I'm not sure if this is the technical term, but I'm asking: At what rate do things scale bigger and smaller as they move toward and away from the human eye?

I know the equation for the linear perspective grids below are,

(size changed by this %) = 5/((distance to object) + 5) for the grid on the left
(size changed by this %) = 3/((distance to object) + 3) for the grid on the right

The '3's and the '5's come from the distance from the distance point to the vanishing point. On the grid on the right the objects smaller at a bigger rate as they move back than they do on the grid on the rate that has a distance point of 5.

for example: a circle with a 1' diameter at (distance to object) = 0
will be 0.5' at 5' away because...
0.5 = 5(5 + 5) = 5/10 = 1/2

I know one point perspective is not nearly the most accurate model, but I'm only interested is calculating the size of objects depending on their size, or vice versa. Maybe you have a better equations that scales objects the same way the eye perceives.

Right now I only know what it isn't. It's not depth of field because that has to do with what's in focus. It's not the range of vision in terms of degree or depth perspective because the eye(s) can see things shrink at the same rate with one eye looking at things going straight away from me.

In both of my grids the hieght from the horizon is 5' but things scale bigger and smaller at the same rates away from you whether you're on the ground or in a hot air balloon.

I suppose whatever the different setting there is between a fish eye and my eye is what I'm asking.

So, please. What is the distance point for the human eye?

Two different distance points


What you are referring to what is usually referred to as the "field of view" angle. The concept of FOV angle approaches the issue from the opposite direction, though. It defines how much width is visible at what distance. At a distance of d, the vertical width which is visible on the screen is d * tan(fov / 2) * 2. That means the screen-width of a circle of radius r would be r / (d * tan(fov / 2) * 2).

You can zoom in and out by changing the FOV angle. Sniper scopes in first person shooters are often implemented that way.

What FOV angle is the "natural" one is a matter of discussion. People's preferences are found anywhere between 45° and 90°. If you want to determine the ideal FOV angle in a physical way, you would need to know the physical width of the player's monitor and the distance between the monitor and the player's eyes.


There isn't one: Human eyeballs are round and heads pivot, so they don't have a flat projection plane.

When rendering, the proper Field Of View (FOV) angle for projecting into the flat screen plane depends on the human eyeballs distance from the screen and the size of the screen.

enter image description here

You can let the users change the FOV angle to match their gaming setup if they wish to and guesstimate the proper default value from the average screen size and average eye distance.

A mobile game may need a different default FOV from a TV console game or a PC desktop/laptop game.

  • \$\begingroup\$ +1 for the illustration. Note that on VR displays, the field of view is measurable and developers are encouraged to use the exact "arc length that the screen covers from your eyeballs" as their fov setting. \$\endgroup\$ – Jimmy Jul 19 '17 at 21:51

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