A 5% grade at 70 mph provides the power that the car needs to maintain that speed, so you would be extracting no power from the battery nor storing any to it. At shallower grades, the motors would have to provide some of the power to maintain 70 mph, but you would be recovering 100% of the potential energy.
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SageBrush
REJECT Fascism
You continuing to deny it doesn't make you correct.
The power from change in height wrt time is m*g*v, where v is the vertical velocity. The roadsters mass is about 1300Kg. Looking at the power vs speed plot, the required power to maintain 70 mph is about 21KW. So 21000 = 1300 * 9.8 * v, which makes v = 1.648 m/s. If the speed along the hypotenuse of the hill is 70 mph (31.3 m/s), the sine of the angle of the hill is 1.648 / 31.3 = 0.0527. The tangent of that angle is 0.0527, so the percent grade is 5.27 %. As I said, about a 5% grade provides the power to keep the car traveling at 70 mph.
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SageBrush
REJECT Fascism
The power to offset air resistance is proportional to the cube of the speed, but that is not a useful model for a car that has significant linear losses from tyre and road friction.
Air friction is 0.5*rho*CdA*v*v
Rolling friction is m*g*RR
Integrate these two frictions together, and remember that you are throwing away ~ half of the KE in regen your way.
This graph models a possible Model 3 with a Cd of 0.21 and area of 2.22 meter squared.
You can plug in speeds if you like, but it is obvious from eyeballing the graph that in this speed range the function is far from cubic. As one example, 90 kph is 1.5x the speed of 60 kph. If your cubic model was correct the power in kW at 90 kph would be 3.96*1.5*1.5*1.5 = 13.36 kW. You have other errors and omissions in your reasoning, but this should be more than enough for you to get a clue.
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I didn't use air resistance. I used the total power curve from this plot for the Tesla roadster. This is from a JB blog. It's probably the data they developed for their epa schedules. I haven't been able to find a plot for the model s, but the results should be similar.
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SageBrush
REJECT Fascism
Good.
5 kW at 30 mph.
15 kW at 60 mph
Is that a ratio of 8 between the two powers in your world ?
No, but I didn't use air resistance in my calculations. I used the data for total power from the plot. If you check my earlier post, you'll see I made a distinction between high speed and low speed sources of drag.
SageBrush
REJECT Fascism
You keep saying that the car speed should be bled off quickly through regen because power is cubic in relation to speed. From your post #40:
Your own graph shows you that the function is far from cubic. What is so hard to understand here ?
You should have said: at 60 mph, you are bleeding off KE 3x as fast as at 30 mph but are only going twice as fast. Continue your calcs from this point, but remember that regen is ~ 50% lossy on average with a range for this scenario of 20 - 60% depending on load.
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It doesn't matter that I said that power loss due to wind resistance is cubic. I didn't use wind resistance in my calculations. What is so hard to understand here? I used the actual power required at any given speed and I took into account regen efficiency. I said I integrated the power curve. It's better to regen quickly even though it's lossy. In coasting down from 70 mph, you end up with about 100 KJ in the battery, if you regen quickly to 33 mph, that would have been dissipated as heat.
SageBrush
REJECT Fascism
Except you did not.
You said the power to move the car is cubic to speed. Here is your post #40 again, since you did not quote it:
Integrating the real power curve is a start. Show your math please
I admit that I was just using those speeds as an example of a cubic relationship. I did not check the actual values off the chart. Here's one for you from the plot:
120 mph->78.9 Kw
115 mph->69.5 Kw
78.9 / 69.5 = (120 / 115) ^3
But again, I used the plot for power requirement. I've already provided several examples. What would you like me to calculate.
Abqnative
Member
I have been lurking following this discussion. While the math is interesting I learned in college physics that some problems are best understood with simple principles. If you go up a hill you need work to gain potential energy and coming down potential energy can become kinetic energy. If you have way to recover 100%, then hills can be disregarded. With regen in a Tesla and speed limits for safety, we will always have losses with hills. If we are on flat roads, we have losses due to friction, drag and non productive waste (heating,A/C). There is no such thing as "free" coasting. Either you are using energy to maintain it, or you are losing it.
The point I got from this thread is don't "aggressively "" use regen or acceleration. Keep your speed (kinetic energy) steady. If you go slower on average (lower kinetic energy & losses) you will save energy. Playing tricks on hills does not avoid the laws of physics.
The point I got from this thread is don't "aggressively "" use regen or acceleration. Keep your speed (kinetic energy) steady. If you go slower on average (lower kinetic energy & losses) you will save energy. Playing tricks on hills does not avoid the laws of physics.
I took a short trip on interstate through town yesterday in my model s. There were lots of overpasses, so lots of relatively steep low hills. With the car on autopilot at 70 mph, the power meter never moved into the green on the downhill sides. Occasionally the orange would disappear leaving only the gray bar in the middle. At 80 mph, there was always some orange showing.
