How Can You Calculate the Area of Light Hitting a Dome?

[laminatedevildoll]

Let's say that there's a dome with some radius and height. Is there a way to calculate the area of the light hitting the dome (half an ellipse) when the sun is near the side of the dome and when the sun is right above the dome? I know that the sun angle is approximately 60 degrees. I'd appreciate any help.

 

[DaveC426913]

How high is the apex? Are we talking about the apex of the dome eclipsing any part of the perimeter from the sun's POV? If the sun is 60 degrees high in the sky, that would be pretty difficult to have, meaning at all times, your dome is merely an ellipse from the sun's POV.

 

[tony873004]

compute the area of a 2-d cross section of the dome from the Sun's vantage. For example, when the Sun was directly overhead, the dome would look like a 2-d circle on the ground. Then the formula becomes pi * r^2.

If the dome were a semi sphere, when the Sun was shining directly on the side of the dome (sunset or sunrise), then it would be 0.5 pi*r^2.

 

[laminatedevildoll]

How high is the apex? Are we talking about the apex of the dome eclipsing any part of the perimeter from the sun's POV? If the sun is 60 degrees high in the sky, that would be pretty difficult to have, meaning at all times, your dome is merely an ellipse from the sun's POV.

Well, the height of the dome is two times its radius. I am interested in determining the area of light hitting the dome. For instance, when the sun is at the top, from the its POV, the area is a circle right? So would it just be the area of the dome the light is hitting? How would I calculate how much light is being absorbed within that area? How about from the side? From the sun's POV, how do I calculate the area? Would it look like a rectangle? Thanks for your help.

 

[laminatedevildoll]

compute the area of a 2-d cross section of the dome from the Sun's vantage. For example, when the Sun was directly overhead, the dome would look like a 2-d circle on the ground. Then the formula becomes pi * r^2.

If the dome were a semi sphere, when the Sun was shining directly on the side of the dome (sunset or sunrise), then it would be 0.5 pi*r^2.

That does make sense. I didn't know it can be as easy as that. Thanks.

 

[tony873004]

Yes, it can be that easy. The easiest way to think of it is to think in terms of what kind of a shadow would the object cast. A round ball 1 inch in diameter will case the exact same shadow as a round disk 1 inch in diameter if the Sun is shining down directly on the disk. If they both cast the same size shadow, then they both intercepted the same amount of sunlight.