A Brief Word on Batteries

[Jason McNeil]

I had a Monster Energy drink this morning & so embarked on the ambitious project of completing the 'Choosing the Right Battery Pack' Blog  article. Euphoria has since worn off, will adding additional text tomorrow... 

https://www.ewheels.com/choosing-an-electric-unicycle-with-the-right-battery-pack-for-you/

There are now dozens of different battery cells being made by the big manufacturers LG, Samsung, Panasonic, & Sony. For use in an application like an Electric Unicycle, the two qualities that make a particular cell more suitable over another are it's power output (Amps) & energy density (Ah).

Until fairly recently there was an inevitable trade-off between either high-powered cells (e.g. Sony VC3/A123) and energy dense cells—there's the example of the Panasonic NCR18650B, which had capacity, but could only deliver 2-3 sustained Amps. 

What has been one of the key drivers in powering the Personal Electric Vehicle revolution, is the introduction of >3Ah (>10Wh) batteries which can ALSO capable of sustaining high currents above 10 Amps.

The Battery cells which are most commonly found in Electric Unicycles are shown above. Although lacking in capacity & somewhat outdated, the Samsung 22Ps are still quite common for many Self-Balancing devices because the cost per cell is very low.

Ideally, a smaller battery pack (<32 cells) would have more powerful cells, like the HG2 (found in our small 12" IPS a130s) or the VC3 (used in the Solowheel) to provide the necessary sustained power even, if the speeds are modest.

Because the cells are nearly all capable of the same 10A power output, the calculation then becomes one of how many battery cells you need for your requirements of speed & Rider weight—hill climbing expectations are also important factor for power, but to keep it simple, it's has not been shown here.

These figures are probably on the conservative side, but for a device where getting enough power is the vital for survival, erring on the side caution is preferable. 

This next illustration shows how the battery cells are configured, their capacity & power outputs in several popular Electric Unicycles. There are two defacto standardized voltages for most Wheels: 67.2v, which is comprised of 16 cells in a series & 84.2v, with 20 cells in series. Good arguments exist for both voltages, but for higher performance Wheels, the increasing trend is converging towards the higher 84.2 voltage. This is mainly on account that motors of  higher voltages are capable of higher speeds with better brown-out safety margins.  

[meepmeepmayer]

Interesting overview of what each wheel uses. The color coding is neat. Guess this post a good way to get people buy bigger wheels from you, that's kind of the conclusion.

2 quick nitpicky questions, just to be sure:

• When computing Wh capacity, people here usually use 3.7V nominal voltage for each cell (e.g. 16*4*3.7V*3.5Ah=828,8 Wh for the usual Kingsong "840Wh" packs). You used 3.6V. Is this an unimportant coincidence or something else going on with the different values?
• Is it 84V or 84.2V? Because 16*4.2 = 67.2, and 20*4.2 = 84. Again, coincidence (and it's not like the difference between 4.2 and 4.21 max voltage matters) or something specific going on?

[esaj]

56 minutes ago, meepmeepmayer said:

Interesting overview of what each wheel uses. The color coding is neat. Guess this post a good way to get people buy bigger wheels from you, that's kind of the conclusion.

2 quick nitpicky questions, just to be sure:

• When computing Wh capacity, people here usually use 3.7V nominal voltage for each cell (e.g. 16*4*3.7V*3.5Ah=828,8 Wh for the usual Kingsong "840Wh" packs). You used 3.6V. Is this an unimportant coincidence or something else going on with the different values?

Most lithium batteries have nominal voltage "around" 3.6-3.8V, I've seen both 3.6 and 3.7V figures used most commonly. The term "lithium-ion" is sort of a (marketing) umbrella term, actually comprising of multiple different chemistries, that have somewhat different characteristics, like nominal and maximum voltages (some cells are claimed to be chargeable up to 4.35V or even above, but I wouldn't suggest trying if yours are such ), energy density, internal resistances and such. Even lithium iron phosphate (LiFePo4, 3.2-3.3V nominal) and lithium titanate (LTO, 2.4V nominal) are "lithium-ion" -batteries, despite very different nominal and maximum voltages vs. the rest.  

