Future EUC Wishlist

[Mono]

6 minutes ago, Nic said:

Interesting ... shouldn't the downward force due to gravity for 10 degree slope be the vertical component of F = m * g * Sin(slope) = 100kg * 9.8m/s² * sin(10^o) = 170N

Therefore P = FV = 170N * 1m/s = 170W that is available for regen.

I'm not sure if this is correct, but it seems more reasonable to me as my bicycle brakes don't catch fire when going down hill.

Consider it the other way around: could you (assuming for a moment you weigh 100kg ;-) drive up a 20% incline at 20km/h using just 170W pedal power? No way. It indeed takes a little over 1kW to do this.

[Nic]

33 minutes ago, Mono said:

Consider it the other way around: could you (assuming for a moment you weigh 100kg ;-) drive up a 20% incline at 20km/h using just 170W pedal power? No way. It indeed takes a little over 1kW to do this.

I couldn't do it on a flat never mind a hill.

You would need to add in the other 180W for drag, so that would be 350W to do 20km/h up a 20% incline.

My best guess. ... 1kw just sounds like a lot. I think my v8 could handle a 20% hill at 20km/h and not use 1kw (its an 800W motor). I'm just speculating of course.

Our grey cells certainly get a workout on this forum.

Edited November 19, 2019 by Nic

[zeke]

Oh dear, we're going to have to untangle some misconceptions about how motors work. I'll save that for another post perhaps.

1 hour ago, xorbe said:

Why don't wheels use resistive forces instead when the battery is greater than 90% full?  (ie, coils open, then close and pull the magnets back after passing.)  Just like changing direction, I mean.

 

Interestingly enough, it already does this. The trick is, you have to go very, very slowly. But that's probably not what you wanted to hear.

It's a bit more complicated, but yes I've thought about having a mode that intentionally dissipates the energy rather than storing it in the battery. On several occasions I've found myself with a fully charged battery, wanting to descend a hill. To avoid walking like a plebian, I burned off the extra energy by zooming around like a maniac to dissipate the energy into air drag.

The energy could be dissipated as heat in a couple ways:

1. Without adding any extra components, the energy could be dissipated as heat into the motor windings and MOSFETs. This would require reprogramming the EUC firmware to make the ripple current very large, so that the RMS current is much larger and increases I_RMS²R losses. Also, the motor driver switching frequency could be increased (perhaps in short, high-frequency bursts) to burn energy into transition losses.
2. With the addition of a resistor that is switched into parallel with the battery, the energy could be dissipated as heat into this resistor. This approach is common in motor drives and class-D amplifiers to avoid "bus pump."

In either of these approaches, the extra power dissipation into the EUC would be tremendous. Normally, when you regen brake, about 90% (assuming the EUC operates with 90% efficiency) of the energy delivered to the EUC is delivered back into chemical energy in the battery, and the remaining 10% is dissipated as heat into the motor windings and driver. It is not uncommon to deliver 2kW into the EUC while descending a hill, so the cooling system is built to handle about 200W while the battery absorbs 1.8kW. If you instead dissipate all of the energy into the EUC as heat, then the cooling system needs to handle 2kW instead. That's a big difference which would require a big fan.

You could constrain the dissipative braking power to 200W or thereabouts, whatever the cooling system can handle, and only allow the rider to descend below speeds that dissipate this much power. Not much fun, but it'd do the trick. Note that cooling becomes more difficult at low speeds, where there isn't much airflow.

Another possibility is to install brakes. Unlike electronics, which must be kept at low temperatures not exceeding 150°C and therefore must have very large heat sinks in order to dissipate a given amount of heat power, brakes are a simple structure that can still function at very high temperatures and therefore don't need extravagant cooling systems. Or better yet, carry chains and drag them behind you when you need to descend on a full charge. Or perhaps, just bomb down the hill so fast that air drag eats all the energy, like a skydiver at terminal velocity. The steeper the hill, the faster you go. 

 

1 hour ago, Mono said:

Or it might be that the controller limits the regen current. Not at all unlikely, IMHO.

Any limits or changes you make to the battery current are directly tied to changes in the motor behavior, such as its torque. You can't just "limit the regen current" without changing the torque the motor produces, unless you're providing something else (besides the battery) to eat the regen energy. And of course, the motor torque must be a certain value, in accordance with the rider's weight and the grade of the slope in order to prevent the rider from simply accelerating down.

