5 ms·

1.3kW/m^2 * 20% efficiency = 260w/m^2 260w/m^2 * 5m^2 = 1.3kW 1.3kW * 4 hours daily sun = 5.2kW/day 265 miles range / 85kW full charge capacity = 3.1 miles/k

1.3kW/m^2 * 20% efficiency = 260w/m^2 260w/m^2 * 5m^2 = 1.3kW 1.3kW * 4 hours daily sun = 5.2kW/day 265 miles range / 85kW full charge capacity = 3.1 miles/kW 5.2kW/day * 3.1 miles/kW = 16.12 miles / day Still sounds too good, what did I miss?

thfuran 7y ago
That initial 1.3kW is ignoring atmospheric absorption. You lose some 300 W even at zenith without clouds. More if you're really far from the equator and/or it isn't currently solar noon. If you're in a city, you may be in shade a lot and certainly don't have the sun at its zenith for 4 hours regardless.
Mvhsz 7y ago
5m^2 is too big for a car, and the panels are heavy and expensive. Toyota used to have a prius with a solar panel roof - it produced enough energy to run the fans (not the AC) while the car was parked on a hot day. Not even in the ballpark of the energy needed to move the car (and it was an expensive option).
txcwpalpha 7y ago
16.12 miles a day is not going to cut it. In your previous comment you suggested 50 miles a day. In various studies, Americans drive >30 miles a day. 16 miles would hardly even cover most American's one-way commute, let alone any other errands or trips.
- NullPrefix 7y ago
  >you suggested 50 miles a day /s/suggested/calculated using data provided by parent comment/
  - txcwpalpha 7y ago
    Nobody in the entire comment chain brought up distance except for you, which you kicked off by suggesting 50 miles daily usage. Regardless, even your drastically lower 16 miles a day isn't going to satisfy the needs of the average driver.
adrianN 7y ago
Affordable solar panels don't get 260W/m^2 peak. The solar panels on a car don't track the sun, so average power output is way below peak. You don't get 1.3kW per square meter for 4 hours a day. 42,339 MW of PV installations produced 39,401 GWh of energy in 2017 in Germany[1], so on average you get peak power for just two hours a day with panels that are angled appropriately. [1] https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statistics https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statist...

thfuran 7y ago

That initial 1.3kW is ignoring atmospheric absorption. You lose some 300 W even at zenith without clouds. More if you're really far from the equator and/or it isn't currently solar noon. If you're in a city, you may be in shade a lot and certainly don't have the sun at its zenith for 4 hours regardless.

Mvhsz 7y ago

5m^2 is too big for a car, and the panels are heavy and expensive. Toyota used to have a prius with a solar panel roof - it produced enough energy to run the fans (not the AC) while the car was parked on a hot day. Not even in the ballpark of the energy needed to move the car (and it was an expensive option).

txcwpalpha 7y ago

16.12 miles a day is not going to cut it. In your previous comment you suggested 50 miles a day. In various studies, Americans drive >30 miles a day. 16 miles would hardly even cover most American's one-way commute, let alone any other errands or trips.

NullPrefix 7y ago
>you suggested 50 miles a day /s/suggested/calculated using data provided by parent comment/
- txcwpalpha 7y ago
  Nobody in the entire comment chain brought up distance except for you, which you kicked off by suggesting 50 miles daily usage. Regardless, even your drastically lower 16 miles a day isn't going to satisfy the needs of the average driver.

adrianN 7y ago

Affordable solar panels don't get 260W/m^2 peak. The solar panels on a car don't track the sun, so average power output is way below peak. You don't get 1.3kW per square meter for 4 hours a day. 42,339 MW of PV installations produced 39,401 GWh of energy in 2017 in Germany[1], so on average you get peak power for just two hours a day with panels that are angled appropriately. [1] https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statistics https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statist...