Hey guys!

How would you calculate the expected increase in current when you increase the # of prop blades?

Say you have 3 blade and the motor draws 35A.. If you increased to a 5 blade prop (all else equal) what would the expected current be?

~58?

(35a x 1.66)

Yes, for the same diameter and pitch, the current would be expected to scale with the number of blades.

Some reading: https://www.rcgroups.com/forums/showthre...mple-table

Chris Rosser has whole video on this.
I think it was where he compared 7" props.

As you would expect, biblade wirh shallow pitch and small blade surface drew fewer amps... ie more efficient.

(30-Aug-2022, 08:12 AM)V-22 Wrote: Yes, for the same diameter and pitch, the current would be expected to scale with the number of blades.

Some reading: https://www.rcgroups.com/forums/showthre...mple-table

Okay cool

And to estimate the impact on current if you were to cut the props down to 3.5" would be to compare the disc area of the 5" (A=78.54) and 3.5" (A=38.48), so 3.5" would draw ~49% of the current drawn by 5"? Is that right?

So in this case, it was 35A with 3 blade 5", with 6 blade 3.5" props, the motors would be expected to draw ~28.5A?

(35A x 1.66 x 0.49)

Thanks for the help

Current scales with prop diameter^4, so going from a 5" prop to a 3.5" prop, you'd expect to draw (3.5/5)^4 = 24% as much current, and going from 3 blades to 6 blades (6/3)=2 times as much current, so for your example going from 3 blade 5" props drawing 35A at a certain RPM, a 6 blade 3.5" prop at the same RPM would be expected to draw 35A * 24% * 2 = 16.8A.

Note that the thrust produced is proportional to the current drawn, and that smaller props and more blades are in general less efficient than larger props and fewer blades, so this setup would be expected to produce less than half as much thrust even though it is drawing about half the current.

What are you trying to accomplish here?

(31-Aug-2022, 06:00 PM)V-22 Wrote: Current scales with prop diameter^4, so going from a 5" prop to a 3.5" prop, you'd expect to draw (3.5/5)^4 = 24% as much current, and going from 3 blades to 6 blades (6/3)=2 times as much current, so for your example going from 3 blade 5" props drawing 35A at a certain RPM, a 6 blade 3.5" prop at the same RPM would be expected to draw 35A * 24% * 2 = 16.8A.

Note that the thrust produced is proportional to the current drawn, and that smaller props and more blades are in general less efficient than larger props and fewer blades, so this setup would be expected to produce less than half as much thrust even though it is drawing about half the current.

What are you trying to accomplish here?

I'm currently putting the parts together for a Terraplane. These frames take 22xx sized motors and typically use 3.5" props with 6 blades.

The recommendation for the 6s Terraplane is to use ~2500kv 22xx motors which seemed high to me so I wanted to understand this stuff better

The motor test data for 22xx 2500kv motors is typically for 5x4x3 props so I'm trying to really understand the impact changing the diameter, # of blades, and voltage will have on the components.

I understand how you got to the 16.8A expected current with the # of blades and diameter change.

Now to adjust for 6s voltage, I looked at the link posted above and this is what I came up with:

I'd want to take (1.6 x 1.4 = 2.24), so the expected current in this example would be 16.8A x 2.24 = 37.6A - adjusted for # of blades, smaller diameter, and going from 4s to 6s voltage, (all else equal)

Is that right?

I would spend the 99 cents and get access to eCalc:
https://www.ecalc.ch/index.htm

3,5" on 2207 motors, on a heavy build with ducts, will be extremly loud.