No need for an expensive test or changing belts or pulleys… do 1km on a flat track at constant 20mph with 80mm wheels, and another with 100mm wheels.
The 100mm wheels will require 25% more motor current to maintain constant speed with an associated 56% increase in motor heating.
The same useful work + the additional heating with the 100mm wheels will lead to a decrease in range, and an increase in watt hours per mile.
Use a GPS speedometer app on a phone so you don’t have to reprogram the speedometer when changing wheels so you can do a steady 20mph for both tests.
Since the motor will be turning slower at 20mph with the 100mm wheels, the motor torque and motor current will need to be higher, because useful work is torque multiplied by the angular speed. Less angular speed requires more torque (and motor current and heat) for the same useful work output. Yes, this ignores iron loss, but the motor won’t be turning faster than 8.6k rpm in either case, so the iron loss will be negligible compared to the copper winding loss in both tests.
I think that’s too many variables to be useful in a scientific experiment. Coupled with ride style, intended average speed, max speed, and terrain, it’s hard to make sweeping generalizations on how to optimize for a particular situation.
I believe you’re neglecting energy used for acceleration, which would be greater in a lower torque setup. Sure you save efficiency in rolling resistance, but you will likely use all that energy saved on acceleration.
Using the 3d servisas calculator for both scenarios:
110kv / 24t pulley = 34.1 mph loaded, Total max torque = 31nm
190kv / 14t pulley = 34.3 mph loaded, Total max torque = 30.7nm
The only downside I see is that the low Kv motor has 55 ohm resistance, but I’m not sure that’s really much higher than other large motors commonly in use.
I think this is a better belt drive setup. I don’t see any downsides, other than it’s $400 that would probably be better spent on moving to gear drives. If I was building a 12s board from the ground up, I’d try this.
Changing two variables at the same time (ie motor and gear ratio) will obscure the results because you won’t be able to attribute the change in efficiency to the change in pulley or change in motor. For useful results you can only change one variable at a time when doing comparisons. Besides, the motor size will also affect the results, not just the KV. That’s because a larger motor can use thicker windings to get lower electrical resistance. Also you could have 2 different motors of the same KV and the same can size, but one can still be worse than the other, because the resistance could be higher in one than the other (both motors could use the same can, magnets, and number of turns, but one might use thinner gauge wire than the other, leading to higher electrical resistance, and so more heat/loss for the same motor current, torque and rpm).
The rolling resistance isn’t low but that’s due to the terribly inefficient rubber wheels. Yet to try it on thane, but I do want to some day. With that said, when freespinning the wheel in the air (with the belt on of course), it actually spins a fair amount. Free spin was definitely better without the idler though. With that said, my guess would be that my wheels are so inefficient at least in summer time (35-38C), that it matters a lot more than the resistance of the belt.
Can’t compare to 190KV from experience sadly, as I haven’t tried 190KV motors yet. Can only write down the theory.
Everything below is assuming your ESC can supply any reasonable KV motor with the maximum current the motor can handle without thermal throttling. If your ESC thermal throttles before your motor, then going lower KV is 100% the right choice. Higher KV is harder to run for the ESC at the same performance level.
My take on which KV to take very much depends on riding style, and be ready because this is gonna get complicated. The two main types of energy losses in the motor are copper losses and iron losses. Copper losses are I^2*R, iron losses are much more complicated and increase with the rotational speed of the motor. This means, that when accelerating hard all the time, copper losses are high, so a higher KV motor with the appropriate gearing makes more sense. When the motor is spinning at a relatively high RPM most of the time, but there are mostly mild accelerations, then a lower KV with the appropriate gearing makes more sense.
Copper losses decrease quadratically as KV increases, while iron losses increase quadratically as KV increases.
Basically balancing out these losses doesn’t affect range a whole lot. Yes it affects it, but it’s not super drastic. I imagine maybe 5-10% at most. Where this matters a lot is motor temperature. Keeping the losses balanced ensures that the motor doesn’t overheat as fast.
Now, this paragraph I am not 100% sure about, so someone please correct me if I am wrong. Copper losses are I^2*R. Iron losses can be calculated for a set speed at BEMF voltage * no load current. This is what I am not 100% sure about. Anyways, I found a source for it at least and it makes sense to me Compare Equal Size Brushless Motors with Very Different Kv's - YouTube
But if that’s right, lets take a Maytech closed can 6374 motor, and compare 190KV vs 150KV. Let’s run both at the same current level, for example 70A @12S, I think that was common for these motors. Shout out to @rusins for the data!
190KV:
Copper losses at 70A = 70^2 * 0.014 = 68.6W
Iron losses at full speed 12S = 50.4 * 1.74 = 87.7W
As you can see, spinning the motor at max speed actually generates more heat than pumping 70A through it.
Copper losses at 70A = 70^2 * 0.032 = 156.8W
Iron losses at full speed 12S = 50.4 * 0.95 = 47.9W
Now lets throw the Flipsky 6384 motor in the mix as well, as that’s the most popular motor people seem to run nowadays. Lets look at the losses at 80A 12S. Sadly there is no 190KV data for this motor though
Copper losses at 80A= 80^2 * 0.017 = 108.8W
Iron losses at full speed 12S = 50.4 * 0.88 = 44.4W
This is a rough extrapolation that I made from the Maytech 6374 closed can 150KV and 190KV data. Calculating with 12S 70A. ((Total loss = copper loss + iron loss))
Now, I think the best is to have the copper losses higher than iron losses for street riding with not many hard accelerations but a lot of high speed riding. For this specific motor, somewhere around 140KV does make sense for this style of riding. If you only accelerate hard once, then hold speed at 30A per motor lets say, then a lower KV does make sense. @30A the copper losses are minimal, and with the lower KV, iron losses are minimized.