As you can see, there are more specific chemistries in lithium-ion (and Li-pos, too, actually Li-pos seem to be not much else than the "usual" chemistries packaged in pouches, both to make them more unusually shaped as well as to lower the internal resistance?), usually adding more/different mixes of metals into the cathodes ... I can't claim to understand why they work differently, probably someone who knows about chemistry and physics could explain that.

This chart states 3.6V nominal voltage for cobalt-based and 3.8V and manganese-based cells:

On top of that, manufacturers use (possibly proprietary) additives and coatings to enhance some characteristics, just as an example, from some some research with coatings ("Modification of Ni-Rich FCG NMC and NCA Cathodes by Atomic Layer Deposition: Preventing Surface Phase Transitions for High-Voltage Lithium-Ion Batteries") :

In this report, we show that atomic layer deposition (ALD) of titania (TiO₂) and alumina (Al₂O₃) on Ni-rich FCG NMC and NCA active material particles could substantially improve LIB performance and allow for increased upper cutoff voltage (UCV) during charging, which delivers significantly increased specific energy utilization. Our results show that Al₂O₃ coating improved the NMC cycling performance by 40% and the NCA cycling performance by 34% at 1 C/−1 C with respectively 4.35 V and 4.4 V UCV in 2 Ah pouch cells. 

Cycling performance (cycle life) from NMC electrodes in 2 Ah pouch cells at (a) low disharge rate (0.3 C/−0.3 C) and (b) high discharge rate (1 C/−1 C) with voltage window 3.0–4.35 V. Rate capability of NMC half coin cells with 3.0–4.8 V vs. Li⁺/Li voltage window from (c) fresh electrodes and (d) electrodes extracted from 2 Ah pouch cells after 1 C/−1 C cycling.

Cycling performance (cycle life) from NCA electrodes in 2 Ah pouch cells at (a) low disharge rate (0.3 C/−0.3 C) and (b) high discharge rate (1 C/−1 C) with voltage window 3.0–4.4 V. Rate capability of NCA half coin cells with 3.0–4.8 V vs. Li⁺/Li voltage window from (c) fresh electrodes and from (d) electrodes extracted from 2 Ah pouch cells after 1 C/−1 C cycling.

https://www.nature.com/articles/srep26532

This kind of research is what makes the "usual" chemistries go up in capacity/discharge capability/lifetime etc. slowly over the years, typically it seems something like a 5% (pure guess) increase in capacity is achieved yearly through small changes and additives etc.

 

TLDR; 3.6V and 3.7V seem to be used interchangeably as typical nominal voltages for most lithium-ion batteries, of which there are actually several different chemistries.

 

 

Quote

• Is it 84V or 84.2V? Because 16*4.2 = 67.2, and 20*4.2 = 84. Again, coincidence (and it's not like the difference between 4.2 and 4.21 max voltage matters) or something specific going on?

Probably just a mistake, not that it makes much of a difference.

[Jason McNeil]

49 minutes ago, meepmeepmayer said:

When computing Wh capacity, people here usually use 3.7V nominal voltage for each cell (e.g. 16*4*3.7V*3.5Ah=828,8 Wh for the usual Kingsong "840Wh" packs). You used 3.6V. Is this an unimportant coincidence or something else going on with the different values?

Official manufacturer nominal voltages are 3.6v; King Song & GW generously rounding up the capacity figures... 

http://www.rc-electronic.com/downloads/pdf/infos/datasheet_LiIo_Panasonic.pdf

https://www.imrbatteries.com/content/sanyo_ncr18650ga.pdf

On the other hand Inmotion have a historic tendency of under-rating their packs (???). Their Scooters (L6/Lively) with 52 cells, use the Samsung 22P, which yields, 52x 2.2Ah x 3.6v = 411.84Wh, yet the battery label & marketing materials have 384Wh. Same with the V5F, originally rated as 288Wh, until someone probably asked why they why cutting themselves short, now with 320Wh capacity label. 

