Hope this helps :o)

March 2nd, 2011

Let the light on a flat surface be a, and the object which forms the primary shadow be bc. The wall de is the surface on which the derivative shadow is received at the place mn. The remainder, dn and me, continues to be illuminated by a, and the light from dn is reflected in the primitive shadow bc, and the light from me does the same. Thus the derivative shadow nm, which is not exposed to the light a, remains dark, while the primitive shadow is illuminated by the lit background which surrounds the derivative shadow. Therefore the derivative is darker than the primitive.

- Leonardo da Vinci
March 2nd, 2011
at the risk of sounding dumb. . .huh?
March 2nd, 2011
There was a drawing, it doesn't help.
March 2nd, 2011
@tolpol Thanks for the info. I had my de's and mn's mixed up before. Makes way more sense now!
;-)
March 3rd, 2011
If you say so!
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