Homework 1
E = F / A
F = 1700 lumens x .30
A = 10’ x 10’ = 100 ft2
E = F / A = (1700 x .3) / 100 = 5.1 footcandles (or lumens/square foot)
To convert to lux, multiply by 10.764 lux/fc.
E = 5.1fc * 10.764 = 54.8964 or about 55 lux
A single bare bulb is usually a source of glare.If it is right next to the ceiling, it might not be too bad, because the ceiling will be pretty bright all around it.The further the bulb hangs down from the ceiling, the worse it gets, because the ceiling gets darker and the bulb is in contrast to the wall behind it.The contrast (and the resultant glare) get much worse if the walls are painted black.
Find E at the upper surface of the diffuser.(This is similar to problem #1.)
E = F / A = (3 x 2400 lumens x .8) / 4’ x 4’ = 360 fc
Find the light leaving the underside of the diffuser (and the resultant brightness or luminance.)
L = E x
One lone fixture on a black ceiling will cause glare, because there is extreme contrast.Many fixtures on a white ceiling will be much more even within the field of view.Many fixtures will also provide more illumination.Glare is not necessarily the result of too much illumination, but rather too much contrast.
Indirect light, coming from the North, is much better for viewing a painting.It casts no shadows and has no "hot spots" (reflections of the sun, itself).Direct light would be more dramatic for a sculpture, because there would be sharp shadows, popping out the three dimensional nature of the form.There should still be some fill light, or the contrast might be slightly high, but this is not a major problem.In fact, this is one reason that you often find sculpture gardens in museums and never find painting gardens.Even paintings which were done outdoors (eg Monet’s gardens or lily ponds, etc.) are usually displayed indoors with diffuse lighting, if possible.)