Warnings should depend from current, not from speed

[marc]

The reason why there are speed warnings and tilt-backs (even tilt-forward for my gotway Msuper) is to avoid that the rider falls forward while accelerating because the engine, the converter cannot deliver that high CURRENT, while he is able to work at that voltage. But strange enough, the warnings, the alarms and tilt-backs depend from speed (which is proportional with the voltage) instead of current.
 You would still say: so what?

Well, if you are climbing a steep hill of 35 degrees and you are 120kg and you accelerate very fast on this hill from 0 to 20km/h. You wouldn't even hear a warning before falling because your speed is OK (less than 20km/h) but the engine and the converter cannot deliver that high current: current is proportional with the load (=weight, inclination and accelaration). voltage is proportional with the speed (motor speed and unicycle speed).

Also for braking, once I had braked heavily and my gotway msuper did a tilt-forward, to make me clear that he cannot brake that fast. Again without warning. since my speed was ok, besides I was braking so of course there will be no warnings for brakings: but it should. I became a little unstable due to tilt forward during braking but again this is a current and load issue not a speed and voltage issue. why don't they focus to current to make the unicycle safer. 

The voltage nearly rests the same for all moments: 52-68 volt:empty, full, high speed, low speed, braking, accelerating. always between 52-68.

But the current can go from minus 50 amperes to plus 50 amperes: a huge difference and the most restricting factor for the capabilities of the product.

If we look at the progression in the car industry in the last 2 centuries, one can give an example for safe braking to compare with the safe accelerating on the euc's. They make an ABS system to make you brake safer. They don't say:  "you are braking, your intention is to stop, and stopping is safe, so you are in the good way. you don't need a ABS. you may brake and slip when you brake from 200km/h to 0km/h in a rainy environment and due to slipping you may make an accident and die."  But the engineers of euc do still need time to make the euc safer, not slower, safer by inventing the current warning system. No?

I think the reason is: current can change instantly (accelerating from zero or hard braking at high speed) and warnings should not have an impact by that time, while speed increases slowly (e.g. 3 seconds from 0 to 30km/h), then it has sense to send an alarm at 25km/h and by the time you hear it and you stop accelerating you ride already 35km/h. If the warning would be at 33km/h, by the time you hear the alarm you would be at 44km/h and you would have fallen.

But for people want to accelerate slowly the alarms at low speed levels is not relevant.

Engineers... am I correct?

[WakefulTraveller]

It would certainly make sense to have a more intelligent alarm / tilt back system, especially when it comes to GotWay. However with my MCM3, I never had an issue with the motor not having enough current. It would always alert me when I should slow down (with the first two alarms disabled) no matter the conditions. Yet I suppose there's a limit to how much torque a motor can supply, even with all of the current that it can take. I can certainly see how a heavy person going uphill on a HS MSuper could cause some forward tilting (and overheating).

[Jeffrey Scott Will]

Someone correct me if I'm wrong, but dont we already get these types of warnings? On recent powerhouses like KS and GW, you have programmable warnings based on the speed you define. But then there is also a built in warning for "low battery" which is based on voltage, not how much estimated distance you have left. Because it's based on voltage, if stressful situations like hill climbing cause the voltage to drop, you're going to get the low battery warning when you're pushing it too hard. This is what happened from time to time to me on my earlier wheels with smaller batteries like airwheel x8 (174wh) and KS 14c 500w (340wh). The battery wasn't actually low, but prolonged hill climbing will cause an alarm from the voltage drop which is interpreted the same as "low battery"

You're always going to put yourself at more risk when you push the speed closer to the point where you have less available torque because any sudden bump cannot be predicted by your wheel, and therefore, it will shut down if said bump requires more torque than you have available. 

[marc]

4 hours ago, Jeffrey Scott Will said:

Someone correct me if I'm wrong, but dont we already get these types of warnings? On recent powerhouses like KS and GW, you have programmable warnings based on the speed you define. ...

