the stated P-51 for example, the weight and airframe are similar yet doubling the power only results in an increase of about 25% in speed. (In actuality, the weight is lower and aiframe is cleaned up quite a bit.) I couldn't find any aircraft with different engine options that came any where near the doubling of speed with twice the HP. The average was arount

25-30%.
Reply to
Paul McIntosh

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