Downhill regen energy/power formula

[Alternator]

Hello,

I'm questioning the use of some formulas (the ones used to calculate energy and power on flat and climbing) in the event of downhill slowdown (not complete stop). I'm not sure about the formulas being adequate for what I use them to, so your help is greatly appreciated.

The idea is to compute how much energy can be harvested from going downhill at a reduced speed vs coasting on a heavy touring bike, and the corresponding power requirements.
It is NOT about the motor efficiency at given RPM or anything electrical. We only focus on mechanical energy. Electrical losses will be computed later.

I compute the energy and power from first calculating force vectors. Then we just have to multiply by the speed to get the power, and the distance to get the work (energy). I use SI and metric units. The system is at equilibrium, constant speed. We won't study the kinetic force of acceleration and stopping (regen braking) as it's not we're focusing on.

At constant speed, the sum of the rolling resistance + air drag + slope force equals zero. Fr+Fa+Fs=0

- Rolling resistance
Fr=Crr.m.g.cos(a)

With Crr the rolling resistance coefficient, g (gravity acceleration) of 9,81 m/s^2. For any reasonable incline (<20%), cos(a) -> 1 so if you feel lazy, get rid of that term (I'll plug it in a spreadsheet so I don't care). Crr for touring bike is around 0,01. I'll use a mass of 120kg.

- Air drag
Fa=(1/2).rho.Cda.(v+vw)^2

With rho density of air of 1,22kg/m^3; Cda the combination of S.Cx of 0,4 for a touring bike, v the speed of the bicycle, and vw the speed of the wind in the axis of motion.

- Slope force
Fs=m.g.sin(a)

Gravity my friends. m the mass (still 120kg), g still 9,81 m/s^2, and a the angle of the incline in ° (or rad). You'll get it from the incline in % by arctan(angle in %). (if you forget this step, the result will be less precise, but in the same ballpark anyway).

I want to climb a hill
So let's imagine a 10 km long, 10% incline, that I want to climb at 10 km/h with no headwind.
Fr=12N, Fa=1,9N, Fs=117N (gasp) so Fresist=131N.
That gives a consumed energy of 1310 kJ and a constant power of 363W

Now I'm on top, exhausted and I come back home.
It's still 10 km long, 10% incline (but this time is downhill). Let's say I go downhill 10 km/h too (for computing sake, no way in real life).
Fr=12N, Fa=1,9N, Fs=-117N. Same thing but with the slope the other way. Fresist=-104N
The energy I'll have to dissipate in the brakes (or harvest via regen) is 1040 kJ at a constant power of 288W given I don't pedal.

The equation allows me to know which speed I'd go without braking (or regen), that is when all the forces cancel out. It gives about 75 km/h
Which sounds coherent given the load and the slope.
It also tells me that if I wanted to go dowhnill at 40 km/h, id' be able to harvest 753 kJ but would need to dissipate or absorb 837W. Less energy because going faster makes wind rob me energy, but more power as I harvest this energy in less time.

If I check with the gravitational potential energy, Egp=m.g.(h2-h1). A 10% slope 10km long has a 1km elevation, m and g still the same, I get 1177 kJ, slightly more than computed at 10km/h which is logical: at 10 km/h I start losing energy by rolling resistance and air drag anyway. So it seems correct.

If I add the cinetic energy, (1/2).m.v^2, I get at 10 km/h I get some 463 Joules more, and at 40 km/h some 7,4 kJ more.

The question is: do these equations apply for regenerating? Are the energy and power computations correct? I would think that, yes, but the power seems quite abrupt to me.
Could you confirm or correct my assumptions?

Thank you,
Nicolas

 

[MadRhino]

Sorry that this reply may seem out of topic, according to math. This is only about my personal experience with regen.

Real world regen current back in the battery is never close to theorical regen. Riding downhill on regen is hard on controllers, hot for the motor, and all the fun of DH riding is lost. I find the range advantage is not worth the reliability inconvenience. The only interesting feature of regen is braking performance. My controllers are set to regen on brake switch only and the switch is independent from the brake lever operation. The more powerful your setup will be, the less often you will use regen because its stronger braking power makes it interesting only for emergency braking situations. A good front brake is more important than any regen setup for braking precision and stopping distance improvement.

 

[John in CR]

I'm pretty sure the guys with heat stress in their systems from regen braking is from low speed regen where the motor is operating in it's very inefficient range, which I take works similarly in forward motion or braking.

