# Building on a hillside

I'm simulating a test structure on a hillside site; the ground drops away dramatically. So a lot of sky is visible below what would be the horizon on a level site. So far as I can tell, the only way to simulate this correctly (if the ground reflectance and sky light is significant) is to give the model an actual elevation, z = 700 feet or whatever. Do I have that right?

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Randolph M. Fritz • [email protected]
Environmental Energy Technologies Division • Lawrence Berkeley Labs

Hi Randolph,

I am not sure wether I get your answer correct. The horizon is (almost) at an infinite distance, and moving up 700m doesn't change that much. In fact, you apply your sky distribution which depends on angles, not carthesian height, to a source object to make it appear infinitely distant in Radiance.

I would try to get a good representation of the horizon line (in degrees from the ideal horizon) to assess shadowing. Reflective properties in the mountains can be interesting, too, e.g. if you have snow for a long period of the year. In general getting climate data for such an exposed location is interesting.

Cheers, Lars.

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Dipl.-Ing. Architect Lars O. Grobe

On Nov 15, 2010, at 22:45, "Randolph M. Fritz" <[email protected]> wrote:

I'm simulating a test structure on a hillside site; the ground drops away dramatically. So a lot of sky is visible below what would be the horizon on a level site. So far as I can tell, the only way to simulate this correctly (if the ground reflectance and sky light is significant) is to give the model an actual elevation, z = 700 feet or whatever. Do I have that right?

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Randolph M. Fritz • [email protected]
Environmental Energy Technologies Division • Lawrence Berkeley Labs

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[email protected]

Hi Randolph,

It's a bit tricky to modify skybright.cal to account for an elevation causing the horizon to be below a 0-degree altitude, but it's possible. First, you should check that it's necessary. The formula you want for the angle below the horizon is:

theta = acos(R/(h+R))

where R is the radius of the Earth (about 6360 km) and h is the distance above the horizon (sea level, presumably). Even from the top of Mt. Everest, the change in the horizon is only about 3 degrees, assuming you could see something at sea level from there, which I'm not sure you could. (Anyone who's been to the top of Everest, please pipe in.)

Cheers,
-Greg

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From: "Randolph M. Fritz" <[email protected]>
Date: November 15, 2010 1:45:55 PM PST

I'm simulating a test structure on a hillside site; the ground drops away dramatically. So a lot of sky is visible below what would be the horizon on a level site. So far as I can tell, the only way to simulate this correctly (if the ground reflectance and sky light is significant) is to give the model an actual elevation, z = 700 feet or whatever. Do I have that right?

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Randolph M. Fritz • [email protected]

Duh. Me dumb.

It seems like a large angle when I'm looking over it...

Thanks, Greg, Lars.

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On 2010-11-15 14:18:27 -0800, Greg Ward said:

Hi Randolph,

It's a bit tricky to modify skybright.cal to account for an elevation causing the horizon to be below a 0-degree altitude, but it's possible. First, you should check that it's necessary. The formula you want for the angle below the horizon is:

theta = acos(R/(h+R))

where R is the radius of the Earth (about 6360 km) and h is the distance above the horizon (sea level, presumably). Even from the top of Mt. Everest, the change in the horizon is only about 3 degrees, assuming you could see something at sea level from there, which I'm not sure you could. (Anyone who's been to the top of Everest, please pipe in.)

Cheers,
-Greg

From: "Randolph M. Fritz" <[email protected]>
Date: November 15, 2010 1:45:55 PM PST

I'm simulating a test structure on a hillside site; the ground drops away dramatically. So a lot of sky is visible below what would be the horizon on a level site. So far as I can tell, the only way to simulate this correctly (if the ground reflectance and sky light is significant) is to give the model an actual elevation, z = 700 feet or whatever. Do I have that right?

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Randolph M. Fritz • [email protected]

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Randolph M. Fritz • [email protected]
Environmental Energy Technologies Division • Lawrence Berkeley Labs