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Need a few good eyes to double check my math, also found a discrepancy I can't explain.


Using a jvc dla-rs48 projector and a 130 inch wide 2.35 screen (55.3 high) with 26 inches of drop and the case is 6 in. This should also yield a 55.3 high by 98.3 wide 16x9 image. All of this via zooming for now (no lense or sled).


Throw distance:


The projectors site say the lense is 1.4:1 to 2.8:1 so:

At 2.35 of 130 wide x 1.4 = 182" or 15'2" minimum


At 16x9 of 97 wide x 2.8 = 271" or 22'7" maximum


So my plan is to put the projector lens 15'5" back


Here is one thing I can't reconcile is that jvc's own website calculations show at 16x9 the min at 11.15 and max at 22.7 kind close. But when I change it to 2.35 it kicks out, on a 130w 55.3h a minimum distance of 14.78 and max of 30.8. Shouldn't my numbers and theirs match up exactly?


Vertical shift:


The projector features a 80% v shift, which in jvc's case really means 30% of the screen height. So rounding the height to 55" you get:


55" x 0.30 = a little more than 16 inches of shift (I'm trying to stay away from using any edges of limits).


So the screen case is 6 in mounted and 26 in of drop make the top of the screen 32" from the ceiling.


32" - 16" shift = 16" projector height.


Is this from center lens, top of lens or bottom?


So if I use a 18 in downtube I should be ok...?


Thank you in advance
 

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Is the JVC specs for TR based on the 16:9 image width (as we use to calculate) or the diagonal?
 
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