Because it's based on voltage,

To avoid misunderstandings: that is what I am saying too: " we have alarms depending on speed and voltage."

But what I say different than you is: "alarms based on the most restricting factor of the euc, namely the CURRENT, would be better for safety, because you can have dangerous situations at low speed also." And there, it will be the current who will predict the danger in advance.

[Mono]

1 hour ago, marc said:

And there, it will be the current who will predict the danger in advance.

Why not voltage? As long as voltage doesn't drop, torque should be in good shape, I would believe. 

[marc]

7 hours ago, Niko said:

Why not voltage? As long as voltage doesn't drop, torque should be in good shape, I would believe. 

voltage drop is too little (max. 1 or 2 volt in 1 second) and difficult to base on to announce when the unicycle is not able to do the asked job, while current drop can vary from -50ampere to +50 ampere in 1 second (change of 100 amperes in 1 second is huge).

[John Eucist]

I agree that warnings should be for all three.  Over-current, over-speed, and under-voltage.  The latter two already standard in most eucs.

[Mono]

To my understanding over-current must be prevented to prevent damage of components, pretty much the same as over-heat, though better predictable. The current is to my understanding limited more or less independently of the driving state and gives an upper limit for delivered power. Over-current seems to be a slight misnomer, it should rather be max-current. Am I wrong? This seems different from under-voltage, indicating a drop of power output compared to normal, or over-speed, indicating a mismatch between available power and speed.

Roughly speaking, under-voltage indicates the battery getting to its limits, and max-current indicates other components getting to their limits, right?

@marc from what I have seen voltage depends dramatically and instantaneously on load. My hunch is that to predict an upcoming face plant, current is usually a worse predictor than voltage, maybe unless you have a large motor and a very large battery (or a different battery type with smaller internal resistance under large loads), but I would certainly be interested to see data on this.

[John Eucist]

5 hours ago, Niko said:

Over-current seems to be a slight misnomer, it should rather be max-current.

Yeah I was trying to make it neat and consistent sounding with the list of "over" and "unders".    What I meant by "over-current" was basically if you demand too much power the euc might shut down resulting in possible faceplant.

[Keith]

6 hours ago, marc said:

voltage drop is too little (max. 1 or 2 volt in 1 second) and difficult to base on to announce when the unicycle is not able to do the asked job, while current drop can vary from -50ampere to +50 ampere in 1 second (change of 100 amperes in 1 second is huge).

So what! The concept that current can change, I.e. Swing from (say) -50 Amps to +50 so fast that the voltage would not drop enough to alarm is a red herring, a short instantaneous high current would not damage the motor or controller, it is the heating it causes, if it continues, that does any damage.

Please define what you mean by over-current in this situation? What exactly are you considering is too much current for the wheel?  Do you mean a fault condition where a short circuit is going to cause an overheat situation in the battery and/or the controller or over current in normal operations?

over current in normal operation should not be possible. With a full battery, wheel standing still, rider leaning hard forward the only current limiting factor is the total resistance of the power train, including internal resistance of the battery. That should be well within the capability of the wheel I.e. The max power it can instantaneously give, I.e. There should NEVER BE a situation where the current is higher than that without an actual fault condition in the electronics. Under fault conditions, I.e. Some form of short circuit, tilt back would be impossible (even more power would be required to tilt you back) and an alarm probably superfluous as the BMS, or at the very least high current fuse should fail at about that point.

Key point, and the reason the tilt back/alarms etc, work that way is that a stack of 16 series cells have some internal resistance, the higher the current, the more the pack voltage will drop, that voltage drop is what triggers alarms/tilt back.if a high current continues for too long, and the pack is full enough that voltage does not get low enough to alarm, then the controller will heat up and alarm for that instead.

