[mjcumming]

I have searched through many threads but I am still unclear how to calculate screen size and projector location from throw distance. In particular, I am looking at the Sharp 9000 with a throw of 1 to 1.3. My viewing distance is about 110 inches and was planning on a 16:9 screen with a width between 73-80 inches (83-91 diagonal). The screen width to view distance would be 1.4 to 1.5 times (perhaps a little large http://www.avsforum.com/ubb/smile.gif ) Given the above how does I calculate projector position. There is a beam in my ceiling that I need to avoid....


Mike

 

[cam]

Mike,


Below is a excerpt that I cut from a forum member's post :


=====================================================

16:9 screen screen width 92 " throw MAx 18'11" Min

13'11"


from the xv-Z9000U throw distance chart here is the formula for your own calculations..16:9 screens

y1(Max) = (.05510x - 0.04593) X 3.28

y2(min) = (.04068x - 0.04369) X 3.28


x= screen Size (diag) (inches)

y= Projection distance (feet)

========================================================


I think in your case :

x = ((y2 / 3.28) + 0.04369) / 0.04068


since your y2 = 9.167 feet, so x = 69.8"


Sorry, it appears you can't get 83-91" screen


Cam

 

[mjcumming]

Cam, thanks for the reply... but I am still having a hard time following where the numbers come from and where to plug them in to get min/max projector locations for a given screen size. ie where does 3.28, .04369 and .04369 come from???


Why does the Sony PJCalc program give much shorter distance for a lens with a 1.2 zoom? Where do you use the 1.3 zoom in the equations?


Thanks Mike

 

[cam]

Hi Mike,


I think the original formulae are for throwing distance in metric, and therefore 3.28 is to convert it from meter back to feet.


About those fractional parameters I think the original poster got it from the Sharp manual.


The things get more complicated with zoom ratio, which is quantified in optical terms, that is for a fixed distance, with minimum zoom, you get the smallest screen size, and with maximum zoom of 1.35, you get the biggest screen size.

This is all reflected in the formulae mentioned. Zoom ratio alone can not determine the throwing distance/ratio. (Sometimes throw ratio is specified as 1.8 - 2.5 for example, which is related to the screen width, that is if you have a 80" 16x9 screen, the screen width will be 69.7", and therefore the minimum projecting distance is 69.7"x1.8 = 10'5", likewise for the maximum distance)

 

[noah katz]

That 1 to 1.2 is not the throw ratio, it's the zoom ratio. The throw ratio is about 1.8 to 2.2 screen widths, so for 100" width the projector can be located between 180" and 220" from the screen.


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Noah

 

[mjcumming]

Noah,


Ok, as I understand then, the throw ratio and zoom are different but surely they are related? Is the throw ratio a fixed number which is then changed by the zoom? I am also guessing the throw ratio always represents a diagonal size of the screen and this number would depend on the panel configuration (ie 16:9 or 4:3). As you can see, I am still not quite clear on all this http://www.avsforum.com/ubb/smile.gif


Mike

 

[noah katz]

Mike,


"Is the throw ratio a fixed number which is then changed by the zoom?"


That's one way of putting it; there's not a lot of consistency in the way mfgr's present specs. Usually they give the ratio of minimum to maximum zoom, like 1 to 1.3, and the corresponding range of throw distances for that range.


Throw ratio is defined as the ratio of throw distance (to the screen) to either screen width or screen height.


Go to http://www.projectorcentral.com/ and on the left is a selection for projector calculator for screen sizes and throw distances.


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Noah