"Peak Efficiency" Control Mode?

General topics and discussions about the VESC and its development.
devin
Posts: 255
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 06 Sep 2017, 05:53

markorman wrote:Stop thinking about controlling the motor with watts. Nobody nowhere is controlling the motor with watts. The only time you care about watts consumed by motor is when you have to limit the power drained from battery.
Almost are motors are in torque control mode (at least in automotive world) since the torque is the force the motor produces and force you feel when accelerating/decelerating. In BLDC and PMSM motor you have to limit the torque because the mechanical limits, and limits due to current in copper windings.

This is typical graph for PMSM motors. You have to limit the torque (motor current) and then limit the power.


@markorman... if i was elon musk, i would instruct my engineers to modify my personal tesla S in the following manner...

1. put a motor temp thermometer in the middle of my dashboard with a "red line" slightly below the winding enamel breakdown temp

&

2. instruct my control algorithm engineers to continue the "peak battery amps" plateau all the way to the left side of the graph to ensure control of constant max battery watt via the electricity pedal leading to increased acceleration:

Image

3. triple the tire width

4. head down to the local drag strip....

Image

markorman
Posts: 14
Joined: 25 May 2016, 11:35

Re: "Peak Efficiency" Control Mode?

Postby markorman » 06 Sep 2017, 06:24

1. put a motor temp thermometer in the middle of my dashboard with a "red line" slightly below the winding enamel breakdown temp

They ARE measuring the temperature of the motor and will limit the current through it if you drive at max speed too long.
2. instruct my control algorithm engineers to continue the "peak battery amps" plateau all the way to the left side of the graph to ensure control of constant max battery watt via the electricity pedal leading to increased acceleration:

You HAVE to limit the torque to the motor or you will brake some mechanics (motor, gearbox, drive train...) before you will overheat the motor.
Also the most powerful model S has acceleration of ~1g all the way to the point where the power is at max. You cannot accelerate faster that 1g.
3. triple the tire width

And this would decrease the efficiency of the car.

devin
Posts: 255
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 06 Sep 2017, 16:33

devin wrote:3. triple the tire width


markorman wrote:And this would decrease the efficiency of the car.


i'm more concerned about tire slippage...

Image

markorman wrote:You HAVE to limit the torque to the motor or you will brake some mechanics (motor, gearbox, drive train...) before you will overheat the motor.
Also the most powerful model S has acceleration of ~1g all the way to the point where the power is at max. You cannot accelerate faster that 1g.


5. add fins

6. double pack volts [1x(1/2)amp hours]

7. add on/off switch to the steering wheel for traction control

8. triple tread depth

devin
Posts: 255
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 07 Sep 2017, 23:51

markorman wrote:You HAVE to limit the torque to the motor or you will brake some mechanics (motor, gearbox, drive train...) before you will overheat the motor.
Also the most powerful model S has acceleration of ~1g all the way to the point where the power is at max. You cannot accelerate faster that 1g.


@markorman I think strengthening the mechanics is a viable alternative to lowering the torque....

for example i'm aware an anonymous skateboard designer has recently upgraded their electric skateboard hub motor axles from 8mm to 12mm after testing 48A batt max / 120A motor max settings @ 36V on a VESC for about 7 days... the failure occurred accelerating at low speed in a flat area.

Image

Image

Image

Image

Image

^this last one is true color but enhanced with additional contrast

my personal analysis is it was a torque-related fatigue based on the visible rotational pattern of failure...

others have suggested it was caused by the leverage of a wider than normal wheel but I doubt this.

compare to this image:

Image
Image Source: http://www.ferret.com.au/articles/in-focus/a-review-of-failures-in-machineries-why-your-machines-fail-and-how-you-can-stop-it-n2509249

devin wrote:3. triple the tire width


markorman wrote:And this would decrease the efficiency of the car.


i'd argue with hub motors it increases efficiency.
Last edited by devin on 08 Sep 2017, 16:10, edited 1 time in total.

rew
Posts: 943
Joined: 25 Mar 2016, 12:29
Location: Delft, Netherlands.

Re: "Peak Efficiency" Control Mode?

