knightmb said:
Makes me wish some of us from here could get in on this "challenges" and surprise everyone.
Nah
You can figure out this optimization problem by solving the first derivative of the time function for zero. **
As speed increases, you only care about the change in aero related work.
So long as you are charging at 50 kW (let alone 70), you want to drive a lot faster than 30 mph.
Unless your route and driving forces you to use L2 charging. Then by all means, slow down !!
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You are right though -- these C&D drivers are pretty representative of clueless EV drivers. In that sense they offer "real world" results
A smarter but still EV clueless driver would have figured out the arithmetic challenged approach to EV racing: drive slower at the start of a leg and if possible, faster for the second half gauged to arrive at the DCFC with a low SoC. The really clever would know about the battery reserve.
[rant]
I find C&D obnoxious. They imply that EV drivers need co-pilots to be hunched over spreadsheets. As if every drive is a 1000 mile race.
[/rant]
**
You can also estimate with a little algebra:
Say 60 mph is 200 Wh/mile and 100 Wh/mile is Aero (this should be measured on the road)
The change in Aero losses are proportional to the square of the speed change so e.g. 70 mph would be (7/6)*(7/6)*100 - 100 = 36 Wh more per mile so 2.52 more kWh per hour. The higher speed saves 10 minutes per hours or about 30 minutes per 3 hours. The overhead involved in going to a charger could be some 10 minutes so net 20 minutes are saved in driving time every 3 hours. That implies that charging has to be faster than 2.52*3*3 = 22 kW to save time.