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Epson HC2040

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  • Epson HC2040

    Hi guys. I just picked up the Epson HC2040 projector but I was wondering what your thoughts on this projector

    The reviews online for the HC2040 look pretty good but I'm wondering about its application for simulator use

    Here are the reviews for this projector that I found: http://www.projectorcentral.com/epso...040-review.htm

    I did the projector calculator and I think it says it should be 4.3m away. That seems pretty far and I think it's because it's not a short throw projector?

    If I'm using this for a simulator, do I need a short throw projector? Would I be standing in the way of the image?

    I read about the projector offset but I'm not sure if this will projector will work. Please help!

    Thanks

  • #2
    Just for some more info here are some of the dimensions for my setup

    Ceiling Height: ~130 inches (10.8 feet)
    Setup Length: 17 feet
    Distance from hitting area to screen: ~6.5 feet

    Screen size: 83" High x 94" Wide (6'9" H x 7'8" W) -- Net Return Pro Series Simulator Screen

    I'm very confused about this projector calculator....what does my attachment mean?
    Attached Files

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    • #3
      Sorry to say, the projector calculator is way off. Firstly, it's showing info for 16:9 and your screen is closer to 4:3

      Start by adjusting the screen dimensions to your screen size, or as close as you can get it. Forget "screen diagonal" it is the most useless piece of information ever and has been popularized by TV manufacturers because it makes their TV sound bigger than it actually is. Use screen width or height - but pay attention to both of these since that is going to be very close to what the image will project at.

      Lastly, this is probably not a short throw projector, so better look for a different one.

      Comment


      • #4
        I know the screen is better off in a 4:3 format but I don't mind just using it in a 16:9 format.
        I'd like to just set it at 16:9 and then I could play movies on it if I wanted. From what I read, it seems like if I wanted to have my golf sim in a 4:3 format and then watch movies in a 16:9 format, I would need to be able to change the distances by setting it on a sliding track in order to achieve that, which is probably too much trouble for me.

        If I wanted to have it set at 16:9 and based on the Epson throw distance calculator, it says I need to mount the projector 116" (9' 8") away from the screen and it also says I need to have the projector mounted 50" from the ceiling. The top of my screen would be at 83" and it would bottom out at 30" (Since the height of the displayed image would 53"....and the width of the displayed image would be 94")

        Based on this information, it seems like the angle that the projector would be using to display the entire screen would be 24.6 degrees. Please correct me if I'm wrong, but if my hitting area is 6.5 feet from the screen, then at 6.5 feet from the screen, then anything that is taller than ~5 feet would block some of what the projector is displaying and then it would cast a shadow.

        Here are my calculations:

        Screen size: 83" High x 94" Wide (6'9" H x 7'8" W) -- Net Return Pro Series Simulator Screen
        Floor to ceiling: 130"
        Screen from bottom to top: 83"
        --> Since I'm using 16:9 and the width of my screen is 94", then the displayed image height would be 53"
        --> The bottom 30" is not used and the image to be displayed would be from 30"-83"

        Since the Epson throw distance calculator says that it needs to be 116" away from the screen it gives me a triangle with these dimensions:
        a^2 + b^2 = c^2
        116" (a) x 53" (b) x 127.5" (c - Hypotenuse) --> Using these dimensions, the angle that the projector needs to display the full image is 24.6 degrees

        If my hitting area is 6.5 feet (78"), then to check if a shadow will be displayed, I need to scale down the triangle.
        116" - 78" = 38"
        38" (a) x 17.4" (b) x 41.8" (c - Hypotenuse)

        Projector is mounted 50" from the ceiling and the height of the room is 130" --> 130" - 50" = 80"
        80" - 17.4" = 62.6" = 5.2 feet

        So basically, anything over 5 feet at 6.5 feet away from the screen will disrupt the image displayed on the screen and cause a shadow

        Is my logic flawed or am I missing anything? I just want to know for future reference.



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