Comparison:

Image Source: https://met-tech.com/race-car-transmission-input-shaft/

Caption: "Fatigue Fracture Of A 300M Ultra-high Strength Steel Race Car Transmission Input Shaft"

Caption: "Fracture of an 8mm Steel Hub Motor Axle @ 90kv @ approximately 80mm outer tire diameter @ ~120A max motor amps setting w/VESC"

^12mm axle upgrade

## "Peak Efficiency" Control Mode?

### Re: "Peak Efficiency" Control Mode?

[post removed by author]

Last edited by devin on 15 Sep 2017, 18:31, edited 1 time in total.

### Re: "Peak Efficiency" Control Mode?

Efficiency Control Mode of AC induction motors?

A=“Synchronous” RPM @ Present AC Frequency

B=120

C=Present AC Frequency

D=Poles

E=85/100=Present RPM Desired % of Synchronous RPM @ Present AC Frequency (example where 85% = 85/100)

C=(AD)/(BE)

A=(BCE)/D

B=(AD)/(CE)

D=(BCE)/A

E=(AD)/(BC)

Theory: By variably adjusting parameter “E — Desired % of Synchronous RPM @ Present AC Frequency”, electrical to mechanical conversion efficiency and acceleration can be variably adjusted in AC induction motors.

references: https://www.quora.com/What-is-the-formula-to-calculate-a-synchronous-speed-of-a-motor

"f = PN/120 where f is frequency of power supply, P is number of poles in the machine and N = speed in rpm.

So, Synchronous speed N = 120f/P

Thus for 50 Hz supply, 2 pole machine; synchronous speed = 3000rpm."

Full throttle = C

&

% Full throttle = % motor amps or watts or torque as defined by C, F or G

&

F = Minimum motor amps or watts or torque at full throttle setting

if C results in less motor amps or watts or torque than F, than full throttle = F & % full throttle = % motor amps as defined by F

&

G = Maximum motor amps or watts or torque at full throttle setting

if C results in more motor amps or watts or torque than G, than full throttle = G & % full throttle = % motor amps as defined by G

&

H = Minimum frequency setting

edit: I may have found some issues with these equations so I've reformulated them and reposted the new versions below, so please take the above equations in this posting with a grain of salt...

Last edited by devin on 16 Sep 2017, 19:09, edited 3 times in total.

### Re: "Peak Efficiency" Control Mode?

Devin, I've been thinking about how to explain things.

The picture that does not apply to BLDC motors that you keep posting is based on the simple assumption that all losses are due to copper losses. In first approximation that's true.

So... We have a motor where I = U / R and U = Uapplied - Ubemf . The losses then are I

From this you can conclude, as R cannot change, that you want to keep I as small as possible. So... My old strategy: "accelerate fast because the extra energy goes into kinetic energy anyway, I know I'm using lots of energy/sec and energy/km, but it's not wasted" is not a good one: Because of the high currents the losses are larger than with a slower acceleration.

So under these assumptions you need to apply the minimal current until you arrive at your destination.

Suppose that applying 20A results in a 200N force allowing me (100kg) to accelerate from 0 to 5m/s in 2.5 seconds. Suppose you propose to apply 10A for twice as long. The force of 100N will accelerate to 5m/s as well, but now in 5 seconds. The losses per-second are however 4x times less, but because you apply them for twice as long, you'll only save 50% of the energy.

Now when we've got the simplest model figured out we can add other effects. For example: Suppose things are not frictionless, but there is a constant drag-force. Now we might want to travel 1km. and the firiction might be 10N. This costs us an extra A in current to overcome. How does this change the strategy of "as little current as possible?

If we need to travel 1km, there is simply 10N * 1000m = 10kJ of energy that needs to come from the battery through the motor.

Suppose I want to take 200s at 5m/s. I'll use 1A at say(*) 50V, or 50W during 200 seconds for 10kJ. Suppose you ride only 2.5m/s for 400s, so you'll still use that 1A but now at only 25V effective. Now you will incur the I

Next you might start to add effects like iron losses or air resistance. Things get complicated.

