Here's an interesting efficiency thought experiment for you. Those without an engineering/geeky mind may wish to move on .
When driving at high speeds, drag consumes the most amount of energy. When driving at low speeds, other effects consume the most amount of energy.
Imagine I'm going 75 MPH on the freeway and I take an off-ramp. Ahead, the ramp merges onto a road with a 55 MPH speed limit, and my goal is to enter that road having consumed the least amount of energy after leaving the highway.
Obviously, braking to the target speed of 55 MPH is not the best thing to do. We all know that regen is more efficient than braking.
So we have two remaining options to slow down efficiently:
1. Press the pedal enough to coast--no regen, no power delivery--until I hit my target speed of 55 MPH.
2. Allow the car to regen by some amount until I reach the target speed of 55 MPH.
In option (1), I don't consume any power (or generate any), but I remain at a higher speed for a longer time. As a result, the car is exposed to higher drag forces for a longer time.
In option (2), I regen at a maximum efficiency of about 85% (so I lose 15% due to regen alone), but I spend less time at the higher speeds where drag forces are higher.
So the question: Which option is more efficient?
In my college days, I (probably?) could have solved this. But with two young kids running around, I'm too lazy and/or time-constrained to make an effort. So maybe someone out there wants to impress us .
Assumptions:
1. Drag force is 1/2 (rho) v^2*Cd*A. (Pick a reasonable density, area, and a Cd of 0.24).
2. Energy consumed by drag force is the drag force times the distance over which it applies (F * d). However, since F is not constant, you must use the integrated version: Energy = the integral of F over the distance d.
3. Speed is reducing from 75 MPH to 55 MPH.
4. Assume regen efficiency of 85%. Either assume a regen rate (like 30 kW), or--for bonus points--leave it as a variable to test the result at different values!
5. Ignore other energy consumption effects--(bearing friction, rolling friction, HVAC, etc.) as I believe they are relatively small compared to these two. Besides, most of those can be considered close enough to be identical in both situations.
6. The road is level.
Anyone want to propose a solution? (This will require some paper and a writing utensil .
So why do I ask?
First, I'm curious. Not only is it not directly obvious to me which is better in this scenario, but I'm fairly certain it changes depending on the speeds involved. If going from 35 to 15 MPH, for instance, coasting is likely MORE efficient than regen, since drag forces are relatively small. At some speed, regen becomes more efficient. WHERE is that boundary?
Second, you'll notice that when in cruise control and you drop the set speed, the car uses regen to reach that speed. Perhaps the engineers at Tesla already considered this--or perhaps it's required by law--but it would be cool if the car considered the current speed and the set speed, and used the most efficient method of getting there. Not a huge energy difference granted, but at the very least something to consider.
OK...now solve, ye bright minds!
Note: I solved this in Post #68 if you want to jump to the good stuff!
When driving at high speeds, drag consumes the most amount of energy. When driving at low speeds, other effects consume the most amount of energy.
Imagine I'm going 75 MPH on the freeway and I take an off-ramp. Ahead, the ramp merges onto a road with a 55 MPH speed limit, and my goal is to enter that road having consumed the least amount of energy after leaving the highway.
Obviously, braking to the target speed of 55 MPH is not the best thing to do. We all know that regen is more efficient than braking.
So we have two remaining options to slow down efficiently:
1. Press the pedal enough to coast--no regen, no power delivery--until I hit my target speed of 55 MPH.
2. Allow the car to regen by some amount until I reach the target speed of 55 MPH.
In option (1), I don't consume any power (or generate any), but I remain at a higher speed for a longer time. As a result, the car is exposed to higher drag forces for a longer time.
In option (2), I regen at a maximum efficiency of about 85% (so I lose 15% due to regen alone), but I spend less time at the higher speeds where drag forces are higher.
So the question: Which option is more efficient?
In my college days, I (probably?) could have solved this. But with two young kids running around, I'm too lazy and/or time-constrained to make an effort. So maybe someone out there wants to impress us .
Assumptions:
1. Drag force is 1/2 (rho) v^2*Cd*A. (Pick a reasonable density, area, and a Cd of 0.24).
2. Energy consumed by drag force is the drag force times the distance over which it applies (F * d). However, since F is not constant, you must use the integrated version: Energy = the integral of F over the distance d.
3. Speed is reducing from 75 MPH to 55 MPH.
4. Assume regen efficiency of 85%. Either assume a regen rate (like 30 kW), or--for bonus points--leave it as a variable to test the result at different values!
5. Ignore other energy consumption effects--(bearing friction, rolling friction, HVAC, etc.) as I believe they are relatively small compared to these two. Besides, most of those can be considered close enough to be identical in both situations.
6. The road is level.
Anyone want to propose a solution? (This will require some paper and a writing utensil .
So why do I ask?
First, I'm curious. Not only is it not directly obvious to me which is better in this scenario, but I'm fairly certain it changes depending on the speeds involved. If going from 35 to 15 MPH, for instance, coasting is likely MORE efficient than regen, since drag forces are relatively small. At some speed, regen becomes more efficient. WHERE is that boundary?
Second, you'll notice that when in cruise control and you drop the set speed, the car uses regen to reach that speed. Perhaps the engineers at Tesla already considered this--or perhaps it's required by law--but it would be cool if the car considered the current speed and the set speed, and used the most efficient method of getting there. Not a huge energy difference granted, but at the very least something to consider.
OK...now solve, ye bright minds!
Note: I solved this in Post #68 if you want to jump to the good stuff!
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