5
$\begingroup$

I've been trying to find the required motor power for a quadcopter. I first calculated drag power and then I made a linear trend line for motor power against motor weight. I made the motor power 1.5 times the drag power and then used the trend line to find its weight but this seems to give numbers that are much too small. What am I doing wrong and how can I fix this?

What I did was take the altitude, the temperature, estimated weight including payloads, rotor radius and chord, and the speed to solve for drag using the drag equation. An example is alt: 400 ft, temperature: 65 F, estimated weight: 3 pounds, rotor radius: 6 inches, rotor chord: 1 inch, and speed: 20 kts. This would give a drag power of 16 watts and this is so small that the motor weight is negative according to the trendline.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ If you can post an example of what you did (including the numbers you used), it would enable us to better help your problem. $\endgroup$
    – kevin
    Commented Oct 30, 2016 at 19:18
  • $\begingroup$ Probably answered in How to calculate the required motor power and rotor size with respect to the weight of quadcopter? $\endgroup$
    – mins
    Commented Oct 30, 2016 at 20:29
  • $\begingroup$ Would I just add the power needed to overcome drag to the amount needed to overcome weight? $\endgroup$
    – user2888499
    Commented Oct 31, 2016 at 2:14
  • $\begingroup$ I think the assumption is to use twice the power required to compensate for weight. Half the power is in excess at sea level, and used to provide for drag, wind, temperature, density, etc, and horizontal and vertical accelerations. I'm not an expert. Drag here seems difficult to evaluate anyway. Indeed this total power has to be split by the number of motors, and the efficiency of the motor / propeller must be taken into consideration. $\endgroup$
    – mins
    Commented Oct 31, 2016 at 9:15

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .