Regeneration on Full Battery, bad news :(

[Gaston]

Europeans sophisticated education and knowledge meets Chinese culture. It's really interesting to see what you measured. What are the next steps on the manufacturers side? Look at the warranty time and the treshholds for a bad battery. Why should they improve anything? 

[cg]

 

... I actually thought that you could to some extent determine whether the wheel is going down or up a hill or straight based on speed & power change, like more power, speed doesn't change or drops -> going up hill, less power, speed doesn't change or raises -> going down hill. If both change, then it's hard to say

 

Yes, I was thinking the same.

but I think this will only indicate the start of going uphill, and the start of going downhill, and if you are looking closely to your own plot, for instance around the first minute, it is difficult to say what is happening, my guess is you are just accelerating normally :), and then, at 2.1, you are decelerating, but it seems there might be an offset between speed and power (again looking around 2.1 in your asfallati data)... could also be i'm to tirred :/.

Anyway, I tried it out with some heavy smoothing, I dare you to prove it wrong

I have massaged all the scales to be in the neighbourhood of the speed. so the real values aren't really readable from the power...

the hills scale, has zero = 5, and then +-5 for the arbitrary scale.

I have done the following

-----

X=real(dlmread("asfaltti_20150915_210726.txt" , ",", "emptyvalue", 0));

# Acceleration defined as speed_sample(n)-speed_sample(n-1)]
S=X(:,2);
ACC = [0; S(2:end)-S(1:end-1)];

P=X(:,3);
D_PWR = [0; P(2:end)-P(1:end-1)];

# smooth acceleration and power
s_size = 25;
S_ACC = conv(ACC, ones(s_size,1)./s_size, "same");
S_DPWR = conv(D_PWR, ones(s_size,1)./s_size, "same");

AX = 1:length(X);
INDICIES_DHILL = S_ACC > 0 & S_DPWR <= 0; # Accelerate and decrease power usage: going downhill?
INDICIES_UHILL = S_ACC < 0 & S_DPWR >= 0; # Decelerate and increase power usage: going uphill?
INDICIES_STRAIT = ~INDICIES_UHILL & ~INDICIES_DHILL; # Or hard to say?

Q = zeros(1,length(X));
Q(INDICIES_UHILL)=1;
Q(INDICIES_DHILL)=-1;

ss=201;
clf;
hold on;
plot(conv(Q, ones(ss,1)./ss).*10+10,'r');
plot(S);
plot(P./100+10, 'g:');
legend("hill?", "speed", "pwr");

 

[Mono]

Going faster requires a great deal more energy in itself (more than what is in linear proportion with speed). 

sure, but this is because of drag (while friction grows pretty much linear in speed). The forward-backward balancing task is invariant under the chosen speed, apart from the wheel spinning at a different speed. Newtonian physics doesn't depend on whether the train you are in is standing still or moving with constant velocity. 

I also see no reason to believe that balancing consumes more energy at lower speeds

That's not what I said or meant. It consumes the same amount of energy per time at any speed, hence more at lower speeds to cover the same distance, because it takes longer to cover the same distance at lower speed.

[Jason McNeil]

The data in the energy consumption tests is that going faster requires more energy per/km on account of air resistance. In steady-state cruising mode, there should be no micro-braking action: the 100s/1000s calculations per second should only inject sufficient power for that point-in-time. Obviously if the rider is making erratic movements, this may not be the case.

[MrBump]

Setting aside the relevance or otherwise of simplistic analogies for the moment - do you have any evidence or any reasoning to back up the claim that speed has no effect on the amount of energy used in balance along the axis of travel (whether positive or negative)?  

Seeing how the unicycle balances a rider by varying speed, separating the two factors doesn't seem to make much sense.

[PedalFaster]

Seeing how the unicycle balances a rider by varying speed

Actually, a unicycle, electric or pedal, corrects balance errors by applying an acceleration or deceleration to the wheel.

The only reason the balance "angle" at which the center of mass leads or lags the axle changes at all as a function of speed is to counteract the amount of force applied by drag - in your basic "frictionless" high school physics model, the rider is balanced directly over the axle regardless of speed, and only departs from there temporarily when desiring to accomplish acceleration or braking.

Absent disruptions from the rider, ground or wind gusts, no physical work needs to be done to balance, because balance is a state of equilibrium.

It is true that for a given angular error in rider position ahead of the axle, the faster you are going the more physical work that will be needed to accelerate the wheel to correct that, because work is acceleration times mass times the distance over which it is applied.  However, what is less clear (and probably depends on efficiency curves of the motor and non-linearity of drag) is the degree to which such work to accelerate the wheel under an overly forward rider, or such omitted work to decelerate it back to a rearwards one, can simply be counted as part of the net "getting you where you were going anyway" and so fungible.

It's also tempting to suspect that rider lean angle errors will be smaller (or corrected before they grow) at higher speeds.  Yet disruptions from bumps in the pavement will probably be larger at higher speed.

 

The forward-backward balancing task is invariant under the chosen speed, apart from the wheel spinning at a different speed. Newtonian physics doesn't depend on whether the train you are in is standing still or moving with constant velocity.

While that is true, beware of incorrect conclusions.  Specifically, the work done by a unicyclist riding a train to correct a balance error is independent of the speed of the train.

However, Newton's laws also demonstrate that when the unicyclist applies force against the floor of the train car to restore their balance, the engine will have to apply a comparable force to maintain the speed of the train, and the work done to accomplish that _will_ be a function of the speed of the train over the tracks.