[Keith]

On 24 May 2016 at 10:36 AM, marc said:

The reason why there are speed warnings and tilt-backs (even tilt-forward for my gotway Msuper) is to avoid that the rider falls forward while accelerating because the engine, the converter cannot deliver that high CURRENT, while he is able to work at that voltage. But strange enough, the warnings, the alarms and tilt-backs depend from speed (which is proportional with the voltage) instead of current.
 You would still say: so what?

Well, if you are climbing a steep hill of 35 degrees and you are 120kg and you accelerate very fast on this hill from 0 to 20km/h. You wouldn't even hear a warning before falling because your speed is OK (less than 20km/h) but the engine and the converter cannot deliver that high current: current is proportional with the load (=weight, inclination and accelaration). voltage is proportional with the speed (motor speed and unicycle speed).

Picking up further on my previous reply, I've realised where you are coming from in your above statement. Conceptually it sounds fine but, actually it is completely wrong.

What you are actually talking about is having insufficient torque, I.e. The rider leans forward too far for the present conditions, the power train cannot deliver enough torque to hold the rider up, rider falls forward. There is no demand for higher current, the power train hits the maximum current it can provide for the present battery voltage and does not/ cannot demand any more. There is no current sensor that could detect that condition as there is no condition to detect, the motor is happily giving all its got and won't ask for any more.

What you would actually require would be a force sensor built into the peddles and tied into your wheel's computer to tell you that you are trying to demand more torque than the wheel can presently give. Actually, of course the wheel would need to alarm when you reached close to max torque to stop you leaning any harder.

At the moment, a faceplant is natures way of telling you you demanded too much torque. It certainly sharpens up the learning curve to assist the rider in instinctively knowing this in future :-)

[Chriull]

38 minutes ago, Keith said:

...What you are actually talking about is having insufficient torque, I.e. The rider leans forward too far for the present conditions, the power train cannot deliver enough torque to hold the rider up, rider falls forward. There is no demand for higher current, the power train hits the maximum current it can provide for the present battery voltage and does not/ cannot demand any more. There is no current sensor that could detect that condition as there is no condition to detect, the motor is happily giving all its got and won't ask for any more.

A little addentum to the "present battery voltage" - the back EMV (the higher the speed the more) - leads to this situation that no more current can be "pumped" into the motor to produce more torque. So as quite often written here - at high speeds there is less torque and vice versa.

38 minutes ago, Keith said:

 There is no current sensor that could detect that condition as there is no condition to detect, the motor is happily giving all its got and won't ask for any more.

I would say that with known speed (rpm), voltage and current for a BLDC with a given characteristic the controller could know that the torque limit is (about to be) reached. (If the firmware programmer knows about this "relation" and is able to express it formaly...)

38 minutes ago, Keith said:

...At the moment, a faceplant is natures way of telling you you demanded too much torque. It certainly sharpens up the learning curve to assist the rider in instinctively knowing this in future :-)

And there imho one can feel a little "softness" starting, once one comes in the region where torque gets "sparse"... So driving a bit carefull and "listening" to the wheel is a very good "Warning indicator" too imho...

[Mono]

56 minutes ago, Keith said:

What you are actually talking about is having insufficient torque, I.e. The rider leans forward too far for the present conditions, the power train cannot deliver enough torque to hold the rider up, rider falls forward. There is no demand for higher current, the power train hits the maximum current it can provide for the present battery voltage and does not/ cannot demand any more. There is no current sensor that could detect that condition as there is no condition to detect,

You could have a max current warning, say integrated over 0.5s or so, no? It would indicate that the power/torque limit is reached, because current cannot be increased and voltage can't be either, naturally. 

[Keith]

2 hours ago, Niko said:

You could have a max current warning, say integrated over 0.5s or so, no? It would indicate that the power/torque limit is reached, because current cannot be increased and voltage can't be either, naturally. 

