You have selected a motor with a peak efficiency of about 30%. That is not realistic. I provided links to a realistic motor with mostly the same specs as yours. You ignore me. I asked for a reference to where you found the motor with the specs you quoted. You ignore me. I told you where your specs are wrong. You ignore me.
What do you expect happens now?
"Peak Efficiency" Control Mode?
Re: "Peak Efficiency" Control Mode?
You have selected a motor with a peak efficiency of about 30%. That is not realistic. I provided links to a realistic motor with mostly the same specs as yours. You ignore me. I asked for a reference to where you found the motor with the specs you quoted. You ignore me. I told you where your specs are wrong. You ignore me.
What do you expect happens now?
I found the specs I used on the manufacturer's website: http://alienpowersystem.com/shop/brushlessmotors/alien5065outrunnerbrushlessmotor270kv2200wa/
http://alienpowersystem.com/shop/brushlessmotors/alien5065outrunnerbrushlessmotor270kv2200wa/ wrote:
MOTOR: 5065
KV: 270
MAX POWER: 2200W
WIRE WINDS: 9
MAX AMP: 60A
ESC: 80/120A
MAX VOLT: 8S
RESISTANCE (Ohm): .42
NO LOAD CURRENT: 1,5
SIZE: 50 x 65 ( without shaft )
WEIGHT (g): 0,380
SHAFT: 8mm with 3mm keyway
devin wrote:4.88 battery amps = 9.35 motor amps * (% 52.28duty cycle/100)
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.
The question then becomes, does any other duty cycle besides 52.28% provide a higher electrical to mechanical wattage conversion efficiency than 86.57% at 6000rpm?
I'll analyze a second motor in the next post...
Re: "Peak Efficiency" Control Mode?
I'm interested in using the efficiency algorithm for human propulsion, therefore for the next analysis I'll use this previous generation long hub motor which was tested and recommended by the manufacturer for performance to use up to 120 motor amps at stall if rider stays aware of temperature & keeps below 250F...
(vesc used with 200 motor amp setting had a coded 120 motor amp limit)


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
36.6V31.111V=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 wrote:I'm telling u I can start coggingfree from standstill with 200 motor and other settings I've had, 60/60 or 70/70 didnt have nearly the same low end torque or smoothness and that's with two motors! To me it's worth finding out the details on.
(vesc used with 200 motor amp setting had a coded 120 motor amp limit)


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
36.6V31.111V=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?
Last edited by devin on 02 Sep 2017, 13:35, edited 2 times in total.

