# Real Height versus Perceived Height, of Objects at a Distance.

1. Aug 14, 2012

### colinven

This thought came from staring at telephone poles and light posts along roads and walkways. I noticed that these objects all had the same height more or less and were regularly spaced. The poles and appeared to reduced in height (following some mathematical relationship) as the distance between me and the object increased. When I first thought about this I would hold my fingers up close to my face, and with my pointer finger and thumb encase the height of a telephone pole between my two fingers. The height between my two fingers might have only been a few centimeters (this is the height that the telephone pole seemed to be from my distance to it), but the real height of the telephone pole was much larger. I thought, how could the perceived height and real height be related; is there some way in which I could simply encase any object at some distance and by measuring the height between my fingers know the real height of the object encased?

What I found was a simple ratio: $\frac{h}{l}$=$\frac{H}{L}$

where h is the height between my fingers or the perceived height, l is the distance my fingers are from my eye, H is the real height of the object, and L is the distance from my fingers to the object.

Using this formula I only had to know my distance to the object, the height between my two fingers and how far my fingers were from my face in order to know the real height of any object.

The perceived height of telephone poles as the fade into the distance is proportional to $1/L$, or in other words the perceived height is inversely proportional to your distance from each pole. This is true if you are standing directly in line with the poles, however we know from experience that if you are standing directly in line you will not observe this telescoping series. You have to stand some distance from the line of telephone poles in order to observe this phenomenon. To account for this I found the following equation using simple geometry for right triangles: $h$=$\frac{Hl}{\sqrt{n^{2}d^{2}+L^{2}}}$

The above equation assumes you are standing parallel to the telephone poles and directly across from the first one in your series. Thus, L is the perpendicular distance to the first telephone pole, d is the pole spacing, and n is the number of telephone pole in the series (the first telephone pole is n=0, the second n=1,...).

The whole denominator in the above equation acts as a way to indirectly measure the distance to each successive telephone pole using the Pythagorean theorem. The above equation has some finite bound, that means as n → 0 h → $\frac{H}{L}$. Whereas the first equation has no bound, that means as L → 0 h → ∞. Therefore as L → ∞ the $\sqrt{n^{2}d^{2}+L^{2}}$ approximates L and the difference between the two equations becomes negligible.

I offer this as a start to a discussion on how our perception forms the space around us. Is it possible that by our observation we impose restrictions on the allowed heights of an object? Does our observation create a potential curve like $1/L$? Where does distance come from, us or is it apriori?

2. Aug 14, 2012

### Staff: Mentor

You have discovered the concept of perspective: http://en.m.wikipedia.org/wiki/Perspective_(graphical [Broken])

Last edited by a moderator: May 6, 2017
3. Aug 14, 2012

### Bobbywhy

colinven, Welcome to Physics Forums!

Russ is exactly right: If Leonardo Da Vinci was alive today he would congratulate you! Just have a look at his "The Last Supper".

4. Aug 14, 2012

### colinven

Thank you Bobbywhy and russ_watters for reading and responding to this post. This subject has been on my mind for a while now, and it feels great to share my ideas.

When we look at telephone poles fade into the distance they do not fade linearly, but instead the relative height of each successive pole reduces proportional to $1/L$; where L is the distance from the observer to each pole. It would be incorrect, then, to draw a straight line extending from a point in the center of a page to the edge and create my telephone poles under this line. I would have to draw the curve $1/x$ and draw my telephone poles under this function in order to create the correct perspective.

What poses this restriction on space; that an object (when observed) must have a relative height fall under some curve approximating $1/L$?

5. Aug 14, 2012

### haruspex

No, you're overlooking the fact that they do not appear to be equally spaced either. Projection (perspective) preserves straight lines.
There's an interesting conserved quantity in perspective called the cross-ratio. E.g. if there are four equally spaced points in a straight line (in reality) then in the projection, when you have positioned three of them, you can calculate where the fourth must go.

6. Aug 15, 2012

### Bobbywhy

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