"Peak Efficiency" Control Mode?

[Hummie]

if your most efficient speed is 25mph and youre starting from 0 but limited in your ability to accelerate you will lull around in inefficiency for a longer time as it takes longer to get to the efficient speed. it seems easier to use your "internal computer" in choosing what amount of acceleration-compromise to make when trying to be as efficient as possible as we have to deal with the necessity of real-world stop-and-go driving.
the most efficient speed might also not be sustainable as it could require too many continuous watts to maintain and then with the increased heat, while still being at a decidedly optimum speed, could end up very inefficient even just in the motor and this would probably be the case with those little motors. As you know if we wanted efficient travel more so than anything else the wind resistance would likely be the biggest factor and we'd be riding at maybe 12mph with a motor that had that as about the top speed. not so practical or fun though.

[devin]

if your most efficient speed is 25mph and youre starting from 0 but limited in your ability to accelerate you will lull around in inefficiency for a longer time as it takes longer to get to the efficient speed. it seems easier to use your "internal computer" in choosing what amount of acceleration-compromise to make when trying to be as efficient as possible as we have to deal with the necessity of real-world stop-and-go driving.
the most efficient speed might also not be sustainable as it could require too many continuous watts to maintain and then with the increased heat, while still being at a decidedly optimum speed, could end up very inefficient even just in the motor and this would probably be the case with those little motors. As you know if we wanted efficient travel more so than anything else the wind resistance would likely be the biggest factor and we'd be riding at maybe 12mph with a motor that had that as about the top speed. not so practical or fun though.

@Hummie ...according to the algorithm, you can get nearly max theoretical 85% peak efficiency at nearly every speed, not just one speed...

simply the controller needs to calculate the most efficient wattage to apply to the motor to achieve peak 85% electrical to mechanical conversion efficiency depending on the kv and present speed.

[Hummie]

seems like it would just be a really low acceleration program as it would always have to keep the applied voltage so close to the back emf and minimal current flow.




http://www.ebikes.ca/tools/simulator.ht ... grade_b=13
comparing two motors with a different voltage, one with 72 volts, other at 36, at the same speed the inefficiency shows up a lot in the bottom right of this link. "performance" shows the huge contrast in temp and efficiency.
why doesn't the esc send the same voltage to each motor regardless of the pack voltage? similar to what you're talking about and if you have control of the effective voltage.

[devin]

seems like it would just be a really low acceleration program as it would always have to keep the applied voltage so close to the back emf and minimal current flow.




http://www.ebikes.ca/tools/simulator.ht ... grade_b=13
doing the half speed with double the voltage shows much more inefficient which relates and I'm wondering why the esc doesn't just send the same voltage in these simulations regardless of what the battery is.
bottom right "performance" shows the huge contrast in temp and efficiency.

@Hummie surprisingly, i don't think it would be a low acceleration algorithm... for example:

with the @Hummie 8mm axle long hub motor, at 26.2mph, according to the "peak efficiency" algorithm, the mechanical output of the motor would be 2057.59W (for 2420W Electrical Input, or 48.407 battery amps @ 50V -- 85% electrical to mechanical conversion efficiency) -- since the mechanical load opposing the rider at that speed would be significantly less than 2057.59w, the rider is in a state of rapid acceleration.

He says he tested the 90kv motor with a motor amp setting of 200A... & the vesc detection was 0.0415ohm...

That version of the vesc has a coded motor amp limit of 120A and the lead to lead is double the detection value, therefore I will use 120A and 0.0830ohm....

Stall 120A @ 0.0830ohm = 1195.2W Electrical at Stall = 9.96V Effective @ Stall = 50V Battery @ %19.92 duty cycle

The 90KV motor is capable of turning @ 50V Battery * 90KV = 4500rpm at no load rpm @ 100% duty cycle

If we use the "peak efficiency" algorithm, and the motor is turning 2800rpm:

A = (20B)/(17C)

A = XX.XXV = Full Throttle Effective Voltage
B = 2800rpm = Present RPM
C = 90kv = Motor KV

therefore

A = (20B)/(17C)

A = ((20)(2800))/((17)(90))

36.6V = ((20)(2800))/((17)(90))

Simply according to the algorithm, in theory, for peak electrical to mechanical conversion efficiency, when the rotor is turning 2800rpm, the applied effective voltage should be 36.6V.

1/90=0.01111... volts per rpm <---- simply 90 rpm per volt is 0.011111... volts per rpm

2800rpm*0.0111111v=31.111V <---- present back emf voltage at 2800rpm

36.6V-31.111V=5.489V <------- Effective Voltage minus back emf voltage equals 5.489 net volts

5.489V=66.13A*0.0830ohm <---- 66.13A Motor Amps @ 2800rpm

90*2=180
180*pi=565.48
565.48/60=9.424r/vs
0.10611=1/9.424 <--------90KV = 0.10611Nm/A

66.13A Motor Amps * 0.10611NM/A = 7.017Nm <----- 7.017 Newton Meters Torque at 66.13A Motor Amps

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

2057.59w = 7.017Nm*2800rpm/9.5488 <------ 2057.59W Mechanical Output @ 2800rpm @ 36.6V Effective Volts

