If you think about it for a moment, Shel, it is obvious that the diameter of the circle of illumination needs to be equal to the diagonal of the negative. Going back to HS geometry, the sum of the square of the two sides is equal to the square of the hypotenuse. I could have used trig, but the battery's dead in my scientific calculator and I couldn't find my side rule. Come to think of it I haven't seen that slide rule since the sixties <grin>.
Ciao, graywolf [EMAIL PROTECTED] ----- Original Message ----- From: Shel Belinkoff <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Tuesday, January 01, 2002 11:16 AM Subject: Re: Questions for Beseler 23CII Users > Hi Tom, > > How did you arrive at this conclusion? What's the math behind it? > Might be worthwhile to know. > > Tom Rittenhouse wrote: > > > > My calculator says a 9x9 negative requires a 12.8cm diameter coverage. I > > don't believe a their comdenser head 23C can cover that. A 6x9 image requirs > > 10.8, so to cover 9x9 the light source needs to be almost an inch bigger > > than to cover 6x9. > > -- > Shel Belinkoff > mailto:[EMAIL PROTECTED] > http://home.earthlink.net/~belinkoff/ > - > This message is from the Pentax-Discuss Mail List. To unsubscribe, > go to http://www.pdml.net and follow the directions. Don't forget to > visit the Pentax Users' Gallery at http://pug.komkon.org . - This message is from the Pentax-Discuss Mail List. To unsubscribe, go to http://www.pdml.net and follow the directions. Don't forget to visit the Pentax Users' Gallery at http://pug.komkon.org .
