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

Posted:

**26 Sep 2017, 15:31**A forum for my open software and open hardware projects

http://vedder.se/forums/

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Posted: **26 Sep 2017, 15:31**

Posted: **28 Sep 2017, 15:11**

I updated the code to account for the fact that max duty cycle is 95% and max motor amps is 120a... by my calculations a single hub motor+vesc could conceivably propel a human to 40mph at more than 90% conversion efficiency with the following assumptions...

Efficiency Control Program

K= 90.00% = Desired Efficiency % Setting

L= 500.00w = Desired Min Watts Available Setting

P= 4500.00w = Desired Max Watts Available Setting

Y= 120.00a = Max Motor Amps Setting

Z= 95.00% = Max Duty Cycle % Setting

M= 100.00% = Throttle % Setting

G= 48.2v = Battery Voltage

F= 0.025ohm = Winding Resistance

I= 120kv = Motor KV

J= 2032.81rpm = Present RPM

A= XX.XXXa = Battery Amps

B= XX.XXXa = Motor Amps

C= XX.XXX% = Duty Cycle

D= XX.XXXv = Back EMF Voltage

E= XX.XXXv = PWM Voltage

H= XX.XXXw = Electrical Wattage

N= XX.XXXw = Desired Full Throttle Wattage

Where:

D=(1/I)*J

&

E=0

&

if D>0 then E=D/(K/100)

&

if E>(G*(Z/100)) then E=G*(Z/100)

&

N=0

&

if E>0 then N=(E*(E-D))/F

&

if N<L then N=L

&

if N>P then N=P

&

H=N*(M/100)

&

A=H/G

&

E=(1/2)(sqrt((4*A*F*G)+(D^2))+D)

&

if E>G then E=G

&

if E>(G*(Z/100)) then E=(G*(Z/100))

&

B=(E-D)/F

&

if B>Y then B=Y

&

E=(B*F)+D

&

C=(E/G)*100

&

A=B*(C/100)

&

H=A*G

Therefore:

20mph @ 84mm tire diameter @ 1:1 gearing:

A= 29.40a = Battery Amps

B= 75.28a = Motor Amps

C= 39.05% = Duty Cycle

D= 16.94v = Back EMF Voltage

E= 18.82v = PWM Voltage

F= 0.025ohm = Winding Resistance

G= 48.2v = Battery Voltage

H= 1417.12w = Electrical Wattage

I= 120kv = Motor KV

J= 2032.81rpm = Present RPM

K= 90% = Desired Efficiency % Setting

L= 500w = Desired Min Watts Available

M= 100% = Throttle % Setting

N= 1417.12 = Desired Full Throttle Wattage

P= 4500w = Desired Max Watts Available

Y= 120a = Max Motor Amps

Z= 95% = Max Duty Cycle %

Wind Watts Assumptions:

1.2 = Standing Human Estimated Drag Coefficient

0.929m^2 = Standing Human Estimated Frontal Area

1.225kg/m^3 = Fluid Density of Air

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

In theory simply, a single hub motor and vesc using efficiency control could conceivably propel a human rider to 40mph at greater than 90% electrical to mechanical conversion efficiency.

Posted: **28 Sep 2017, 19:25**

Posted: **29 Sep 2017, 19:59**

looking at the grin tech motor simulator you can see when the load is increased the efficiency drops with every motor on there regardless of the power applied. having the esc programmed to only apply a voltage within a close proximity of the back voltage, which would limit the power, still has the motor at much lower efficiency than 90 percent with a big load

http://www.ebikes.ca/tools/simulator.ht ... se&grade=9

http://www.ebikes.ca/tools/simulator.ht ... se&grade=9

Posted: **29 Sep 2017, 20:40**

Posted: **29 Sep 2017, 22:01**

@hummie in the first picture the assumption is that you are at 42% throttle going a steady 15.4mph at 0% grade & this steady speed requires 3.4a battery amps (most likely they’ve factored wind drag at 0% grade 15.4mph). in the second picture the assumption is that you are at 42% throttle going 8.3mph and accelerating at 2.66mph/s up a 12% grade. Traveling at 8.3mph and accelerating at 2.66mph/s at 42% throttle up a 12% grade would require 19.3a battery amps for the load, according to the simulator.

if you want to have identical conversion efficiency at 8.3mph & 15.4mph, you have to draw less battery amps at 8.3mph than at 15.4mph, not more, and this fact means you won’t be going up a 12% grade at 8.3mph & accelerating at 2.66mph/s if you are achieving identical electrical to mechanical conversion efficiency at both speeds, and drawing only 3.4a battery amps at 15.4mph, simply.

(but you can achieve identical conversion efficiency at both speeds by drawing some amount less than 3.4a battery amps at 8.3mph assuming you draw exactly 3.4a battery amps at 15.4mph)

if you want to have identical conversion efficiency at 8.3mph & 15.4mph, you have to draw less battery amps at 8.3mph than at 15.4mph, not more, and this fact means you won’t be going up a 12% grade at 8.3mph & accelerating at 2.66mph/s if you are achieving identical electrical to mechanical conversion efficiency at both speeds, and drawing only 3.4a battery amps at 15.4mph, simply.

(but you can achieve identical conversion efficiency at both speeds by drawing some amount less than 3.4a battery amps at 8.3mph assuming you draw exactly 3.4a battery amps at 15.4mph)

Posted: **30 Sep 2017, 07:35**

The ebikes.ca simulator is great !

By default, it tells you what speed you get with 0 acceleration. When you slide the speed cursor, it will tell you the grade it would require to reach that speed OR the acceleration you would get if the grade was to stay at 0%: notice that 12.1% grade is the same as 2.66mph/s

2.66*1.609/3.6/9.81=0.121 or 12.1%

By default, it tells you what speed you get with 0 acceleration. When you slide the speed cursor, it will tell you the grade it would require to reach that speed OR the acceleration you would get if the grade was to stay at 0%: notice that 12.1% grade is the same as 2.66mph/s

2.66*1.609/3.6/9.81=0.121 or 12.1%

Posted: **01 Oct 2017, 15:34**

Posted: **01 Oct 2017, 17:43**

Posted: **01 Oct 2017, 18:21**