[meepmeepmayer]

Thanks for the info, thought I was missing something besides the natural spec uncertainty of batteries. @esaj your technical drawing/diagram/table per post ratio must be insane

[Mono]

I am still not sure I understand why max speed capabilities are related to the battery. The amount of current flowing decreases linearly with increasing speed. That is, the maximal load/demand on the battery we see at 0km/h. What am I missing?

EDIT: I would believe that max speed is a result of the motor characteristics/specifications, like the operating voltage of the motor is a specification for the motor. The job of the battery is to provide the (ideally constant) voltage source with the specified value under varying operating conditions. Operating the motor at higher voltages may work and lead to higher speeds, but also risks to cause motor failure. Please correct me if I am wrong. 

[Chriull]

10 minutes ago, Mono said:

I am still not sure I understand why max speed capabilities are related to the battery.

The motor (the coils moving in the magnetic field) generates a voltage proportional to the rotational speed. 

For 1p vs 2p vs 4p configurations the big difference is imho the lowered internal resistance, which reduces the voltage sags and by this avoiding cut outs/overleans if one tries to accelerate at higher speeds.

@Jason McNeilhas a detailed voltage/current log from a 9be+ (15s2p) with the extrem voltage sags. With a 4p configuration the would be halved.

[Mono]

30 minutes ago, Chriull said:

The motor (the coils moving in the magnetic field) generates a voltage proportional to the rotational speed. 

Right. This is called back EMF and is exactly the reason which prevents any deep drag from the battery at high speed (for acceleration). It increases the resistance of the motor when spinning fast, hence decreases the maximal possible current/battery drag. What am I missing?

[esaj]

2 hours ago, Mono said:

Right. This is called back EMF and is exactly the reason which prevents any deep drag from the battery at high speed (for acceleration). It increases the resistance of the motor when spinning fast, hence decreases the maximal possible current/battery drag. What am I missing?

"Same" acceleration (as in, accelerate as much as fast) at higher speeds requires more energy (and thus more current), even without taking air drag into account. Using this: https://physics.stackexchange.com/a/105286  as starting point:

Let a be acceleration; F force; m mass; v0 initial velocity; vf final velocity; P power required; x distance travelled and t time taken. Hence,

P = F/t= (m*a*x)/t

Then a=(vf²-v0²)/2x and t=(2x)/(v0+vf)

I figured that it finally gives out something like (just to prevent mixing the travelled distance into that)...

P  = (m *(vf²-v0²)) / (2*t))

I think... I probably should practice some algebra again and just check that on paper, but assuming it's correct

Using example values of m = 100kg, t = 5s, v0 = 0m/s (0 km/h) and vf = 2.77777...m/s (10km/h), ie. a 100kg mass (rider + wheel) accelerates from 0km/h to 10km/h over 5 seconds. Required average power is

(100kg * ((2.77777m/s)² - (0m/s)²)) / (2 * 5s) = 77.1562W

Same weight, accelerating from 10km/h (2.77777m/s) to 20km/h (5.5555..m/s) over the same amount of time (5s) will require

(100kg * ((5.55555m/s)² - (2.77777m/s)²)) / (2 * 5s) = 231.481W

 

On top of that become at least the power to over come rolling friction & airdrag

But I'm not sure if my calculations were correct...

[Mono]

2 hours ago, esaj said:

"Same" acceleration (as in, accelerate as much as fast) at higher speeds requires more energy

Right, but it is not like that we can make any acceleration actually happen independently of speed. The maximal acceleration depends on the speed. On the electric side the available power is quite simple to calculate: it's current times voltage and the maximal current decreases linearly with increasing speed while the voltage is essentially constant, unless for very large currents. From this, the maximal possible acceleration is given, but not the other way around. As-fast-as-possible acceleration hence requires less power with increasing speed, simply because the motor allows less current to flow. What am I missing?

[esaj]

I'm not even sure I understand what you're asking here...  

48 minutes ago, Mono said:

Right, but it is not like that we can make any acceleration actually happen independently of speed. The maximal acceleration depends on the speed. On the electric side the available power is quite simple to calculate: it's current times voltage and the maximal current decreases linearly with increasing speed while the voltage is essentially constant, unless for very large currents. From this, the maximal possible acceleration is given, but not the other way around. As-fast-as-possible acceleration hence requires less power with increasing speed, simply because the motor allows less current to flow. What am I missing?

