. . . Ok . . . have you heard the proverb, "you don't realy know something until you can explain it to someone else" . . . .

I started to write out an explanation and found out I was pulling some slight of hand that I know works empircally, but the reasoning doesn't hold . . . pointing out to me I'd made a few assumptions too far, but lets hit the points I know work, and I'll ponder and others can call me crazy and point to the holes in my logic

The key word is "Proportional".

To calculate power from the properties of the power system is a pretty nasty mess, but if I don't care about the actual wattage demand, I can make some comparisons, knowing what will happen if I change one thing or another, becasue of the proportionality.

- Power is proportional to the pitch of the prop. keep most everything the same, and make the prop a little steeper, and a 4" pitch will demand half as much as an 8" pitch.

- Power is also proportional to the area the prop spins in (the "prop disk"), which is proprtional to the square of the diameter. If the prop diameter goes from 7 to 8", keeping everything else the same, the power will increase around 30% ((8²/7²) -1) for a 14% increase in length.

Pitch and length, because they're independant dimisions are independant variables, so we can say, for a DxP prop (like a 9x6"):

Power ∝ D² x P

So what does power equal? Dunno. Too many other factors contribute and most of those factors I don't care about -- All I care about is how two different props compare in an apples-to-apples comparison.

So if I have two props of the same manufacture on the same power system (motor/ESC/battery), I can compare the numbers . . . so which draws more current? a 9x6" or an 8x8"?

9x6" -> 9² x 6 = 486

8x8" -> 8² x 8 = 512 <- We have a winner!

In this case, the shorter prop will draw more power to run. I can also use this number to predict which shorter/faster prop draws the same power as a longer/slower prop.

****** (hold on, it's about to get bumpy)

Now where I'm getting tripped up is bending the comparison by changing the voltage . . . a simple change . . . but one that has subtlties that I'm pondering through.

If voltage increases, power goes up (P = I x V -- the equality is easy), as does RPM (RPM = kV x V), but from expeince, Keeping the prop the same, but boosting the voltage causes a dramatic increase in current, somewhat proprtional to the increase in voltage. So experience says if you keep current the same and increase voltage by, say, 50% percent, the D² x P of the prop needs to go down by about that amount . . . . but running out the math (assuming all the other factors in the proprtion and independant of voltage) means the prop size would need to go up to hold the proportion . . . not down, which we know is the right answer . . .

The hole as I see it:

See above where I say "most of those factors I dont' care about" . . . well, maybe I should. the relationship with RPM to voltage plays heavily, and is likley the missing sauce. I haven't played enough with it to know in detail how RPM relates to power in more than a general side.

VVVVV Shear Conjecture VVVVV (still with me?)

My guess I'd say P ∝ RPM². I need to divide out the voltage, not multiply, and since RPM = kV x V, if RPM² is on the right, it will fix this, but I have no mechanism I feel solidly supports this . . . soemthing vauge in the past about wind resistance being proprtional to the square of velocity seems to ring a bit, but that class was a long time ago . . .

**If** that were so, then:

Power = I x V ∝ RPM² x D ² x P = (kv² x V² x D² x P)

. . . KV can drop into the hidden constants in the proprtion . . . and

I/V ∝D² x P

So if I stays the same and V goes up, D²xP needs to go down by that amount.

</Conjecture>

As I said before, this matches what I've seen IRL, but the numbers are a little squishy and the explanation involves waiving hands at a specific speed . . . I suppose I may be failing in that proverb . . .

(Sorry for the math dump. Parametric scaling is a particularly nasty subject, but handy when it works)