Acc to the link, the highway speeds are from 62 to 75 mph. Midpoint is 68.5 so I'll use that as the "actual" speed.

22 mph tail wind reduces speed-into-wind to 68.5 - 22 = 46.5 mph. So I expect drag to reduce to (46.5/68.5)**2 = 0.46 of what it was, so roughly halved. Power saved was about 1.5 kW.

22 mph headwind increases speed-into-wind to 90.5 mph so drag should increase to 1.745 of what it was. Extra power needed was about 3 kW.

Acc to ev-database.uk Tesla 3 LR 73.5 kWh does 255 miles highway mild weather, so at 68.5 mph 255 miles takes 3.723 hrs, so to drain 73.5 kWh in that time the consumption must be about 19.75 kW continuous.

If the *only *load on the car is due to wind resistance, I'd expect the tailwind to reduce the load to half what it was, so I'd hope for a saving of 10kW. We see actual 1.5 kW saving.

If the *only *load on the car is due to wind resistance, I'd expect the headwind to increase the load to 1.75 times what it was, so I'd expect an increase of 0.75 from existing load, so extra 15 kW needed. We see actual 3 kW extra load.

These extra loads are about a fifth of what the aerodynamic-drag-only theory predicts. So I think this is saying that the fixed loads due to tyre rolling resistance, viscous drag in gears/bearings, inefficiency in converting the 73.5 kWh to mechanical power at 68 mph, heating loads, air-conditioning loads, anything else you can think of that's consuming power at a steady rate, these contribute about 4/5 of the car's loads, and the wind-drag is contributing around 1/5.

Does this sound about right? Or did I get my estimates badly wrong somewhere? Would be nice to have an actual figure of instantaneous Model 3 power consumption at 68 mph in windless "average" day on the flat!