Postby rew » 08 Sep 2017, 07:36

devin wrote:-earlier you proposed a scenario of an e-bicyclist at 25kmh with 100kv motor, 0.1ohm resistance, 50V and no load = 50kmh (@25kmh, the present speed is half of no load speed which on the objectionable graph shows this rpm point should supply peak power
BULLSHIT.

Your graph shows peak power at that point. But your graph does not apply to this motor. That is why it is not relevant and you should stop referring to that graph. The graph predicts we can put 250A through the motor to achieve peak output power. That is not relevant because the motor can only handle 15A.

devin
Posts: 255
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 08 Sep 2017, 12:54

rew wrote:Your graph shows peak power at that point. But your graph does not apply to this motor. That is why it is not relevant and you should stop referring to that graph. The graph predicts we can put 250A through the motor to achieve peak output power. That is not relevant because the motor can only handle 15A.


@rew -- An important part of the graph i think remains accurate (the part pertaining to this thread)-- the column showing 85% electrical to mechanical conversion efficiency when the present RPM equals 85% of the applied effective voltage times the KV (before factoring iron & bearing friction losses.) Also I didn't make the graph, I linked to it so if it isn't accurate please don't blame me.

rew wrote:The graph predicts we can put 250A through the motor to achieve peak output power. That is not relevant because the motor can only handle 15A.


@rew Test Question: There is a contest with a prize of $1000 USD for whoever achieves the highest mechanical power output from a particular BLDC motor for 100 milliseconds. My entry fee costs $100, therefore the $1000 prize would cover 10 times the entry fee so the longevity of the motor is of little concern for winning this contest. I have a controller capable of 100% duty cycle at any RPM in bursts of 100 milliseconds and a dynamometer. Assuming all contestants have the same voltage lab supply, if I want to win the prize, at what % of no load rpm should I apply my 100 millisecond 100% duty cycle burst for the greatest chances of victory? What is the approximate electrical to mechanical % conversion efficiency during this 100 milliseconds?
Last edited by devin on 08 Sep 2017, 15:50, edited 15 times in total.

markorman
Posts: 14
Joined: 25 May 2016, 11:35

Re: "Peak Efficiency" Control Mode?

Postby markorman » 08 Sep 2017, 13:11

devin wrote:
rew wrote:Your graph shows peak power at that point. But your graph does not apply to this motor. That is why it is not relevant and you should stop referring to that graph. The graph predicts we can put 250A through the motor to achieve peak output power. That is not relevant because the motor can only handle 15A.


@rew -- An important part of the graph i think remains accurate (the part pertaining to this thread)-- the column showing 85% electrical to mechanical conversion efficiency when the present RPM equals 85% of the applied effective voltage times the KV (before factoring iron & bearing friction losses.) Also I didn't make the graph, I linked to it so if it isn't accurate please don't blame me.


Graph you posted have nothing to do with 3-phase PMSM motors, but with brushed DC motors. How many times to you need to be told that?

devin
Posts: 255
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 08 Sep 2017, 13:15

markorman wrote:Graph you posted have nothing to do with 3-phase PMSM motors, but with brushed DC motors. How many times to you need to be told that?


@markorman pertinent to this thread is the part of the graph showing 85% conversion of electrical to mechanical watts when the present RPM equals 85% of the applied effective voltage times the KV (before factoring iron & bearing friction losses.) -- according to my calculations this part of the graph is still accurate with 3-phase pmsm motors, i think.

please have a look at the formulas i posted not the graph.

devin wrote:1297.275W Electrical = 25.9455 battery amps * 50V <-------- 1297.275W Electrical In

1102.756w = 4.212Nm*2500rpm/9.5488 <------ 1102.756W Mechanical Output @ 2500rpm @ 29.411V Effective Volts @ 100kv @ 0.1ohm @ %58.82 duty cycle

(1102.756W/1297.275W)*100=85.00% <----------- 85.00% conversion efficiency of electrical to mechanical watts.


