ThosEM
Space Weatherman
Ok, here's my take on this question, based on a modest amount of analysis, but assuming superchargers are always sufficiently close together to avoid running out:
The total trip time is the sum of the driving time plus the charging time:
T = D/V + Dr/Vc
where Dr is the rated distance used for the trip (to be restored by charging) and Vc is the average rate of charging (in rated distance units).
Let's assume that Dr = D * V/Vr, where Vr is the speed at which one gets rated range. This says the energy consumed (rated range units) is proportional to the speed driven, which is pretty consistent with published Range vs speed plots.
So now we have:
T = D/V + D*V/(Vr*Vc)
If you plot T vs. V here, or do a bit of calculus, you’ll find that the minimum time for the trip occurs for:
Vopt^2 = Vr*Vc
Too slow and the driving time increases and dominates; too fast and the charging time increases and dominates. Hit it right and you have the minimum trip time.
If Vr is 100 kph (it might be a bit higher in good weather) and Vc is 300 kph (typical average for a supercharger session starting out at 600 kph and tapering down), this says the minimum trip time is obtained by driving at an optimal average speed of:
Vopt = √(100*300) = 175 kph or 108 mph
On the other hand, with L2 charging at a rate of perhaps 45 kph / 30 mph:
Vopt = √(100*45) = 67 kph or 42 mph
Bottom line: your charging rate does determine the optimal (minimum time) speed for road trips, but if you drive that fast with superchargers, you'll spend as much time charging as you do driving, which will not feel optimal and probably get you a speeding ticket. Or if traveling with L2 charging, your gains from driving faster than optimal will indeed be eaten up charging and more so. With moderate CHAdeMO charging, you'll need to spend about the same time charging as driving and it won't feel optimal.
A good compromise is to speed up as much as possible as the next supercharger approaches.
The total trip time is the sum of the driving time plus the charging time:
T = D/V + Dr/Vc
where Dr is the rated distance used for the trip (to be restored by charging) and Vc is the average rate of charging (in rated distance units).
Let's assume that Dr = D * V/Vr, where Vr is the speed at which one gets rated range. This says the energy consumed (rated range units) is proportional to the speed driven, which is pretty consistent with published Range vs speed plots.
So now we have:
T = D/V + D*V/(Vr*Vc)
If you plot T vs. V here, or do a bit of calculus, you’ll find that the minimum time for the trip occurs for:
Vopt^2 = Vr*Vc
Too slow and the driving time increases and dominates; too fast and the charging time increases and dominates. Hit it right and you have the minimum trip time.
If Vr is 100 kph (it might be a bit higher in good weather) and Vc is 300 kph (typical average for a supercharger session starting out at 600 kph and tapering down), this says the minimum trip time is obtained by driving at an optimal average speed of:
Vopt = √(100*300) = 175 kph or 108 mph
On the other hand, with L2 charging at a rate of perhaps 45 kph / 30 mph:
Vopt = √(100*45) = 67 kph or 42 mph
Bottom line: your charging rate does determine the optimal (minimum time) speed for road trips, but if you drive that fast with superchargers, you'll spend as much time charging as you do driving, which will not feel optimal and probably get you a speeding ticket. Or if traveling with L2 charging, your gains from driving faster than optimal will indeed be eaten up charging and more so. With moderate CHAdeMO charging, you'll need to spend about the same time charging as driving and it won't feel optimal.
A good compromise is to speed up as much as possible as the next supercharger approaches.