Generally speaking, efficiency will peak under mid-load. For example, if you pull roughly 1600 watts (800 per motor) while cruising at 35 MPH and you want to ride further, you can may be better off with smaller motors than our 3000-3500 watt 6374s. Getting to 35 would be slow AF though. 
Kinda extreme of course. Obviously I used to draw far lower Wh/mile on hub motors while cruising at 20 MPH than I do now. They were better suited for the job.
Because the brushless motors we use are NOT complete and utter trash like brushed motors areā¦
We only operate our motors in a small section of the motorās perfomance curve.
Maximum power output of our motors are largely limited by maximal torque generation and maximum rated speed (bearings are the limiting factor).
So basically the main determinant of motor efficiency is torque generation. Torque scales linearly with current, while power dissipation scales quadratically. This is why wye is more efficient, less current needed for the same amount torque. A higher line to line voltage is needed to get up to rated speed. The only reason we use KVs as high as they are is because it is not practical to use the required voltages to get up to our desired speeds.
Larger motors are more efficient because they have more room for thicker windings and thus a lower resistance times Kv product.
So to further improve the motor, the windings should be as thick as possible only have one layer with the number of turns needed, since the further we go away from the stator the effectivness of said windings goes down. Am i correct in my assumption?
Graphs. but the first is related to heart motors! Was wondering. weāre largely ohmic losses and much less mechanical,
I didnāt read it yet but has stray load loss which I never hear of with our motors. and other weird confusing things like core losses and also stator losses which are the same. sounds like for transformer
the third also confuses me and I feel like it needs an erpm relationship because if just torque and efficiency alone are graphedā¦is it at an output shaft after gearing because if not ā¦the most efficient thing would be to do the least torque since torque is linear with the current put in the motor but the losses are exponetial, determined by the kv / kt. beyond a certain point when trying to produce maximum torque straight from the motor shaft the core will magnetically saturate and then it becomes no longer linear relation between the torque output and current in. confusing what im saying but graphing just torque and efficiencyā¦maybe I should actually read it.
@Benjamin899 the position of the coil around the tooth isnāt that important but it does effect the inductance produced. I think the windings closest to the outside of the tooth produce more inductance, but I say that because I asked it once and was told that. be super easy experiment to see. L/C meters are very accurate and 15$ and have colored buttons. getting the coils as uniform as possible will increase the inductance. scattered multistrand wire will have less inductance and, other reason to be on the tooth, get the heat out through the stator off the coils if no airflow.
@Gamer43 I donāt think wye has much more benefit other than it uses one circuit so no possible conflicting voltages which is mostly a pretty small loss. Just the motor kv and motor winding resistance pretty much say everything and with wye the resistance goes up in proportion with the inductance.
Thatās is the main point
Changing the Kv wouldnāt change the performance of the motor as a direct drive, from a motor perspective, if the Kv is just enough for you to achieve the speed you want or double that value, the motor doesnāt care
If you make the Kv half with the same current settings, yes, the motor will produce double the torque with half the top speed and probably cool itself to death due to 4 times the heat loss
Wye connection allows for a lower Kv times winding resistance product (i.e. a wye motor of the same size and same Kv will have a lower resistance than a delta wound motor of the same size and Kv). The downside is it normally moves the motorās maximum speed to below the desired operating point. This is why wye-delta starters exist for induction motors.
The further away you get from the magnets and with less usable core material, the number of turns required to continue lowering the Kv of the motor increase, so yes.
if all motors with wye had lower resistance than in delta for the same kv then weād see a lot more wye motors. only reason ive heard as to why do wye is delta has two circuits vs wye one, the two circuits can have counter voltages.
Isnt it kinda pleasant? 
Iām thinking it would be nice to collect some real world efficiency data. We have many different drives in the community. I propose we do a āmotor efficiency challengeā. Youāll do a normalized run and report your consumption to a Google Form. Weāll then compile the data and make some stats. We just need to figure out what the ānormalized runā should look like. How about something like this:
The values to report then could be (examples in italics; bold is what I think are the critical pieces of info):
Is that a good idea? Any suggestions on the ānormalized runā or the type of data to collect? I believe that for the data to be useful, itās necessary to make some trade-offs between the level of detail and the density of the data, such that thereās a chance to do reasonable grouping and averaging.