^85% electrical to mechanical converison efficiency @ 50% of no load rpm

devin wrote:The cold logic of the algorithm is "whenever I am pushing full throttle, my motor is at [edit: desired] electrical to mechanical conversion efficiency, whatever amount of power that happens to be at the time*"

*except at very low rpms-- full throttle equates to a rider-specified minimum wattage-- otherwise the formula will produce insufficient electrical and mechanical wattage to propel a human rider. (less than full throttle, throttle % equates to % max available motor amps as defined by the algorithm)

@ full throttle:
if according to the algorithm at full throttle G < E, then full throttle = E or max duty cycle, whichever is lower wattage, & %throttle = %max motor amps as defined by E or max duty cycle, whichever supersedes based on lower wattage. (E supersedes & is lower wattage when rpms are below the useable max efficiency power threshold,and max duty cycle supersedes & is lower wattage when Back EMF at high rpms prevents obtaining E).

if according to the algorithm G > F, then full throttle = F & %throttle = %max motor amps as defined by F

B = (D)(AC)

where:

A= present effective voltage
B= present rpm
C= motor kv or rpm per volt
D= 85/100 [17/20]
E= minimum full throttle wattage [edit: or torque or motor amps] setting
F= maximum wattage [edit: or torque or motor amps] setting
G= present wattage [edit: or torque or motor amps]

^by variably changing D the (85/100) ratio, the rider could change the power curve, for example changing this ratio to (90/100) would give lower acceleration, or changing the ratio to (80/100) would give greater acceleration.


rew wrote:You are suggesting that we find a control method that is more efficient at driving a motor than "standard".

Let's try this with a theoretical motor first. Let's try this first "by hand".

Easy numbers. 50V max, about 100KV: 10 rad/sec/V, 0.1 Ohm resistance. Max 50A.

The 10 rad/sec/V means that this motor produces 0.1 Nm/A . So at 50A, a maximum of 5Nm.

So I've mounted this on my bike (or skateboard if you like) and the gearing is such that at no load and 50V on the motor, the board/bike goes 50km/h. So now I'm cruising along at 25km/h


devin wrote:Assuming 50V, 100kv, 0.1ohm, no load = 5000rpm = 50km/h, present speed = 2500rpm = 25km/h = [edit: 50% no load rpm]:

The 100KV motor is capable of turning @ 50V Battery * 100KV = 5000rpm at no load rpm @ 100% duty cycle

If we use the "[edit: desired] efficiency" algorithm, and the motor is turning 2500rpm:

A = (20B)/(17C)

A = XX.XXV = Full Throttle Effective Voltage
B = 2500rpm = Present RPM
C = 100kv = Motor KV

therefore

A = (20B)/(17C)

A = ((20)(2500))/((17)(100))

29.411V = ((20)(2500))/((17)(100))

Simply according to the algorithm, in theory, for [edit: desired] electrical to mechanical conversion efficiency, when the rotor is turning 2500rpm, the applied effective voltage should be 29.411V.

1/100=0.01 volts per rpm <---- simply 100 rpm per volt is 0.01 volts per rpm

2500rpm*0.01v=25V <---- 25V present back emf voltage at 2500rpm

29.411V-25V=4.411V <------- Effective Voltage minus back emf voltage equals 4.411 net volts

44.11A=4.411V/0.1ohm <---- 44.11A Motor Amps @ 2500rpm

100kv*2=200
200*pi=628.3185
628.3185/60=10.4719r/vs <-------------- 100kv is 10.4719 radians per second per volt
0.09549=1/10.4719 <--------100kv = 0.09549Nm/A

44.11A Motor Amps * 0.09549NM/A = 4.212Nm <----- 4.212 Newton Meters Torque at 44.11A Motor Amps

Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488
Source: http://www.wentec.com/unipower/calculat ... torque.asp

1102.756w = 4.212Nm*2500rpm/9.5488 <------ 1102.756W Mechanical Output @ 2500rpm @ 29.411V Effective Volts @ 100kv @ 0.1ohm

Assuming 50V battery pacK

(29.411V Effective V/50V Pack V)*100= 58.82% duty cycle <------------ 29.411V Effective Volts is 58.82% duty cycle w/ 50V Battery Pack

&:

Battery Amps = Motor Amps x (Duty Cycle/100)

therefore:

25.9455 battery amps = 44.11 motor amps * (%58.82 duty cycle/100) <------------ 44.11 motor amps @ 58.82% duty cycle is 25.9455 battery amps

1297.275W Electrical = 25.9455 battery amps * 50V <-------- 1297.275W Electrical In