Assuming 50V battery pacK

(36.6V Effective V/50V Pack V)*100= 73.20% duty cycle <------------ 36.6V Effective Volts is 73.20% duty cycle w/ 50V Battery Pack

&:

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

therefore:

48.407 battery amps = 66.13 motor amps * (%73.20 duty cycle/100) <------------ 66.13 motor amps @ 73.20% duty cycle is 48.407 battery amps

2420W Electrical = 48.407 battery amps * 50V <-------- 2420W Electrical In

2057.59w = 7.017Nm*2800rpm/9.5488 <------ 2057.59W Mechanical Output @ 2800rpm @ 90kv @ 36.6V Effective Volts @ 73.20% duty cycle

(2057.59W/2420W)*100=85.02% <----------- 85.02% conversion efficiency of electrical to mechanical watts.

------------------------

In conclusion, using the peak efficiency algorithm, efficiency is near the theoretical maximum 85% @ 2800rpm (ground speed 26.2mph or 42.22kph).

The question then becomes, does any other duty cycle besides 73.20% provide a higher electrical to mechanical wattage conversion efficiency than 85.02% conversion efficiency at 2800rpm?

[Hummie]

maybe it was mentioned and the wattage posted is a the high end of the speed spectrum and how much power it would have going from 3 to 5mph when the voltages are so low? if that's what youre saying...always keeping the applied voltage just a hair above the back emf.

[devin]

maybe it was mentioned but that would be the easy end of the speed spectrum and how much power it would have going from 3 to 5mph when the voltages are so low. if that's what youre saying...always keeping the applied voltage just a hair above the back emf.

@Hummie I think by having parameter "E" minimum full throttle wattage setting in the control algorithm, you could have as much wattage as you want from 3-5mph. (at the sacrifice of efficiency)

The cold logic of the algorithm is "whenever I am pushing full throttle, my motor is at peak 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 < [is less than] 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
E= minimum full throttle wattage setting
F= maximum wattage setting
G= present wattage

^by variably changing D the (85/100) ratio, the rider could change the power curve (at the sacrifice of near-peak conversion efficiency), for example changing this ratio to (90/100) would give lower acceleration, or changing the ratio to (80/100) would give greater acceleration.

[rew]

Bullshit

[devin]

Bullshit

@rew how many more examples are necessary to prove?

4.88battery amps * 50V = 244W Electrical <-------- 244W Electrical In

207.79W = 0.3307Nm*6000rpm/9.5488 <------ 207.79W Mechanical Output @ 6000rpm @ 26.14V Effective Volts

(207.79W/244W)*100=86.57% <----------- 86.57% conversion efficiency of electrical to mechanical watts.

2420W Electrical = 48.407 battery amps * 50V <-------- 2420W Electrical In

2057.59w = 7.017Nm*2800rpm/9.5488 <------ 2057.59W Mechanical Output @ 2800rpm @ 90kv @ 36.6V Effective Volts @ 73.20% duty cycle

(2057.59W/2420W)*100=85.02% <----------- 85.02% conversion efficiency of electrical to mechanical watts.

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.

[rew]

Bullshit

[devin]

Bullshit

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

Assuming 50V, 100kv, 0.1ohm, no load = 5000rpm = 50km/h, present speed = 2500rpm = 25km/h:

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

If we use the "peak 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 peak 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.

------------------------
Conclusion

Using the peak efficiency algorithm, during acceleration, electrical to mechanical conversion efficiency is near the theoretical maximum 85% @ 2500rpm (1102W mechanical output for 1297W electrical input at ground speed 15.53mph or 25kph).

Since the mechanical load opposing the rider (caused by wind, friction, eddy currents, etc) at 15.53mph or 25kph is significantly lower than the 1102W of mechanical output at peak efficiency, the rider is in a state of rapid acceleration. If they remove their thumb from the throttle they will be coasting close to constant speed (with deceleration caused by wind, etc).

The question then becomes, does any other duty cycle besides 58.82% provide a higher electrical to mechanical wattage conversion efficiency than 85.00% conversion efficiency at 2500rpm?

Now suppose that your optimal setting suggests max efficieny at e.g. 5A when moving slowly. Do you think that results in a usable system? People will want to accelerate quicker even if that costs a few joules.

The cold logic of the algorithm is "whenever I am pushing full throttle, my motor is at peak 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
E= minimum full throttle wattage setting
F= maximum wattage setting
G= present wattage

^by variably changing D the (85/100) ratio, the rider could change the power curve (at the sacrifice of near-peak conversion efficiency), for example changing this ratio to (90/100) would give lower acceleration, or changing the ratio to (80/100) would give greater acceleration.

^@rew by incorporating E the minimum full throttle wattage setting, the rider can choose a minimum wattage that corresponds to full throttle for use at low rpms (in other words a necessary-for-usability sacrifice of efficiency at low rpms).

Also, by incorporating F, a maximum wattage setting, if the wattage suggested by the algorithm would exceed the hardware capabilities of the system, the F setting prevents exceeding the hardware capabilities (at the sacrifice of efficiency to prevent system damage) -- or the rider can partially release the throttle to lower the watts (again, at the sacrifice of efficiency).

Bullshit

@rew -can you be more specific i.e. where is the math wrong?