May 16, 2002, 08:04 AM Registered User United States, VA, Warrenton Joined Apr 2001 1,148 Posts How does prop RPM affect power? I just got AstroFlight's Electric Motor Handbook and I'm trying to puzzle out something regarding the power you get from a motor. As a for instance, on 10 V you could prop a motor to draw a certain current, say 20 amps. Now, you can then use a gear box to swing a bigger prop at the same 20 amps, but get more thrust. The system you have is still generating 200 Watts, right? But now your RPMs are reducing. Theoretically, you could continue to increase the gear ratio and keep propping for 20 amps and your system is still a 200 W system, right? But your RPMs will go down, down, down... What is the effect of RPMs on performance? Your system may be using 10 volts and 20 amps, but as you increase the gear ratio, are you really getting that 200 W in the air? What RPM range should you shoot for for good flight? Thanks, Dan Nash
 May 16, 2002, 08:52 AM Member Edina, MN, USA Joined Oct 1999 11,592 Posts Yes, thrust increases with prop diameter; but for your fixed input power, prop wash velocity decreases. It takes both thrust and prop wash velocity to fly a plane. Prop wash velocity is a function of propeller pitch and rpm. One rule thumb, and method for motor/prop selction, is to select you prop pitch 1st, based on 2.5 to 3 time stall speed -- 20 mph stall speed them pitch speed should be 50 to 60 mph. A 10" pitch at 5400 rpm is a velocity of 54 mph. Down load a copy of MotoCalc ( www.motocalc.com ) and learn a lot about pitch speed, thrust, rpm, etc.
 May 16, 2002, 11:46 AM Registered User Williamsville, NY, USA Joined Sep 2000 4,553 Posts A quick estimate of the power required to turn the prop can be found by: P = ((dia/12)^4)*((RPM/1000)^3) * (pitch/12) So, for a 12 x 8 prop turning at 5000 RPM, the power required is: ((12/12)^4) * ((5000/1000)^3) * (8/12) = 83 watts The most critical term is the diameter. Double the diameter ((24/12)^4)*((5000/1000)^3)*(8/12) = 1333 watts or 16 times the power
 May 17, 2002, 11:50 AM Registered User Dallas,TX, USA Joined Mar 1999 834 Posts Ninjak2k, Larry Dudeck procided the power formula;from that you could compute your 200 watt example and how RPM decreases with increase in prop diameter. Static thrust, Ts=CT0*Rho*(RPM/60)^2*(Dia/12)^4. Where Ts is static thrust in lbs,CT0 is the static thrust coefficient,Rho is the air density in slugs/ft^3,RPM is the prop speed in revs per minute and Dia is the prop diameter in inches. The CT0 is measured imperically for a family of props that depends on power coefficient. Anyway, the static thrust varies as a function RPM^2 times Dia^4; you can play with that relationship keeping a constant 200 watts. Yours, Dick Huang
 May 18, 2002, 07:23 AM Registered User Williamsville, NY, USA Joined Sep 2000 4,553 Posts Dick, Any feel for the magnitude of 'CTO'? General info for any who might be interested... the density of air in 'slugs/ft^3' at sea level, 68 F is aprroximately equal to .075/32.17 or .00233
 May 18, 2002, 12:12 PM Registered User Dallas,TX, USA Joined Mar 1999 834 Posts Larry, CT0=CP0*1.727*(1.97-PTD) where PTD is the pitch/diameter and CP0 is the power coefficient=(hp*550)/(rho*(RPM/60)^3*(Dia/12)^4. Most of this work was performed in the 1920's by NACA using 2 blade wood propellers. For an APC 10x4 running at 10000 RPM and SL, I get CP0=0.03775, CT0=0.1023 and static thrust=3.26lbs. Dick Huang
 May 19, 2002, 12:32 PM Motors beat engines! Milwaukee Wisconsin, United States Joined Feb 2001 4,564 Posts Basically you're looking for the biggest prop that you can still get the 2.5 to 3x stall speed pitch speed number with. The other variable is props. You have to work within what's available, and its also best to not get too close to the 1:1 diam/ pitch ratio, as the prop will be stalled untill the plane gets some speed up reducing thrust, reducing the gain expected fro the whole excercise. That being said, I have had best luck going with closer to the 3x figure than the 2.5x, as you can throttle back more giving longer flights. Every plane I've geared flew better geared than DD. No exceptions. My best example is my 48" span, 500 in/sq pattern plane at 3lbs. I'm only using about 240 input watts into a 12x8 prop, but the plane goes better than 50 mph and can hang on its prop! Dean in Milwaukee
May 19, 2002, 01:27 PM
Boffin
Joined Apr 2001
3,404 Posts
Re: How does prop RPM affect power?

Quote:
 Originally posted by Ninjak2k What is the effect of RPMs on performance?
The simple answer is that a big slowly moving propeller is more efficient at moving air. So the higher the gear ratio, the more static thrust you get for the same power. What you sacrifice is speed and some mechanical loss in the gearbox.

Indoor flyers can use very tiny motors geared 16:1 and huge props. They can only fly very slowly. They couldn't fly at all with the same motor direct.
If you want top speed for racing you ordinarily use a direct drive motor with a small prop turning very quickly. The choice depends on how you want to fly.

A complicating factor is voltage and current. The direct drive motor might be 10 volts at 20A to get your 200W. The same motor with a gearbox might be 20 volts and 10 amps (for example). The motor will turn much faster and the current will be lower. Both lead to higher electrical efficiency.

So a well chosen geared setup can be more efficient in almost all ways which helps explain why they have become so popular in most planes.

Rick.