RC airplane required engine power 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
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