No, you aren't getting it yet, @Chriull just said it, back EMF reduces the apparent voltage at the wheel. Let us say you lean forward the maximum you can so the motor produces the maximum power it can, the highest current occurs at zero RPM. As the wheel gets faster the back EMF reduces the apparent voltage at the motor and that in turn reduces the current. This is why, the faster you are going the less torque there is available. Measuring current tells you absolutely nothing about how close you are to a faceplant.

Actually, if you could measure that apparent voltage, and I cannot think how you would do it because it is inside the motor, then that would indeed indicate how close to not enough torque you were.

there are three common characteristics to ALL electric motors. If you fit the motor into a dynamometer, apply full power to the motor and restrict the RPM with the dynamometer you will find:

Maximum current & torque is at Zero RPM

Maximum Power is at 50% of no load speed

When the motor has zero load, power (obviously) is zero, torque is zero, current is a tiny trickle to cover friction and windage losses, the RPM (no load speed) is equal to the motors kV value x the applied voltage.

So, for example a motor with a kV of 100 will spin with no load and an applied voltage of 60V at 6000RPM (or just under due to losses) maximum power will occur at 3000RPM. The motor will have best efficiency at around 70-80% of no load speed and at zero RPM maximum current will be limited only by the impedance/ resistance of the motor windings plus any other resistance in the power train.

[Mono]

@KeithI don't see how what you are saying prevents, as @Chriull suggested, to measure speed and current (and voltage given that there isn't unlimited supply) to know whether torque could still be increased, as the characteristics of the motor are known. Sure, max current (as max power and max torque) depends on RPM, but that doesn't mean we can't compute the supposed limit as a function of RPM. I still believe that voltage drop is more worthwhile to look at, but that's a different question.

[John Eucist]

1 hour ago, Keith said:

Actually, if you could measure that apparent voltage, and I cannot think how you would do it because it is inside the motor, then that would indeed indicate how close to not enough torque you were.

Gotway and King Song apps show live monitoring of voltage and current.  I have noticed that when I do hard leans (hard pressure on front or back of pedals) the current would spike up while the voltage would suddenly drop (assuming the app readings were correct).  On a slightly different yet related topic, I was just looking at the Ohm's Law formula (I=V/R) which states that voltage and current are proportional to each other.  Doesn't that contradict the phenomenon I just described or am I just not understanding something correctly?  Could someone explain?

[Tomek]

20 minutes ago, John Eucist said:

On a slightly different yet related topic, I was just looking at the Ohm's Law formula (I=V/R) which states that voltage and current are proportional to each other.  Doesn't that contradict the phenomenon I just described or am I just not understanding the formula correctly?  Could someone explain?
 

ideally the voltage would be constant and when resistance of the mottor decreases (=it increases speed or torque) the current flowing through the motor increases. However, when the motor draws a lot of current from the battery, that battery's voltage proportionally drops (because of the internal resistance phenomenon). high drain batteries have less internal resistance, hence less voltage drop than normal ones.

PS

to my understanding A-brand (KS, Gotway at least) wheel warnings are derived from speed, current, voltage (battery level/strain) and temperature.  e.g. the 3 beep warning on my MCM4 activates on much lower speeds when on low battery or riding up a hill.

also, as mentioned by others, max speed of the motor is the property of the motor (depends on how it's wound) and cannot be exceeded or the electronics will fry - that's why wheels shut down, to prevent frying which would mean a shutdown anyway AND a likely fire (I assume).

 

[Keith]

@Niko, Yes, I absolutely agree what @Chriull said is spot on and nothing I said prevents it but you said: "have a max current warning".

If you measure RPM, instantaneous voltage and instantaneous current, then it should be possible to calculate how close to maximum torque you are for that instant. Interestingly, most EUC apps capture all of that from the wheel already I.e. report speed (which is basically RPM X circumference, ), voltage and current. 

However I suspect the algorithm needed to estimate available torque could be challenging, and, of course, the weight of the rider would need to be input as well.

