### "Peak Efficiency" Control Mode?

Posted:

**28 Aug 2017, 17:17**What if there was a control mode where full throttle always equates to peak electrical to mechanical power conversion efficiency from the motor?

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Summary:

Assuming 100% duty cycle, a motor gets maximum (about 85%) conversion of electrical to mechanical power at about 85% of no load rpm.

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Logic:

full throttle applies an effective voltage (A) such that the present rpm (B) is 85% of (C) motor KV times (A) the present effective voltage.

B = (85/100)(AC)

therefore:

A = (20B)/(17C)

assuming the motor is 100kv, battery 50V, present rpm is 2500rpm:

A = XX.XXV = Full Throttle Effective Voltage

B = 2500rpm = Present RPM

C = 100kv = Motor KV

D = 50v = Battery Voltage

B = (85/100)(AC)

2500 = (85/100)(A*100)

therefore:

A = (20B)/(17C)

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

29.41V = (20*2500)/(17*100)

therefore:

A = 29.41V = Full Throttle Effective Voltage

B = 2500rpm = Present RPM

C = 100kv = Motor KV

D = 50v = Battery Voltage

E = XX.XX% duty = Duty Cycle

&:

(A/D)*100 = E

(29.41/50)*100=58.82

therefore:

A = 29.41V = Full Throttle Effective Voltage

B = 2500rpm = Present RPM

C = 100kv = Motor KV

D = 50v = Battery Voltage

E = 58.82% duty = Duty Cycle

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Theory: If the motor KV is 100kv, the battery voltage is 50V, and the duty cycle is 58.82%, and the present RPM is 2500rpm, then the motor is running very close to peak electrical to mechanical conversion efficiency.

Simply if the duty cycle is always maintained such that the present rpm is always 85% of present effective voltage times the KV, then in theory in this control mode the motor should always be running at peak electrical to mechanical conversion efficiency.

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

Summary:

Assuming 100% duty cycle, a motor gets maximum (about 85%) conversion of electrical to mechanical power at about 85% of no load rpm.

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

Logic:

full throttle applies an effective voltage (A) such that the present rpm (B) is 85% of (C) motor KV times (A) the present effective voltage.

B = (85/100)(AC)

therefore:

A = (20B)/(17C)

assuming the motor is 100kv, battery 50V, present rpm is 2500rpm:

A = XX.XXV = Full Throttle Effective Voltage

B = 2500rpm = Present RPM

C = 100kv = Motor KV

D = 50v = Battery Voltage

B = (85/100)(AC)

2500 = (85/100)(A*100)

therefore:

A = (20B)/(17C)

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

29.41V = (20*2500)/(17*100)

therefore:

A = 29.41V = Full Throttle Effective Voltage

B = 2500rpm = Present RPM

C = 100kv = Motor KV

D = 50v = Battery Voltage

E = XX.XX% duty = Duty Cycle

&:

(A/D)*100 = E

(29.41/50)*100=58.82

therefore:

A = 29.41V = Full Throttle Effective Voltage

B = 2500rpm = Present RPM

C = 100kv = Motor KV

D = 50v = Battery Voltage

E = 58.82% duty = Duty Cycle

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

Theory: If the motor KV is 100kv, the battery voltage is 50V, and the duty cycle is 58.82%, and the present RPM is 2500rpm, then the motor is running very close to peak electrical to mechanical conversion efficiency.

Simply if the duty cycle is always maintained such that the present rpm is always 85% of present effective voltage times the KV, then in theory in this control mode the motor should always be running at peak electrical to mechanical conversion efficiency.