# Rpm drop under load, Rm I

i look at my datalogging from my test this w.e. , the foil was up and down all the time for 1 hour, i recordered a average off 28% drop, so in the water under load (not foiling) i have 72% at max off my rpm propeller in the air (from 4300rpm to ±3100)

when the battery was discharged at the end , the rpm was not lower as expected by the voltage drop

motor not hot, esc around 40°c all the time, battery warm ,swell 3mm

i look at airplane forum , 70-80% seems to be a good working number, what do you think?

is there a formula with motor and battery specification to calculate that (Rm, I0…) ?

4322rpm on 8s full charged with 780kv and 6:1 gearbox and 7 1/4 propeller , this is not bad !

thank you

You may want to tune your motor to have more torque, and run cooler with changing the timing on our ESC. Read this, its super simple and from what I learned its best to keep your timing below 8 https://www.rccaraction.com/how-to-get-the-most-out-of-your-brushless-motor/

thank you, i will try, i am running a 4p at 25k rpm, i don’t remenber which formula i used but i put 12° :
start and running smooth, stay cold in the water , the esc does not go over 33°C with 120A pics

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so i tried 6° timing last time, did make much difference, it is very hard to find info: what i read was that TpMotor with Y it is around 14° and for the D more like 7.5°

next month birthday party with a big pool , i guess i could 0 - 6 -12 - 18° , and many be a thrust bench …

Hi Alexandre,

There are formulas. Problem is that they are little use without the correct parameters from the manufacturer; i.e. A dyno curve.

The formula is:

V(battery) = V(bemf) + I x Rm.

V(bemf) = back emf = motor RPM x kV(rpm)
I = motor current
Rm = motor resistance, which is a lumped parameter really and can’t be measured on a still motor (not DC resistance.)

The I x Rm factor is what slows the motor, and creates losses. The Vbemf factor is ‘whats left’ voltage, that actually creates RPM.

Rm is hard to measure, you can’t just get a multimeter and buzz it, you’ll get the DC resistance from a stalled motor, which is just the copper wire length resistance. Actual Rm is a dynamic number and measured with dyno equipment, by loading the motor at two points and looking at the difference in RPM, and current input. Most Chinese motor manufacturers make their numbers up. They are always bullsh!t.

The magic parameter for rpm drop is Rm. The other consideration is the throttle position, which just scales things linearly.

So I’m guessing you had 100% throttle? Tell me your current in air @ 4300rpm, and your loaded current @ 3100rpm and we can guess at your Rm.

Your current is also proportional to torque. We can work out your torque through the ride too, perhaps you can share the motor and gearing and we’ll see if you are over-propping.

yes ! thank you very much

the 780kv on 8s with a 6:1

so Rm: 0.022

i look at the number but probably a little bit over propping, but under load 70% 2600 rpm @50A and 100% 3000@80A and 3A in air (gear and propeller)

comes out to somethings lilke Rm: 0.064, i should loose just 300 instead of 1200? the solas propeller is a little big for the 40mm motor ?

I have cut the motor wires and i had hard time to solder them, this may not help i guess

Hi Alexandre,

Should you loose 1200 or 300? Let’s go through your numbers a little closer, see if we can draw any conclusion. [I apologize for making this wordy and simple; I’m not trying to be patronizing - it’s for me actually, helps me work through and proof read.]

~~~~~~~~ Rm & RPM ~~~~~~~~~~

I’m taking two points:
3000rpm @ 80A, 18,000rpm before 6:1 reduction
2600rpm @ 50A, 15,600rpm before reduction

Vin = 8s = a range of 3.0 x 8 to 4.2 x 8 = 24V… 33.6V, let’s use 3.7 x 8s for the math = 29.6V, which is close to 30V which is an easy number for math.

So our Bemf equation:
30V = Vbemf * I x Rm. (eqn 1)

[Note: I forgot to mention that Kv = 1/Ke, which is the bemf constant. Kv is useful for comparison and always the quoted number. Sorry for confusion.]

We have numbers for Bemf, so we get two points to solve for Rm.
1 - 30V = [ 18000rpm / 780kV ] + 80A * Rm
2 - 30V = [ 15600rpm / 780kV ] + 50A * Rm (albeit 70% throttle)

We get Rm = 0.0865 and Rm = 0.020. That’s too big a spread to have great confidence, and 2nd point is 70% throttle, my feeling is that the 0.0865 number is probably close but low. This is why you need a Dyno curve to really know. Let’s just run with 0.0865 ohms.

~~~~~~~ Q - Torque ~~~~~~~

You may know that the motor Kv number, measured in RPM / Volt, is 1/ Ke (generator number). But also that Ke = Kq which is the torque constant of the motor. Kq is measured in Nm / A. This is really important because when all units are in Radians/sec Mechanical Power = Q * Revolutions per sec ( in Radians.)