The maximum possible current should indeed decrease when the back-EMF starts to rise near the battery voltage at high speeds, but the resistance in the circuit itself is relatively low, and in reality the back-EMF varies wildly all the time. Neither voltage (battery voltage sags, back-EMF changes) nor resistance (battery and mosfet internal resistances go up with temperature) are actually constant, and the current is pulsed in the motor through the mosfets with PWM. The inductance of the coils will slow down the rate the current can rise and drop, so it's not only about the (DC) resistance and the voltage difference. The basic Ohm's law power equation P = U*I gives theoretical ballpark figures, but the reality is much more complex...

I've learned a bunch over about the past year and a half fiddling with electronics, but I still have no means to build a simulation of an actual 3-phase motor, because I have no idea on the amount of inductance in the coils or how much effect the permanent magnets have on the coils. Plus I've kept mostly to DC-side of things anyway, and motors are AC-devices  Somebody well versed in electric motor-drives probably can explain those better. 

The coils act as inductors, so they also have inductive reactance, that's relative to (sine-wave) frequencies. It's an imaginary component (yes, imaginary, as in the 2nd component of complex numbers, not "imaginary" like I made it up ), a sort of a "frequency-dependent AC resistance" that with the resistance forms the impedance, so certain parameters also change with the PWM-frequency (which, again to make things more hairy, is not sine-wave frequency, because it's square wave, but comprises of multiple summed sine-wave harmonics over the fundamental frequency with different amplitudes). The back-EMFs of the of coils change along with the rate the current's changing, not with immediate current value (something like u = L* (di/dt), that reads out something like "small signal voltage equals inductance times the rate of small-signal current change over time"), as they actually try to keep the current from changing (ie. doesn't matter if the current is trying to go down or up, the back-emf changes to try to keep the same amount of current flowing). Capacitors have a similar effect, except they change the current in relation to change of voltage, the rule of thumb is something like "capacitors oppose change in voltage, inductors oppose change in current"... So the real-world situation is much more complex than the simplified models let you believe, although they're not completely wrong either.

Probably I misunderstood what you were even asking. If you really want to start breaking your head, you can go google something like "PMSM motor acceleration equations" or such. There's a reason that many large automation companies have "[electric] drives" -divisions that focus entirely on this stuff. Don't say I didn't warn you and hope you're much more well versed in differential calculus than I am, I forgot most of it after polytechnic, since I never needed it (before now) 

[Rehab1]

Professor @esaj.....well stated! And that is precisely why you have my ACM board to dissect!

[esaj]

18 minutes ago, Rehab1 said:

Professor @esaj.....well stated!

 Professor of what? Nonsense?  I'm not trying to sound like a smart-ass, as I always say, I'm just a hobbyist on this stuff. (I think) what I'm trying to say with the above is that a lot depends on the "level of detail" you want to look at the motors and their drives, and giving "absolute" values for things becomes very tricky (like you really think that 1500W nominal motor is capable of exactly 1500W continuous? Ball park figures ). Or how much power you have available for acceleration... Things like simplifying back-emf and the PWM output to some "average" (RMS or whatever) DC-like voltage and such make looking at the whole thing a lot more simpler, and you can make working stuff just based on simplified models (well, maybe not something like an EUC, but get the motor turning at least with some speed-control etc), but diving into the more detailed stuff has a tendency to make me want to curl up in a dark corner   Or then I just cling too much into the details and miss the whole big picture...

 

 

[Mono]

12 hours ago, esaj said:

So the real-world situation is much more complex than the simplified models let you believe, although they're not completely wrong either.

Either the (simple) model of the motor which connects speed with current is completely wrong, or the maximally possible current/torque decreases with increasing speed and hence the battery should not be the limiting factor at high speed. Make your pick. I tend to be hesitant in considering a seemingly established simple model (which also looks consistent) to be wrong, but what do I know.

