The maximum daily energy density of sunlight in sunny Los Angeles is about 6.4 kWh / m2 [1] (assuming perfect, moving angle of panels to sun).
If we can turn the entire footprint of a Model X into solar panels, that gives us about 10 m2 (big car!).
The US DoE reports the Model X gets 100 mi / 31 kWh [2]. Or 12.4 kWh for 40 mi.
So those panels would need to get 1.2 kWh / m2 of solar power. Which is about 18% efficiency and pretty reasonable for good consumer panels [3].
But it assumes the car is in sunlight all day at the perfect angle, there is no loss (eg due to weight), in a locale as sunny as LA (eg Seattle gets half of the sunlight as LA), and can be completely coated in efficient panels (the model S solar roof is <1 m2 in comparison). Bumping efficiency to 30% gives some headroom but it still seems pretty impractical.
The maximum daily energy density of sunlight in sunny Los Angeles is about 6.4 kWh / m2 [1] (assuming perfect, moving angle of panels to sun).
If we can turn the entire footprint of a Model X into solar panels, that gives us about 10 m2 (big car!).
The US DoE reports the Model X gets 100 mi / 31 kWh [2]. Or 12.4 kWh for 40 mi.
So those panels would need to get 1.2 kWh / m2 of solar power. Which is about 18% efficiency and pretty reasonable for good consumer panels [3].
But it assumes the car is in sunlight all day at the perfect angle, there is no loss (eg due to weight), in a locale as sunny as LA (eg Seattle gets half of the sunlight as LA), and can be completely coated in efficient panels (the model S solar roof is <1 m2 in comparison). Bumping efficiency to 30% gives some headroom but it still seems pretty impractical.
[1] https://globalsolaratlas.info/detail?c=34.270738,-116.929301...
[2] https://www.fueleconomy.gov/feg/Find.do?action=sbs&id=41196
[3] https://news.energysage.com/what-are-the-most-efficient-sola...