MikeNoon
Electrical
- Jan 24, 2003
- 3
I need to design a solar testing chamber that my client is basing on MIL-STD-810F, where they prescribe a range of illuminance from 55 to 1140W/m^2,delivered to three different sides of a 4' cube.
Using the conversion formula of 683 x 1140W/m^2 would give us a delivered amount of 778,620 lux, or roughly 77,800 foot-candles, per side of the cube. This would be roughly seven times the level of natural sunlight on a cloudless day, which is approximately 10,000 fc. Is this conversion and requirement correct, or is my customer looking for too high a level of illumination?
I can't figure out how I can fit enough luminaires around a 4' cube object to deliver that much light, in the required spectral mix of HPS and MV. Any ideas, or experience with these testing chambers to offer? Thanks in advance.
Mike Noon LC
Palindrome Lighting Design, Inc.
Using the conversion formula of 683 x 1140W/m^2 would give us a delivered amount of 778,620 lux, or roughly 77,800 foot-candles, per side of the cube. This would be roughly seven times the level of natural sunlight on a cloudless day, which is approximately 10,000 fc. Is this conversion and requirement correct, or is my customer looking for too high a level of illumination?
I can't figure out how I can fit enough luminaires around a 4' cube object to deliver that much light, in the required spectral mix of HPS and MV. Any ideas, or experience with these testing chambers to offer? Thanks in advance.
Mike Noon LC
Palindrome Lighting Design, Inc.