FTR: the simple model that computes the maximal current c as a function of the velocity v is

c(v) = c_0 * (v_max - v) / v_max

where v_max is the maximal velocity of the motor and c_0 is the maximal current at velocity 0.

[esaj]

2 minutes ago, Mono said:

Either the (simple) model of the motor which connects speed with current is completely wrong, or the maximally possible current/torque decreases with increasing speed and hence the battery should not be the limiting factor at high speed. Make your pick. I tend to be hesitant in considering a seemingly established simple model to be wrong, but what do I know.

Ah, now I at least understand what you were asking   From what I know, the input voltage limits the speed, precisely because of the back-EMF (average/RMS/whatever) voltage going up with the motor rotational speed, and thus causing it to not produce enough torque to accelerate more when no high enough potential difference can be caused for enough current to flow. Why do you think the battery voltage would not limit the speed?

[Chriull]

9 hours ago, Mono said:

Either the (simple) model of the motor which connects speed with current is completely wrong,

The speed is not connected with the current. Current is just proportional to the torque.

one common diagramm shows max possible current (hence torque) over speed (assuming a constant supply voltage)

Quote

 or the maximally possible current/torque decreases with increasing speed

Yes

Quote

and hence the battery should not be the limiting factor at high speed.

No. The battery voltage minus the back emf limits the maximum possible current/torque. (The motor model for this is a voltage source proportional to the speed with a fixed internal resistance + 1) )

by decreasing the battery voltage the whole max torque over speed limit shifts nearer to 0. (max no load speed and max torque (at 0 speed) decreases)

edit: 1)

+ an inductance, which should be neglectible once the system got stable (no current changes).

[esaj]

I was supposed to be doing something else tonight, but instead ended up simulating (DC 1-phase) motor in LTSpice. I found a nice article describing how to simulate the back-EMF and motor parameters:  https://www.precisionmicrodrives.com/application-notes/ab-025-using-spice-to-model-dc-motors

Having no much real values to use, I used Kt/Kemf and inductance parameters in the ball park of some ebike-data I found, and then pretty much pulled the inertia and loss-parameters from thin air (basically testing values that seemed to give "reasonable" behavior until I had a working model). I don't claim this to prove beyond doubt that the back-EMF & battery voltage limit the maximum speed of the motor, but it does seem so

Electrical Variable description, symbol, unit	Mechanical Equivalent description, symbol, unit
Voltage, V, [V] ≡ [kg⋅m2⋅A^-1⋅s^-3 ]	Torque, τ, [N⋅m] ≡ [kg⋅m2⋅s^-2]
Current, I, [A]	Angular velocity, ω, [rad⋅s^-1]
Resistance, R, [Ω] ≡ [kg⋅m2⋅A^-2⋅s^-3]	Coefficient of viscous friction, μ, [kg⋅m2⋅s^-1]
Inductance, L, [H] ≡ [kg⋅m2⋅A^-2⋅s^-2]	Moment of Inertia, I, [kg⋅m2]

The details how this works (or at least is supposed to work) to model the motor and the physical properties are in the article.

 

60V pulse given at 5kHz, R1 depicts motor winding resistance (0.3 ohms), L1 is the inductance (100uH), Back_emf is a behavioral voltage source that uses the physical model (made by V_torque, L_inertia and internal resistance of V_SENSE_2) to simulate the back-EMF from the motor.  The last bit on the right is a circuit that can track the rotor position, but it's not relevant.

Probably my inertia / loss -values for the physical model aren't exactly correct, but the end result does give the kind of behavior I'd expect to see from a motor.

At 50% duty cycle, the motor starts from 0RPM and goes up to close to 300RPM (the lower graph, the factor 9.5493 comes from the fact that the current in physical model depicts the motor speed in radians per second). The upper graph shows back-EMF average voltage (the thin red line in the middle, around 30V, which makes sense for 50% duty cycle at 60V), the green "bar" is the voltage pulse from the input voltage source V1 and the blue is the voltage measured between R1 and L1 (back-EMF + extra noise from the inductor).

