estimating power needed for level flight

[flieslikeabeagle]

Hi all, I'm trying to figure out how to estimate the amount of power needed to maintain level flight.

For starters, there's a rule-of-thumb for parkflier foamies that says that roughly 2 watts per ounce will probably fly the plane. Its too general to be accurate for every model, obviously, but it seems to be reasonably close to the mark for most slow-flier foamies.

I had an idea to use some basic physics; assuming the plane is gliding at a uniform speed and has a (minimum) glide angle of theta, then the drag force equals the component of gravitational acceleration along the planes thrust line, which is m*g*sin(theta).

Since power = force x velocity, the power needed to move this plane at this speed, with this drag force, is mg sin(theta) V, where V is the speed. Knowing the rough glide ratio of the plane (power off) gives you theta; knowing an approximate flying speed gives you V; so I can now calculate the power needed to keep it flying.

Well, that seems straightforward enough, but if I put in 1 oz (0.0283 kg), a glide ratio of 5:1, and a speed of 5 m/Sec (about 11 mph), I come up with the result that it only needs 0.28 watts to fly one ounce of plane! That number is about seven times lower than the rule-of-thumb of 2 watts/oz, which is based on experience! I can believe a factor of two (50% efficiency in the motor/gearbox/prop), but seven???

What am I missing?? :confused:

-Flieslikeabeagle.

[Andy W]

Your trying to put too much faith into generalizations that are too simplistic to effectively model flight dynamics.
..a

[vintage1]

No, he is close.

I did all these calcs as well, and you are bang on theory wise.

.28 watts per oz is 4.5 watts per lb. Which is consistent with my calcs of about 3W per pound at 15mph and a 10:1 glide angle.

Motor is 50% efficient, but the prop? make that 50% efficient as well, and shove in 80% on the gearbox and you get about 22W/lb. For level flight.

But your parkflyer can't just potter around on the edge of stalling - it needs to climb as well.

Lets dump the efficiences, and say you have the drag and glide angles correct, and stay with your flying speed - which is very low in my opinion - I'd say nearer 15mph but hey, lets not argue over detail.

From the web converter prgrams..

1 foot pound-force/second = 1.3558179 watt

so for every oz raised at one foot per second it takes about .084 watts A measly 60 feet per MINUTE climb rate. We need at LEAST 100fpm to really cope with just minor turbulence and about double that is better, so let say 200fpm is a decent outdoor parkflyer, so that adds 0.2824620625 watts per once to your figure

About double to .56 watts per ounce. This is what teh prop needs to produce more or less, in order to either climb at 200fpm at stall speed or to fly level at maybe 1.5 times stall speed...cos drag will go up as speed increases, and power goes up with drag AND speed...

Its stacking up.

So now we stick in 50% for the motor efficiency, which gives us 1.12W/oz for a decent plane.

80% efficiency for a GWS gearbox, takes us up to 1.4W/oz.

Which is nearly 70% of your 2W/oz ...you can lose the alst 30% either in a better climb rate and more top speed or a nasty orange propellor, or both :D

My IPS powered Peter Rake Sperry monoplane was 15W input, weighed 7oz but was capable of better than 300fpm and over 22mph.

I estmated motor eficiency about 55%.

Throttled back I am sure it drew a lot less, and would wafle around on 1/2 throttle 1/4 power qute nicely.

So I think your thinking is spot on, as fars as it goes, it just needs to go a little further.
:D


.

[flieslikeabeagle]

Andy, thanks for the feedback. Just to clarify, my goal was to do a quick back-of-the envelope estimate, without worrying about extreme accuracy. I find this kind of order-of-magnitude estimate often helps me understand the basic physics concepts behind the problem. Once the concepts are understood, one can always add detail, but if I start with too much detail, then I can't see the wood for the trees. :)

Vintage1, thanks very much for your feeback! Very informative!

You're right about needing more power to climb, of course, but I wanted to start by understanding the bare minimum amount of power needed to hold level flight. Any extra power beyond that would be used to climb, so the next step I was after was to estimate a climb rate (as you did, equating potential energy increase per second to power), and then climb angle, knowing the climb rate and flying speed.

I understand the drag increasing with speed (proportional to V^2), and consequently the power requirements going up even more steeply (V^3). Which is why a 70 HP car can often do over 100 mph, but it takes some five or six hundred horsies to get to 200 mph...

The 11 mph figure came from the estimated stall speed of my Wattage Camel. You obtained that number for me using MotoCalc during an earlier discussion we had, and subsequently I found an equation in an issue of "Fly RC" connecting wing loading to estimated stall speed, and it returned the same number. And sure, flying around at the edge of a stall wouldn't be much fun (!), so I can throw in a 50% increase over stall speed and the appropriate increase in drag force and power requirement.

All this is to help me develop a little JavaScript program to be used to help design parkflyers...with the behind-the-scenes goal of helping me learn both a little more about the physics of flying, and coding in JavaScript. The program exists in fledgling form, but at present it uses the "watts/oz" rules of thumb rather than any actual physics equations.

Thanks for the feedback!

-Flieslikeabeagle

[vintage1]

Sounds like you are replicatiing Motocalc.

PM me if you need to toss any ideas around. I'm vaguely considering teh cost benefit of writing articles (for publication) to basically explore this very territory.

I.e. to answer the simple question 'what power train do I need for plane X' with the very long and complex answer...