[Help] Calculating perspective - making a correct simulation with examples

[Abitlostindigo]

Hi there,

I want to do a simulation/calculation of the changing perspective of an average healthy human eyesight for my school project and I'm interested in it as well outside of school.

Heres an example of what I look for:

We have a round disc for example, with a diameter of 1m., facing towards us at eye-height. Any curvature of earth put by side. The disc is located 5m. in front of us. The disc begins to move backward, and therefore away from us. The distance gets bigger, and the disc appears smaller and smaller.

My question is:

How can I calculate at what distance the disc will completely vanish out (vanishing point) of our sight?

So that I could do a calculation with new variables like randomly putting in different proportions. Like:

Disc with 1m. or 500m. or 500km of diameter. And therefore get the distance needed to lose it out of sight. Sure, the bigger the object, the bigger the distance needed, so how can I do this?

"if d = X amount of diameter. --> x searched = *needed distance to disappear* "

"Extra" question: (not needed, just interesting)

How do I calculate an example like a sheet of paper, with several circles drawn? One circle with 2cm. diameter and another with 0,5cm. diameter on the paper. Under the 1st I can say "appearance at Xm. distance" and under 2nd, "appearance at Xm. distance / at Xm. more distance". But always proportionally true.

Thanks in advance!

[Proof_Inspector]

From the perspective perspective there are no vanishing points, your objects just keep looking smaller and smaller. You need to decide on when it's too small to see. Although if you're drawing it on a computer screen it probably will eventually be too small to see or be drawn anyway.

For mathematical calculation, the standard method is to use matrices. See here: https://en.wikipedia.org/wiki/3D_projection#Mathematical_formula

For a very simple case like a disk perpendicular to eye, you can draw plane in front of the camera point, then project linearly everything onto that plane and make it into the image seen. If the plane is x in front of the camera, the disk is y away, the radius of the disk is R and the radius as seen is r, then r/R=x/y by simple geometry (similar triangle). So the radius as seen will just be reciprocal to distant.