A close up of the upper graph shows the voltage pulses, the back-EMF and the voltage ripple caused by the inductance itself:

 

Start up current is really high here, as the motor starts from standstill, but drops fast and ripples around almost 0A after reaching the full speed, after pretty much steady speed is reached near 2 seconds, the average is around 30mA, yet the RMS-value is around 8.5A. Probably my parameters aren't exactly correct vs. real world?  Or could there actually be that high ripple...

Any way, setting the duty cycle to 99%, the speed pretty much doubles, and the back-EMF (again the red line) climbs very near the maximum input voltage of 60V:

Since the duty cycle can't be raised higher (well, yeah, could be raised to 100%, but I'm not sure if the wheels can actually do that, and there's at least some switching on/off delays in the real things), it would certainly seem that the battery voltage will limit the maximum speed of the motor (at least as far as this model can be trusted).

Added bonus: what happens if the voltage source is dropped to 0V? Braking... but this is not actual regenerative braking, but basically short circuiting the motor phase. After 10000 pulses (2 seconds at 5kHz), the voltage source stops pulsing:

The blue line in the upper graph is the (average) back-EMF, dropping off as the motor starts to brake. The lower graph shows the RPM (the I(L2)*9.5493 -graph, I'm not sure what that color is called, cyan?) and the current in green. Slamming the brakes causes a nice 200A backwards current at first that dies off over the next second or so...

But like said, since I pulled the values pretty much out of my a$$, I can't claim this to be exactly correct, and it's said to model a single-phase DC motor, not a 3-phase BLDC/PMSM or such.

 

 

[Carlos E Rodriguez]

Battery voltsge let say 67vdc (VBAT).

Wheel not spinning. Motor could back emf (VEMF)is 0vdc. 

Max current = (VBAT - VEMF) * PWM duty cycle. Let use 50% just for getting some numbers. By the way even at 100% duty cycle one phase will be turned of for every two magnet pole transitions. 

So (67-0) .50 = 33.5 volts. Let say coil resistance is 2 ohms. You get 16.525 AMPS Maximum produced and no more. 

So let's say we now spin the wheel and because the coils are going pass magnets they produce a voltage. Let's say the wheel is spinning at what is determined 50% of a theoretical maximum RPM. At that Rpm the motor coils are producing a VEMF of 33.5 volts. 

Now do the same calculation at that speed. 

(67-33.5) .50 = 16.525volts. So now you are only able to put 16.525/2 = 8.25amps. 

So now because torque is amps you basically only have 1/2 the torque available to push. 

Next let say we double the rpm so now the motor could produce a VEMF of 67 volts. 

Do the math. 

(67-67) .50 = 0 volts., 0 amps available. 

The battery has 67 volts but the back emf is also 67 volts so there is no more voltage potential to produce current. 

And that is the story. And there is no way around it other than have voltage supply able to produce voltages higher than the VEMF the wheel is producing. 

That is why the new wheels are coming out with 87volts. The motor design is the same so it produces the same back emf profile but because we have more VBAT the controller can still put current to go faster. 

[Mono]

18 hours ago, esaj said:

Why do you think the battery voltage would not limit the speed?

I don't think that 

16 hours ago, Chriull said:

The speed is not connected with the current.

The simple model of the motor that connects the maximal current c as a function of the velocity v is

c(v) = c_0 * (v_max - v) / v_max

where v_max is the maximal velocity of the motor (for the given voltage) and c_0 is the maximal current at velocity 0 (for the given voltage). 

There may be a misunderstanding: I don't say we don't need the battery to reach or keep up high speed. I am saying that when comparing limit demands at different speeds we have the larger demands on the battery, surprisingly, at lower speeds. That is, if the battery can satisfy the current-demands of the motor at lower speeds (i.e. maximal acceleration at low speed), it shouldn't have any problem at higher speeds.

[Chriull]

1 hour ago, Mono said:

The simple model of the motor that connects the maximal current c as a function of the velocity v is

c(v) = c_0 * (v_max - v) / v_max

where v_max is the maximal velocity of the motor (for the given voltage) and c_0 is the maximal current at velocity 0 (for the given voltage). 

