I How can I tell how much an object will shrink when it moves away?

  • Thread starter JeffCyr
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How can I find the size an object will appear (Such as on a photo with a ruler) When it is in the distance by knowing how far it is and it's size when closer
This is a weird question that I'm not quite sure how to phrase so I'll do my best.

Reason for the question: I'm trying to learn perspective in art but guides and tutorials aren't helping so I thought of using geometry.

Scenario/question: I'm looking to have a formula I could use to see how far an object would appear in the distance compared to a closer "origin" point. For example how wide a road is 100 meters away base on it's size up close "X" or how tall a building's wall is 10 meters away when it's other end is visible at 0 meters at a height of "Y"

Does this make sense? Sorry I don't know which prefix to use for this I don't know how advanced this question is so I went with undergrad to be safe since I don't think that was covered in highschool 10 years ago


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The ratio of the apparent linear size (eg width or height) of an object k metres away to that of a same-sized object m metres away is approximately m/k. When the angle subtended at the eye by the object (the angle between a line from your eye to one extreme of the object - eg top - and a line to the opposite extreme - eg bottom) is large, we need to make a trigonometric correction to this. But for objects that don't fill most of the field of view, that approximation should be fine.
Thank you, since it's for drawing I was planning on using the outer most lines for measurements. So it should be fine for the main shape the rest I'll tweak as need be. Thank you very much! I knew I should have gone to scientists rather than artists to learn perspective!


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It's a matter of "similar triangles". Suppose you have a vertical stick, of length L, at distance D from you. You look at it through a sheet of glass at distance d from you. The lines of sight from the ends of the stick to you form a triangle with base length L and altitude D. The lines of sight from the points at which those first lines cross the glass form a triangle with exactly the same angles (so similar to the first triangle). Call the distance between the endpoints of the image on the glass L'. The altitude of this triangle is the shorter distance D-d Then [tex]\frac{L'}{L}= \frac{D- d}{D}[/tex] or [tex]L'= \frac{D- d}{D}L[/tex].

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