"Peak Efficiency" Control Mode?

[devin]

If you keep talking about kinetic energy in electrons, one day I'll calculate that kinetic energy for you. It's going to be less than you expect. For all practical purposes the kinetic energy in the electrons is zero. You should also stop talking about 100% duty at stall.

The hobbyking motor I linked simply allows 1.5V of effective voltage above the BEMF. So if the rotor is stalled, you're allowed 1.5V or 3% duty cylce (at 50V battery voltage). If the rotor is doing 150RPM, (1V BEMF) you're allowed to do 2.5V or 5% dutycycle.

@rew If we take the number of copper free electrons in a typical motor winding, and send them on a straight line trajectory relative to the earth at their fermi velocity (1570km/s), how much kinetic energy is this?

@rew In theory, the stored kinetic energy in the free electrons of 10 cubic centimeters of copper is equivalent to the kinetic energy at the muzzle of about 63 x .50 caliber rifle bullets or about 1688 x .45 caliber handgun bullets. This same amount of kinetic energy is also equivalent to about 264 watt hours, or about 5AH from a 50V battery pack.

——————————

0.0000007725 kgs free electrons in 10 cubic centimeters copper (nearly a milligram or 1/1000th of a gram or roughly equivalent to a single grain of sugar)

1570000 meters per second = V = fermi velocity of copper free electrons

KE=(1/2)*(0.0000007725*(1570000*1570000))

KE = 952067.625 joules

—————

Math:

10 cubic centimeters copper

1,000,000 cubic centimeters per cubic meter

10 cubic centimeters is 1/100,000th of a cubic meter

8.49x10^28 free electrons per m^3 copper

(1/100000)*(8.49x10^28)=8.49x10^23 free electrons in 10 cubic centimeters copper

8.49x10^23 free electrons * 9.10*10^-31 kg mass per electron

(8.49*10^23)*(9.10*10^-31) = 7.725*10^(-7) kgs free electrons in 10cc’s copper

7.725*10^(-7)= 0.0000007725 kgs free electrons in 10cc’s copper

fermi velocity of copper free electrons = 1570000 meters per second = V

M = 0.0000007725 kgs

KE = 1/2 (M * (V * V))

KE=(1/2)*(0.0000007725*(1570000*1570000))

KE = 952067.625 joules

——————————

952067 joules / 15037 joules per .50 cal rifle bullet = 63.31 x .50 cal rifle bullets equivalent Kinetic Energy

952067 joules / 564 joules per .45 handgun bullet = 1688 x .45 bullets equivalent Kinetic Energy

Source: https://en.wikipedia.org/wiki/Muzzle_energy

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Image Source: https://www.survivalmonkey.com/threads/short-range-media-penetration-by-50-bmg-ap-rounds.49556/

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*Also, the kinetic energy which is in theory stored in the roughly single-sugar-grain-sized mass of free electrons in 10 cubic centimeters of copper, 952067 joules, is the same amount of energy transferred by a human-sized 80kg object striking a brick wall at 345.10 miles per hour, 144.277 meters per second, or 555.39 kilometers per hour.

[Addy]

. For all practical purposes the kinetic energy in the electrons is zero.

This is the most important part. Devin, you can keep spouting calculations but what would actually be interesting is if you put some of that effort towards implementing your "peak efficiency" control mode with a custom VESC app or something.

[rew]

Devin, your calculations/simulations of a while back show your "max efficiency" method reducing the throttle value at certain points in the trip. Thus the guy riding "your algorithm" will be at work later than the guy riding "full trhottle". I propose adding a third virtual rider to the comparison. Mine gets a reduced "full throttle" (i.e. max motor amps) setting, so that in the end he arrives at the next stop sign at the same time as your "max efficiency" rider.

Intuition tells me that he'll use only very little more juice than your rider, or even less.

[devin]

Devin, your calculations/simulations of a while back show your "max efficiency" method reducing the throttle value at certain points in the trip. Thus the guy riding "your algorithm" will be at work later than the guy riding "full trhottle". I propose adding a third virtual rider to the comparison. Mine gets a reduced "full throttle" (i.e. max motor amps) setting, so that in the end he arrives at the next stop sign at the same time as your "max efficiency" rider.

Intuition tells me that he'll use only very little more juice than your rider, or even less.

@rew — both riders are using “full throttle” (different algorithms), but, yes, in the previously discussed scenario, “my rider” arrives about 4 seconds later at the 400 meter mark, traveling approximately the same speed at this line as the other rider. if they are headed towards a stop light they might arrive about the same time at work, assuming each red light lasts longer than 4 seconds. the scenario we discuss requires both riders to use full throttle the entire 400 meters, and both riders to be capable of 30mph on flat ground.

according to the wind drag and vehicle thrust equations, maintaining 30mph on flat ground with the 4 motors in the scenario requires no less than ~11a motor amps @ 30mph.

