-
Want to remove ads? Register an account and login to see fewer ads, and become a Supporting Member to remove almost all ads.
Regen vs. Coasting
- Thread starter Todd Burch
- Start date
-
- Tags
- Driving Dynamics Model S
Thank you - I hoped that was the case though until my own car arrives am too distant from the nearest Model S to check it out.
the principle still applies as the losses when acting as a motor and as a regen unit may well differ.
Does anybody know what the regen on the dash is actually measuring? Is 60kW the power going into the battery? Or is that the power coming out of the motor and into the charging system? Or the power at the axle going into the motor? I'm guessing it's the DC power into the battery but don't know.
> My point was that at higher target speeds, the speed you have to hold is higher, and therefore the energy usage increases significantly. [Todd Burch]
When I hit an exit ramp at high speed my strategy is to immediately regen down to that magic speed that will coast me to 40 mph as I hit the first intersection (where no stop required). A commuter doing this daily could get really good at it. If you DO need to stop then a final full regen down to the 7 mph cut off at which point you brake. (But this might not exactly be Todd's point).
> all is serene in the middle [lane]. [stevezzzz]
My 'diesel vertigo' only allows me enough tether to enjoy going down this massive descent in regen, where I'm free to modulate speed within its accel/decel range. I.e. you accelerate by reducing regen, decelerate by increasing regen. Semis are prone to pop into your center lane so being OFF CC gives you immediate full regen as opposed to waiting for the brake light to come on (small point admittedly). But mostly just the fun of working one's way down the 12 miles just using regen - worth the trip from anywhere to enjoy such a decline. :smile:
--
When I hit an exit ramp at high speed my strategy is to immediately regen down to that magic speed that will coast me to 40 mph as I hit the first intersection (where no stop required). A commuter doing this daily could get really good at it. If you DO need to stop then a final full regen down to the 7 mph cut off at which point you brake. (But this might not exactly be Todd's point).
> all is serene in the middle [lane]. [stevezzzz]
My 'diesel vertigo' only allows me enough tether to enjoy going down this massive descent in regen, where I'm free to modulate speed within its accel/decel range. I.e. you accelerate by reducing regen, decelerate by increasing regen. Semis are prone to pop into your center lane so being OFF CC gives you immediate full regen as opposed to waiting for the brake light to come on (small point admittedly). But mostly just the fun of working one's way down the 12 miles just using regen - worth the trip from anywhere to enjoy such a decline. :smile:
--
This post confuses me a bit. I guess the idea is to maximize efficiency by slowing down (which begins regen) but then giving the car enough juice to cancel regen while you coast until it's safe to pick up speed again, or alternatively, to slowly coast to a stop when necessary.
Is this any different than the way I was taught to drive efficiently, which was to always looks ahead and use the brake as little as possible?
Is this any different than the way I was taught to drive efficiently, which was to always looks ahead and use the brake as little as possible?
I have wondered that myself. But in addition, I have also wondered what the energy usage is measuring - kW per what? I assume kW per mile? But the math doesn't seem to really line up with my 'average' kWh, or with the averages in the Energy screen.
The units are not kW per anything. It's a measure of power, not energy. It's just kW. Power is the rate at which you are converting energy while kWh is a measure of energy. For example, if you could regen at 60 kW for one hour, you would put 60 kWh back into your battery. But if you only did it for a minute, you would only put 1 kWh back into your battery.
What part of the math doesn't line up with the averages in the Energy screen?
The "kW" (kilowatt) unit is already a rate. What your asking is akin to saying "horsepower per what"?
That is a good question. I can't find a discussion of this on the Tesla website. But they do have this page on the Roadster efficiency:
Energy Efficiency of Tesla Electric Vehicles | Tesla Motors
The plot is interesting for the various pieces it contains, but the numbers seem fundamentally wrong to me. For example, they show power losses from the tires losses as a constant independent of speed which is clearly impossible. It seems like they should have said the force required to overcome tire rubber hysteresis is a constant independent of speed, so that the power loss is linear in speed.
It would be nice to have a correct version of this plot for the Model S, including the net gains from regenerative braking.
The vertical axis is in Wh/mile, not Wh. Therefore if the power loss is linear in speed, the graph should be a flat line.
