# Thread: size of magnified image

1. ## size of magnified image

If something that is originally seen across 1 degree of arc of sight is magnified to 2 degrees, exactly how much larger would the area of the image appear to be, what ratio? What is the formula for that?

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As area is the square of length, if you double the linear dimension of something then the area must be 4 times the original,, 3 > 9, 4 > 16, etc.
this remains true regardless of the shape, square circle ellipse etc.

3. Originally Posted by Senor Molinero
As area is the square of length, if you double the linear dimension of something then the area must be 4 times the original,, 3 > 9, 4 > 16, etc.
this remains true regardless of the shape, square circle ellipse etc.
But I'm not curving it around with the arc, just magnifying it from the same distance. So at 180 degrees of sight, wouldn't it appear infinitely large?

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Not sure what you're getting at here. If we're talking about small angles, ie, through a telescope, if the apparent image has twice the dimension then it must have four times the area. With the optics of a fish-eye lens (180 degree field of view), if you magnify that image twice then you're only seeing the central 90deg spread out to 180deg.
please explain the basis of your query.

5. Originally Posted by grav
But I'm not curving it around with the arc, just magnifying it from the same distance. So at 180 degrees of sight, wouldn't it appear infinitely large?
I kind of get what you might be getting at, and that might be equating what you are seeing with the sine of the angle.
The problem is that what you are looking at is not flat, so the triangle/sine thing doesn't apply.
Your field of view is going to be the same at any angle, therefore what you can see does not change between the adjecent and hypotonuse of the triangle that you are looking at. Your shape is a wedge with an arced opposite end instead of a triangle with a flat side.
That arc is proportional to the angle of view therefore the area is 4x proportional.

6. Originally Posted by NEOWatcher
I kind of get what you might be getting at, and that might be equating what you are seeing with the sine of the angle.
The problem is that what you are looking at is not flat, so the triangle/sine thing doesn't apply.
Your field of view is going to be the same at any angle, therefore what you can see does not change between the adjecent and hypotonuse of the triangle that you are looking at. Your shape is a wedge with an arced opposite end instead of a triangle with a flat side.
That arc is proportional to the angle of view therefore the area is 4x proportional.
Yes, that is what I'm asking, except I was thinking it would be the tangent of the angle. Are you sure about the proportion to the angle? I know the arc would be twice as great, but if we see the cross section of an object some distance 'd' away that covers an angle of 1 degree, how would an object at the same distance covering two degrees compare? Wouldn't an object at the same distance that covers 180 degrees appear infinitely large, since that would mean one cannot then see any boundaries or edges of the object, so it wouldn't be just a direct ratio of the angles involved? Otherwise, that would be like saying that a galaxy that is a billion light years away covering our whole field of view is only 180 times wider than one that covers 1 degree of arc, wouldn't it?

7. Originally Posted by grav
Yes, that is what I'm asking, except I was thinking it would be the tangent of the angle. I know the arc would be twice as great
Good, we're at least on the same wavelength here.

Originally Posted by grav
but if we see the cross section of an object some distance 'd' away that covers 1 degree of arc, how would an object at the same distance with two degrees of arc compare?
yes, tangent is the one because the distance would be the adjecent side.
tan(theta) = opposite / adjecent
Now your viewing area is most likely circular since you have a fixed angle.

So, the radius of that circle ends up being the opposite side of the triangle at half the viewing angle.

Radius = distance * tan(theta/2)

And now, of course, Area = pi*r^2

And you get Area = pi * (distance * tan(theta/2))^2

Now plug in an arbitrary distance (1 works well ) and compare the result from two different thetas. The ratio will depend on the sizes of the angles and ends up not being linear.

Originally Posted by grav
Wouldn't an object at the same distance that covers 180 degrees of arc appear infinitely large, since that would mean one cannot then see any boundaries or edges of the object, so it wouldn't be just a direct ratio of the angles involved?
Yes; but only if you are assuming the object is flat, and the distance to the object is not infinately small.