(*) The 50V follows from the 10N/A that I assumed before.

The picture that does not apply to BLDC motors that you keep posting is based on the simple assumption that all losses are due to copper losses. In first approximation that's true.

So... We have a motor where I = U / R and U = Uapplied - Ubemf . The losses then are I

^{2}.RFrom this you can conclude, as R cannot change, that you want to keep I as small as possible. So... My old strategy: "accelerate fast because the extra energy goes into kinetic energy anyway, I know I'm using lots of energy/sec and energy/km, but it's not wasted" is not a good one: Because of the high currents the losses are larger than with a slower acceleration.

So under these assumptions you need to apply the minimal current until you arrive at your destination.

Suppose that applying 20A results in a 200N force allowing me (100kg) to accelerate from 0 to 5m/s in 2.5 seconds. Suppose you propose to apply 10A for twice as long. The force of 100N will accelerate to 5m/s as well, but now in 5 seconds. The losses per-second are however 4x times less, but because you apply them for twice as long, you'll only save 50% of the energy.

Now when we've got the simplest model figured out we can add other effects. For example: Suppose things are not frictionless, but there is a constant drag-force. Now we might want to travel 1km. and the firiction might be 10N. This costs us an extra A in current to overcome. How does this change the strategy of "as little current as possible?

If we need to travel 1km, there is simply 10N * 1000m = 10kJ of energy that needs to come from the battery through the motor.

Suppose I want to take 200s at 5m/s. I'll use 1A at say(*) 50V, or 50W during 200 seconds for 10kJ. Suppose you ride only 2.5m/s for 400s, so you'll still use that 1A but now at only 25V effective. Now you will incur the I

^{2}.R losses for twice as long and you'll expend twice as much energy in the motor as I. So now going fastest is best....Next you might start to add effects like iron losses or air resistance. Things get complicated.

(*) The 50V follows from the 10N/A that I assumed before.

### Re: "Peak Efficiency" Control Mode?

rew wrote:Devin, I've been thinking about how to explain things.

The picture that does not apply to BLDC motors that you keep posting is based on the simple assumption that all losses are due to copper losses. In first approximation that's true.

So... We have a motor where I = U / R and U = Uapplied - Ubemf . The losses then are I2.R

From this you can conclude, as R cannot change, that you want to keep I as small as possible. So... My old strategy: "accelerate fast because the extra energy goes into kinetic energy anyway, I know I'm using lots of energy/sec and energy/km, but it's not wasted" is not a good one: Because of the high currents the losses are larger than with a slower acceleration.

So under these assumptions you need to apply the minimal current until you arrive at your destination.

Suppose that applying 20A results in a 200N force allowing me (100kg) to accelerate from 0 to 5m/s in 2.5 seconds. Suppose you propose to apply 10A for twice as long. The force of 100N will accelerate to 5m/s as well, but now in 5 seconds. The losses per-second are however 4x times less, but because you apply them for twice as long, you'll only save 50% of the energy.

Now when we've got the simplest model figured out we can add other effects. For example: Suppose things are not frictionless, but there is a constant drag-force. Now we might want to travel 1km. and the firiction might be 10N. This costs us an extra A in current to overcome. How does this change the strategy of "as little current as possible?

If we need to travel 1km, there is simply 10N * 1000m = 10kJ of energy that needs to come from the battery through the motor.

Suppose I want to take 200s at 5m/s. I'll use 1A at say(*) 50V, or 50W during 200 seconds for 10kJ. Suppose you ride only 2.5m/s for 400s, so you'll still use that 1A but now at only 25V effective. Now you will incur the I2.R losses for twice as long and you'll expend twice as much energy in the motor as I. So now going fastest is best....

Next you might start to add effects like iron losses or air resistance. Things get complicated.

(*) The 50V follows from the 10N/A that I assumed before.

