[Radiance-general] Re: Building on a hillside

[Randolph M. Fritz]

Duh.  Me dumb.

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

Thanks, Greg, Lars.

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" <RFritz at lbl.gov>
>> 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?
>> 
>> --
>> Randolph M. Fritz • RFritz at lbl.gov


-- 
Randolph M. Fritz • RFritz at lbl.gov
Environmental Energy Technologies Division • Lawrence Berkeley Labs