Is this what you want to see?
speed = 70 mph
StartingKE = 1/2 * mass * speed ^ 2
DO
kW= f(speed) -- fit a third order polynomial to the total power vs speed plot
dDissipatedE = dDissipatedE + 1000 * kW * dt -- accumulated energy lost to drag forces
distance = distance + speed * dt;
dHeight = distance * sine(theta) -- theta is the angle of the hill from horizontal
dPE = dHeight * mass * g -- potential energy due to change in height
dMotorEnergy = dMotorEnergy + MotorWatts * dt -- accumulated energy added to the car by the motor
KE = StartingKE - dDissipatedE - dPE + dMotorEnergy -- current kinetic energy
speed = sqrt(2 * KE / mass) -- solve KE = 1/2 * mass * speed ^2 for the new speed
time = time + dt
WHILE speed is greater than 5 mph, for instance
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SageBrush
REJECT Fascism
Cute, but the example being discussed was 60 mph down to 30 mph via early regen followed by (low) power or coasting followed by some lesser amount of regen.
Here are the relevant data points from your Roadster graph:
30 mph: 5 kW
60 mph: 15 kW
Personally, I think the integration of the power curve is clumsy but you are stubborn and want to use your method. So go ahead, but use the correct power curve and not the cubic figment of your imagination.
Here is a simpler, semi-quant way to reason the problem:
A quick regen (of a much suspect 50% regen estimate) puts 50% of the KE bled off between 60 and 30 mph into the battery and then consumes it at 5 kW while travelling at 30 mph
Coasting results in speeds greater than 30 mph on average, at less than double the 5 kW of 30 mph but with twice the available KE.
Figuring out exactly how much coasting down is more efficient than regen is a chore, but that it is more efficient is obvious from the above.
Oh, yes it's much better to hand wave than use actual data.
What figment? Power losses are cubic at high speeds where aerodynamic forces dominate as I said. I showed you that, but you are either unwilling or unable to comprehend it.
Damn. How many times do I have to tell you that I have been, from the beginning, using the total power vs speed curve data to calculate losses.
You just keep harping on this because I've shown that your assumptions were incorrect, and you keep diverting to avoid that.
It's very easy using the clumsy power vs speed curve. Computers are very happy doing clumsy things.
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SageBrush
REJECT Fascism
Yeah, I saw the discrete math and the nebulous f(some polynomial)
You might as well use a spreadsheet then, plugging in numbers from the drag equation
*cough*
And I showed you the power at 30 and 60 mph steady state from your own graph is a factor of 3 and not 8 as you thought.
And I showed you with some trivial arithmetic that your conclusion that regen should be used aggressively and early is wrong -- but by all means, do as you please.
Thats not what I thought. I told you the the 30 and 60 mph were just two numbers that were a factor of two different. Some people don't know what cubed means. I was just providing a concrete example. If you prefer, use 100 mph and 200 mph if you want it to work out to 8.
You didn't show anything. You just made a bunch of unsubstantiated statements and then said see. Where as I actually calculated the energy to both coast and drive constantly at 33 mph. It take less to drive at constant speed. Any difference in energy has to go to the battery and heat lost due to regen efficiency.
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Uncle Paul
Well-Known Member
Basics.
Going uphill uses more power than going downhill
Going faster uses more power than going slower.
Going into a wind uses more power than going with the wind. (More wind equals more difference)
Trying to compute all these (and more) to obtain the least possible current drain can be fun, but kind of an endless battle.
Bet Elon would say,,,just enjoy your fantastic car, and let the electronics sort out the details.
So, if you have enough juice to reach your next charging opportunity, just enjoy the ride.
This is not always about winning, or how many leaves you can make grow on your display.
Compared to depreciation, the electricity saved by trick driving might not be all that much.
Going uphill uses more power than going downhill
Going faster uses more power than going slower.
Going into a wind uses more power than going with the wind. (More wind equals more difference)
Trying to compute all these (and more) to obtain the least possible current drain can be fun, but kind of an endless battle.
Bet Elon would say,,,just enjoy your fantastic car, and let the electronics sort out the details.
So, if you have enough juice to reach your next charging opportunity, just enjoy the ride.
This is not always about winning, or how many leaves you can make grow on your display.
Compared to depreciation, the electricity saved by trick driving might not be all that much.
Thank you for the very informative post. I think I was looking at it all wrong when I first started the thread. I thought that the way the motors were designed they always had high resistance present which one had to overcome by pressing the go pedal and once we let go of the go pedal it would slow down the car and regenerate energy. I guess best analogy would be those toy cars you have to repeatedly push to get the motor going and let go and it takes off but slows down quickly because of the constant resistance (vs a free wheeling toy car you just push and it goes much further)
The true version you explained makes a ton more sense (there is no inherent drag on the tesla until the circuit completes by lifting off pedal)
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