23 minutes ago, John Eucist said:

On a slightly different yet related topic, I was just looking at the Ohm's Law formula (I=V/R) which states that voltage and current are proportional to each other.  Doesn't that contradict the phenomenon I just described or am I just not understanding something correctly?

No, no contradiction. Remember batteries have internal resistance (R) so, rearranging Ohms law to V=IR and seeing the battery as a system in itself then the 'V' is the voltage drop across the battery terminals due to the internal resistance 'R' and current flowing 'I'. Thus the higher the current the lower the battery voltage.

For example: if a single 18650 cell has (say) 25milliohms resistance and the pack is 16s2p then total pack resistance will be (16x0.025)/2 = 0.2 Ohms. Since V=IR, if the current is now 10Amps then Voltage drop across the battery V=10x0.2 = 2Volts. I.e. If, at rest the pack is currently at (Say) 60V, when the current reaches 10Amps the voltage will drop to 58V.

 

 

 

 

 

 

[Keith]

28 minutes ago, Tomek said:

also, as mentioned by others, max speed of the motor is the property of the motor (depends on how it's wound) and cannot be exceeded or the electronics will fry - that's why wheels shut down, to prevent frying which would mean a shutdown anyway AND a likely fire (I assume).

Arrrrrrrrrgggggggggggggghhhhhhhhhhhhhhhhh.......... @Tomek, I think I've gone into a parallel universe. The motor cuts out at max RPM (I.e. If lifted up) For the safety of the user, imagine what would happen if you put it back on the ground whilst spinning that fast. Nothing will fry! the motor, as has been stated umpteen times now, is drawing virtually no current at that speed, no power, no torque, no electrical problem whatsoever!!!!!!!!!!!!!!!!!!!!

It cannot be exceeded because it is a physical law, No Load Speed is the point where back EMF, I.e. The motor acting as an alternator/Dynamo equals the applied voltage, no current flows so no further increase in RPM is physically possible.

[John Eucist]

41 minutes ago, Keith said:

No, no contradiction. Remember batteries have internal resistance (R) so, rearranging Ohms law to V=IR and seeing the battery as a system in itself then the 'V' is the voltage drop across the battery terminals due to the internal resistance 'R' and current flowing 'I'. Thus the higher the current the lower the battery voltage.

For example: if a single 18650 cell has (say) 25milliohms resistance and the pack is 16s2p then total pack resistance will be (16x0.025)/2 = 0.2 Ohms. Since V=IR, if the current is now 10Amps then Voltage drop across the battery V=10x0.2 = 2Volts. I.e. If, at rest the pack is currently at (Say) 60V, when the current reaches 10Amps the voltage will drop to 58V.

Well if V (in the formula) stood for the "voltage drop" (the volts to subtract from the value previously), as you used in the example, then that would make sense.  I was thinking along the lines of V being the actual voltage (e.g. what's shown on the GW or KS app).  In that case, (using V=IR) if the total voltage (V) dropped while the current (I) increased then the resistance (R) would have to suddenly fall dramatically for that equation to hold true?  So I guess my question is why would the V in the formula be representing the "voltage DROP" instead of the actual voltage?  I'm sorry for being so ignorant. 

[Tomek]

18 minutes ago, Keith said:

Arrrrrrrrrgggggggggggggghhhhhhhhhhhhhhhhh.......... @Tomek, I think I've gone into a parallel universe. The motor cuts out at max RPM (I.e. If lifted up) For the safety of the user, imagine what would happen if you put it back on the ground whilst spinning that fast. Nothing will fry! the motor, as has been stated umpteen times now, is drawing virtually no current at that speed, no power, no torque, no electrical problem whatsoever!!!!!!!!!!!!!!!!!!!!

