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Question about perspective

  1. Sep 27, 2005 #1
    Hi there everyone,

    I'm troubled with a certain problem. I'm trying to locate the circles of view in perspective using math rather than using a right angle tool to manually locate the circles of view. I've summarized my problem in the picture below:

    [​IMG]

    So, if line #1 is equal to say, 1. What would line #2 be? 1.__% larger?

    Thanks everyone.
     
  2. jcsd
  3. Sep 27, 2005 #2

    HallsofIvy

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    Call the length of the perpendicular to the image plane L. Then #1/L= tan(15)
    and (#1+ #2)/L= tan(30). That is, #1= L tan(15) and so #2= Ltan 30- Ltan 15
    #2/#1= (tan 30- tan 15)/tan 15.
     
  4. Sep 27, 2005 #3

    DaveC426913

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    Uh yeah. Wot he said goes double for me. (He just types faster.)
    :tongue2:
     
  5. Sep 27, 2005 #4
    hmm... okay, I don't understand the 'tan' or the reasoning behind this explanation, but I'm doing some research now on it. Thank you very much.
     
  6. Aug 24, 2007 #5
    Well, I'm still working on this. It's been quite some time since I asked my initial question. When I first posted this question, I barely understood how perspective worked. Since then, I've struggled very hard for nearly 2 years (just about) to visually understand how perspective worked and how it could be roughly applied with complex arrangements of parallelepipedic shapes, resulting in inaccurate, but sufficient representations of said structures. However, I'd like to see what it takes to place a cube without any faces or outlines parallel to the projection plane in perspective using trigonometry.

    I've recreated the initial graphic I supplied with the original post. It is located here (old link for it is dead).
    http://mywebpages.comcast.net/unitedtiles/perspective03.jpg

    The goal of the original post was to understand how to properly represent something simple like a cube with 2 faces parallel to the projection plane (0 degree cube or '1 point cube), then 'link' that representation to a field of view size. By this, I mean that cameras have constant angles of vision emitting from them. If then, I was able to use simple math to represent this '0 degree cube' on the projection plane, I'd be able to simultaneously understand how that camera would 'experience' the representation relative to its inherent angle of vision. I've demonstrated this in this next illustration.
    http://mywebpages.comcast.net/unitedtiles/perspective01.jpg

    On the right of the illustration, I've shown how the 'camera' would interpret the '0 degree cube' if it had 2 different angles of vision emitting from its viewpoint. The one where the square of the cube appears smaller is due to a larger angle of view emitting from the viewpoint. This would be acquired through the usage of the equation supplied to me 2 years ago.

    NOW then, I've strengthened up a bit. I'm ready to start playing with cubes with faces at complex arrangements to the projection plane. However, I have no idea how I should go about this. I've created a rough orthogonal perspective illustration of the 'perspective happening' where we have a cube with all its planes at complex angles to the projection plane. I know we will need to know the angle of one set of planes of the cube relative to the projection plane to do anything. I also know we will need to know where the centerpoint of the cube or any 'corner point' of the cube is located relative to an established 3dimensional numerical grid system in space. Beyond this, I'm completely lost. I do know one thing though -- it would be somewhat tough to complete this operation. However, I'd like to know how it would be done.

    Here's the illustration demonstrating the new problem:
    http://mywebpages.comcast.net/unitedtiles/perspective02.jpg
     
    Last edited: Aug 24, 2007
  7. Aug 24, 2007 #6
    Incidentally, I know that I simply need to understand the distance between each of the corner points of the cube and the direction of view (line connecting the viewpoint with the projection plane at a right angle). Once I know that distance, I can use the law of similar triangles (or, as I call it, 'distance triangular proportions') to place the points in perspective. So the issue is understanding the distance between each of the corner points of the cube and the direction of view.

    Here's the method I'm using once I know the distance between points and the direction of view:

    http://www.handprint.com/HP/WCL/perspect2.html#distance

    So essentially, I'm not terribly familiar with trigonometry, but think it should be possibly in theory to start with a given -- the location of either the center point of the cube at a specific distance from either the direction of view or projection plane, then somehow use trig to understand the distance between the 'corner points' of the cube from the direction of view. In the case of using distance triangular proportions (as I linked to), you must understand the distance between that point and the direction of view (line connecting veiwpoint with projection plane at right angle and going on to infinity on opposite side of projection plane) such that the distance is that of a line at a right angle to the direction of view. This allows usage of the law of similar triangles.

    Essentially, I just want to know if this would actually be possible and how hard it would be.. and if someone could give me an example of how to do it. ;)
     
    Last edited: Aug 24, 2007
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