@rew I think the rider decides the speed they'd like to go and the algorithm in oversimplified terms ensures a "minimum electrical to mechanical conversion efficiency" the entire time while accelerating (for example in start and stop traffic -- close to 85% electrical to mechanical conversion efficiency throughout most of the entire operating range or more if desired -- not only at 85% of max rpm), with lower desired efficiency leading to more rapid acceleration, and higher desired efficiency leading to more measured acceleration.

If the rider chooses a higher final cruising speed, more mechanical watt hours per kilometer will be expended moving air, and then kilometers per electrical kilowatt hour will decrease, while electrical to mechanical conversion efficiency during acceleration is preserved at some desired minimum, for example 85% electrical watts to mechanical.

### Re: "Peak Efficiency" Control Mode?

I may have found some odd results with the AC Induction Motor Efficiency Control theory previously posted, so here's round 2:

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

Efficiency Control Mode of AC induction motors?

Example Values:

J=Present RPM = 2500rpm

E=Present RPM Desired % of Synchronous RPM @ Present AC Frequency = 0.85 = 85/100 = 85%

^Theory: By variably adjusting parameter “E — Present RPM Desired % of Synchronous RPM @ Present AC Frequency”, electrical to mechanical conversion efficiency and acceleration can be variably adjusted in AC induction motors.

C=Desired AC Frequency at Present RPM= XX.XX

A=“Synchronous” RPM @ Desired Present AC Frequency = XX.XX

B=120

D=Poles = 2

F = Minimum motor amps or watts or torque at full throttle setting

G = Maximum motor amps or watts or torque at full throttle setting

H = Minimum frequency setting

I = Maximum frequency setting

If C>F & C<G & C>H & C<I, then Full throttle = C

If C<F then Full throttle = F

If C>G then Full throttle = G

If C<H then Full throttle = H

If C>I then Full throttle = I

&

% Full throttle = % motor amps or watts or torque as defined by C, F, G, H or I

therefore:

2500 present rpm [J]/ 0.85desired % sync rpm [E] = 2941.17 sync rpm @ desired frequency[A]

J/E=A

2500/0.85=2941.17 <---present 2500rpm is 85% of a synchronous rpm of 2941.17rpm

therefore:

2941.17rpm = A = “Synchronous” RPM @ Desired Present AC Frequency

&

references: https://www.quora.com/What-is-the-formula-to-calculate-a-synchronous-speed-of-a-motor

"f = PN/120 where f is frequency of power supply, P is number of poles in the machine and N = speed in rpm.

So, Synchronous speed N = 120f/P

Thus for 50 Hz supply, 2 pole machine; synchronous speed = 3000rpm."

N = 120f/P = 2941.17[A] = ((120[B])(C))/2[D]

A=2941.17

B=120

D=2

C=XX.XX <——need to determine the desired AC Frequency at 2500rpm

A=((B)(C))/D

therefore:

C=(AD)/B

therefore:

C=((2941.17)(2))/120

C=49.01HZ

therefore:

C=Desired AC Frequency at Present RPM = 49.01hz <—— desired 49.01hz frequency at 2500rpm @ 2 poles @ desired 85% of synchronous RPM

therefore:

J=Present RPM = 2500rpm <——Present RPM

A=“Synchronous” RPM @ Present Desired AC Frequency = 2941.17rpm

B=120

C=Desired AC Frequency at Present RPM= 49.01hz <— Desired AC Frequency

D=Poles = 2

E=Present RPM Desired % of Synchronous RPM @ Present AC Frequency = 0.85 = 85/100 = 85%

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

Simply, in theory, If the rotor is turning 2500rpm, and there are 2 poles, and we want the present RPM to be 85% of the "Synchronous" RPM, the desired AC Frequency should be set to 49.01hz. At full throttle, if wattage control is used the electrical wattage should be equivalent to Setting F - minimum wattage setting, and % throttle is % motor amps as defined by F.

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

Efficiency Control Mode of AC induction motors?

