View Full Version : Question Calculate required power to stay airborne?
skogsvargen
Sep 29, 2008, 06:29 AM
I am trying to design an electric long endurance airframe. It will basically be a glider which i try to optimize to stay up as long as possible without relying on thermals.
The question is how do i calculate how much power that i will need to keep the plane flying at a given speed?
I know the airfoil and can calculate lift and drag from the polars. I know the weight of the airframe and of course the wing diemsions. Calculating the drag from the fuselage and the stabilizers is harder, but i just need an approximate number.
/Magnus
biber
Sep 29, 2008, 07:13 AM
Level flight will require a Power = Airspeed*Weight/(Lift/Drag).
Seek for the speed of minimum sink as optimum speed, as the required power can also be written as Power = Weight*Sinkrate.
Note that the power requirement calculated above is what the propeller has to effectively deliver.
That has to be divided by the prop efficiency to get the engines mechanical power required,
which again has to be divided by the motor efficiency to get the required electric power.
biber
vintage1
Sep 29, 2008, 07:37 AM
This is incredibly simple.
What you need is the L/D ratio, and the flying speed.
watts per pound needed is the speed divided by the lift to drag ratio. With suitable constants.
You can get the speed from the wing loading..genarlly about 1.25 times stall..others will know better.
A good slow revving propelleor (think rubber model if you know what I mean) seems to be around 70% efficient: if you motor is similar, then simply double the watts per pound needed to sustain flight, and that's your input power.
I calculated a clean airframe should need as little as 3W/lb to stay up. Output. So 6W/lb input.
An example calculation. Let's say you are flying at 20mph wih an L/D ratio of 20.
So te drag is 1/20h of the weight, in lb, time steh speed, which is 20mph.
20 mph is 20x44/30 feet per second 29.33333 feet per second..divide that by the ratio and its 1.4666 feet per second..
That times the factor that google turns up (1.355818) gives you 1.9998 watts/lb.
So, if we take the two contants - mph to feet per second, and pound feet per second to watts, the overall formula is
watts/lb = ((Speed in mph/ (L/D)) * C) where C = 1.998. Call it two for simplicity.
That's how to derive it from those figures: to derive it from the actual airframe means you have to find its glide angle essentially, or calculate drag somehow.
What you should realise is this is JUST enough to keep it in the air : to get any sort of climb out of sink, or raw climb, you need an excess of power added to that which is rate of climb in feet per minute divided by 60 times 1.355818: so a 150 fpm climb needs say an excess of three and half watts per pound. Roughly.
Looking again at your post, you seem to need an approximate figure for profile drag.
ISTR there is a formula that relates to the frontal area and a Cd figure. I suggest about 0.1 of the 'flat plate of similar area' which you will have to look up.
But it may be simpler to fly the model and set it up in still air and a altitude recorder, and simply see what its rate of descent is. The watts per pound to keep it up is almost the sink rate in feet per minute times .00226
So that's another way, with an approximate doubling for motor and prop inefficiencies.
I cannot emphasise enough how much a very big slow prop will improve efficiency by the way. Rubber models seem to operate sub 600 RPM..if you can gear that deep you are in good territory. The prop will be MASSIVE...and very coarse pitch probably.
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