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Discussion Starter · #1 ·
I am getting ready to install a Barco 808 on my ceiling and was playing around with the lens program. For a 52X92" wide screen I am getting that it will have a "negative a" of -7.6" but the picture shows the top of the image this much higher than the projector, not lower than the projector. I am confused as changing the projection angle by just .1 inverts the picture so that the top of the projected image is about 7.7" below the projector. (which is what I think it should be). What is the deal here? Per the way the picture comes out I would have to mount it about 16" below the ceiling to get an image that stops 8" or so from the top of the wall. Given that the unit is at least 14" deep that would be quite an obstruction.


Brian
 

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When it says negative it means that the picture will be past the top of the projector. Meaning it will be shining on the ceiling. Look at the picture again. Remember that the flattest part of the "projected image" is along your ceiling and it shows that it will project past the top of the wall. Get it?
 

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Discussion Starter · #3 ·
Why does it show the opposite if I change the projection angle by just 0.1 degree (an insignificant amount)? Brian
 

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The LENS program picture really doesn't show the projector in the proper relative position when it gives the output diagram. They just have several standard pictures and you have to adjust based on the numbers. The little red arrow pointing up with "A Negative" means the 7.6 inch dimension goes upward from the top edge of the image. The picture just shows it as a positive offset.


For a negative 7.6 inch offset, that basically means that if you mounted your projector flush with the ceiling, the top of your image would be 7.6 inches down from the ceiling.


For a comparison, input a 4:3 92x69 screen and you'll see the offset is basically zero meaning the image edge would also be flush with the bottom of the projector. For the 16:9 screen you're really still pointing to the exact same place on the wall with the smaller vertical image dimension.


Hope that helps...


Mark
 
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