For best usefulness on hills you need a fairly low regen current limit setting. It will of course vary by winding, but I have my cargo bike set that way. It has a high Kv winding and the regen limit produces between 20A and 25A braking at speeds above 25mph or so. My battery is typically at near 80V, so that's 1600-2000W of energy added back to my battery. That bike pushes a pretty good load about 275lb all up, often another 50lb in cargo, and it's low slung so definitely requires some braking even on a 8% downhill grade on the highway where I commonly stay on the regen for long periods to keep speed below 50mph. On really steep stuff mechanical braking is needed too.

If your lowest setting is still too forceful, then (at least with the Xiechang controllers) simply increase shunt resistance to fool the controller into thinking current is higher than it is, the opposite of the typical shunt mod for more power. Then program the forward current limits higher to get back to where you were for forward motion.

John

 

[adrian_sm]

The question is: do these equations apply for regenerating? Are the energy and power computations correct? I would think that, yes, but the power seems quite abrupt to me.
Could you confirm or correct my assumptions?

Nicolas,

Looks about right. Don't need to get any more accurate as all the regen efficiency will swamp that later.

I noticed in my tests that my controller heats up a lot more under regen, than under full power delivery. Same with the motor. So this may limit how much you can rely on the regen for looooooong downhill braking depending on what motor/controller you have.

BTW I measured the regen torque curve for various speeds, and settings on an EB306 contoller. You may find the results useful. Your actual regen will obviously depend on your system voltage, motor, and controller. But I was surprised by the shape of the curve.

http://www.endless-sphere.com/forums/viewtopic.php?f=30&t=46210&start=15#p675415

 

[Ken Taylor]

Using the same model in a spreadsheet at https://docs.google.com/spreadsheet/ccc?key=0AnRMBzgiNpwGdDVwOFhiT2tyMlFKWDFaMFhyczhVQVE#gid=0 with your numbers gives the same results for 10km/hr downhill and 75 km/h but differs from your numbers in the bit that says:-

It also tells me that if I wanted to go dowhnill at 40 km/h, id' be able to harvest 753 kJ but would need to dissipate or absorb 837W.

Here you have 840W available for harvesting.

Also you add:-

If I add the cinetic energy, (1/2).m.v^2, I get at 10 km/h I get some 463 Joules more, and at 40 km/h some 7,4 kJ more.

But if you start at zero speed and end at zero speed then change in kinetic energy = 0. If you start stationary and end at 40km/h you subtract rather then add the kinetic energy.

You ask:-

Could you confirm or correct my assumptions?

I hope the above is useful.

On your model parameters 0.4 CdA is probably too low for someone that weighs 120kg and the Crr value is probably too high. I use 0.004 and there is some estimates at http://www.industrializedcyclist.com/tiretest.pdf . I've been wondering how much power is ideal for motor assist at http://blog.urremote.com/2012/12/what-would-it-take-to-cycle-fast-when.html .

 

[Alternator]

Excellent!
Thank you very much

Ken, your article made me really laugh. So true what you say about electrical power being highly addictive!
My connection has some problems so your spreadsheet doesn't load, I'll check it again later. I don't fully understand what difference you are talking about? Or is it that I computed 837W while you computed 840W? (by "absorb" I meant harvest).

Thank you very much for the reminder on kinetik energy, you're absolutely right.

Re. the values, I used 0,4 Cda and 120 kg because I'm considering heavily loaded panniers... I would need to measure the frontal area of the bags but as they are quite in the shadow of my legs I don't know how that sums up. It might quite possibly be higher. For rolling resistance, it will be for large touring tires ala Schwalbe Marathon on backcountry roads... not the best rolling stuff.

Adrian, your experiment is absolutely brillant and will help me tremendously. Thank you very much!

John and MadRhino, thank you for reminding me to mind the gap between theory and practice xD

 

[Ken Taylor]

My connection has some problems so your spreadsheet doesn't load, I'll check it again later.

I hope it worked.

I don't fully understand what difference you are talking about? Or is it that I computed 837W while you computed 840W? (by "absorb" I meant harvest).

Well I didn't understand the difference between "harvest and dissipate". What I meant was there is 840W available not 753 kJ + 837W. The difference between 837 and 840 is immaterial.

Re. the values, I used 0,4 Cda and 120 kg because I'm considering heavily loaded panniers... I would need to measure the frontal area of the bags but as they are quite in the shadow of my legs I don't know how that sums up. It might quite possibly be higher. For rolling resistance, it will be for large touring tires ala Schwalbe Marathon on backcountry roads... not the best rolling stuff.

We are guestimating. Cda. 0.4 is reasonable and it would need measurement, like that described at http://www.trainingandracingwithapowermeter.com/2011/04/estimation-of-cda-from-anthropometric.html to do any better.
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