Solar Geometry and Projection
modelling Level 3

Solar Geometry and Projection

A south-facing slope receives more solar energy than a north-facing slope at the same latitude. Why? Because the angle between the sun's rays and the surface normal determines the energy per unit area. This model introduces trigonometry and vector geometry to model solar radiation on tilted surfaces.

Prerequisites: trigonometry, vectors, dot product, projection

Updated 11 min read

1. The Question

Why does the same incoming solar radiation produce different heating rates on slopes of different orientations?

A horizontal surface in Denver receives a certain amount of solar energy per square meter. A south-facing slope (tilted toward the equator) receives more energy per unit area. A north-facing slope receives less. The total radiation from the sun hasn’t changed — but the effective area intercepting the light has.

This is a projection effect, and it’s fundamental to:

  • Topographic climate (microclimates in mountainous terrain)
  • Seasonal energy balance (why summer is warmer than winter)
  • Solar panel placement (tilt angle matters)
  • Glacier mass balance (north vs. south aspects)

The mathematical question: How do we quantify the relationship between surface orientation and energy receipt?

2. The Conceptual Model

Projection: The Core Idea

Imagine a flashlight shining on a wall:

  • When the flashlight is perpendicular to the wall, the beam creates a small, bright circle.
  • When you tilt the flashlight, the beam spreads out into an ellipse — same total light, larger area, lower intensity.

The intensity (energy per unit area) depends on the angle between the light direction and the surface orientation.

Key principle: Energy flux is maximized when the light hits the surface perpendicularly and decreases as the angle becomes more oblique.

Surface Normal Vector

To describe surface orientation, we use the normal vector — a unit vector pointing perpendicular to the surface, away from it.

Examples:

  • Horizontal surface: normal points straight up (0, 0, 1)
  • Vertical wall facing east: normal points east (1, 0, 0)
  • 30° south-facing slope: normal tilts south and up

The angle between the sun’s direction and the surface normal determines the projection factor.

3. Building the Mathematical Model

Trigonometry of Projection

Consider a surface tilted at angle $\beta$ from horizontal, receiving light from an angle $\theta$ from the surface normal.

The effective area intercepting the light is:

\[A_{\text{eff}} = A \cos \theta\]

Where:

  • $A$ is the physical area of the surface (m²)
  • $\theta$ is the angle of incidence (angle between light direction and surface normal)

Energy flux density (W/m² on the surface):

\[S = S_0 \cos \theta\]

Where $S_0$ is the incoming solar flux density (W/m² perpendicular to the sun’s rays).

Special cases:

  • $\theta = 0°$ (sun perpendicular to surface): $\cos 0° = 1$ → full intensity
  • $\theta = 45°$: $\cos 45° = 0.707$ → 70.7% of perpendicular intensity
  • $\theta = 90°$ (sun parallel to surface): $\cos 90° = 0$ → no energy received

The Dot Product

We can express $\cos \theta$ using the dot product of two unit vectors:

\[\cos \theta = \hat{\mathbf{n}} \cdot \hat{\mathbf{s}}\]

Where:

  • $\hat{\mathbf{n}}$ is the unit normal vector to the surface
  • $\hat{\mathbf{s}}$ is the unit vector pointing toward the sun

The dot product of two vectors $\mathbf{a} = (a_x, a_y, a_z)$ and $\mathbf{b} = (b_x, b_y, b_z)$ is:

\[\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z\]

Geometric interpretation:

\[\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\]
For unit vectors ($ \hat{\mathbf{a}} = \hat{\mathbf{b}} = 1$):
\[\hat{\mathbf{a}} \cdot \hat{\mathbf{b}} = \cos \theta\]

The dot product directly gives us the cosine of the angle between the vectors!

4. Worked Example by Hand

Problem: A solar panel is tilted 35° from horizontal, facing due south. At solar noon on the equinox, the sun is at an elevation angle of 50° above the horizon (also due south).

(a) What is the angle of incidence $\theta$ (angle between sun and surface normal)?
(b) If incoming solar radiation is $S_0 = 1000$ W/m², what is the effective flux on the panel?

Solution

(a) Angle of incidence

Surface normal: The panel tilts 35° from horizontal toward the south. The normal tilts 35° from vertical toward the south.

In spherical coordinates (measuring from vertical):

  • Normal elevation from horizontal: $90° - 35° = 55°$

Sun direction: Elevation 50° above horizon, so zenith angle $= 90° - 50° = 40°$

Both the sun and the surface normal are in the same vertical plane (the north-south meridian). The angle between them is:

\[\theta = |55° - 40°| = 15°\]

(b) Effective flux

\[S = S_0 \cos \theta = 1000 \times \cos(15°)\] \[\cos(15°) \approx 0.966\] \[S \approx 966 \text{ W/m}^2\]

The panel receives 96.6% of the perpendicular solar flux — very nearly optimal!

Note: If the panel were horizontal, the flux would be:

\[S_{\text{horiz}} = 1000 \times \cos(40°) = 1000 \times 0.766 = 766 \text{ W/m}^2\]

The tilted panel receives 26% more energy than a horizontal surface.