If y'all want, I can make a post about how to think about motor drives. But rest assured, when you go down a hill the energy is going into your battery. Unless you fall off, in which case it goes into your skin.

[Mono]

20 minutes ago, Nic said:

I think my v8 could handle a 20% hill at 20km/h and not use 1kw

For sure, but only without the rider. Anyways, in case you are interested: I think the problem in the calculation was that you account two times for the grade, once in the force and once in the speed, which is one time too many. That is, you need to push 170N with 5m/s to get up the grade at 18km/h, or 980N with 1m/s to get the 100kg up 1m/s in lift form.

[Aneta]

50 minutes ago, Nic said:

I couldn't do it on a flat never mind a hill.

You would need to add in the other 180W for drag, so that would be 350W to do 20km/h up a 20% incline.

My best guess. ... 1kw just sounds like a lot. I think my v8 could handle a 20% hill at 20km/h and not use 1kw (its an 800W motor). I'm just speculating of course.

Our grey cells certainly get a workout on this forum.

1kW is nothing when going up steep slopes. As previously shown, going up at 1m/s (vertical speed) for 100kg total weight is equivalent to rate of change of potential energy 980W. This is pure mechanical power, that must be 980W after losses for friction and heat in the motor/controller/battery. This simulation demonstrates that on 18% slope at 20kph, battery power is about 1500W, while mechanical power is about 1150W.

https://www.ebikes.ca/tools/simulator.html?batt=cust_84_0.2_10&cont=cust_40_100_0.03_V&wheel=16i&frame=cust_1_0.01&hp=0&grade=18&autothrot=true&throt=65.2&motor=M3525

So, about 1150-980=170W of battery power goes into overcoming motion resistance, about 1500-1150=350W goes into heating of motor/controller/battery, and 980W goes into potential energy.

800W is just motor's rating, it just says that at 60V the motor can take up to 800/60 = 13.3A current indefinitely, without overheating.

Edited November 19, 2019 by Aneta

[Mono]

37 minutes ago, zeke said:

Any limits or changes you make to the battery current are directly tied to changes in the motor behavior, such as its torque. You can't just "limit the regen current" without changing the torque the motor produces, unless you're providing something else (besides the battery) to eat the regen energy. And of course, the motor torque must be a certain value, in accordance with the rider's weight and the grade of the slope in order to prevent the rider from simply accelerating down.

or falling on their butt. I agree, either the motor produces the torque without regen or it limits the produced torque (as if this would be unheard of. While I didn't look at or reverse engineer any controller software, judging from measurements and behavior I think I have seen wheels on both of these possibilities.

37 minutes ago, zeke said:

If you instead dissipate all of the energy into the EUC as heat, then the cooling system needs to handle 2kW instead. That's a big difference which would require a big fan.

There is more than this one possibility. For example, the system heats up and when it is too hot braking becomes squishy (goes back to limited regen).

Quote

If y'all want, I can make a post about how to think about motor drives. But rest assured, when you go down a hill the energy is going into your battery.

Several independent sources and independent measurements though seem to suggest that hard braking is not always regen. AKA and discussed under "power braking" in this forum. But we are always happy to learn, that's what we are here for.

 

Edited November 19, 2019 by Mono

[Aneta]

41 minutes ago, zeke said:

If y'all want, I can make a post about how to think about motor drives.

Please.

[zeke]

3 minutes ago, Aneta said:

Please.

Okay! Give me a few days though; I have a project deadline to hit.

[Aneta]

55 minutes ago, zeke said:

Or perhaps, just bomb down the hill so fast that air drag eats all the energy, like a skydiver at terminal velocity. The steeper the hill, the faster you go. 

The problem with this is that the air drag will be mostly compensated by g-thrust (mg*sin(alpha)), so the motor has little work to do, and taking into account that motor thrust is nearly nonexistent when approaching no-load speed, the speed won't go much past terminal speed for that hill (without motor).

Perhaps, we should have something like a steam iron? A small water boiler. 1ml of water takes ~2300J to evaporate, or 0.6Wh, so a glass of water (200ml) can help us burn off about 120Wh, or about 10% of average EUC battery.

[zeke]

35 minutes ago, Aneta said:

The problem with this is that the air drag will be mostly compensated by g-thrust (mg*sin(alpha)), so the motor has little work to do, and taking into account that motor thrust is nearly nonexistent when approaching no-load speed, the speed won't go much past terminal speed for that hill (without motor).