Thanks - got it! (It might have been clearer if they had labeled the graph 'Energy Consumption per Mile' rather than 'Power Consumption' )
Do you know if a similar graph exists for the Model S?
But if kinetic energy and drag are both proportional to the square of speed, then the reduction in drag per second due to the regen would actually be independent of speed, right?
Not quite. Ignore drag for a minute and consider 100% efficient regen for simplification. See the below picture for a proof (sorry for image quality).
Assume you apply regen at 60 kW for a second. For that second, regen is able to remove a fixed amount of energy from your kinetic energy.
As I show, as your speed increases toward infinity, the contribution of a fixed amount of regen decreases to zero.
So for a second of regen, the faster you start out, the more slowly regen slows you down. Since regen slows you more slowly, more energy is lost to drag.
Sorry if you have to turn your head...I'm at my iPad right now...
Last edited:
Not exactly. As speed increases, the loaded radius of the tire decreases (less bending of the tread area and sidewasl) and the pressure in the tire increases (because the tire heats up). Both of these reduces the rolling resistance of the tire so that it's really close to being a constant.
Yes, I get that it decelerates less, but v^2-u^2 is a constant, then if drag is proportional to speed squared, drag reduces at a constant rate.
But then the energy lost to drag (= work done by drag) would be proportional to the cube of the speed
Let u be initial speed and k be the power of the deceleration per 1/2 kilo
The after 1 second, we have speed v = sqrt(u^2-k)
Then reduction in drag energy is proportional to u^3 - (u^2-k)sqrt(u^2-k).
= u^2(u-sqrt(u^2-k))+ksqrt(u^2-k)
which increases as u increases.
So, although the regen causes less deceleration at higher speed, it would actually cause a larger decrease in drag work. Which, thinking about it, makes sense because it's just a twist on the idea that acceleration gets more difficult as you get faster: small decreases in speed make a bigger difference at high speed.
Thanks Jerry. I misinterpreted the graph assuming from the title that it was a graph of power rather than energy per mile (i.e. a force in some odd units). Doug_g caught my error. That the rolling resistance force is basically constant is plausible to me because to roll the tire a little bit requires that you deform - push in - the front of the tire (and the road) a little bit and that takes a certain amount of force applied over a certain distance. Some of the energy you put into deforming the front of the tire (where it meets the road) will be given back as the squished area moves past the vertical (directly under the car) and instead of pushing against you, this part of the tire will push the car forward as it restores its shape. But just as with a bouncing rubber ball, you don't get back everything you put in - some is lost as heat - so there is a net force pushing against the direction of travel. I don't see offhand why there should be much of an effect on the area that is squished with increasing speed (it's squished/unsquished faster, but why a smaller area?). I agree that the increase in air pressure with speed caused by higher tire temperatures should reduce resistance a bit, maybe a couple of %?
physicsfita
Member
The most straightforward way to see that would be to boost to the non-inertial frame of the tire. In this frame, the tread of the tire would be experiencing an outward fictitious force (the centrifugal force) that depends on the rotation rate. (This is the same effect that makes planets bigger around their equators than pole-to-pole.) This makes the tire bigger and rounder, reducing the contact patch.
If want to do it from an inertial frame, the argument is a little more involved: as you speed up the rotation rate, the speed of a portion of the tread must also increase. Boosting to a frame co-moving with the axle, we see that the tread portion would have to experience a greater centripetal force. This force is provided by an increased electric-force interaction with other parts of the tire, which happens because we've increased average intermolecular bond lengths. Thus, the tire gets bigger and rounder, reducing the contact patch.
JRP3
Hyperactive Member
If the tire gets bigger, meaning the diameter increases, the contact patch would also get larger.
That effect must be very small in practice. If you raise a car on a mechanics lift and then step on the gas so that the wheels are spinning fast in the air, you're not going to see the tires getting bigger. For rolling resistance to be basically independent of speed, (which it more or less is empirically), the same area of tire must be squished (and unsquished past the vertical, leading to some restoring force, but less than the force you had to use to squish, because of hysteresis in the rubber - just like bouncing a rubber ball) by the same amount when you roll forward a distance dx independently of the speed. You need to supply power that increases linearly with speed to overcome rolling resistance simply because when you double the speed, you are squishing every patch on the tire twice as often per unit time, and so generating twice as much heat. At least, that's how I understand it.
Similar threads