1102.756w = 4.212Nm*2500rpm/9.5488 <------ 1102.756W Mechanical Output @ 2500rpm @ 29.411V Effective Volts @ 100kv @ 0.1ohm @ %58.82 duty cycle

(1102.756W/1297.275W)*100=85.00% <----------- 85.00% conversion efficiency of electrical to mechanical watts.


devin wrote:B = (D)(AC)

where:

A= present effective voltage
B= present rpm
C= motor kv or rpm per volt
D= 85/100 [17/20]
E= minimum full throttle wattage [edit: or torque or motor amps] setting
F= maximum wattage [edit: or torque or motor amps] setting
G= present wattage [edit: or torque or motor amps]

^by variably changing D the (85/100) ratio, the rider could change the power curve, for example changing this ratio to (90/100) would give lower acceleration, or changing the ratio to (80/100) would give greater acceleration.


devin wrote:in a couple mathematical examples I've done, I've found that changing variable D to a fraction higher than 85/100 (such as 90/100) can provide even higher conversion efficiencies than 85% at the sacrifice of acceleration. On that basis, as mentioned, I propose calling the control mode "Desired Efficiency Control Mode" rather than "Peak Efficiency" control mode.

rew
Posts: 943
Joined: 25 Mar 2016, 12:29
Location: Delft, Netherlands.

Re: "Peak Efficiency" Control Mode?

Postby rew » 08 Sep 2017, 19:18

devin wrote:@rew Test Question: There is a contest with a prize of $1000 USD for whoever achieves the highest mechanical power output from a particular BLDC motor for 100 milliseconds. My entry fee costs $100, therefore the $1000 prize would cover 10 times the entry fee so the longevity of the motor is of little concern for winning this contest. I have a controller capable of 100% duty cycle at any RPM in bursts of 100 milliseconds and a dynamometer. Assuming all contestants have the same voltage lab supply, if I want to win the prize, at what % of no load rpm should I apply my 100 millisecond 100% duty cycle burst for the greatest chances of victory? What is the approximate electrical to mechanical % conversion efficiency during this 100 milliseconds?

Well... If the iron in the motor cannot saturate (it can) and if the permanent magnets are infinitely strong, then at 50% of the max RPM. You happy now?

But the iron that helps increase efficiency DOES saturate. I expect not very much above the rated max current. And the magnets will become demagnetized if you apply too big magnetic forces on them.

The VESC is quite capable of driving say a DT750 motor. It has a rated current of 18A and 61mOhms of winding resistance. That would mean 180A at stall and 90A at your max-power point. Do you have a VESC6? That will do a short burst of 90A no problem. But maybe, to be on the safe side, this experiment should be done with a motor that is just a little bit smaller. One with say 5A max current and we drive it with 50-70A for that short burst of yours.

devin
Posts: 255
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: "Peak Efficiency" Control Mode?

Postby devin » 09 Sep 2017, 02:58

devin wrote:@rew Test Question: There is a contest with a prize of $1000 USD for whoever achieves the highest mechanical power output from a particular BLDC motor for 100 milliseconds. My entry fee costs $100, therefore the $1000 prize would cover 10 times the entry fee so the longevity of the motor is of little concern for winning this contest. I have a controller capable of 100% duty cycle at any RPM in bursts of 100 milliseconds and a dynamometer. Assuming all contestants have the same voltage lab supply, if I want to win the prize, at what % of no load rpm should I apply my 100 millisecond 100% duty cycle burst for the greatest chances of victory? What is the approximate electrical to mechanical % conversion efficiency during this 100 milliseconds?


rew wrote:Well... If the iron in the motor cannot saturate (it can) and if the permanent magnets are infinitely strong, then at 50% of the max RPM. You happy now?

But the iron that helps increase efficiency DOES saturate. I expect not very much above the rated max current. And the magnets will become demagnetized if you apply too big magnetic forces on them.

The VESC is quite capable of driving say a DT750 motor. It has a rated current of 18A and 61mOhms of winding resistance. That would mean 180A at stall and 90A at your max-power point. Do you have a VESC6? That will do a short burst of 90A no problem. But maybe, to be on the safe side, this experiment should be done with a motor that is just a little bit smaller. One with say 5A max current and we drive it with 50-70A for that short burst of yours.


@rew A+ ;)


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