Exactly. If the motor current is zero, then it's not absorbing any regen power into the battery. The problem of over-charging your battery is solved. 

35 minutes ago, Aneta said:

Perhaps, we should have something like a steam iron? A small water boiler. 1ml of water takes ~2300J to evaporate, or 0.6Wh, so a glass of water (200ml) can help us burn off about 120Wh, or about 10% of average EUC battery.

If the electronics were better waterproof, this could be an interesting idea. So long as you don't fall and spill boiling water on yourself.

Edited November 19, 2019 by zeke

[Aneta]

25 minutes ago, zeke said:

Exactly. If the motor current is zero, then it's not absorbing any regen power into the battery. The problem of over-charging your battery is solved. 

If the electronics were better waterproof, this could be an interesting idea.

I'm also thinking about ways of dissipating regen energy because I recently found myself in this limbo situation - I was camping somewhere in the mountains and needed to go down on my wheel, but it was charged to 100%. It was not the top of the hill, so I could go up, too. So I rode about 1km up the modest ~5% slope, then down, in hopes that I can continue going down after that. To my amuzement, I returned to initial point with the same 100% voltage as I started, so again I couldn't go down. Of course, there was no violation of conservation of energy, it's just the battety is "hot" right after charging with high current, so its voltage is not resting voltage. But it's a serious problem. Of course, better plan ahead, but in this case, it was charged fully to ascend a mountain, but plans changed and I needed to descend it.

I found that riding down slowly (walking speed) burns off the regen energy predominantly in the motor, not in the battery, as I kept a close eye on the voltage reading in Wheellog to make sure it does not exceed 4.2V/cell.

 

Edited November 19, 2019 by Aneta

[Nic]

10 hours ago, Nic said:

Interesting ... shouldn't the downward force due to gravity for 10 degree slope be the vertical component of F = m * g * Sin(slope) = 100kg * 9.8m/s² * sin(10^o) = 170N

Therefore P = FV = 170N * 1m/s = 170W that is available for regen.

I'm not sure if this is correct, but it seems more reasonable to me as my bicycle brakes don't catch fire when going down hill.

I slept well last night and the solution came to me ... my answer was incorrect, so is what I believe to be the correct answer ...

1) 100kg person (in. EUC) going DOWN a 10^o (20%) slope at 20km/h

Downward force due to gravity for 10 degree slope be the vertical component of F = m * g * Sin(slope) = 100kg * 9.8m/s² * sin(10^o) = 170N

Therefore P = FV = 170N * 1m/s = 170W max available for regen.

I will use Zeke's value for drag at 20km/k as 180W

This means that when going down a 10^o (20%) slope at a speed of 20km/h you would use 10W (180W - 170W = 10W) of power and there would be no regen!

In order to get regen you would need to be braking by going slower than this speed.

2) 100kg person (in. EUC) going UP a 10^o (20%) slope at 20km/h

In this case you are using the slope to gain mechanical advantage to reduce the required force to move up the slope, so we still have 170N required to do this (not 980N as we are not moving vertical direct lift).

Therefore P = FV = 170N * 1m/s = 170W to move up the slope.

We also have the 180W to overcome drag at 20km/h, so moving up the 10⁰ (20%) slope will require 350W (170W + 180W = 350W) of power required.

3) Conclusions

Regen power is minimal on a 10⁰ (20%) slope and for an 84V wheel can generate a MAX of approximately 2A (170W / 84V = ~2A) under the conditions assumed in these calculations.

To get regen you need to be going slow by apply braking force, preferably on a steep hill. Obviously you will get more regen power when decelerating rather than maintaining 20km/h speed down the slope, but that is not the subject of this analysis with the example provided.

I did a quick search and found this article on this topic and it seems to confirm these conclusions...

https://electrek.co/2018/04/24/regenerative-braking-how-it-works/

Quote

For smaller EVs such as personal electric vehicles, the numbers aren’t quite as optimistic. On multiple electric bicycles with regenerative braking options, I’ve generally averaged around 4-5% regeneration, with a maximum of around 8% in hilly areas. Other personal electric vehicles including electric scooters and skateboards have similar results, usually in the lower single digits. Again, keep in mind this isn’t the raw efficiency of the system (as in how much braking energy is lost in the energy transfer), it’s the effectiveness (as in how much further your range increases due to the use of regenerative braking).

...