8. Actually, now that I think I've got it straight in my head what it is exactly I am trying to find here, I think you had it right the first time, with the sine of the angle. I am trying to find out how much larger an object would appear to be if viewed at a larger angle for the line of sight at the same distance. Now, thinking about this in terms of the cross sectional area of a flat plane across the line of sight, the two sides of a right triangle are the distance and radius, so we use the tangent of the angle to compare the two directly. But for a three dimensional object such as with a sphere, the "edge" of the line of sight would fall tangent to the surface of the sphere, putting that and the radius at right angles to each other, making the distance the hypotenuse. So in this case, the distance and radius of the object can be compared directly by using the sine of the angle. And since it is the size of the object I am really trying to find, its width would be in proportion to its radius, of course, so we use the sine of the angle.

The thing is, though, a formula I came up with shows a tremendous difference for the result than what one might otherwise think, so I am trying to find the angles as precisely as possible now to compare results, but even at a distance of just a few times greater than the radius, the tangent and sine of an angle really wouldn't make that much difference at all, so that's not it. In fact, the formula shows a greater deviation with greater distance, and therefore at smaller angles, while using a ratio of the tangent of angles or the sine, or even a direct ratio of the angles themselves for the normal object size and magnified image, at small angles converge to almost precisely the same thing very quickly, so it really wouldn't make too much difference which one we used at very large distances.

Thank you both for your replies.

9. Originally Posted by grav
Actually, now that I think I've got it straight in my head what it is exactly I am trying to find here, I think you had it right the first time, with the sine of the angle. I am trying to find out how much larger an object would appear to be if viewed at a larger angle for the line of sight at the same distance.
Somehow I got lost on this...I have a fairly good idea what you are saying, but, instead of addressing this particular point, I think the following may bring a bit more "perspective" to the situation.
Originally Posted by grav
But for a three dimensional object ... So in this case, the distance and radius of the object can be compared directly by using the sine of the angle.
I almost thought of that in the beginning, but with perspective, the distance is the closest point. Viewing is in relation to a plane that contains that point, and is perpendicular to the viewing angle. All other points of the object are then projected onto that plane to determine the size.

So, taking a sphere for example, the distance would be the distance to the center of the sphere, less the radius of the sphere.
The limit of the sphere (outside viewable edge) would be the tangent (as you said).
This point now creates a right triangle with the radius of the sphere as the opposite side, and the line of sight to the point is the adjacent side.
To project that point to the plane, it becomes a ratio of its distance, therefore the size at the distance of the plane appears as
projectedradius * (planedistance / (planedistance - radius))

3D projections start to get a little messy. Wiki has some formulas there.

Now; without knowing where the point lies (in 3d space, or distance to the plane) it's going to be impossible to compute without having two different viewing locations, which is where parallax comes in.

10. Thanks again. I guess I'll rephrase the question, though, so you can be sure what I was asking to begin with, since I wasn't even sure at first what I was looking for precisely, struggling with the best way to think about it, as you now appear to be doing. For a sphere with its center at some distance from an observer, how large would it appear to be if viewed at some angle between its tangents? Seems simple now, doesn't it? Basically, the distance as I have defined it is just that from the center of the sphere to the observer. The tangent would form a right angle to the radius, making a right triangle with the distance as the hypotenuse. I'm looking for the size of the sphere, so it is the radius I want. That is just the sine of the angle times the distance, just as you said before.

11. Originally Posted by grav
For a sphere with its center at some distance from an observer, how large would it appear to be if viewed at some angle between its tangents?
I think it's the word "appear" that was confusing me, because, in my mind, that implies not knowing a distance, and bringing in all sorts of transformations. "Appearances are decieving."

So; you know the distance to the center of the sphere? Yes; the sine of the angle...
Radius = Distance * Sin(theta)

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