That c(v) is the maximum possible current at speed v. The real needed current at speed v is just the current which is needed to "create" enough torque to overcome all the frictions, air drag, accelerations (balancing), etc,... As long as the real needed current is below this max current c(v) everything is ok - if not a crash is likely to happen/happening.

1 hour ago, Mono said:

There may be a misunderstanding: I don't say we don't need the battery to reach or keep up high speed. I am saying that when comparing limit demands at different speeds we have the larger demands on the battery, surprisingly, at lower speeds. That is, if the battery can satisfy the current-demands of the motor at lower speeds, it shouldn't have any problem at higher speeds.

The maximum possible current decreases with increasing speed. The current demand goes a bit up with higher speeds for friction and air drag, which lies normaly nicely below the max current at this speed. But the gap to the max possible current gets lower and lower, so with some extra acceleration one can reach more easily the max possible current at this speed -> overlean/crash/etc. The battery could theoretically deliver the needed power and also the needed current. But in this closed system with the motor generating the back-emf there is not enough voltage difference, that this "possible" current could flow. To "repair" this the battery needs a higher voltage -> the difference between battery voltage and motor back-emf gets bigger -> more current can flow! (I= Delta U/R, Delta U=Ubattery-Uback-emf, R is constant (once current has "settled"), Uback-emf is proportional to the speed so the only thing left to "influence" I is to change Ubattery.)

[esaj]

2 hours ago, Mono said:

I don't think that 

The simple model of the motor that connects the maximal current c as a function of the velocity v is

c(v) = c_0 * (v_max - v) / v_max

where v_max is the maximal velocity of the motor (for the given voltage) and c_0 is the maximal current at velocity 0 (for the given voltage). 

There may be a misunderstanding: I don't say we don't need the battery to reach or keep up high speed. I am saying that when comparing limit demands at different speeds we have the larger demands on the battery, surprisingly, at lower speeds. That is, if the battery can satisfy the current-demands of the motor at lower speeds, it shouldn't have any problem at higher speeds.

This is how I understand it: the theoretical maximum current is limited by the battery voltage, battery internal resistance (which I left out in the simulation), winding resistance / impedance and back-EMF from the motor (disregarding things like connector resistances, mosfet Rds(on)-resistances, switching losses, changes due to heating etc. "minor details"). The cells themselves are limited by their voltage, internal resistance and the resistances of the circuit they're attached to (for DC-currents). If an LG MJ1 cell charged to 4.2V and with 37 milliohm internal resistance was shorted over a super conductor (basically 0-resistance outside the cell itself), it could theoretically produce a maximum current of 4.2V / 0.037ohm = about 113.5A as long as the voltage doesn't start to drop due to discharging. In a more realistic scenario, if the wire & connections short circuiting the cell had a resistance of 13 milliohms (a value picked just to get a total resistance of 50 milliohms), the theoretical maximum current (before the voltage starts to drop due to discharging / resistances start to change due to heating) would be 84A. Both cases would probably lead to a thermal runaway in real world, but it's a theoretical maximum for single cell.

At steady duty-cycle, high current should occur at start to get the motor moving, and less current is "available" the higher the back-EMF goes with the speed (in DC-terms, the voltage U of I = U / R becomes smaller, as it's the difference between the battery / PWM average voltage and the back-EMF => I = (U_bat - U_backemf) / R). The 99% duty-cycle simulation is basically accelerating the motor nearly as fast as it can to as high speed as it can (100% would give that one final notch more, but that's it). The battery itself probably should have no problem supplying the (usually) lower currents at steady higher speeds on flat ground, but if you look at the speed-curve in the simplified LTSpice-model, you notice that with constant losses in the physical side and no changes in resistances due to heating, acceleration is still not constant but slows down the faster the wheel is turning as the back-EMF goes up dropping the current flow but everything else stays the same. So at least in that case, the "available" torque/current needed for acceleration is limited by the back-EMF getting closer to the battery voltage (or battery voltage * PWM duty cycle), and thus it will reach a point where it cannot accelerate further, not because the battery couldn't give out higher current by itself, but because the voltage difference is too small to cause as high current flow as would be required to keep accelerating. This is of course a simplified model, because it assumes steady battery voltage etc., whereas in reality if you were to lean forward a lot at high speed and there wouldn't be enough voltage difference to cause high enough current to produce the required torque to accelerate to pull the wheel back upright, and/or the battery internal resistance could cause the battery voltage to drop enough as the current goes up, you would likely end up overleaning the wheel. Also the physical losses probably go up with speed too, like air drag and such, making the power required to accelerate go higher with speed.

