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Homework Help: Converting real life dimensions of a canvas into video file dimensions.

  1. May 30, 2012 #1
    Hey people,

    1. The problem statement,

    I'm doing a Uni Art assignment, and I'm going to be projecting a video onto a canvas (which I don't know the exact dimensions of yet).
    Video file dimensions are stated as 1080 x 980 and so forth, whereas the real life canvas will be something like 50 inches x 100 inches.
    I'm going to need to edit the video file dimensions in order to project it onto the canvas without letting any of the image spill.

    Can someone please help me in figuring out a formula (or whatever you call it) for converting these size/measurement dimensions to be compatible.

    I don't know what SI units a canvas is measured in, but just say the canvas is 70 inches by 150 inches, what resolution would I change my video file to?

    Thank you in advance.

    Oh, and I'm terribly sorry if I made this thread in the wrong category. I honestly don't know what calculus is, so I assumed Measurement would fit under per-calculus... Maths C drop out here. I would really appreciate the help!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 30, 2012 #2


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    Gold Member

    Welcome to PF!,

    The ratio of the width to the height is 2:1 (two-to-one) for the canvas because:

    100 inches / 50 inches = 2

    I don't think that there are too many standard video resolutions that have this sort of aspect ratio. Most of them are closer to square. For instance, for the resolution that you stated (1080x980), the aspect ratio is more like 1.1 : 1

    Wait, it's a canvas, right? That usually means that it is taller than it is wide. I was thinking of it being the other way around (wider than tall). I realize that you can orient it however you want, I was just thinking of the typical case.

    Either way, you can make the size of the projected image as large or as small as you like by varying the distance between the projector and the canvas. (I think that some projectors also have adjustable lenses so that you can change the image size without varying the distance).

    So, if the canvas is taller than it is wide, then all you have to do is adjust the size of the projected image until the width of the image matches the width of the canvas. (The height of the image will be smaller than the height of the canvas).

    If the canvas is wider than it is tall, then all you have to do is adjust the size of the image until the height of the image matches the height of the canvas. (The width of the image will be smaller than the width of the canvas).
    Last edited: May 30, 2012
  4. May 31, 2012 #3
    Thanks a million Cepheid! :smile:

    I got the final dimensions of the canvas, 1.25 x .90 meters. So I worked the aspect ratio to be 1 : .72

    I wish I knew it was as simple as just dividing. :blushing:
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