# Rate at which things appear smaller and establishing a baseline

1. Jul 29, 2012

### LykosPF4

Hi, everyone!

To be honest, I don't know that this is a math question or what field of science and math can help understand this, so this is my first try to ask others on the subject. Here it goes.

For a while now, I've been trying to understand visual perspective. Things get smaller as they move away, that's true, but what is the rate at which they appear smaller? One thing I felt was necessary was creating a baseline such as holding a 1 meter long object 1 meter away from my eyes. From there, you could say that

1 meter appears to be "l" meters long, where l<1
at d meters away from the baseline where d>0.

The only problem with that on paper is that I realize I really have no way of establishing visual angles. If I draw out my possible baseline, I cannot determine accurately measurements or field of vision a 1 meter long object takes up. 20°, 30°, 40°? I don't know.

One thing that comes out of this is calling out apparent lengths or how long something looks to be at a certain distance away. Like if you looked at the sun (with PROPER radiation blocking equipment!) it appears like a very tiny circle even though it is gargantuan. But if I hold out a penny in front of my face 2 feet away, how much smaller is the sun to the penny?

I don't know really if this has any function in real life, but it seems like it has some astronomical function that may already be in place, else we couldn't be determining the incredibly massive sizes of stars that are nearly as large as our our solar system. So this is why I'm wondering if this belonged in physics or astronomy as well.

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Does anyone have an idea what I'm talking about? Surely someone has attempted such concepts before me. If I need to draw any of this out on paper, let me know.

2. Jul 29, 2012

### awkward

Hi LykosPF4,

I'm not sure what you are asking. Are you asking how to measure small angles, like the width of a penny at a distance of a mile? If so, the astronomers (I am not one) have been working at this for a long time and have developed techniques for measuring really small angles. For example, these guys claim to be able to measure small angles to a resolution of 2.2 x 10^-6 radian: http://www.opticsinfobase.org/ao/abstract.cfm?uri=ao-43-29-5438
I am sure if you asked this question in an astronomy forum someone could give you a much better answer than I. However, the disk of a distant star makes far too small an angle for us to measure directly from earth, so estimates of its size must be based on some other means of measurement.

With respect to your question about the size of a penny at two feet versus the apparent size of the sun, it may help to know about solid angles: http://en.wikipedia.org/wiki/Solid_angle

3. Jul 29, 2012

### awkward

One more note, which you may already know. The length s of an arc subtended by an angle of $\theta$ radians on a circle of radius r is $s = r \theta$; so $\theta = s / r$. You can use this to compute the angle subtended by a penny at a distance of a a mile, for example.

4. Aug 5, 2012

### LykosPF4

Thank you. I'll see where this can go in the Astronomy forum as well as the equation you gave me. What I'm trying to determine is the rate at which things appear smaller as they move further away. the length of a 12" ruler never gets smaller, but it appears smaller as it moves away. I guess if I created a word problem, it may make more sense! Haha all those years in math courses and now I'm the one created word problems.

Let's say my arms are 2 feet long. I have 2 rulers that are 12 inches long each. Let's place one ruler propped up at eye level 4 feet from my eyes. I then hold up the other ruler 2 feet from my eyes and just far enough below the propped up ruler using my arms such that I can see both rulers simultaneously. How many inches would the propped up ruler take up on the ruler in my hands?

One must assume the propped up ruler takes up less than 12 inches on my in hand ruler, but I don't know how to do that yet. I'll see what I can do with your equation.