So putting more batteries in parallel should help with the high current requirements at start up and during hill climbing/acceleration, as well as to lower the voltage sag due to internal resistance... And of course give more range. The highest possible speed is still limited by the maximum battery voltage / duty cycle. I don't know the real values for the currents, but if climbing a hill on a powerful wheel can cause the solder to melt in connectors, it must be high. One of the ecodrift.ru dyno-tests showed Gotway Monster producing up to steady 4.7kW at 58% battery before the power output starts to drop, at a voltage around 75-80V, that would mean around 60A for output power alone (and more due to internal losses everywhere, like heating up those motor connectors ).

[Chriull]

Some more pointers in regard to max speed versus battery:

Lift cut-off speed measured with half empty and full batteries.

http://forum.electricunicycle.org/topic/5624-ks16-lift-cut-off-speed/

 

The following images are links to http://airwheel.ru/test-monokoles-na-dinostende/  (dynamometer test of different wheels):

( Hi @EcoDrift, i hope it is ok to link to images of your website and reuse two of your images? If not, please tell and i'll remove this post! )

Load test (dynamometer) with an Inmotion V8 (so more or less max load):

and the corresponding battery voltage / current:

There is roughly a voltage sag of 20V at full load (~45A) and this is not at a really low speed (~20 km/h).

Imho the Inmotion has a 20s2p configuration - with a 20s4p configution the voltage sag in the same situation would only be 10V -> higher max speed, more torque at higher speeds, less risk of cutout.

 

Here one sees the comparison of a "normal" Inmotion V8 versus one with better/stronger battery cells (V8 Vitaliy): More torque at the same speed available! (Also this v8 from vitaliy burned the mb at the second test run...;( ) Also one sees nicely the max-torque vs speed limit shifted upwards

Don't know why the max-torque vs. speed limit is not a straight - could be that ecodrift did not fully accelerate at the end, or this is a sign of voltage sag due to the high load?

Another comparison: KS16 340Wh (16s2p with ~2850 mAh cells) vs. KS16 840 (16s4p with 3500mAh cells):

 

[Mono]

The "corresponding battery voltage / current" graph looks pretty weird (mostly no current flow, huge V-sags without current flowing...) and the correspondence is not quite clear to me. Where do you get the correspondence "at ~20km/h" from?

[US69]

@esaj

Fantastic explanation!!

I have learned from the e-bike forums to say the complicated EMF things in a one short term:

Speed is Voltage related

Torque is Amp related

:-)

what also has helped me in vaping to understand how each variable has its factor is this little tool:

http://www.sengpielaudio.com/Rechner-ohm.htm

Give in a Voltage (67volt full/or 53Volt near empty) and a needed wattage (800watt) and you will get shown the amperage which has to be delivered by the batteries:

67/800: 11,94Amps

53/800: 15,09Amps

Thats why i try to avoid high speeds, hills and hefty accelerations on half or going empty batteries...the amperage can get a lot higher to do the same thing...and may be to much for the batteries designed envelope 

[Chriull]

6 hours ago, Mono said:

The "corresponding battery voltage / current" graph looks pretty weird (mostly no current flow, huge V-sags without current flowing...) and the correspondence is not quite clear to me. Where do you get the correspondence "at ~20km/h" from?

@EcoDrift wrote on his page that This current voltage graph is from three measurements - so everything inbetween ( no current flowing) is to be ignored.

the correspondance with the speed comes from the power over speed graph. The max current happens somewhere near the max power, so both graphs can be "combined"

edit: the power over speed graph should be recorded at one of the three current peeks of the voltage and current graph.

Ps.: If you describe a bit more more why you think battery voltage does not inflict max speed it would be easier for us all together to refine and formulate correctly the relation.