How to calculate apparent size of an object based on distance

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I would like to know what the size of my monitor would appear to be if I were to move further away from it. It is a 22 inch monitor, and let's say right now I am 24 inches away from it. If I were 30 inches away from it, what would the size appear to be? I have an eye tracking device mounted on my monitor and it needs to know the size of my screen. If I move the device closer to me but keep the monitor in the same place, I will need to give it different measurements, because it will be as if I have a smaller screen.
 
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dmehling said:
If I were 30 inches away from it, what would the size appear to be? I

You have to say precisely what you mean by the size appearing to be something. Are you asking the question:

"A 22 inch monitor at a distance of 30 inches from the eye appears to be the same size as X inch monitor that is 24 inches from the eye. Find X."
 
In other words, if I moved further away, it would get smaller. So, it would appear to be a certain size 24 inches away. I would like to know how big it would seem if I moved away another 6 inches, to be a distance of 30 inches away.
 
Or maybe I should make it more specific to my situation. I have to tell the software the size of my monitor. If I move the eye tracking sensor further away from the screen, I will have to give new measurements, otherwise the eye tracking will be completely an accurate. So, if I have a 22 inch monitor, and I move the sensor further away from the monitor, it would be like using it on a small monitor.
 
No, it will be like using it on the same monitor. The monitor has not changed size.
 
No, it would be different. It is the way the sensor is designed. The sensor is tracking my eyes so that I can control the mouse cursor with my eyes. So it needs to know what the size of the monitor is. If the sensor is moved closer to me, but the monitor stays in the same place, the position of my eyes will have changed in relationship to the sensor. Then, the cursor will not be where I'm looking at.
 
[itex]\frac{22}{30} = \frac{X}{24}[/itex]

[itex]X = 17.6[/itex]
 
So it's that simple? What about angular size? Does that not play a role in this kind of problem?
 
The tangent of the angle that the diagonal of the screen subtends is approximately given by the ratio [itex]\frac{22}{30}[/itex]. Setting the two angular sizes equal implies setting the two ratios equal. You can solve the problem without computing the angles.