 Posts: 86
 Joined: 09 Aug 2017, 11:10
Re: "Peak Efficiency" Control Mode?
devin wrote:
I found the specs I used on the manufacturer's website: http://alienpowersystem.com/shop/brushlessmotors/alien5065outrunnerbrushlessmotor270kv2200wa/
Hi Devin,
Please pay attention to the given specs
I've found many errors even in reknowned manufacturers like TMotor.
Hobbyking sells perhaps good products but is the "king" of the wrong specs.
More, VESC detection delivers sometimes R and L values far from reality, but I don't know why.
So, don't trust all what you see.
Have a Nice WE.
Thierry
Re: "Peak Efficiency" Control Mode?
devin wrote: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?
Devin, the losses are easily calculated:
* resistive losses: motor resistance * current squared.
* mechanical losses: the motor has internal friction that requires some power to overcome. The power is (almost) linear with the speed.
* eddy current losses. These go up with the square or the cube of the speed.
For most motors I've tested the friction is almost constant, leading to a power loss that is linear with the speed.
Now those eddy current losses are VERY annoying. the constant that defines those losses should be low enough that they are still negligible at the top speed.
Then... Mechanical output power is linear with the rotation speed. With our modern motors that come to within 3% of their max RPM at max voltage under max load (torque/amps) that means that they will get their max power output simply at the max amps and max volts.
Anyway. 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 and that requires 2Nm of torque from the motor. What would the normal control algorithm give? What would you suggest we try to improve efficiency? Note that you can't tell the rider to speed up to 30 km/h because that's more efficent for the motor. Similarly, you can't tell him to change the gearing on the bike/board because a 20% higher RPM would be more efficient in the motor.
By my math, you need 20A of current for the 2Nm of torque. 20A of current requires 2V of voltage across the windings, so the PWM would make the effective voltage across the motor 27V, or 54% PWM. That's it. No choices available.
If you've commuted to work 100 times and determined that you often ride at 25km/h and 2Nm of torque you can adjust the gearing for optimum efficiency at that point. But you'll lose either in maxtorqueonthewheels or in maxspeed. If you like the maxtorque and top speed of your setup, then there is not much you can do.
To improve efficiency, get a bigger motor. Say one with 0.05 Ohm resistance, One that allows 100A of current, but that you'll only load upto the 50A max of your ESC. This reduces the losses almost by a factor of two.
Re: "Peak Efficiency" Control Mode?
rew wrote:Devin, the losses are easily calculated:
* resistive losses: motor resistance * current squared.
* mechanical losses: the motor has internal friction that requires some power to overcome. The power is (almost) linear with the speed.
* eddy current losses. These go up with the square or the cube of the speed.
For most motors I've tested the friction is almost constant, leading to a power loss that is linear with the speed.
Now those eddy current losses are VERY annoying. the constant that defines those losses should be low enough that they are still negligible at the top speed.
Then... Mechanical output power is linear with the rotation speed. With our modern motors that come to within 3% of their max RPM at max voltage under max load (torque/amps) that means that they will get their max power output simply at the max amps and max volts.
Anyway. 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 and that requires 2Nm of torque from the motor. What would the normal control algorithm give? What would you suggest we try to improve efficiency? Note that you can't tell the rider to speed up to 30 km/h because that's more efficent for the motor. Similarly, you can't tell him to change the gearing on the bike/board because a 20% higher RPM would be more efficient in the motor.
By my math, you need 20A of current for the 2Nm of torque. 20A of current requires 2V of voltage across the windings, so the PWM would make the effective voltage across the motor 27V, or 54% PWM. That's it. No choices available.
If you've commuted to work 100 times and determined that you often ride at 25km/h and 2Nm of torque you can adjust the gearing for optimum efficiency at that point. But you'll lose either in maxtorqueonthewheels or in maxspeed. If you like the maxtorque and top speed of your setup, then there is not much you can do.
To improve efficiency, get a bigger motor. Say one with 0.05 Ohm resistance, One that allows 100A of current, but that you'll only load upto the 50A max of your ESC. This reduces the losses almost by a factor of two.
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
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
29.411V25V=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?
Re: "Peak Efficiency" Control Mode?
(I haven't checked if your math is correct)
Most people have a throttle like control that dictates what the torque is that the motor is supposed and allowed to deliver. In my example that throttle was set to "equivalent of 2Nm", or about 20A. If your control algorithm suddenly decides to apply 4.4Nm, then the rider will protest that he didn't order that much power.
If for example, you're trying to optimize energy use and you're accelerating at 25km/h but have a target of 30km/h that you want to reach as quickly as possible, then optimizing motor efficiency is not relevant to the problem. In my example, 2Nm was enough to stay at 25km/h. If reaching 30km/h as energyefficient as possible requires you to apply MORE than the 44A that you calculated: Only 2.4 Nm of the 4.4 can be used to go faster. If you apply say 2A more current, you'll accelerate 0.2/2.4 = 8% faster for only 4.5% more energy use.
But this: "reach 30km/h as energyefficient as possible" is seldom the target: you're going to want to get somewhere. So once you're at 30km/h very quickly and very energy efficent, you're going to have to travel longer than when you accelerate more slowly. So the strategy changes again!
The question is moot.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?
Most people have a throttle like control that dictates what the torque is that the motor is supposed and allowed to deliver. In my example that throttle was set to "equivalent of 2Nm", or about 20A. If your control algorithm suddenly decides to apply 4.4Nm, then the rider will protest that he didn't order that much power.
If for example, you're trying to optimize energy use and you're accelerating at 25km/h but have a target of 30km/h that you want to reach as quickly as possible, then optimizing motor efficiency is not relevant to the problem. In my example, 2Nm was enough to stay at 25km/h. If reaching 30km/h as energyefficient as possible requires you to apply MORE than the 44A that you calculated: Only 2.4 Nm of the 4.4 can be used to go faster. If you apply say 2A more current, you'll accelerate 0.2/2.4 = 8% faster for only 4.5% more energy use.
But this: "reach 30km/h as energyefficient as possible" is seldom the target: you're going to want to get somewhere. So once you're at 30km/h very quickly and very energy efficent, you're going to have to travel longer than when you accelerate more slowly. So the strategy changes again!
Re: "Peak Efficiency" Control Mode?
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 riderspecified 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 [or torque or motor amps] setting
F= maximum wattage [or torque or motor amps] setting
G= present wattage [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.
*except at very low rpms full throttle equates to a riderspecified 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 [or torque or motor amps] setting
F= maximum wattage [or torque or motor amps] setting
G= present wattage [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.
Last edited by devin on 06 Sep 2017, 01:24, edited 11 times in total.
Re: "Peak Efficiency" Control Mode?
So if you can manually figure out the max efficiency setting... does it differ with speed?
So now the only parameter for the function is speed: we must assume the user commands full throttle.
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. Note that most transportationoptimizations are not optimal if you optimize against time or distance. You have to optimize the whole. All transportation optimizations allow more joules for earlier arrival.
So now the only parameter for the function is speed: we must assume the user commands full throttle.
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. Note that most transportationoptimizations are not optimal if you optimize against time or distance. You have to optimize the whole. All transportation optimizations allow more joules for earlier arrival.
Re: "Peak Efficiency" Control Mode?
rew wrote:So if you can manually figure out the max efficiency setting... does it differ with speed?
So now the only parameter for the function is speed: we must assume the user commands full throttle.
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. Note that most transportationoptimizations are not optimal if you optimize against time or distance. You have to optimize the whole. All transportation optimizations allow more joules for earlier arrival.
rew wrote:So if you can manually figure out the max efficiency setting... does it differ with speed?
yes, the most efficient wattage applied to the motor varies with speed. (more wattage at higher speeds and less wattage at lower speeds)
rew wrote: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.
devin wrote: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 riderspecified 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, 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 necessaryforusability 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).
Last edited by devin on 05 Sep 2017, 13:48, edited 1 time in total.
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