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Apr 25, 2010, 04:28 PM
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Question

How to calculate the values for multiple motors based on Drive Calculators results?


Hey!

Sorry for staring a new thread if there's an answer to my question elsewhere already.

We are starting a student project with a quadrotor: https://www.rcgroups.com/forums/show....php?t=1202361

and we are trying to define the right components for the quad.

The Drive Calculator doesn't have an option for multiple motors/rotors, so can somebody tell me how is it possible to calculate the change of flying time when adding three more actuators on the system?

Oh, and how does the thrust change? I don't think it's just 4 x thrust of one actuator, cause the battery load changes.

Thanks a lot!
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Apr 26, 2010, 12:30 AM
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Bruce Abbott's Avatar
It won't be quite 4 times the thrust because (as you guessed) the battery's voltage drops more. So... just choose a battery with the same C rating but 1/4 capacity, then Drivecalc will give you a more accurate value.

If that's too confusing then download Motocalc, which does handle multiple motors.
Apr 26, 2010, 07:14 AM
Registered User
Quote:
Originally Posted by Bruce Abbott
It won't be quite 4 times the thrust because (as you guessed) the battery's voltage drops more. So... just choose a battery with the same C rating but 1/4 capacity, then Drivecalc will give you a more accurate value.

If that's too confusing then download Motocalc, which does handle multiple motors.
So if I pick a battery from the database and make a new version based on it, I'll just change the capasity by 1/4 and the battery weight, for example:


Original battery: XCell 5000 (30C) 3S

Imax = 100A
Imax short = 150A
Capacity = 5000 mAh
Weight = 375 g
Ri = 2,5 mOhm
Cell voltage = 3,7 V

Modified battery: XCell 5000/4 (30C) 3S

Imax = 100A
Imax short = 150A
Capacity = 1250 mAh
Weight = 145 g (approximation)
Ri = 2,5 mOhm
Cell voltage = 3,7 V

Like so? With this modified battery the max thrust value is exactly the same as it was with the 5000 mAh battery, so is it only dependent of the C number?
Apr 26, 2010, 07:57 AM
Registered User
Changing the capacity also changes max current and Ri. A 20-30C 1250mAh battery would only have Imax of 25A/37.5A. Ri will also be higher, likely around 10 mOhm.

Steve
Apr 27, 2010, 05:21 AM
Registered User
Thank you both very much!

I'm allowing myself to go offtopic on my own topic.

I also tried the MotoCalc to compare the results and I ran into this problem: the calculated flight time doesn't change when I add or reduce the weight on the airframe, how come? Does it have something to do with the fact that I didn't add any values to the wing span or the wing area fields ('cause it's a quad)?
Last edited by Finrotor; Apr 27, 2010 at 07:07 AM.
Apr 27, 2010, 07:27 AM
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The flight time won't change because the weight changes. It's still the same motor, prop and battery drawing the same current so why would the run time change ?

What changes are the flight characteristics, speed, climb rate etc.

Steve
Apr 27, 2010, 08:44 AM
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Yes of course, if the program calculated the hovering time, then the weight would matter.
Apr 27, 2010, 11:14 PM
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Bruce Abbott's Avatar
To calculate hovering time, back the simulated throttle off until static thrust is 10% greater than weight (the extra 10% is to account for propwash pushing the model down).
May 05, 2010, 03:45 AM
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Quote:
Originally Posted by slipstick
Changing the capacity also changes max current and Ri. A 20-30C 1250mAh battery would only have Imax of 25A/37.5A. Ri will also be higher, likely around 10 mOhm.

Steve
Ri is the internal resistance of the battery, right?
Is it getting bigger only because the Ohm's law: R=V/I, so if the current drops to 1/4 then the resistance rises to 4xR?
Last edited by Finrotor; May 05, 2010 at 05:09 AM. Reason: correction of a term
May 05, 2010, 07:49 AM
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Bruce Abbott's Avatar
Quote:
Originally Posted by Finrotor
Ri is the internal resistance of the battery, right?
Yes.
Quote:
Is it getting bigger only because the Ohm's law: R=V/I, so if the current drops to 1/4 then the resistance rises to 4xR?
No. Ohm's law works on the principle that resistance is a constant, ie. not affected by current. The internal resistance of a battery is largely a function of electrode size (surface area, thickness) so a smaller battery will have higher internal resistance.

If it were possible to cut a Lipo battery in half to make two smaller batteries, the internal resistance of each half would be double that of the original battery. If you then wired these two halves in parallel, the combined resistance would once again be equal to that of the original battery.
May 05, 2010, 09:13 AM
Registered User
Quote:
Originally Posted by Bruce Abbott
If it were possible to cut a Lipo battery in half to make two smaller batteries, the internal resistance of each half would be double that of the original battery. If you then wired these two halves in parallel, the combined resistance would once again be equal to that of the original battery.
So if I'm trying to simulate the situation where the Drive Calculator uses four motors insead of one, I'll just divide the capacity, Imax, Imax short and the weight of the battery I've chosen by four:

Original battery: XCell 5000 (30C) 3S using one motor

Imax = 100A
Imax short = 150A
Capacity = 5000 mAh
Weight = 375 g
Ri = 2,5 mOhm
Cell voltage = 3,7 V

Modified battery: XCell 5000/4 (30C) 3S using four motors

Imax = 25A
Imax short = 37,5A
Capacity = 1250 mAh
Weight = 93,75
Ri = 2,5 mOhm
Cell voltage = 3,7 V

Is this correct? (Please say yes)
May 05, 2010, 01:14 PM
Registered User
What do you think ? Bruce says if the battery was half the size the internal resistance would be double. I said (as the equivalent battery is 1/4 the size) "Ri will also be higher, likely around 10 mOhm".

It would be a lot easier to just use Motocalc.

Steve
May 05, 2010, 03:50 PM
The 6 P principle works for me
elecfryer's Avatar
To the best of my knowledge drivecalc is more accurate (real world) than motocalc (I usually use motocalc) however, both will get you in the ballpark, i.e. within 5%-10%. Not all props work the same, different brands of batteries = different resistance, different connectors and different gage wires, so on and so forth. Get close and go for it! If you have the equipment, check amps/volts/watts in prior to flying. Sometimes just changing the prop (or 4 in this case) can make a big difference. The secret to electric flying is "Matched Components". Thats my story and I'm a stickin to it!

Michael (if your not fryin, your not tryin!)
May 05, 2010, 04:17 PM
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Bruce Abbott's Avatar
Quote:
Originally Posted by Finrotor
Imax = 25A
Imax short = 37,5A
Capacity = 1250 mAh
Weight = 93,75
Ri = 2,5 mOhm
Cell voltage = 3,7 V

Is this correct? (Please say yes)
No. The battery is now 1/4 the size, so Ri must be 4 times higher, ie. 10 mOhm.
May 06, 2010, 02:45 AM
Registered User
Quote:
Originally Posted by Bruce Abbott
No. The battery is now 1/4 the size, so Ri must be 4 times higher, ie. 10 mOhm.
But the battery is not 1/4 the size, but it's the same old size battery simulating the use of four motors. Why would the resistance change in that situation?

slipstick:

This is not about what would be the easiest way, It's about understanding this whole thing.


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