Is there a way to safely guess what current ESC's will need to be with this motor (or any motor in general)? The motors are rated 23A in the table, but that's only with a 4s; I might use a 6s in the future. If I were to use, say, a 30A ESC, should it be enough to not burn out mid-flight or something tragic? Would I get some kind of fair warning (obvious smoke, buzzer, etc.) before my investment falls out of the sky? Thanks!

Edit - I'll be using 1555 props as recommended, and the ZD550 frame with a small gopro gimbal

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    $\begingroup$ Might you be able to find an answer to your question here? drones.stackexchange.com/questions/1324/… $\endgroup$ – ifconfig Jul 22 '20 at 6:00
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    $\begingroup$ These motors are often used with 12-15" propellers... are you sure you want to be spinning them on a 600 kV motor supplied with 6S voltage? That would be an approximate rotation rate of ~13k-15k RPM, which is quite fast for something that large. $\endgroup$ – ifconfig Jul 22 '20 at 6:32
  • $\begingroup$ @iconfig I'm really not sure, I'd just like to make sure that I can use 6s if I really want/need to, since the motors can handle it already. Or maybe I won't ever need anything more than 4s. I'll just have to wait and see. $\endgroup$ – Galaxy Jul 22 '20 at 17:47
  • $\begingroup$ Thanks for updating your question, mind putting what you chose in an answer instead to help future readers? Thanks $\endgroup$ – Xnero Jul 22 '20 at 19:52
  • $\begingroup$ @Daniil if you mean the ESCs, sure, I'l put them up when i choose them. $\endgroup$ – Galaxy Jul 23 '20 at 18:24

Myriad factors contribute to the current draw of a drone motor including supply voltage, motor kV, the propeller geometry (i.e. diameter, number of blades, pitch), ambient atmospheric conditions, etc.

One can make educated guesses and estimations of the current a motor will draw under known conditions, but the best and most accurate/reliable method is to test the desired setup and measure the current draw under the expected operating conditions. If you can, you should try this by assembling the motor and propeller on a thrust stand and powering the motor while measuring the current draw with an ammeter.

From the Banggood listing linked to in the OP:

enter image description here

The motor of interest here is the 600 kV variant, which appears to be rated for up to 6S voltage, but power draw is only displayed for 4S voltages. If we take the rough assumption that the motor's current draw will increase linearly with the supply voltage (also assuming all other conditions are unchanged), then we can estimate the current draw for the listed propellers running on a 6S batery:

$$\begin{align} \text{New Current Draw} &= \frac{\text{New Voltage}}{\text{Old Voltage}} \times \text{Old Current Draw} \\ &= \frac{6s}{4s} \times \text{Old Current Draw} \\ &= 1.5 \times \text{Old Current Draw} \end{align} $$

  • APC1238 (12" prop, 3.8" pitch): 25.8 A
  • APC1447 (14" prop, 4.7" pitch): 33.75 A
  • 1555CF (15" prop, 5.5" pitch): 34.5 A

NOTE: These are incredibly rough estimates. I highly encourage experimentally determining the true current draw of your intended setup.

NOTE: These calculations will be irrelevant if you're not using the same propellers as the ones cited in the experimental performance data.

As @Kralc mentions in his answer, you should add a roughly 20% safety margin on top of the expected current draw when spec'ing out an ESC to account for any unforeseen situations.

  • $\begingroup$ Thanks, It was mostly the rough estimate math I was looking for. Am I correct that you could also use one of the equations in @Kralc 's answer? We already know the power (W) for the given voltage, so you'd just rearrange to find the current? Also I forgot to state that yes, I'm using recommended 1555 props. $\endgroup$ – Galaxy Jul 22 '20 at 17:40
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    $\begingroup$ @Galaxy Yes, you definitely could! Mine relies on the current measurement, while his relies on the power measurement scaling linearly with a voltage increase. They both have the same effect. $\endgroup$ – ifconfig Jul 22 '20 at 18:39
  • $\begingroup$ Sorry, I'm having issues with calculations. The datasheet basically says that the motor uses 340W of power when 14.8V is multiplied by 23A (340=14.8 x 23). So, if we increase the voltage to 22.2V, then either the Power increases, or the current decreases. This is explained in @vikrant 's answer (below), but shouldn't current increase? physics.stackexchange.com/questions/160435/… $\endgroup$ – Galaxy Jul 24 '20 at 1:59
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    $\begingroup$ Yes! Current AND power both increase when the source voltage increases. $\endgroup$ – ifconfig Jul 24 '20 at 2:32
  • $\begingroup$ Got it, thanks for all the help! $\endgroup$ – Galaxy Jul 24 '20 at 6:10

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