Example Values:

J=Present RPM = 2500rpm

E=Present RPM Desired % of Synchronous RPM @ Present AC Frequency = 0.85 = 85/100 = 85%

^Theory: By variably adjusting parameter “E — Present RPM Desired % of Synchronous RPM @ Present AC Frequency”, electrical to mechanical conversion efficiency and acceleration can be variably adjusted in AC induction motors.

C=Desired AC Frequency at Present RPM= XX.XX

A=“Synchronous” RPM @ Desired Present AC Frequency = XX.XX

B=120

D=Poles = 2

F = Minimum motor amps or watts or torque at full throttle setting

G = Maximum motor amps or watts or torque at full throttle setting

H = Minimum frequency setting

I = Maximum frequency setting

If C>F & C<G & C>H & C<I, then Full throttle = C

If C<F then Full throttle = F

If C>G then Full throttle = G

If C<H then Full throttle = H

If C>I then Full throttle = I

&

% Full throttle = % motor amps or watts or torque as defined by C, F, G, H or I

therefore:

2500 present rpm [J]/ 0.85desired % sync rpm [E] = 2941.17 sync rpm @ desired frequency[A]

J/E=A

2500/0.85=2941.17 <---present 2500rpm is 85% of a synchronous rpm of 2941.17rpm

therefore:

2941.17rpm = A = “Synchronous” RPM @ Desired Present AC Frequency

&

references: https://www.quora.com/What-is-the-formula-to-calculate-a-synchronous-speed-of-a-motor

"f = PN/120 where f is frequency of power supply, P is number of poles in the machine and N = speed in rpm.

So, Synchronous speed N = 120f/P

Thus for 50 Hz supply, 2 pole machine; synchronous speed = 3000rpm."

N = 120f/P = 2941.17[A] = ((120[B])(C))/2[D]

A=2941.17

B=120

D=2

C=XX.XX <——need to determine the desired AC Frequency at 2500rpm

A=((B)(C))/D

therefore:

C=(AD)/B

therefore:

C=((2941.17)(2))/120

C=49.01HZ

therefore:

C=Desired AC Frequency at Present RPM = 49.01hz <—— desired 49.01hz frequency at 2500rpm @ 2 poles @ desired 85% of synchronous RPM

therefore:

J=Present RPM = 2500rpm <——Present RPM

A=“Synchronous” RPM @ Present Desired AC Frequency = 2941.17rpm

B=120

C=Desired AC Frequency at Present RPM= 49.01hz <— Desired AC Frequency

D=Poles = 2

E=Present RPM Desired % of Synchronous RPM @ Present AC Frequency = 0.85 = 85/100 = 85%

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

Simply, in theory, If the rotor is turning 2500rpm, and there are 2 poles, and we want the present RPM to be 85% of the "Synchronous" RPM, the desired AC Frequency should be set to 49.01hz. At full throttle, if wattage control is used the electrical wattage should be equivalent to Setting F - minimum wattage setting, and % throttle is % motor amps as defined by F.

### Re: "Peak Efficiency" Control Mode?

You're not reading my posts, but only quoting them in full, I'm going to stop reading your posts too.

### Re: "Peak Efficiency" Control Mode?

rew wrote:Suppose things are not frictionless, but there is a constant drag-force. Now we might want to travel 1km. and the firiction might be 10N. This costs us an extra A in current to overcome. How does this change the strategy of "as little current as possible?

devin wrote:If the rider chooses a higher final cruising speed, more mechanical watt hours per kilometer will be expended moving air, and then kilometers per electrical kilowatt hour will decrease, while electrical to mechanical conversion efficiency during acceleration is preserved at some desired minimum, for example 85% electrical watts to mechanical.

rew wrote:You're not reading my posts, but only quoting them in full, I'm going to stop reading your posts too.

@rew long story short (in reply to your question), I don't think adding drag force changes "the strategy" because whether the rider decides to go faster or slower, the electrical to mechanical conversion efficiency can remain effectively the same at most operating rpms, and at faster speeds, more joules per km are necessary for overcoming air resistance (which is by far the predominant oppositional force).

simply electrical to mechanical conversion efficiency can remain effectively constant regardless of speed, but higher speeds require more mechanical energy per km to maintain, when factoring wind resistance.

