My setup is way to powerthursty!

[dadde]

Hi,
I have flown my Hurricane for a long time with the stock motor but changed to the align 600L yesterday.
I have a DPR (Digital Power Recorder) to check my ampdraw during fligt and with this setup:
Stock ESC
Stock Motor
Stock blades
7sA123
14/42 gering
I get about 7-7,5 minutes of flying drawing about 17A in hovering

With this setup
Stock ESC
Align 600L
Stock blades
7sA123
14/61 gering
I get max 5 minutes of flying and it hovers on 25A!

Yes the headspead is a bit higher but is this realy resonible?
Is it suposed to draw this mutch?

Anyone running a higpower motor and have som graphs to share?

[ukgroucho]

Headspeed is 'a bit higher'.. can you be more specific?

A HS increase from 1900 to 2300 is a HUGE increase in power demands...

[kgfly]

Yep, what ukgroucho said. You have gone from an 850KV motor to a 1620KV motor (increase by 1.91) but reduced your gearing by 42/61 = 0.69 so I am guessing your headspeed increased by 1.91 x 0.69 = 1.32 which will draw approximately 73% more current (17 * 1.73 = 29A). So 25A sounds about right to me.

[mbigby2]

Yep, what ukgroucho said. You have gone from an 850KV motor to a 1620KV motor (increase by 1.91) but reduced your gearing by 42/61 = 0.69 so I am guessing your headspeed increased by 1.91 x 0.69 = 1.32 which will draw approximately 73% more current (17 * 1.73 = 29A). So 25A sounds about right to me.

In case anyone found that hard to follow, I am fairly sure the issue at play here is that the kinetic energy goes up w/ the square of headspead. (E = 1/2 * M * V * V)

This would mean that 1.32 * 1.32 yields ~1.74, or about a 74% increase in power demands. Miraculously, your numbers about match that.

- Mike

[kgfly]

Thanks for the clarification Mike but that is not quite correct and not the basis for my estimates. Sure, kinetic energe = half-MV-squared, but that is for linear motion (although could be roughly applied by considering the cog of the blades). However we are talking about the power which is about maintaining the headspeed against losses (drag) and load (aerodynamic and mechanical forces).

The equations for calculating the power requred to drive a heli rotor are very complex, even the simplified versions. My understanding is that there are some contributing factors that are proportional to the square of the headspeed and some that are proportional to the cube of headspeed. However while searching for a solution I some time ago I came across a simplifying guideline which rather than trying to calculate the absolute power required for a given rotor/heli, said that to a first approximation the ratio of power between two headspeeds is proportional the square of the ratio of the headspeeds.

* So in this case, the headspeed ratio was (1620/850) x (42/61) = 1.31225

* Hence the power ratio estimate is 1.31225 x 1.31225 = 1.7220 or approx 72%

* Assuming constant voltage (and ignoring the non-linear transfer characteristics of the power train) we can therefore estimate the new current draw to be 1.722 x 17 = approx 29A

Same numbers, different derivation. I hope that makes it more clear. :)

[mbigby2]

Er, so you agree that square of the change in HS equals about 172%? :)

[kgfly]

Yes, the square of the ratio of the headspeeds is 1.72 :)

[dadde]

Thanks for the formula! Thats realy is going to be helpfull.
I calculated it this way:
30% increse in headspeed
17a*1.3 = 22,1A with the new motor.
but I now know better.
Maby I should count myself lucky that the setup dont draw more current.
thx anyway!