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Discussion Starter #1
Here's a question for you.

If you have a hill to descend, what is better ( and if possible with an equation prove it) driving into the hill at 30 mph and descending at same speed (gradient for this experiment would be one steep enough to maintain 30 mph but not accelerate) getting the benefit of x kWh from the regeneration, OR using more power to arrive at top of the hill before descent at a higher speed so you descend faster and therefore create more kWh for same distance travelled?

Can you tell I'm a programmer by trade?! (Although don't do much programming these days being the boss!)

I'm betting it is better to descend faster, but the equation is mightily complex!
 

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I don't have equations but you never get back in regeneration what you expended to get the speed up in the first place - my observation is very (very) roughly 1/2. Also speed always costs - you never get something for nothing.
 

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Discussion Starter #3
I agree that's true on the flat, but I believe there is a tipping point when there's a gradient involved.

The equation will involve extra energy spent versus increase in g from a higher speed descent.
 

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Fairly simple answer. Avoiding the need for equations...

Electric motors are not 100pc efficient, and so the more you accelerate the more energy you lose getting to the top because you're applying more force. If you go up gently you use (say) 110pc of the minimum energy required to reach the top. If you accelerate up the hill you use 110pc of a much bigger number.

Regeneration is way less than 100pc efficient, so you only recover a small percentage on the way back down regardless of how much that is or how fast you go.

The most efficient way up and down a hill is to be going at the exact speed to coast up just reaching the top before stopping, and using nothing but gravity down the other side. That might just be energy positive for the cycle of the hill alone, but isn't overall because you've already supplied the energy to reach the initial speed.
 
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Due to losses everywhere in the system (heat, internal friction, chemical) there will never be a tipping point.
Using X kWh to ascend a hill and regenerating Y kWh descending, X will always be much higher than Y.
The faster the ascent the more air resistance (drag) comes into play, causing even greater losses (on the descent too).
The best case scenario is probably ascending at 1mph (practically no air resistance, less heat generated in the motor and batteries, less loss (heat) in the chemical/electrical conversion, etc) and then descending at 1mph with practically no regeneration, but then again very little loss in the system.
Worst case is storming up the hill at 90mph, loads of air resistance (using energy from the battery), motor heating up (using energy from the battery), battery heating up (using energy from the battery). On the descent, you get lots of regen but loads of air resistance slowing the car (wasting kinetic energy). However, regen is horribly inefficient, the motor heats up (wasting kinetic energy), the actual kinetic energy to electrical energy conversion is poor and stuffing lots of energy back into the battery over a short space of time causes the battery to heat up wasting the electrical energy being (re)generated by the motor*.
So, in summary (sorry that went on longer than planned), probably best at 30mph.

*This is way the sail/coast mode is being talked about on the Volkswagon XL1.
 

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Discussion Starter #7
I actually never mentioned anything about a hill ascent in my op, as I'm guessing anyone would leverage that because of all the things already mentioned and more importantly gravity, it's blindingly obvious that you can't even maintain the energy uphill from down.

No, my point was if you start on a flat road.

I appreciate you'll have to back up the hill to get home!
 

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I'm not so sure, hence why I asked the question, and would need an equation as proof!
The Drag equation is:

Drag Force = 0.5 * Fluid Density * Velocity^2 * Drag Coefficient * Surface area in direction of travel

Given that the Fluid Density, Drag Coeff and Surface area are constant for the same car in the same air, the Drag Force goes with the Square of the Velocity. If you go twice as quick it's four times as hard to get there.

The extra losses from the heat generated within the powertrain at speed also rise, but these are secondary effects and are smaller by comparison.
 

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Discussion Starter #9
Or put it another way, if no one can write the equation (I couldn't!)

When you arrive at the top of a hill and examine the gradient and distance to next car, do you A) simply maintain speed into the hill B) take your foot off the accelerator and have to apply the gas pedal half way down the hill as the regen slows you more than gravity pulls you or C) pick up speed then take your foot off the accelerator when you believe you have enough momentum to get to the bottom/car in front.

I am a C man!
 

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Discussion Starter #10
Don't mean to be picky, but the equation is complex enough without the ascent being included as well, so can we leave ascent out for now?

Drag is one small part of the energy created/diminished in this complex model, but I appreciate your googling for me ;)
 

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The killer is air resistance.
In terms of energy, with no energy being transferred to the motor and coasting up the hill:
Kinetic energy is transferred into
1. potential energy due to it gaining height
2. work done against the air and ground resistance
3. sound and heat
Now, focussing on the air resistance
Drag = 0.5 x ro x v squared x S x CD
S is the frontal area (big flat area bad)
CD is the drag coefficient
Ro is the air density
V squared is the velocity squared. If you treble the velocity, v squared is multiplied by 9.
Speed is the enemy to be avoided at all costs as the work done against the air is never regained.
 

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Hi Dominic, :).
Whatever you do you can't avoid the physics, ignoring the ascent, if you find yourself at the top of a hill, the most efficient way down is still at 1mph, due to the losses incurred. At 30mph you'll get more air resistance, leading to slowing down (or using more energy to stay at 30mph) or if it's steeper, some lossy regen. At 90mph you'll almost certainly be using energy to stay at 90 (unless it's very steep) due to massive air resistance, or getting lots of v. lossy regen if it's really steep.
Essentially the faster you move the worse efficiency gets...
That's why a Bugatti Veyron needs 1000bhp to get to 250mph... but a 109bhp Leaf could do 100mph (if it was wasn't restricted).
 

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Drag is one small part of the energy created/diminished in this complex model, but I appreciate your googling for me ;)
Drag is not part of the energy created - it's what the energy applied to the system has to overcome to achieve the desired outcome I.e. Motion, and it's by far the biggest factor in what you're trying to discover. I did indeed Google the equation to make sure I got the term definitions right, because it's been two decades since I was a rocket scientist and used the thing in anger.
 

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Or put it another way, if no one can write the equation (I couldn't!)

When you arrive at the top of a hill and examine the gradient and distance to next car, do you A) simply maintain speed into the hill B) take your foot off the accelerator and have to apply the gas pedal half way down the hill as the regen slows you more than gravity pulls you or C) pick up speed then take your foot off the accelerator when you believe you have enough momentum to get to the bottom/car in front.

I am a C man!
I would go with D)- use variable regen levels using the throttle pedal to adjust my speed to reach the car/bottom of hill at the pace I desire. It probably means I miss out on some speed or regen on occasions, but my aim is always to carry as much speed as I need to complete the next segment of the journey as effortlessly as possible.
 
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