Solar Power on a Surface
Sunlight power hitting a tilted surface.
Formula
Variables
Example
G=1000, A=2, θ=30°: P ≈ 1732 W
How it works
This formula tells you how much solar power a flat or tilted surface receives from sunlight. The key idea is that the same beam of sunlight spreads over a larger effective area when the surface is tilted, so only the component of the area facing the Sun counts; that is why the cosine of the angle from the normal appears. When the Sun’s rays hit straight on, cos θ = 1 and the surface receives the maximum power for that area.
When to use it
Use this when estimating solar energy on panels, roofs, or any inclined surface exposed to sunlight. It is useful in basic solar design, physics problems about radiation, and comparing how tilt affects captured sunlight.
Worked example, step by step
Given: G = 800 W/m², A = 3 m², θ = 60° Step 1: Write the formula: P = G × A × cos θ Step 2: Substitute the values: P = 800 × 3 × cos 60° Step 3: Calculate cos 60° = 0.5 Step 4: Multiply: P = 800 × 3 × 0.5 Step 5: P = 1200 W Final answer: The solar power hitting the surface is 1200 W.
Common mistakes to avoid
- Using the angle from the surface instead of the angle from the normal
- Forgetting that cos θ is unitless and must be applied to the whole area contribution
- Plugging in degrees without checking the calculator is set to degrees
Did You Know?
Peak sunlight delivers about 1000 W/m² — the '1 sun' benchmark used to rate panels.
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