### Re: "Peak Efficiency" Control Mode?

Efficiency Control Via Watt Control Via Duty Cycle Control Without Motor KV Knowledge?

Where:

K= 90% = Desired Electrical to Mechanical Conversion Efficiency Setting

L= 500w = Desired Minimum Available Wattage Setting

P= 2000w = Desired Maximum Available Wattage Setting

&

P>=L

G= 50V = Pack Voltage

F= 0.1ohm = Winding Resistance

I= 100kv = Motor KV

J= 2200rpm = Present RPM

D= 22V = Present Back Emf V

M= 100% = Throttle Setting

A= XX.XXXA = Battery Amps

B= XX.XXXA = Motor Amps

C= XX.XXX% = Duty Cycle

E= XX.XXXV = Effective PWM Voltage

H= XX.XXXw = Throttled Wattage

N= XX.XXXw = Full Throttle Wattage

D=(1/I)*J <---------Calculating Back EMF Voltage from Motor KV and Present RPM

&

E=D/(K/100) <---------Calculating Effective PWM Voltage from Present Back EMF Voltage and Desired Efficiency Setting

^If Back EMF V is known - Motor KV & RPM knowledge is not necessary

&

if E>G then E=G <---------Ensuring Effective PWM Voltage is less than or equal to Battery Pack Voltage

&

if E>(G*(0.95)) then E=(G*(0.95)) <---------(Optional) Ensuring Effective PWM Voltage results in duty cycle less than or equal to VESC max duty cycle 95%

&

C=(E/G)*100 <---------Calculating Duty Cycle % from Effective PWM Voltage and Pack Voltage

&

B=(E-D)/F <---------Motor Amps From Effective PWM Voltage, Present Back EMF Voltage & Winding Resistance

&

A=B*(C/100) <---------Calculating Battery Amps from Motor Amps and Duty Cycle

&

N=A*G <---------Calculating Full Throttle Desired Wattage from Battery Amps and Pack Voltage

&

if N<L then N=L <---------Ensuring Full Throttle Desired Efficiency Wattage is greater than or equal to Minimum Wattage Setting

&

if N>P then N=P <---------Ensuring Full Throttle Desired Efficiency Wattage is less than or equal to Maximum Wattage Setting

&

H=N*(M/100) <---------Calculating Desired Wattage From Throttle Position and Full Throttle Desired Wattage

&

A=H/G <---------Re-calculating Battery Amps From Desired Wattage and Pack Voltage

&

E=(1/2)(sqrt((4*A*F*G)+D^2)+D) <---------Re-calculating Effective PWM Voltage From Battery Amps, Winding Resistance, Pack Voltage & Present Back EMF

&

C=(E/G)*100 <---------Re-calculating Duty Cycle % from Effective PWM Voltage & Pack Voltage

&

B=A/(C/100) <---------Re-calculating Motor Amps from Battery Amps & Duty Cycle

&

repeat

therefore:

K= 90% = Desired Efficiency Setting

L= 500w = Desired Minimum Available Wattage Setting

P= 2000w = Desired Maximum Available Wattage Setting

G= 50V = Pack Voltage

F= 0.1ohm = Winding Resistance

I= 100kv = Motor KV

J= 2200rpm = Present RPM

D= 22V = Present Back Emf V

M= 100% = Throttle Setting

A= 11.95061A = Battery Amps

B= 24.44444A = Motor Amps

C= 48.88888% = Duty Cycle

E= 24.44444V = Effective PWM Voltage

H= 597.5305w = Throttled Wattage

N= 597.5305w = Full Throttle Wattage

&

597.5305W Electrical = 11.95061 battery amps * 50V <-------- 597.5305W Electrical In

537.78673w = 2.33419Nm*2200rpm/9.5488 <------ 537.78673W Mechanical Output @ 2200rpm @ 24.44444V Effective Volts @ 100kv @ 0.1ohm

