[–] rlad 1y ago ↗ Is it snake oil? The amount of power that would be delivered to the ground seems quite minuscule based on that small orbital mirror size.In fact it should be over three orders of magnitude lower than that of normal sunlight on the solar panel, which is roughly 1000 W per square meter.Here are the calculations:---Assumptions:Solar constant: 1366 W/m²Mirror area: 100 m² (10 m x 10 m)Reflectivity of aluminized Mylar: 90%Atmospheric attenuation: 70% of reflected sunlight reaches Earth’s surfaceSpot diameter on Earth: 500 metersSpot area on Earth: π × (250 m)² ≈ 196,350 m²Calculation:Total incident power = 1366 W/m² × 100 m² = 136,600 WReflected power (after reflectivity) = 136,600 W × 0.90 = 122,940 WPower reaching Earth’s surface (after atmospheric attenuation) = 122,940 W × 0.70 = 86,058 WPower per square meter actually delivered at Earth’s surface = 86,058 W ÷ 196,350 m² ≈ 0.438 W/m²
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[ 2.9 ms ] story [ 15.3 ms ] threadIn fact it should be over three orders of magnitude lower than that of normal sunlight on the solar panel, which is roughly 1000 W per square meter.
Here are the calculations:
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Assumptions:
Solar constant: 1366 W/m²
Mirror area: 100 m² (10 m x 10 m)
Reflectivity of aluminized Mylar: 90%
Atmospheric attenuation: 70% of reflected sunlight reaches Earth’s surface
Spot diameter on Earth: 500 meters
Spot area on Earth: π × (250 m)² ≈ 196,350 m²
Calculation:
Total incident power = 1366 W/m² × 100 m² = 136,600 W
Reflected power (after reflectivity) = 136,600 W × 0.90 = 122,940 W
Power reaching Earth’s surface (after atmospheric attenuation) = 122,940 W × 0.70 = 86,058 W
Power per square meter actually delivered at Earth’s surface = 86,058 W ÷ 196,350 m² ≈ 0.438 W/m²