5. Vector Calculation Example

Problem: A surface has normal vector $\mathbf{n} = (0.5, 0, 0.866)$ (pointing northeast and upward). The sun direction is $\mathbf{s} = (0.3, 0.3, 0.9)$ (pointing slightly south and upward). Both are unit vectors.

(a) Compute the angle of incidence.
(b) If $S_0 = 1200$ W/m², what is the flux on the surface?

Solution

(a) Angle of incidence

\[\cos \theta = \mathbf{n} \cdot \mathbf{s} = (0.5)(0.3) + (0)(0.3) + (0.866)(0.9)\] \[= 0.15 + 0 + 0.779 = 0.929\] \[\theta = \arccos(0.929) \approx 21.6°\]

(b) Effective flux

\[S = 1200 \times 0.929 = 1115 \text{ W/m}^2\]

6. Computational Implementation

Below is an interactive model showing solar flux on a tilted surface.

Try this:

  • Default (30° south-facing slope, sun at 45° elevation due south): Near-optimal alignment, high flux
  • Change aspect to 0° (north-facing): Flux drops dramatically
  • Set slope to 0° (horizontal): See how flux varies only with sun angle, not aspect
  • Solar elevation to 90° (sun directly overhead): Horizontal surface receives maximum; all slopes receive equal flux
  • Explore the polar plot: shows how flux varies with aspect for the current slope and sun position

7. Interpretation

Why South-Facing Slopes Are Warmer (Northern Hemisphere)

In the Northern Hemisphere:

  • The sun is always in the southern part of the sky at midday
  • South-facing slopes tilt toward the sun → smaller angle of incidence → more energy
  • North-facing slopes tilt away from the sun → larger angle of incidence → less energy

This creates topoclimate effects:

  • South-facing slopes: warmer, drier, different vegetation (e.g., grassland vs. forest)
  • North-facing slopes: cooler, moister, more shade-tolerant species
  • Snowpack persists longer on north-facing slopes

Seasonal Variation

The sun’s elevation changes with season:

  • Summer solstice: High solar elevation → even horizontal surfaces receive high energy
  • Winter solstice: Low solar elevation → south-facing slopes receive much more than horizontal surfaces

This is why winter heating in buildings benefits more from south-facing windows than summer cooling suffers from them.

Optimal Solar Panel Tilt

For maximum annual energy collection at latitude $\phi$:

  • Tilt angle $\approx \phi$ (equal to latitude)
  • Face south (Northern Hemisphere) or north (Southern Hemisphere)

For maximum winter energy (heating season):

  • Tilt angle $\approx \phi + 15°$

For maximum summer energy (if needed):

  • Tilt angle $\approx \phi - 15°$

8. What Could Go Wrong?

Confusing Slope and Aspect

Slope is how steep the surface is (0° = flat, 90° = vertical).
Aspect is which direction the surface faces (0° = north, 180° = south).

A 30° north-facing slope and a 30° south-facing slope have the same slope but opposite energy receipts.

Forgetting the Cosine Can Be Negative

If $\theta > 90°$, then $\cos \theta < 0$. This means the surface is tilted away from the sun (backside).

In reality, a backside surface receives zero direct solar radiation. The model should set negative values to zero:

\[S = S_0 \max(0, \cos \theta)\]

Ignoring Diffuse Radiation

Our model assumes all radiation is direct (straight from the sun). In reality:

  • Diffuse radiation (scattered by atmosphere and clouds) comes from all directions
  • On cloudy days, diffuse radiation dominates (50–100% of total)
  • Diffuse radiation is less sensitive to surface orientation

Full solar models separate direct and diffuse components.

Neglecting Terrain Shading

A slope may be oriented toward the sun but still in shadow due to adjacent ridges or peaks. This requires viewshed analysis (Model 9 previews this).

9. Math Refresher: Trigonometry and Vectors

Right Triangle Trigonometry

For a right triangle with angle $\theta$:

\[\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\] \[\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\] \[\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\]

Unit Circle

On the unit circle (radius = 1):

  • $\cos \theta$ is the x-coordinate
  • $\sin \theta$ is the y-coordinate
  • $\theta$ is measured counterclockwise from the positive x-axis

Dot Product Properties

\(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\) (commutative)

\(\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}\) (distributive)

$$\mathbf{a} \cdot \mathbf{a} = \mathbf{a} ^2$$ (magnitude squared)

If $\mathbf{a} \cdot \mathbf{b} = 0$, the vectors are perpendicular.

Summary

  • Solar flux on a surface: $S = S_0 \cos \theta$, where $\theta$ is the angle of incidence
  • $\theta$ is the angle between the sun direction and the surface normal
  • The dot product $\hat{\mathbf{n}} \cdot \hat{\mathbf{s}} = \cos \theta$ gives us the projection factor
  • South-facing slopes receive more energy than north-facing slopes (Northern Hemisphere)
  • Optimal solar panel tilt ≈ latitude for annual energy maximization
  • Real surfaces also receive diffuse radiation and may be shaded by terrain

References