When towing with an EV (especially something large like an RV) it is well understood that the #1 greatest drain on available range is wind resistance. Driving more slowly helps to mitigate this. This is of course true even when not towing, but the effect is much more pronounced when you have a giant sail of a trailer behind you.
Then there is the energy lost due to gravity when climbing hills. You can of course get a fair amount of that back on the backside of the hill through regen, but regen is not 100% efficient.
It requires a certain amount of energy just to counteract gravity when your vehicle is on a hill. Imagine that you are on a hill and hold the car still just by feathering the accelerator. You’re getting zero mi/kWh. You’ll drain the entire battery after some time (probably a few hours) without moving an inch. That energy doesn’t just disappear once you start moving. It’s there the whole time you’re on a hill, a parasitic load on your battery that grows linearly with the amount of time you spend on the hill.
So based on that, there is clearly some advantage to completing the ascent of a hill faster, so as to spend less time on the hill (and thus spend less of that parasitic drain). However, this has to be balanced by the wind resistance. It doesn’t make sense to go 70mph towing an RV up a grade, as the additional losses due to wind resistance would likely exceed the gains from spending less time on the hill. Conversely, driving 20mph up the hill would also not make sense, as the parasitic drain from gravity would almost certainly exceed the gains from less wind resistance.
There’s two curves here and they surely intersect at some optimal speed to climb a hill. So given your vehicle’s frontal area, Cd, angle of the grade, length of the grade, and probably a few other parameters, it must be possible to determine the optimal speed to ascend a hill. It’s surely also possible to factor in the descent, assuming something like an 70% efficient return of energy on the backside through regen.
Does anybody know if anybody has worked out such a formula? Maybe the wizards at ABRP?
Speed also matters because going up a 25 mile long grade at 45mph takes 33 minutes, but taking that same grade at 60mph takes just 25 minutes. You’re fighting gravity for an extra 8 minutes at slower speed.
I understand what you are saying about how the work required to climb the hill is the same regardless of how fast you climb it. But that looks at the problem in a “spherical cow on a frictionless plain” kind of way. If it really was true that it’s all the same no matter how fast you go, then it should be possible to climb a grade at some ridiculously slow speed – say 1mph, and it would use the same amount of energy as climbing it at 50mph?! No…that’s not right. That super slow 1mph climb may consume the same amount of power to climb the elevation required ast the 50 mph climb, but the overall energy consumption of the entire system is going to be a lot higher for the 1mph climb.
That’s the crux of my question. And the article you linked about the Tesla Semi (and EV trucks in general) is more about “can they drive steep grades at the speed limit”, not “what is the optimal speed to climb a grade”. The computations done in the blog post are all assuming driving at the speed limit, not comparing the different consumption amounts for climbing at different speeds. It also completely hand-waves away the effects of wind resistance, which is the crux of what I’m asking.