At the end of the day, regenerative braking will never be as effective in smaller vehicles as it is in larger ones simply due to physics. Because of this, the lack of regen in e-bikes and other PEVs isn’t a deal breaker. However, the benefits of regenerative braking outside of simple energy recapture can’t be ignored. And hey, I’ll take a free 5% range increase any day!

 

Edited November 19, 2019 by Nic

[zeke]

 

1 hour ago, Nic said:

Downward force due to gravity for 10 degree slope be the vertical component of F = m * g * Sin(slope) = 100kg * 9.8m/s² * sin(10^o) = 170N

Therefore P = FV = 170N * 1m/s = 170W max available for regen.

  

You are correct in thinking that, if you pushed a 100kg object up a 10.4° incline at 20km/h without drag, that the force you would be applying in the direction up the incline (at a 10.4° angle) is 176N. This is how wedges work: they allow force reduction, similar to a lever, pulley, or gear. Instead of having to push 980N, you only need to push 176N. You forget, however, that you would be pushing the mass in a direction up the incline with a velocity of 5.6m/s, which requires 176N×5.6m/s ≈ 1kW. Levers, pulleys, gears, and wedges don't allow you to use less energy; they just change the ratio of force to velocity. Less force, more velocity equals same power.

The result is the same if you don't even consider the wedge action, and just assume you pushed the 100kg object vertically. A 100kg mass has a 980N downward force due to gravity. Traveling up a 10.4° incline at 20km/h has a vertical upwards component (in the direction of gravity) of 1m/s. So if you just pushed the 100kg vertically with velocity 1m/s, this would require a power of 980N×1m/s ≈ 1kW. Same result.

Check out some physics vids on YouTube sometime, they've gotten pretty good.

 

1 hour ago, Nic said:

I did a quick search and found this article on this topic and it seems to confirm these conclusions...

https://electrek.co/2018/04/24/regenerative-braking-how-it-works/

We're convoluting the discussion by comparing apples to oranges. That article is trying to describe how much extra range you get from regen braking under typical riding conditions, rather than how much of the regen braking energy makes it into the battery. This does not justify incorrect math.

Quote

Again, keep in mind this isn’t the raw efficiency of the system (as in how much braking energy is lost in the energy transfer), it’s the effectiveness (as in how much further your range increases due to the use of regenerative braking).

Simple thought experiment. Imagine you're driving an electric car. If you're driving in a residential neighborhood with a stop sign on every corner, you will be using your brakes frequently, and as a consequence, the fact that you have regen braking will extend your range dramatically (because instead of dissipating all your kinetic energy into heat every time you stop, you restore it). If, instead, you drive on the freeway, then you will never use your brakes and the fact that you even have regen braking will do nothing to extend your range. It is the interval between starts and stops that determines how beneficial regen braking is: with a short interval regen braking is awesome, and with a long interval regen braking is pointless.

The purpose of this article is to highlight that lightweight vehicles don't carry much kinetic energy, so the interval at which you'd have to be stopping and starting for regen to have a major impact on range would need to be much shorter than for a heavier vehicle like a car. Stated differently, if a lightweight vehicle is starting and stopping at the same interval as cars, then its range benefit due to regen braking (as a percentage) will be smaller. A deeper discussion would involve the ratio of mass to drag, but let's not get into that.

 

TL;DR: going up and down hills really does require lots of power.

Edited November 19, 2019 by zeke

[mike_bike_kite]

Better security : They could use a class 1 bluetooth on the wheel which would give a 100 meter link to your phone and warn you that the wheel is being moved. I guess you could set whether to have silent, audible and visual alarms. If you don't turn it off with your phone then the wheel won't work for an hour. That way, if your phone dies, you can still ride it home eventually. Once the alarm has been triggered though (and not turned off by your phone) after a set time It will require linking to the original app on your phone to continue working. This will then broadcast the location of the wheel continually via the app. The wheel could also have a warning mode just to tell people it's armed if they touch it. The only extra hardware cost would be the upgraded bluetooth unit.

It would also be nice if all wheels had a lockable point ie through the wheel. It would also be nice if the major components of the wheel (battery, motor and motherboard) had mac-addresses or similar) then all the parts can be logged as stolen. I know there isn't much theft of wheels but ...

[Nic]

2 hours ago, zeke said:

TL;DR: going up and down hills really does require lots of power.

Good analysis @zeke ... makes sense now.

