I know from my experience flying gliders cross-country that to optimize cross-country average speed you should drive at a speed that results in a battery drain equivalent to the rate of charge you'll see when you stop, 90 kW. What's that speed?
This is a decent rule of thumb, but not necessarily exact. For instance, if you're charging off super-slow 110v/12a (1.3kW), which takes 4 days to charge, your ideal speed will still be about 25mph, which drains about 5kW continuous while driving. The "sweet spot" is rather the speed at which going one extra mile per hour "breaks even" as far as extra recharge time required.
Breaking out the math: let's define the function p(v) to be the power (in kW) required to maintain velocity v in MPH.
Then assuming 90kW continuous supercharge, superchargers 90 miles apart (becaues it makes the math work out cleaner), and no overhead time for stopping to plug in, the "cycle time" t(v) from leaving one supercharger to leaving the next supercharger can be calculated as:
t(v) = (time spent driving) + (time spent charging) = (90 / v) + ((p(v) / 90) * (90 / v)) = (90 + p(v)) / v.
Using calculus (yes we're going there) to solve this equation, we want to set the derivative of the function t(v) equal to zero:
t'(v)= p'(v) / v - (90 + p(v)) / v^2 = (v * p'(v) - 90 - p(v)) / v^2 = 0.
From here, we can simply crunch the power vs speed data to find the approximate speed v for which this is true.
For the Roadster, I found this graph:
http://kilowatt-age.com/yahoo_site_admin/assets/images/image003.323212833_std.jpg
and let's assume the Model S uses 25% more power than the Roadster at any given speed.
So from the graph, we can eyeball some data points:
70mph: Wh/mi = 300 * 1.25 = 375, continuous power p(70mph) = 375 * 70 = 26.25 kW
80mph: Wh/mi = 360 * 1.25 = 450, continuous power p(80mph) = 450 * 80 = 36 kW
90mph: Wh/mi = 425 * 1.25 = 530, continuous power p(90mph) = 530 * 90 = 47.7 kW
100mph: Wh/mi = 500 * 1.25 = 625, continuous power p(100mph) = 625 * 100 = 62.5 kW
110mph: Wh/mi = 570 * 1.25 = 712, continuous power p(110mph) = 712 * 110 = 78.3 kW
120mph: Wh/mi = 660 * 1.25 = 825, continuous power p(120mph) = 825 * 120 = 99 kW
and from these numbers we can numerically approximate p'(v), which is simply the marginal increased power required to go one extra mph:
p'(80mph) ~= (p(90) - p(70)) / 20 = 1.073 kW / MPH
p'(90mph) ~= (p(100) - p(80)) / 20 = 1.325 kW / MPH
p'(100mph) ~= (p(110) - p(90)) / 20 = 1.53 kW / MPH
p'(110mph) ~= (p(120) - p(100)) / 20 = 1.825 kW / MPH
So recalling that t'(v) = (v * p'(v) - 90 - p(v)) / v^2, we can plug these in and calculate:
t'(80mph) = (80 * 1.073 - 90 - 36) / 6400 = -0.00628
t'(90mph) = (90 * 1.325 - 90 - 47.7) / 8100 = -0.00228
t'(100mph) = (100 * 1.53 - 90 - 62.5) / 10000 = 0.00005
t'(110mph) = (110 * 1.825 - 90 - 78.3) / 12100 = 0.00268
and we have our clear winner:
100mph is extremely close to optimal. (Drat, I so wanted it to be
88mph!)
The t'(v) calculations above can be interpreted as "number of hours saved for each drive/charge cycle for a 1mph increase in speed", so e.g. at 80mph, you can calculate that driving an extra 1mph faster would save you 0.00628 hours = 23 seconds per drive/charge cycle. However, increasing your speed from 110mph to 111mph would penalize you 0.00268 hours = 10 seconds per cycle. What's your time worth?
The continuous power draw at 100mph is roughly 62.5kW, which means that for optimal cross-country speed, you'll spend about 60% of your time driving and about 40% charging.
By comparison, I had similarly calculated a few years ago that the "sweet spot" for the Roadster, using 17kW charging (70a/240v), was about 55mph.
Someone check my arithmetic?