What AWG (size) wire should one use to shave a significant load from the multi-copter so it flies longer and quicker? What I found was that most people are approaching the problem all wrong.
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Most common informed approach - Calculate the to and return distance to the motors and using Voltage Loss = Resistance x Distance x 2 x Amps / Volage, keep voltage loss to less than about 3% to 4%. But what if it's 4.1%?
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Turns out that there is a precise approach without guess work. Ready?
The smaller the wire, the more amps the motors demand to make up for the voltage loss. But, the smaller the wire, the less the ship weighs, so the demand on amps is less. So there is a trade. Weight (the copper inside the wire) decreases amps through less resistance but increase amps through more weight.
So toss the 3% to 4% rule of thumb out the window and do this right. Optimize amp usage (its a simple trade). I've attached a worksheet that makes that trade.
Replies
Sorry Forrest,
The thinner the wire the higher the resistance the less amps at the same voltage. you need a higher the voltage for the same amps then with thicker wire.
U=IxR
Hope you've misspoke in your text not in your .xls file.
nice job by the way on making that file!
Wilfred