It cannot be exceeded because it is a physical law, No Load Speed is the point where back EMF, I.e. The motor acting as an alternator/Dynamo equals the applied voltage, no current flows so no further increase in RPM is physically possible.

i'm not an expert, so I don't want to argue, but to my understanding it is more complex than that. if the motor spinning at top speed gets pushed e.g. by going downhill, a gust of wind or inbalance of the rider to exceed that speed, the back EMF can damage the electronics. I experienced that several times with brushed motors - switching the direction of the motor too fast literally blew up two of my (properly oversized) h-bridges into pieces, so I assume it can also do serious damage when pushed beyond its limits on a brushless motor.

PS

also, at close to top speed under load the current draw is pretty high, meaning that same goes for back EMF, meaning that when the rider would exceed that speed the current from EMF would start flowing backwards.

[Keith]

2 minutes ago, John Eucist said:

Well if V (in the formula) stood for the "voltage drop" (the volts to subtract from the value previously) as you used in the example then that would make sense.  I was thinking along the lines of V being the actual voltage (e.g. what's shown on the GW or KS app).  In that case, (using V=IR) if the total voltage dropped while the current increased then the resistance would have to suddenly fall dramatically for that equation to hold true.

I'm beginning to think I'm being wound up ;-). The voltage measured, IS the voltage supplied by the battery, so the voltage measured DOES include the voltage drop within the battery. That (and obviously the drop in voltage as the pack discharges) is the ONLY THING that will cause the voltage to drop as current gets higher.

Do not lose sight of the fact that the current is being controlled by the controller. Whether the wheel is drawing 1 Amp or 10 Amps  has nothing to do with resistance per ce, that plus the applied voltage only limits maximum current.

In crude terms you could say that the controller is effectively a variable resistance and it is that which is controlling the current, in practice it is PWM (Pulse Width Modulation) to reduce current the controller send shorter pulses of current, to increase it longer pulses. The current being read by an App is the average current from those pulses.

[John Eucist]

@Keith I have edited my previous reply with a question at the end which I hope you can answer.  I will quote my edited version here (with the question in bold) for convenience:
 

1 hour ago, John Eucist said:

Well if V (in the formula) stood for the "voltage drop" (the volts to subtract from the value previously), as you used in the example, then that would make sense.  I was thinking along the lines of V being the actual voltage (e.g. what's shown on the GW or KS app).  In that case, (using V=IR) if the total voltage (V) dropped while the current (I) increased then the resistance (R) would have to suddenly fall dramatically for that equation to hold true?  So I guess my question is why would the V in the formula be representing the "voltage DROP" instead of the actual voltage?  I'm sorry for being so ignorant. 

 

Also, could you clarify what "That" (the first [of two] "that" which I made BOLD) is referring to in the following reply from you?  In other words, WHAT "is the ONLY THING that will cause the voltage to drop as the current gets higher"?
 

48 minutes ago, Keith said:

The voltage measured, IS the voltage supplied by the battery, so the voltage measured DOES include the voltage drop within the battery. That (and obviously the drop in voltage as the pack discharges) is the ONLY THING that will cause the voltage to drop as current gets higher.

I appreciate your insight to help me learn.

 

[Tomek]

3 minutes ago, John Eucist said:

WHAT is the only thing that will cause the voltage to drop as the current gets higher?

 

voltage is always measured across two points. for explaining the (theroretical) internal resistance (r) the voltage drop is what happens "within" the battery, as if it was an ideal battery with a resistor(r) attached to it and the voltage drop was measured across that theoretical resistor. in this model the resistance of the entire circuit (motor+battery) is the internal resistance of the battery (r) + the resistance of the motor/controller (R). so, V (as measured when the battery is not under load) = I * (R+r)

[Mono]

45 minutes ago, Keith said:

but you said: "have a max current warning"

and it goes without saying that max current would depend on RPM, doesn't it 

43 minutes ago, Keith said:

the weight of the rider would need to be input as well.

Why is that? The demand should be reflected in the current and to know whether it is outside the critical margin of maximum we don't need to know the weight of the vehicle. In other words, to know that I can only push up to 10% (or 20% or whatever) further should be equally useful irregardless of weight.