Join Date: Feb 2009
Location: EasternShore Maryland
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Sorry guys I searched, I really did, but I am still a little confused. I will be using an "A" lens.
I have found all sorts of calculations [.5625] to get the 16x9 screen size image area inside of the 2.35 screen. I have no issue there. Screen is 117.5 wide, 2.35.1, CIH is 50. 16x9 is 89 wide. All looks good and clear right? Now when I start to calculate the throw distance, is it based on the 16x9 screen width, or the 2.35 screen width?
I think the screen width of the 16x9 is the distance I use to set the projector up. Before I calculate what the "A" lens will do.
Example 89 x 2.2 195.8/12=16.3 feet.
This is where I put the front of the projector lens correct?
Or is it based on the 117.5x 2.0= 235/12=19.58 feet?
If this is the case I have no room to put the projector there. The room is only 21.5 from screen to back wall.
Scenario #1 works perfect.
Scenario #2 Well cut a hole in the wall into the hallway...WAF. Non existent.
Projector will be either Runco LS-5 or Q-650i. Both have no room fro scenario #2 unless I get a short throw lens which will cause pincushion issues, Right?
Anyway thanks for looking.