(537.78673W/597.5305W)*100=90.00% <----------- 90.00% conversion efficiency of electrical to mechanical watts.

proof:

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

2200rpm*0.01v=22V <---- 22V present back emf voltage at 2200rpm

24.44444V-22V=2.44444V <------- Effective Voltage minus back emf voltage equals 2.44444 net volts

24.44444A=2.44444V/0.1ohm <---- 24.44444A Motor Amps @ 2200rpm

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

24.44444A Motor Amps * 0.09549NM/A = 2.33419Nm <----- 2.33419 Newton Meters Torque at 24.44444A Motor Amps

537.78673w = 2.33419Nm*2200rpm/9.5488 <------ 537.78673W Mechanical Output @ 2200rpm @ 24.44444V Effective Volts @ 100kv @ 0.1ohm

Assuming 50V battery pacK

(24.44444V Effective V/50V Pack V)*100= 48.88888% duty cycle <------------ 24.44444V Effective Volts is 48.88888% duty cycle w/ 50V Battery Pack

&:

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

therefore:

11.95061 battery amps = 24.44444 motor amps * (%48.88888 duty cycle/100) <------------ 24.44444 motor amps @ 48.88888% duty cycle is 11.95061 battery amps

597.5305W Electrical = 11.95061 battery amps * 50V <-------- 597.5305W Electrical In

537.78673w = 2.33419Nm*2200rpm/9.5488 <------ 537.78673W Mechanical Output @ 2200rpm @ 24.44444V Effective Volts @ 100kv @ 0.1ohm

(537.78673W/597.5305W)*100=90.00% <----------- 90.00% conversion efficiency of electrical to mechanical watts.

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

Simply in theory, without knowing the motor's KV or rpm, it should still be possible to implement efficiency control based only on Back EMF readings, possibly by utilizing watt control via duty cycle control.

Where:

K= 90% = Desired Electrical to Mechanical Conversion Efficiency Setting

L= 500w = Desired Minimum Available Wattage Setting

P= 2000w = Desired Maximum Available Wattage Setting

&

P>=L

G= 50V = Pack Voltage

F= 0.1ohm = Winding Resistance

I= 100kv = Motor KV

J= 2200rpm = Present RPM

D= 22V = Present Back Emf V

M= 100% = Throttle Setting

A= XX.XXXA = Battery Amps

B= XX.XXXA = Motor Amps

C= XX.XXX% = Duty Cycle

E= XX.XXXV = Effective PWM Voltage

H= XX.XXXw = Throttled Wattage

N= XX.XXXw = Full Throttle Wattage

D=(1/I)*J <---------Calculating Back EMF Voltage from Motor KV and Present RPM

&

E=D/(K/100) <---------Calculating Effective PWM Voltage from Present Back EMF Voltage and Desired Efficiency Setting

^If Back EMF V is known - Motor KV & RPM knowledge is not necessary

&

if E>G then E=G <---------Ensuring Effective PWM Voltage is less than or equal to Battery Pack Voltage

&

if E>(G*(0.95)) then E=(G*(0.95)) <---------(Optional) Ensuring Effective PWM Voltage results in duty cycle less than or equal to VESC max duty cycle 95%

&

C=(E/G)*100 <---------Calculating Duty Cycle % from Effective PWM Voltage and Pack Voltage

&

B=(E-D)/F <---------Motor Amps From Effective PWM Voltage, Present Back EMF Voltage & Winding Resistance

&

A=B*(C/100) <---------Calculating Battery Amps from Motor Amps and Duty Cycle

&

N=A*G <---------Calculating Full Throttle Desired Wattage from Battery Amps and Pack Voltage

&

if N<L then N=L <---------Ensuring Full Throttle Desired Efficiency Wattage is greater than or equal to Minimum Wattage Setting

&

if N>P then N=P <---------Ensuring Full Throttle Desired Efficiency Wattage is less than or equal to Maximum Wattage Setting