[mike_bike_kite]

Remove the switches: The IPS i5 allowed you to just tilt the device to turn the lighting on and off (around 8:10 in Ian's video). You could also perhaps rotate the wheel clockwise to turn it on and anti clockwise to turn off. Maybe again to set the alarm. The advantage of not having switches is there's fewer things to go wrong, few places for rain to get in, the case would be stronger without the holes for switches and less stuff would be needed to manufacture the wheel. You could sense the movement using the standard hardware. It would also be quite cool to use.

[atdlzpae]

@mike_bike_kite I think it would be frustrating to use. Waterproof switches are not rocket science and are very reliable.

[Mono]

9 hours ago, Nic said:

Therefore P = FV = 170N * 1m/s = 170W max available for regen.

This computes the potential regen power when going down the slope at 3.6km/h = 1m/s. You already computed the force in direction of the slope, therefore power is simply force times speed. If the speed is 20km/h, the power will be 5.6 times larger, that is, 944W.

[Nic]

27 minutes ago, Mono said:

This computes the potential regen power when going down the slope at 3.6km/h = 1m/s. You already computed the force in direction of the slope, therefore power is simply force times speed. If the speed is 20km/h, the power will be 5.6 times larger, that is, 944W.

Its been a long time since I had to do any physics ... you can get the same result by calculating the amount of potential energy change for 1m increase in height, which over one second gives power (W = J/S) and there is only losses such as drag to then account for. If you don't use it you lose it (grey cells).

This is what is good about community ... we can share knowledge and experience and learn from each other.

[Mono]

16 minutes ago, Nic said:

you can get the same result by calculating the amount of potential energy change for 1m increase in height

exactly 

[Aneta]

1 hour ago, Nic said:

Its been a long time since I had to do any physics ... you can get the same result by calculating the amount of potential energy change for 1m increase in height, which over one second gives power (W = J/S) and there is only losses such as drag to then account for. If you don't use it you lose it (grey cells).

This is what is good about community ... we can share knowledge and experience and learn from each other.

It's actually why it's called potential energy - gravity is a potential field that in physics means that the work done to move the object between points A and B is only determined by the difference PE(B)-PE(A), not by the path taken between A and B. So, gaining 1m of elevation straight up or via an incline along any path requires the same amount of work (in the absence of friction). Even if one at 100kg went to the Moon and then returned to the point 1m higher than the start, apart from huge losses for other things and trillions of dollars spent, the energy difference is the same 980J.

[Mono]

19 minutes ago, Aneta said:

It's actually why it's called potential energy - gravity is a potential field that in physics means that the work done to move the object between points A and B is only determined by the difference PE(B)-PE(A), not by the path taken between A and B. So, gaining 1m of elevation straight up or via an incline along any path requires the same amount of work (in the absence of friction). Even if one at 100kg went to the Moon and then returned to the point 1m higher than the start, apart from huge losses for other things and trillions of dollars spent, the energy difference is the same 980J.

exactly

[zeke]

1 hour ago, Aneta said:

It's actually why it's called potential energy - gravity is a potential field that in physics means that the work done to move the object between points A and B is only determined by the difference PE(B)-PE(A), not by the path taken between A and B. So, gaining 1m of elevation straight up or via an incline along any path requires the same amount of work (in the absence of friction). Even if one at 100kg went to the Moon and then returned to the point 1m higher than the start, apart from huge losses for other things and trillions of dollars spent, the energy difference is the same 980J.

To be technical, what you are describing is a conservative field. Within conservative fields, such as gravitational, electric, and magnetic fields, the energy required to traverse from point A to point B is the same irrespective of the path taken. Said otherwise, if you start at point A, take some random meandering path, and return to point A, then the difference in energy between the start and end is zero: then you say that energy is conserved.

Edited November 20, 2019 by zeke

[Aneta]

1 hour ago, zeke said:

To be technical, what you are describing is a conservative field. Within conservative fields, such as gravitational, electric, and magnetic fields, the energy required to traverse from point A to point B is the same irrespective of the path taken. Said otherwise, if you start at point A, take some random meandering path, and return to point A, then the difference in energy between the start and end is zero: then you say that energy is conserved.

Oh, yes, my mistake, didn't know the proper term in English.

[Nic]

This doesn't work in space. It depends on gravitational field, which varies with altitude.

Its an approximation, which works well enough for most practical purposes.

At least that is what I am thinking ... or maybe it does, but potential energy is just zero ... and everything else is just losses. 

Oops ... we are off-topic again.

Edited November 20, 2019 by Nic