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I'm going through the process of designing an RC airplane. I know that probably most calculations on scale models might be non-precise ones, but I'm trying to do it as correct as possible (it's a school project). I was trying to check how much shaft power I would need for the aircraft, so I did an excel table and used some pessimist assumptions in order to check if my 46 cubic inches, which deliveries 1,63HP, had enough power. These are the aircraft parameters: Chord= 0,3m Wingspan = 1,8m Wing Area= 0,54m2 AR=6 W= 3,7kg (assumed) Cp = 0,048 (assumed) Propeller efficiency= 0,6 (assumed) Air density= 1,225 kg/m3 (Sea Level - where I intend to operate)

I used W = 1/2 * rho * S * V2 * CL to identify the required CL for level flight at any given airspeed, then Cdi = CL2 / (pi * AR * e) and using Ri = 1/2 * rho * S * V2 * Cdi and Ri = 1/2 * rho * S * V2 * Cdi to find out both resistances I summed them up and multiply them by speed to get power, then I divided that power by propeller efficiency to get the needed shaft power.

For my surprise, the maximum speed that I could achieve with that 1,63HP it's between 39 and 40 m/s or 144km/h (89,4775 MPH), it seems like a lot. Because of these results, I decided to contact a friend who is into RC Planes, he told me that that engine was actually too little for a 1,8m wingspan plane, it looks like I'm doing something wrong, but I don't know what. L/Ds actually make sense.

This is the link for the excel file

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  • $\begingroup$ Do you mean 0.46 cubic inches? 50W/lb is normally considered the minimum for RC planes but you have 3 times that, which should be plenty of power for aerobatics. For reference, here's a similar size model with a 10cc (0.60cubic inch) motor - alshobbies.co.uk/rc-planes/planes-fuel/fuel-sports-planes/… $\endgroup$
    – Robin Bennett
    Jul 9, 2020 at 8:06
  • $\begingroup$ Yeah that's a 0.46 cubic inch engine, I spoke to someone with some experience, who recommended a 0.60 cubic inch for 1.8m wingspan, he told me that I can fly with a smaller engine, but I will have problems on windy days or taking off from a grass runway. He also told me that my model will probably end up weighting more than these 4kg $\endgroup$
    – Lucas Delgado
    Jul 9, 2020 at 18:42
  • $\begingroup$ I think I'd agree with that. If you can build it light a 0.46 would be OK but most models that size use a 0.60 (10cc) 2-stroke or 0.90 4-stroke. You could try asking over at drones.stackexchange.com $\endgroup$
    – Robin Bennett
    Jul 10, 2020 at 8:43
  • $\begingroup$ I did not realized I got a specific page for that haha. Anyway, thank you for your help Robin. $\endgroup$
    – Lucas Delgado
    Jul 10, 2020 at 15:29

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Resistance is a term used in politics and electric circuits. We here talk of drag.

Your induced drag equation uses speed where the lift coefficient belongs: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

What is missing now is the zero-lift drag which is the dominant contributor to overall drag at high speed. From your top speed result this seems to be $$c_{D0}=\frac{2\cdot P\cdot\eta_{Prop}}{v_{max}^3\cdot S_{ref}\cdot\rho} = 0.040$$ assuming a propeller efficiency of 70% and neglecting induced drag. That figure is not unreasonable. Is the landing gear retractable?

Now we can write for the total drag as a function of speed: $$D = \frac{\rho}{2}\cdot v^2\cdot S_{ref}\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)$$ $$D = \frac{\rho}{2}\cdot v^2\cdot S_{ref}\cdot c_{D0} + \frac{2\cdot (m\cdot g)^2}{\rho\cdot v^2\cdot S_{ref}\cdot\pi\cdot AR\cdot\epsilon}$$

When sizing an engine the most reasonable parameter is power loading: How many Kilowatts are available per square meter of wing area and per Kilogram of airplane mass. Wingspan is a poor basis for comparison, since a larger span at the same area means a higher aspect ratio and lower demands for power. When using your figures, your power loading is 1.21 kW for 0.54 m² or 2.25 kW per m² or 0.608 kW per kg. Compare these figures to similar model aircraft to see whether you are on the right track.

Nomenclature:
$P\;\;\;\;\;\;\:$engine power
$v\;\;\;\;\;\;\;$flight speed
$\rho\;\;\;\;\;\;\;$air density
$c_{Di}\;\;\;\;\;$induced drag coefficient
$c_{D0}\;\;\;\;\:$zero lift drag coefficient
$S_{ref}\;\;\;$reference area
$AR\;\;\;\;$aspect ratio (= 6 in your case)
$\epsilon\;\;\;\;\;\;\;$span efficiency, use 0.85 for a rectangular wing
$m\;\;\;\;\;\;$airplane mass
$g\;\;\;\;\;\;\:$gravitational acceleration

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  • $\begingroup$ Hey thank you for your answers you are being really helpful, I really appreciate that. $\endgroup$
    – Lucas Delgado
    Jul 8, 2020 at 22:02
  • $\begingroup$ About resistance instead of drag that's my Spanish playing its part, we use the same word for both resistance and drag, and that word is similar to resistance hahaha. $\endgroup$
    – Lucas Delgado
    Jul 8, 2020 at 22:03
  • $\begingroup$ I did use Cl in Cdi equation, I made a mistake when typing it. Is that zero lift drag the same that parasite drag or should I sum both of them up? $\endgroup$
    – Lucas Delgado
    Jul 8, 2020 at 22:06
  • $\begingroup$ I did realise induced drag did almost nothing compared to parasite drag at high speeds $\endgroup$
    – Lucas Delgado
    Jul 8, 2020 at 22:06
  • $\begingroup$ @LucasDelgado. The zero lift drag coefficient is all the drag the airplane produces without creating any lift, referenced to dynamic pressure and reference area. So it sums up everything except induced drag, parasitic drag included. $\endgroup$
    – Peter Kämpf
    Jul 8, 2020 at 22:09

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