So if the third rider reduces their motor amp setting below 11a motor amps in an attempt to improve efficiency, they’d be disqualified from the contest for being unable to demonstrate that their board is able to reach 30mph...

[remember in the scenario the customer states his requirements include at least 30mph on flat ground, and greatest range while repeatedly accelerating full throttle in start and stop traffic.]



^notice both boards use identical wattage and motor amps at 30mph, both rider’s “mechanical watts” are just “a hair above” the wind watts at 30mph, but efficiency control uses less watt hours per mile by: A) converting a greater % of electrical watts to mechanical watts than the classical algorithm & B) taking a little more time to accelerate to the same top speed (the difference in times between the 2 riders at 400 meters is about 4 seconds, but they both cross the 400 meter mark at approximately the same speed— 30mph)

The 30mph-capable rider with “efficiency control” gets 148.25% as much range in start and stop traffic with stop signs placed 183.5 meters apart compared to the 30mph-capable “classical algorithm” rider, while both use full throttle acceleration for the first 150 meters of each acceleration cycle, followed by mechanical braking.

Efficiency Control: 51.62 kilometers = 32.07 miles per kilowatt hour

Classical Algorithm: 34.81 kilometers = 21.63 miles per kilowatt hour

^In the 183.5 meter stop-separation distance scenario, “efficiency control” gives “my rider” nearly 50% greater “guaranteed range” without sacrificing any of her top speed.

[rew]

[remember in the scenario the customer states his requirements include at least 30mph on flat ground, and greatest range while repeatedly accelerating full throttle in start and stop traffic.]

So, your rider is allowed to press full-throttle, but the motor controller will, in the interest of efficiency reduce the throttle setting, but I'm not allowed to suggest another strategy that also reduces the throttle a bit. Weird. It's difficult to argue with you when you always change the rules in your favor for each argument that is provided.

Similarly you introduce new bullshit with every posting you do. I said so a while ago, and didn't follow through, but i'm giving up now. really.

[devin]

So, your rider is allowed to press full-throttle, but the motor controller will, in the interest of efficiency reduce the throttle setting, but I'm not allowed to suggest another strategy that also reduces the throttle a bit.Weird. It's difficult to argue with you when you always change the rules in your favor for each argument that is provided.

Similarly you introduce new bullshit with every posting you do. I said so a while ago, and didn't follow through, but i'm giving up now. really.

@rew it isn’t correct and is a bit misleading to say my algorithm reduces the “throttle” ... both algorithms ultimately constantly adjust the duty cycle, but in this scenario the throttle setting is identical. the throttle i’m talking about is the remote the riders hold in their hands and I model a scenario where both are pushing the throttle to 100%. if your rider is allowed to make throttle position adjustments, then mine is allowed to make identical throttle adjustments and I think my rider will still use less watt hours per mile if permitted to use the same throttle positions as your rider in the test. For example if both riders use 50% throttle position, why would the outcome be any different?

@rew Test Question: I'm an electric skateboard vendor, and a previous customer asks me what options they have to achieve greatest possible range and efficiency on their electric skateboard when commuting in start and stop city traffic. The customer's route to work features many stop signs and stop lights, so they start and stop very frequently, but they live in a completely flat area and don't expect to encounter any hills. The board we are discussing has a battery which typically runs at 45.98V, (4) 81.42kv hub motors which are 0.136ohms and has 83mm diameter tires. The customer states their only requirements are they want to ensure the board is capable of 30mph top speed on flat ground, and aside from that requirement, they also want highest possible range and electrical to mechanical conversion efficiency while repeatedly accelerating at full throttle during their start-and-stop morning commute. Should I recommend "efficiency control" or the "classical algorithm" to achieve this customer's requirements (at least 30mph top speed on flat ground and greatest possible range & conversion efficiency while repeatedly accelerating at full throttle in start and stop city commuter traffic.)

@rew in the mathematical scenario we discussed the customer is requesting “capable of 30mph top speed on flat ground, and aside from that requirement, they also want highest possible range and electrical to mechanical conversion efficiency while repeatedly accelerating at full throttle during their start-and-stop morning commute...”