There's no such thing as a free lunch, but there is such a thing as getting someone else to pay for it.

[Mono]

 

However, Newton's laws also demonstrate that when the unicyclist applies force against the floor of the train car to restore their balance, the engine will have to apply a comparable force to maintain the speed of the train, and the work done to accomplish that _will_ be a function of the speed of the train over the tracks.

I don't think so. It is exactly the work put in by the unicyclist, independently of the speed of the train.

 

[PedalFaster]

I don't think so. It is exactly the work put in by the unicyclist, independently of the speed of the train.

 

No.  Work is force times distance, or acceleration times mass times distance.

The unicyclist only applies acceleration over a small distance on the floor of the car, so they do little work.

But during the same time, the train covers a great distance over the tracks, so in applying equivalent force it does many times as much work.

Another way to look at it is to evaluate the work as the change in kinetic energy - which varies as the square of velocity.  The unicyclist's kinetic energy relative to the train changes little; however, the engine must do a lot of work to change the unicyclist's kinetic energy relative to the earth, as that change happens at a much higher base velocity.

[Mono]

Seeing how the unicycle balances a rider by varying speed, separating the two factors doesn't seem to make much sense.

I believe someone made the claim that the balancing takes large amounts of energy, which seems to suggest that it can be separated. I am sure it wasn't me though, I just questioned the claim. Yet, I think the separation becomes particularly tricky for uneven surfaces, where keeping constant speed over bumps might or might not be interpreted as part of the balancing task.

[Mono]

Absent disruptions from the rider, ground or wind gusts, no physical work needs to be done to balance, because balance is a state of equilibrium.

the problem is that it's an unstable state. However you have a point: it means a perfect rider will use negligible additional energy for balancing, apart from compensating external disruptions.

[Mono]

No.  Work is force times distance, or acceleration times mass times distance.

The unicyclist only applies acceleration over a small distance on the floor of the car, so they do little work.

But during the same time, the train covers a great distance over the tracks, so in applying equivalent force it does many times as much work.

Another way to look at it is to evaluate the work as the change in kinetic energy - which varies as the square of velocity.  The unicyclist's kinetic energy relative to the train changes little; however, the engine must do a lot of work to change the unicyclist's kinetic energy relative to the earth, as that change happens at a much higher base velocity.

The kinetic energy argument depends on the reference frame. I am not so sure what the reference frame for considering the balancing task should be. As you said before, acceleration is not only part of balancing but also serves the original goal to move forward, which seems to suggest that the right reference frame might be the unicycle (in which case my invariance point would be valid). Additionally, as balancing seems symmetric, one should also get back a similar amount of energy due to balancing decelerations.

[PedalFaster]

The kinetic energy argument depends on the reference frame. I am not so sure what the reference frame for considering the balancing task should be.

 

The reference is whatever it is doing work against or colliding with.  In the real-world case, the reference for the unicycle's work is the ground.   

In your train thought experiment, the reference for the unicycle is the train, but the reference for the train engine (which is doing most of the work) is the ground.  Of course if the unicycle rides out the open door of a box car, the energy for the subsequent collision with the ground or trackside signage is the kinetic energy with reference to one of those stationary objects, rather than the train...

 

Additionally, as balancing seems symmetric, one should also get back a similar amount of energy due to balancing decelerations.

 

Yes, that is indeed already raised as one of the interesting questions.  The most obvious hesitation is that the efficiency of the mechanism certainly varies with speed, so you can't perfectly bank bursts (or letups) of energy to resolve balance issues against more time-average transportation goals.

With a decent rider on a smooth surface, you shouldn't be able to observe the balancing scheme making corrections, so deviations from time-average power are probably quite small, and any resulting loss to differential inefficiency even smaller.  But get a wiggly rider or a bumpy road and it might be a different story.

[PedalFaster]

Just to put some bounding numbers on things, for shallow grade angles we can find the power liberated (to all mechanisms) while dropping down a hill at a given speed by

watts = 2.7 * mass * speed (km/hr) * grade (percent as fraction)

So for example, a 48 kilo rider on a 12 kilo wheel descending a 10 % (.1) slope at 20 km/hr liberates 324 watts of power.  Obviously some of that goes to air drag, rolling resistance, bearings, and electrical losses - but it puts an upper bound on the steady state regenerative power which could be flowing into the battery under those conditions.

Or for a given loss of altitude, the energy liberated can be calculated by

watt hours = .0027 * mass * drop (meters)

So for example an 80 kg rider on a 12 kg wheel descending 1400 meters down the Mt. Washington auto road would theoretically produce 347 watt hours of power - on the same order of magnitude as the total capacity of some battery packs, though charging efficiency would be far from perfect.

The calculations work the other way too - these are the required power and total energy to climb a hill, assuming perfect efficiency without looses to other causes.

 

[MaxLinux]

On 9/14/2015 at 8:44 PM, Cloud said:

Usually happens to me when i cross the Brooklyn Bridge on my kingsong

Riding EUC across the Brooklyn Bridge must be amazing! I'd love to see some video footage of that!

[Cerbera]

On 14/09/2015 at 9:48 PM, MrBump said:

Seems like a lot of energy goes into keeping the rider upright.  Has anyone here got to the bottom of a hill with more lights than they did at the top?

Yes. I went from 4 lights to 1 when I made my Airwheel X5 go up the tallest hill in Hampshire, but then got 1 light back going back down it.