[Anthony1]

I know this question must sound strange, but here is the deal. I'm trying to figure out exactly where I'm going to locate my X1 projector in my Garage. I'm using my Garage as the viewing room, because I don't have any other room that is large enough, and/or that my Wife would let me use for this. Plus there is the issue of ambient light.


Anyways, my Garage is going to be the place that I put it.



At this point I have determined that I'm going to have a fixed screen on the left wall of my Garage. The screen is going to be 16:9 approximately 97.906 inches diagonally. It will be 48 x 85.333. I'm using the Parkland Plastics 4 x 8 panel that you can get at Home Depot. I'm going to attatch the screen to my wall in my Garage.


Ok, now, to be able to have a screen of this size, according to the infocus projector calculator, my X1 projector needs to be 13.44 to 16.142 feet away from the screen. The median between those two figures is 14.791. So I figured that I would have my X1 ceiling mounted just under 15 feet from the screen.



Now here is where my question comes into play. I have determined a spot on my ceiling that I think would be ideal to mount the X1. But there is an issue of an obstruction that is on the ceiling, that is about 3.5 feet to the right and 6 feet forward of where they X1 would be located. It's the Garage Door Opener Unit. This unit cannot be moved and will have to stay where it is.


Now, I'm guesstimating that the beam of the projector will miss this unit on the ceiling, and it won't be an issue. But I'm not sure if that is true. I'm thinking that the beam of light coming out of the projector is like a triangle, with one point coming out of the lens and the other two points going to the extremes of the screen's width. If this is somewhat correct, then I'm almost sure that the beam will completely miss this Garage Door Opener unit, and I will be o.k.

 

[technut Canada]

Quote:

Originally posted by Anthony1 I'm thinking that the beam of light coming out of the projector is like a triangle, with one point coming out of the lens and the other two points going to the extremes of the screen's width. If this is somewhat correct, then I'm almost sure that the beam will completely miss this Garage Door Opener unit, and I will be o.k.

You are correct. Light travels in straight lines except in unusual circumstances (eg. around strong gravimetric forces).


So if you can clear it with a piece of string from where the lens would be to the edge of the screen, then you should be OK.

 

[Anthony1]

Quote:

So if you can clear it with a piece of string from where the lens would be to the edge of the screen, then you should be OK. [/b]

Hey, that's a really good idea. Cool! Thanks for the tip on that.

 

[sonick182]

I dont know if the string idea would work, since you won't know how much of an angle it shoots the image on the screen without the projector...


IE. The angle of the bottom of the image to the bottom of the lens is smaller than the angle from the top of the image to the top of the lens...


Kinda hard to describe, but just let you know that don't take the string idea too strongly (argh, my brains gone dead, i cant describe what i want to say...)

 

[technut Canada]

If he has correctly calculated his screen size (projector throw) and the vertical offset, then there's no reason that the string method shouldn't be valid within an inch or two.

 

[sonick182]

ah, yeah the vertical offset, thats what i was talkin bout... or at least tryin to

 

[craig john]

Quote:

Originally posted by Anthony1 The screen is going to be 16:9 approximately 97.906 inches diagonally.

Approximately???


Craig

 

[Al Kaseltzer]

From Technut's FAQ:
Quote:

When the projector is in 16:9 mode, the image offset is 132.9%. This means that if you have an image 10 feet high and you are projecting from a table top, the bottom of the image will be 3.29 feet above the center of the lens. If you are projecting from the ceiling with the projector installed upside down then the top of the image will be 3.29 feet below the center of the lens.

So for a 48" high screen, the offset will be about 16" ( or approximately 15.792" if you like).


However, your image is going to be just a bit more than 7' wide at a throw distance of 14'. An obstruction 3.5' to the side at a throw distance of 6' shouldn't be a problem.


Don't take my word for it though, double check all your measurements and plan everything now. It'll save you from a lot of problems later on.