October 12

If the sun is in the sky above a solar panel at azimouth Z1 (0 to 2pi radians) and altitude A1 (-pi/2 to pi/2 radians), and the panel is tilted by altidude A2 (0 to pi/2) and the tilt is facing direction Z2 (0 to 2pi radians), then how can you work out incident light as a protion of light if it was directly facing the sun?

I know that relative to the plane of the panel the altitude of the sun will be A1+A2, so the incidence light will be sin(A1+A2) portion of direct light when A1>0 and zero A2<0. But I am not sure how to now account for the azimouth difference?

At first I thought I simply take the angle between the azimouths and if its in the range -pi/2 to pi/2 it will be visible but this isn't true. The sun could be behind the panel and still shine on it if its high enough.--Dacium (talk) 23:27, 12 October 2011 (UTC)[reply]

The relative portion of incident light is cos θ where θ is the angle between the light source and the plane. That's assuming θ is less than π/2, otherwise the light source would be shining on the back of the plane. The formula to work out θ is the same as the formula to find Great-circle distance, see the formula there and replace the latitude values with altitudes and the longitude values with azimuths.--RDBury (talk) 00:22, 13 October 2011 (UTC)[reply]

Intersting, so the incident angle would be = arccos(sin(A1)cos(A2) + cos(A1)cos(A2)cos(Z1-Z2))?--Dacium (talk) 02:32, 13 October 2011 (UTC)[reply]

You've got a typo so it should be arccos(sin(A1)sin(A2) + cos(A1)cos(A2)cos(Z1-Z2)) and the relative portion of incident light is sin(A1)sin(A2) + cos(A1)cos(A2)cos(Z1-Z2). If both altitudes are 0 then the result is cos(Z1-Z2) and if A1 is π/2 then the result is sin(A2), so the formula works on easy special cases; always a good idea to check.--RDBury (talk) 04:03, 13 October 2011 (UTC)[reply]