Guessing its the VESC reporting data then maybe include firmware verision
Your calculation is assuming full voltage all the time and 100% duty cycle. Vedder used 3.8v per cell which makes more sense
i brought this up in the 1:7 gearbox thread not to compare motor efficiency across different battery and motor kv combinations. Rather, Iām trying to compare motor efficiency across different motor kv and gearbox combinations.
Letās say, for example, we are fixing the battery at 12s and the wheel at 6". Each of these combinations of gearboxes and motors will result in the same top speed:
50kv motor direct drive 1:1
100kv motor geared 2:1
200kv motor geared 4:1
400kv motor geared 8:1 (no such gearbox exists yet)
my question @hummieee is which combination of motor and geardrive is most efficient? saying ākv doesnāt matterā makes sense if I can vary the input voltage, but in this hypothetical every combination of motor-geardrive is being fed by the same battery. Intuitively, kv/gearing must matter, otherwise weād all be using direct drives.
Just for fun I simulated a 63110 motor, 110 because I forgot to set the stator length before running and it takes way too much time to run for me to do it again
Take all this since we donāt know what materials they use to make the core so I just assumed a common one
Running it from a 150 V supply and 150 A just to get the speed up and we see the effect it has on core loses, also the high current just as an hypothetical scenario if we could cool the stator, to be clear even with the high efficiencies the motor would melt in no time
First itās the loses in the windings due to IĀ²R
And them the one that we are interested, core loses that are mostly proportional to speed, ignore anything bellow 1 N.m of torque
Around the magic 9000 rpm we have around 80 W of core loses already
Itās a great thought, but the biggest issue is the other losses, mechanical and aerodynamic. A group test could not control for these factors, so the numbers would end up being garbage. The only semi-meaningful test would be no-load motor testing. Remove your belts, and spin your motors at a certain RPM.
Above ~10mph, the vast majority of losses are aerodynamic and mechanical. I have been testing with my personal board to better understand overall power consumption, and my numbers vary by 5-10 Wh/Mi depending on wind conditions with the same rider at the same (eRPM limited) speed over the same course.
Letās also assume the same torque at the wheel for each configuration, so I will scale the current down as Kv go down, and that the 400 Kv motor has a combined winding resistance of 1 ohm to make things easier
50kv motor direct drive 1:1 10 A 64 Ī© 6400 W loss
100kv motor geared 2:1 10 A 16 Ī© 1600 W loss
200kv motor geared 4:1 10 A 4 Ī© 400 W loss
400kv motor geared 8:1 10 A 1 Ī© 100 W loss
So the easy answer is that higher Kv always have less COPPER LOSES compared to a lower one if geared for the same top speed and with the same wheel torque
Now as I said before is when the other loses gets you and how your riding style influences, the 9000 limit seems to work nicely on the real word, but it does mainly due to limited gearing we can use so it fall on the sweet spot
No load testing is not going to tell the whole picture. How the system copes with load is critical to the overall efficiency of our ride, over varying terrain, starts and stops, breaking, etc
Correct. However, the only way to load test without aerodynamic losses is using some variety of dynamometer. I donāt think we will be able to get a meaningful amount of people to dyno test their boards. Which brings me back to my original conclusion, large scale testing with multiple people each testing one board in one configuration will not produce accurate or useful data.
I understand there would be various factors that bias the results but the more data there is and the more averaging is done, the less of a problem these variables would be. The trends would still be apparent from the data analysis.
I disagree. We are looking for a efficiency difference of maybe 1 - 2 Wh/Mi. A light 5mph tailwind can produce that difference. Standing straight up vs riding in a slight tuck can produce that difference. Being much taller or shorter than average can produce that difference. Air pressure in pneumatic tires can produce that difference. Urethane wheel type can produce that difference in certain circumstances. Belt issues can probably produce that. Rusted bearings can produce that.
You are looking for a specific needle in a pile of needles.
If you want to really test this, you will need to get a couple of boards configured differently and dyno test them in a lab.
