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View Full Version : Physics Question: How Does the Distance to an Object Affect Its Apparent Size?



Weimann
2013-07-25, 09:45 PM
This is a thing me and a friend of mine talked about the other day, and I've been trying to google it but can't find quite what I'm looking for.

The original problem is this: we have an object of a certain size (such as, for example, the sun) which we look at from a certain distance. If we were to move the object so that the distance to it was halved, how would that affect the apparent size of the object? It would get bigger, of course, but how much? By what function?

Is there any engineer around who can shed some light on this matter?

faircoin
2013-07-25, 09:55 PM
http://en.wikipedia.org/wiki/Perspective_%28visual%29#Linear_perspective

The light of the object's profile subtends at a smaller angle to our eyes. Apparent height of object (h), distance (d), and actual size (a) are related like so: h=a/d.

Rockphed
2013-07-25, 09:58 PM
To expand on that:

If the sun were half as far away, it would have twice the apparent radius and 4 times the apparent area. This jives with electromagnetic radiation falling off as the square of the radius as it expands in a sphere.

Siosilvar
2013-07-25, 10:10 PM
To expand on that, it's basically because of how triangles work. Imagine a right triangle, with you at one of the acute corners. The adjacent side is the distance between you and the center of the object, and the opposite side is half the object's width. You can scale the triangle to whatever distance or object size so that it looks the same as the first object; since the two sides vary linearly, twice the distance requires an object with twice the width/radius to look the same size. It follows that twice the distance with the same object would appear half as wide across your vision.

...

That made sense when I first wrote it, but not quite so much when I reread it. Oh well.

kinem
2013-07-25, 10:11 PM
Think of it this way: If you halve the radius of a circle, you halve its circumference (the distance around it). It would only take half as many suns to fill up the circle all the way around (the same angle: 360 degrees), so each one would take up double the angle compared to what it would have at the distance you started with.

factotum
2013-07-26, 01:29 AM
Think of it this way: the apparent size of the Moon and the Sun in our sky is about the same (has to be, or else solar eclipses couldn't happen). Moon is 2,200 miles in diameter and around 240,000 miles away. Sun is 880,000 miles in diameter and 93 million miles away, so it's 400 times larger and 400 times further away (approximately, but then, the figures I've listed for size and distance are themselves approximate). So, that tells you that the apparent size of the object varies linearly with distance.

Jimorian
2013-07-26, 01:39 AM
Think of it this way: if the lineality of the objectitude matches the inversal functiative, then the obtusal archetime rectomates across multidecimal projectinates, thereby reversionating spheritoid diagonicities. Got it?

Nerd-o-rama
2013-07-26, 01:47 AM
Look at it this way: if you half the distance to something, its apparent size doubles.

In case the other answers were confusing.

(This is barring refraction, which is a major issue with looking at the sun through our atmosphere, especially at an angle near the horizon.)

TuggyNE
2013-07-26, 08:15 AM
Think of it this way: if the lineality of the objectitude matches the inversal functiative, then the obtusal archetime rectomates across multidecimal projectinates, thereby reversionating spheritoid diagonicities. Got it?

Imma quotebox this, kkthxbai.

shawnhcorey
2013-07-26, 08:33 AM
The angular width is arctangent of the width of the object divided by the distance.

θ = acrtan( width distance )

If the distance is halved, the angular is not doubled.

θ = arctan( 2 width distance )

RCgothic
2013-07-26, 08:42 AM
Think of it like this:

http://img.photobucket.com/albums/v723/rcgothic/apparentSize3_zps61639961.jpg (http://smg.photobucket.com/user/rcgothic/media/apparentSize3_zps61639961.jpg.html)

An object at 2x must be twice as tall (2h) as an object of size h at x to have the same apparent size H. It therefore has four times the actual area 4A.

An object h at 2x only has half the apparent size 0.5H. It therefore appears half as tall and 1/4 the size of the same object at x.

Weimann
2013-07-26, 07:09 PM
That's quite a few ways to think about it, that is.

Well, I sure got informed. :smalltongue: Thanks for your help.

shawnhcorey
2013-07-27, 02:17 PM
An object h at 2x only has half the apparent size 0.5H. It therefore appears half as tall and 1/4 the size of the same object at x.

No, the angle is not halved; it is slightly larger than half.

RCgothic
2013-07-27, 02:38 PM
It depends whether the object is perpendicular to the viewer or not.

I've already demonstrated that an object held perpendicular to the viewer occupies the same angular width if it is half the size at half the distance. That line on the diagram isn't bent.

But if the object isn't perpendicular to the viewer - for instance two buildings of identical size, one at half distance viewed from ground level - in that instance you're right.

shawnhcorey
2013-07-27, 03:02 PM
But there's a line missing from the diagram. Draw one from the apex to the disc at twice the distance. That angle is not 1/2 of the other angle, H.

Consider a disc at a distance equal to its radius. The angle it forms from its centre to its edge is 45 or 90 across its diameter. If you move the disc to half the distance, the angle is not doubled to 180. If it was, it would take up half the sky. That would only happen if you moved it to zero distance.

If the object is far away and move half the distance, then double the angle is a good estimate but not exact. The exact angle would be slightly less than half.just

Manga Shoggoth
2013-07-27, 03:14 PM
To add something further to the mix - the apparant (percieved) size of an object isn't always explained by the maths - for example, the moon illusion (http://en.wikipedia.org/wiki/Moon_illusion).

factotum
2013-07-27, 03:26 PM
The angle it forms from its centre to its edge is 45 or 90 across its diameter. If you move the disc to half the distance, the angle is not doubled to 180. If it was, it would take up half the sky. That would only happen if you moved it to zero distance.


However, at any sort of reasonable distance the approximation of saying that twice as far equals half the size is a pretty reasonable one--for instance, if the object subtends 10 degrees of sky, and you move it twice as close, the arctan method will give you the object's new apparent size as 19.42 degrees rather than the 20 degrees the simple approximation yields. For most purposes that's not a significant enough error to be worried about, and the error gets less as the angular size of the object decreases.

Malacandra
2013-07-31, 11:03 AM
However, at any sort of reasonable distance the approximation of saying that twice as far equals half the size is a pretty reasonable one--for instance, if the object subtends 10 degrees of sky, and you move it twice as close, the arctan method will give you the object's new apparent size as 19.42 degrees rather than the 20 degrees the simple approximation yields. For most purposes that's not a significant enough error to be worried about, and the error gets less as the angular size of the object decreases.

Correct - for any angle q, sin q < q < tan q (where q is in radians), but for small values of q the difference is also small.

e.g. sin 0.1c = 0.0998, tan 0.1c = 0.1003. 0.1 radians is a non-trivial angle, almost 6 degrees.

Mal, B Sc (Mathematics) :xykon: