Question

The relative radiance value of a facet of a Triangulated Irregular Network (TIN) can be computed using: \[ R_f=\cos(A_f-A_s)\sin(H_f)\cos(H_s)+\cos(H_f)\sin(H_s) \] Where, $R_f$ is the relative radiance value of a facet, $A_f$ is the facet’s aspect, $A_s$ is the sun’s azimuth angle, $H_f$ is the facet’s slope and $H_s$ is the sun’s altitude. Suppose a facet of a TIN has a slope value of $10^\circ$ and an aspect value of $297^\circ$ and sun’s azimuth of $315^\circ$. For sun’s altitude angle of $65^\circ$, the relative radiance value of this facet is ................. (Rounded off to 2 decimal places).

Hint:

The first term modulates the slope- and aspect-dependent part with the sun–facet azimuth difference; when the sun is near the facet’s normal (large $H_s$ and small $H_f$), the second term dominates.

Answer 1

Insert the angles in the formula (all in degrees): \[ \begin{aligned} R_f &= \cos(297^\circ - 315^\circ)\sin(10^\circ)\cos(65^\circ) + \cos(10^\circ)\sin(65^\circ) \\ &= \cos(-18^\circ)\sin(10^\circ)\cos(65^\circ) + \cos(10^\circ)\sin(65^\circ). \end{aligned} \] Evaluate the terms: \[ \cos(-18^\circ) = 0.95106, \quad \sin(10^\circ) = 0.17365, \quad \cos(65^\circ) = 0.42262, \quad \sin(65^\circ) = 0.90631, \quad \cos(10^\circ) = 0.98481. \] \[ \Rightarrow R_f = (0.95106)(0.17365)(0.42262) + (0.98481)(0.90631) = 0.0695 + 0.8928 = 0.9623 \approx \boxed{0.96}. \]