Ok, as you’ve done the maths: what charge do you get? Not the ~10% I got from a back-of-the-envelope calculation?
Model 3 is about 5m by 2m, and is apparently rated for 241 Wh/mile
4m * 2m * 1kw/m^2 * 50% * 20% = average power 800 W
(50% because the panels are flat, 20% because cell efficiency)
241 Wh/mile * 60 miles/day = average usage 602 W
I’m not sure what fraction of the day people drive for given that I’m not a driver, but I’m eyeballing 5-10%. I acknowledge professional drivers — taxis etc. — can’t possibly rely on PV alone, that PV can only supply a fraction of what they need (my 10% guesstimate), but I still think this should help with the general public. Or are my assumptions way off?
First you missed a decimal point or two here and scrambled the units of measure-
“ 241 Wh/mile * 60 miles/day = average usage 602 W”
The correct math is
241 Wh/mile * 60 miles/day = average usage 14,460Wh, or 14.46kWh
Further answer - The consensus from people who know this better than you & I, have these cars, and in some cases have tried.. is basically - it won’t charge much, and it’s way more expensive than the electricity it is going to generate.
Note there are AC-DC inverter losses of 10-20%. plus input->battery charge losses which are non-linear and very bad at the low end. For example a Tesla won’t even take a charge if the input is below the ~300-500W range in good weather. In cold weather say Northeast US winter, the floor is closer to a 1kW input as there is a heating system to get the battery put to temperature for charging that is going to eat almost all of that.
So just taking the parent example, even if we assume plastering the car in PV will generate 800W peak
1) This will probably not translate into any charge in winter weather, but possibly allow you to keep the car battery from being fully cold soaked, best case
2) In good weather you are probably looking at post-inverter input to charger at 700W, with charger losses meaning about 400-500W making it to the battery. So that is, in an efficient Tesla about 2 miles of range for every hour of peak sun. Depending on your location, orientation and time of year you might expect peak sun hours of 3-6 hours/day. So grand total 6-18mi/day of range added making a lot of happy assumptions and not moving your car during lunch. This amount of charge per day could be acquired in 1-2 minutes at a supercharger and worth about 30-75cents. Or charge at a L2 charger in your own garage in 12-36 minutes.
I don’t understand your math, I’m sorry. Why would you divide by 24 hours? There is not 24 hours of sun for your PV to capture and put into the car. Peak solar generation is 3-6hrs/day depending on region and time of year.
4m length * 2m width * 1kw/m^2 insolation * 50% loss due to the panel area being calculated by ground area and it not tracking the sun and therefore not getting peak output * 20% cell efficiency = average power 800 W
My BOTE calculation above should use 25% instead of 50% for day-night average of PV panels horizontal to the ground. Can’t edit now, though. 25% is the planet-wide average for day-night and seasonal variation because that’s the ratio of the surface area of the Earth to the area of a disk intersecting the same flux of sunlight at 1AU (4πr^2 : πr^2).
With that correction, that’s 400 watts average over 24 hours (as in: no not merely the peak at noon); which means 24 h * 400 W = 9.6 kWh per day.
If you drive 60 miles per day, and each mile consumes 241 Wh of energy, then you consume 14.41 kWh of energy per day.
Model 3 is about 5m by 2m, and is apparently rated for 241 Wh/mile
4m * 2m * 1kw/m^2 * 50% * 20% = average power 800 W
(50% because the panels are flat, 20% because cell efficiency)
241 Wh/mile * 60 miles/day = average usage 602 W
I’m not sure what fraction of the day people drive for given that I’m not a driver, but I’m eyeballing 5-10%. I acknowledge professional drivers — taxis etc. — can’t possibly rely on PV alone, that PV can only supply a fraction of what they need (my 10% guesstimate), but I still think this should help with the general public. Or are my assumptions way off?