&

H=N*(M/100) <---------Calculating Desired Wattage From Throttle Position and Full Throttle Desired Wattage

&

A=H/G <---------Re-calculating Battery Amps From Desired Wattage and Pack Voltage

&

E=(1/2)(sqrt((4*A*F*G)+D^2)+D) <---------Re-calculating Effective PWM Voltage From Battery Amps, Winding Resistance, Pack Voltage & Present Back EMF

&

C=(E/G)*100 <---------Re-calculating Duty Cycle % from Effective PWM Voltage & Pack Voltage

&

B=A/(C/100) <---------Re-calculating Motor Amps from Battery Amps & Duty Cycle

&

repeat

therefore:

K= 90% = Desired Efficiency Setting

L= 500w = Desired Minimum Available Wattage Setting

P= 2000w = Desired Maximum Available Wattage Setting

G= 50V = Pack Voltage

F= 0.1ohm = Winding Resistance

I= 100kv = Motor KV

J= 2200rpm = Present RPM

D= 22V = Present Back Emf V

M= 100% = Throttle Setting

A= 11.95061A = Battery Amps

B= 24.44444A = Motor Amps

C= 48.88888% = Duty Cycle

E= 24.44444V = Effective PWM Voltage

H= 597.5305w = Throttled Wattage

N= 597.5305w = Full Throttle Wattage

&

597.5305W Electrical = 11.95061 battery amps * 50V <-------- 597.5305W Electrical In

537.78673w = 2.33419Nm*2200rpm/9.5488 <------ 537.78673W Mechanical Output @ 2200rpm @ 24.44444V Effective Volts @ 100kv @ 0.1ohm

(537.78673W/597.5305W)*100=90.00% <----------- 90.00% conversion efficiency of electrical to mechanical watts.

proof:

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

2200rpm*0.01v=22V <---- 22V present back emf voltage at 2200rpm

24.44444V-22V=2.44444V <------- Effective Voltage minus back emf voltage equals 2.44444 net volts

24.44444A=2.44444V/0.1ohm <---- 24.44444A Motor Amps @ 2200rpm

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

24.44444A Motor Amps * 0.09549NM/A = 2.33419Nm <----- 2.33419 Newton Meters Torque at 24.44444A Motor Amps

537.78673w = 2.33419Nm*2200rpm/9.5488 <------ 537.78673W Mechanical Output @ 2200rpm @ 24.44444V Effective Volts @ 100kv @ 0.1ohm

Assuming 50V battery pacK

(24.44444V Effective V/50V Pack V)*100= 48.88888% duty cycle <------------ 24.44444V Effective Volts is 48.88888% duty cycle w/ 50V Battery Pack

&:

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

therefore:

11.95061 battery amps = 24.44444 motor amps * (%48.88888 duty cycle/100) <------------ 24.44444 motor amps @ 48.88888% duty cycle is 11.95061 battery amps

597.5305W Electrical = 11.95061 battery amps * 50V <-------- 597.5305W Electrical In

537.78673w = 2.33419Nm*2200rpm/9.5488 <------ 537.78673W Mechanical Output @ 2200rpm @ 24.44444V Effective Volts @ 100kv @ 0.1ohm

(537.78673W/597.5305W)*100=90.00% <----------- 90.00% conversion efficiency of electrical to mechanical watts.

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

Simply in theory, without knowing the motor's KV or rpm, it should still be possible to implement efficiency control based only on Back EMF readings, possibly by utilizing watt control via duty cycle control.

### Re: "Peak Efficiency" Control Mode?

here's a graph of the output of the efficiency control algorithm assuming 90% desired conversion efficiency, 500 min available watts, 2500 max available watts, 50v, 0.05ohm winding, 84mm tire diameter, 90kv, 1:1 gear ratio (hub motor), 100% throttle, 95% max duty cycle & 120a max motor amps:

### Who is online

Users browsing this forum: Baidu [Spider], Bing [Bot] and 2 guests