...so the outlined scenario specifies full throttle...

you can clearly see in my graphs that at all times both rider’s throttle positions are “100%”

@rew do you believe that if both riders use 50% throttle that the outcome will be any different?

also please note that at no time in the scenario does my algorithm “decrease” the electrial wattage. If you look at the plot of the “electrical watts” my algorithm puts out its either increasing or remaining steady, which is also why I think it’s misleading to characterize my algorithm as performing a “throttle reduction” — if you think this is not correct show me where in the 400 meter scenario does my algorithm ever decrease either electrical watts or duty cycle? i think it doesn’t happen.

if by “throttle reduction” you simply mean my algorithm uses less watts hours to go the same distance without sacrificing any top speed, & without ever “lowering” electrical wattage or duty cycle, while both riders physically push their remote control adjusters to the 100% position, then it is correct.

[ThierryGTLTS]

I would be curious to test your algorithm on a real track.

Have a Nice Day.

Thierry

[devin]

Where:

M= 100 = Throttle % Setting
K= 90 = Desired Efficiency % Setting
L= 500 = Desired Min Watts Available Setting
P= 4500 = Desired Max Watts Available Setting

G= 48.2 = Battery Voltage
D= 16.94 = Back EMF Voltage
F= 0.025 = Winding Resistance Ohms
Y= 120 = Max Motor Amps
Z= 95 = Max Duty Cycle %

N= XX.XXXw = Desired Full Throttle Wattage
C= XX.XXX% = Duty Cycle

As a side note, if we wish to convert the C=Duty Cycle algorithm output to E= XX.XXXv = PWM Effective Voltage

We take:

C= 39.049 = Duty Cycle %

and add the following line before the "repeat":

E=G*(C/100)

Therefore:

E=G*(C/100)

E=48.2*(39.049/100)

18.821618=48.2*(39.049/100)

E = 18.821618 = PWM Effective Voltage <--39.049% Duty Cycle is 18.821618 PWM Effective Voltage

N=L
&
if D>((sqrt(F)*K*sqrt(L))/(10*sqrt(100-K))) then N=(-1)*((100*(D^2)*(K-100))/(F*(K^2)))
&
if N>P then N=P
&
if Y<((sqrt((D^2)+(4*F*N))-D)/(2*F)) then N=Y*(D+(F*Y))
&
if Z<((50*(sqrt((D^2)+(4*F*N))+D))/G) then N=(G*Z*(G*Z-(100*D)))/(10000*F)
&
C=10*((sqrt((25*(D^2))+(F*M*N))/G)+((5*D)/G))
&
E=G*(C/100)
&
repeat

Therefore:

E = 18.821618 = PWM Effective Voltage

Throttle Response Update:

In theory this update to the code improves the dynamic throttle response produced by the algorithm at times when full throttle would result in the motor amp limit setting or duty cycle limit setting being reached.

Where:

M= 100 = Throttle % Setting
K= 90 = Desired Efficiency % Setting
L= 500 = Desired Min Watts Available Setting
P= 4500 = Desired Max Watts Available Setting

G= 48.2 = Battery Voltage
D= 16.94 = Back EMF Voltage
F= 0.025 = Winding Resistance Ohms
Y= 120 = Max Motor Amps
Z= 95 = Max Duty Cycle %

N= XX.XXXw = Desired Full Throttle Wattage
C= XX.XXX% = Duty Cycle

N=L
&
if D>((sqrt(F)*K*sqrt(L))/(10*sqrt(100-K))) then N=(-1)*((100*(D^2)*(K-100))/(F*(K^2)))
&
if N>P then N=P
&
if Y<((sqrt((D^2)+(4*F*N))-D)/(2*F)) then N=Y*(D+(F*Y))
&
if Z<((50*(sqrt((D^2)+(4*F*N))+D))/G) then N=(G*Z*(G*Z-(100*D)))/(10000*F)
&
C=10*((sqrt((25*(D^2))+(F*M*N))/G)+((5*D)/G))
&
repeat


Therefore:

M= 100% = Throttle % Setting
K= 90% = Desired Efficiency % Setting
L= 500w = Desired Min Watts Available Setting
P= 4500w = Desired Max Watts Available Setting

G= 48.2v = Battery Voltage
D= 16.94v = Back EMF Voltage
F= 0.025ohm = Winding Resistance Ohms
Y= 120a = Max Motor Amps
Z= 95% = Max Duty Cycle %

N= 1417.10w = Desired Full Throttle Wattage
C= 39.049% = Duty Cycle

[devin]

Where:
D= 16.94 = Back EMF Voltage

E = 18.821618 = PWM Effective Voltage

Simply because 16.94V Back EMF Voltage is 90% of the applied 18.821618V PWM Effective Voltage, it follows that instantaneous electrical to mechanical conversion efficiency is also 90%.

K= 90 = Desired Efficiency % Setting

M= 100 = Throttle % Setting