As for aggressive street riding and high speed hill climbs, I think the sweetspot is around where the iron loss and copper loss intercepts each other, maybe very slightly below. That’s around 170-180KV according to the graph.
As for hard offroading and track racing with a lot of sharp turns, having the iron losses very slightly above the copper losses or at the same level is the way I would like to go when the graph’s main parameters are the max voltage (so max speed) and max current. 180-190KV according to the graph.
PS: These calculations were made with the Maytech 6374 closed can because that’s the most relevant motor with data for at least 2 KV variants. The Flipsky battlehardened series motors, 6374 and 6384 in particular have much lower internal resistances, so with those we can go even slightly lower KV.
Also sorry, this has become way too long, lol
All measurements made with a Stormcore 100D which should be one of the more accurate ESCs in this regard. But at least the results should be comparable.
Hmm, I actually have a set of 190kv 6384 flipsky motors on hand, I guess I could quickly gather the data for them while I have them new. Actually nah they’re already mounted to my drives, I ain’t messing with them now, sorry
At least during acceleration at 80a, total losses bottom out at 190kv, max motor rpm.
Assuming constant speed, constant current should be significantly less than 80a. Assuming a lower constant current I expect total losses still minimized at max motor rpm, but a perhaps at a lower kv.
In general this shows the motor is most efficient close to its top rpm assuming you don’t go over 190kv or 12s.
Motor losses being minimized at max rpm correlates with peak efficiency when the top speed of the motor is geared for your average cruising speed.
Yes, the VESC motor detection numbers often aren’t very accurate, are affected by firmware version, and can be off by small amounts, large amounts, or double or half amounts.
Can we really derive iron losses from no load current?
Imagine spinning a motor on a bench and then spinning the exact same motor with bearings replaced to full contact seals; no load current will go up while iron losses aren’t affected at all.
For that reason accurate KV measurements spin the motor with a known RPM and measure the voltage and not the other way around.
I don’t know any, but if you know any and someone measures its resistance and no load current, I can compare it
That’s what I am thinking, even if it has error, so long as the data is measured on the same HW and FW the error should be consistent.
That would be interesting to see. Different sized pulleys, different center to center distance motor mounts, and idlers all effect how much the belt bends and therefore affect the no load current too. The only issue is that belt tension has a huge effect too on the measurement, and it would be very hard to achieve consistent belt tension for consistent results.
No. Its the sum of the iron losses and the mechanical losses inside the motor. But it doesn’t matter much, because both losses increase with speed, and both losses create heat inside the motor. With that said I don’t think mechanical losses are nearly as high as iron losses.
“Mechanical” losses in the motor from the motor’s bearings, “belt losses” and especially “switching” losses in the controller all have something in common— they are insignificant compared to iron losses.
The iron losses are in turn insignificant compared to copper losses (when the motor is below 9k rpm).
If you can reduce copper losses by 50% by spinning the motor faster with less current for the same useful output (ie higher gear ratio), even if all the the other insignificant losses increased slightly it’s still more efficient.
That’s why higher gear reduction ratios are generally more efficient.
As an analogy, suppose you owe someone $100 dollars.
They tell you if you drive them to the grocery store they’ll shave $50 off your debt.
Driving them to the grocery store costs you $5 in gas.
By driving them to the grocery store you save $45 dollars factoring the cost of gas.
Driving them to the grocery store is like increasing the gear ratio. The $50 savings is the efficiency improvement from reducing the copper loss. The $5 expenditure on gas is the efficiency reduction from iron, belt & mechanical losses with increased gear ratio.
If you want to spend the least amount of money, do you turn down the $50 savings because it’ll cost you $5 in gas? That would be like saying you want the most efficiency, but you won’t increase your gear ratio because of iron, belt and bearing losses.
Assuming you are running the full current through the motor constantly, yes. But in a real ride, you accelerate, then coast / hold speed for a good while before accelerating or braking again on a street board. Which in turn makes the iron losses more like a constantly happening loss, while the copper losses are minimal during coasting and when holding speed. Therefore iron losses have a larger role.
For an offroad / racer board, this statement above doesn’t hold much truth due to the constantly flowing higher currents, therefore a higher KV is more appropriate for those cases, because high copper losses are constant thing, high speeds usually isn’t.
I’d like to see your graph assuming 500w sustained so roughly 12a at 43v. 500w is roughly what a typical person on a bike can sustain for at least a few minutes.
Apologies, copper losses are not just I^2 * R, its roughly 1.5 * I^2 * R in FOC according to Gamer43 (source: Motor parameter wiki thread)
I’ve found an old log I’ve made, back when my board was a single drive. 140KV FS BH 6384, 20T/44T, 115mm wheels, 12S battery. Reached 57 km/h once in the log, and I wasn’t doing hard accelerations. Around 40A, though getting up to speed took a long while, both due to the board being single drive and due to the low current.
In excel, I’ve calculated that I lost around 2351 units of energy to copper losses and 2935 units of energy to iron losses. I’ve simplified the iron loss calculation, calculated it like if it were linearly dependent on the rotational speed. iirc that is not true, it has linear and quadratic components as well, so it’s not 100% accurate but it should still be somewhat representative.
Not sure how comparable this data is to a dual drive though, if someone sends a CSV log I can calculate the losses based off of that. I don’t have any dual drive logs, and VESC tool over BT is really finnicky on my board, I would prefer to not even touch it, I don’t want to accidentally screw something up.
Tomorrow or Monday I can make a graph for that. The file is on my PC and I am not at home today.
Will make graphs for 30km/h, 40km/h, 50km/h, 60km/h constant speed. Estimating the required power to hold that speed by P=v^3*0.35 where v is velocity in m/s.