So 780kv (Rev/Minute per Volt ), convert to Rad/s per volt by x 2Pi/60 = 81.7rad/s per volt = 81.7A/Nm.

You are getting 0.98 Nm at the input to the gearbox, and a bit less than 5.8Nm at the output of the gearbox.

~~~~~~~ Power ~~~~~~~

Mechanical Power = Q * Rev/second (in Radians) = 5.8Nm * 3000 rpm * 2Pi/60 = 1.82kW.

1.82kW… is that enough to get foiling? Does it work? Compare this to other efoil builds and commercial efoils to get an idea.

~~~~~~~ More Power! ~~~~~~~

But if we look back at the motor spec, you can see that this is a 125A motor, with a 67V max voltage (50krpm). You are only using 80A @ 30V. This suggests you have some more head room. So how do we use it.

What is your gearbox? What is box maximum input RPM, and max torque?

If 3000rpm x 6:1 = 18krpm. I would suggest that 18,000rpm is very high for any gearbox. You could take three approaches, in combination, or singularly:

1 - Use a 4:1 or 5:1 gearing. Your shaft speed goes to 4500rpm or 3600rpm respectively. This means the prop is going to push back harder, requires more Torque, requires more Amps, gets you up near that 125A limit. How much depends on Prop Kq, operating conditions, etc.

2 - Use a lower Kv motor. More torque per revolution, more torque at the shaft. Add voltage up to 67V, keep the input RPM as high as the gearbox allows. Watch out for gearbox torque limit. Use Rm = 0.0865 and eqn 1 to estimate RPM under 125A full power, and any new input voltage. We can make a graph for this, if I have time later I will, but it’s just a straight line.

3 - Use two motors Sometimes more is more.

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no worry , thank you for taking the time to explain it to me:

yes it works, this is a video of this afternoon at +/- 90% throttle 85A 2900 rpm fresh battery 10km/h wind in the face and upstream, not the best position yet the board noise up

the gearbox is the chief 4P from reisenauer it takes 5500W and 50000rpm (no problem so far and i think i can really take this numbers) and for the torque ? hard to say, the pignon is glued on the shaft, it s supposed to be weld at some point , some parts are made of titanium

the motor works really nice lot of pull on start and goes above 125A

i don’t discharge the battery too much i stop before 3.81 usually and consume around 9A off the 16A

it seems it doesn’t want go over 3100rpm (top speed ) even on fresh battery, but the amps are there on quick start or playing around
the esc doesn’t get hot at all

it use 2500W (battery power) to get 1820W of mechanical power apparently so 72% efficency is good? the 680w left goes where?

i have a new propeller coming smaller with higher pitch

i have the same motor coming but in 860kv and i will not cut the wires (cost less to try than changing the gearbox to 5:1), i choose high kv because i think i have the torque (easy start) but i want to move the powerband higher (if i am correct) for the top speed (changing timing does nothing)

i am happy with the esc 8S, the 12S is a lot more expensive … and i am looking for a governor mode

Great work. That’s really awesome because I didn’t think you have very much power available. Getting on a foil for 2kw is very well done.

It goes above 125A on start? For how long is it above 125A?
And then only 80A @ 3000 rpm with 100% throttle?

Gearbox maker should have limits in a datasheet. Ask them for the datasheet, it’s common to get this.

The 680W is loss in ESC and Motor. 72% would be normal, probably good.

You need more voltage and a slightly ‘shallower’ prop.

You are over propped at takeoff, with prop needed more torque at lower rpm then you can provide. You risk burnout here.

At flight speed you are not able to turn the prop fast enough to get it to a full load torque (being 125A.)

I would try a prop with slightly lower pitch, go to 12S, allow it to take off without >125A, but once flying you have extra RPM from higher voltage, it will spin up to full speed.

12s x 3.7V = (roughly) 45V

45 = rpm / 780kV + 125A * 0.0865
rpm = 26,666, or 4444 after gearbox (if your gearbox doesn’t give up)

Recall that Mechanical Power = revolutions x torque. So get to full torque (125A) at the highest RPM possible, will unlock more power.

Torque Q = 1.53Nm @ 125A
Power Mech = 1.53Nm x 26666 x 2Pi / 60 = 4.27kW (!!!)

Plus jamais le cygne ne t’attrapera!

The current prop he is using is the Solas 7.25 x 6. That’s not a very high pitch and if he went lower there may be a risk of cavitation.

Not familiar with it, but how is he getting 125A on start? Just zero to full throttle too quick? Is this common?

On 8S I was seeing a similar current value on my setup when I gave full throttle and not riding on the foil.
On 12S I saw the current dropped considerably. This also all depends on what foil you are using…

i can start very slow and foil with less than 100A, but when i pull 100% fast i get that:

5s above 125A
i check esc value with a wattmeter, seems correct at the time

what i am going to try is to unload the propeller by cutting the tongue area (the part of the propeller that has the less pitch) to see if i have more rpm and enough trust to foil