(Probably) A Vision Related Question
I have been wondering this for some time now, and have no idea under what topic this question falls, but I figure you guys would (as always) have the answer.
Okay assume I am looking at a 3m long object from 10m away, how long would the object appear to be, and how would I work it out?
Okay assume I am looking at a 3m long object from 10m away, how long would the object appear to be, and how would I work it out?
Measure it with a ruler that you hold close to your face. Then you realize you will measure different lengths depending on how close you hold the ruler.
So my boring answer would be: you see a 3m long object from 10m away and that's that.
So my boring answer would be: you see a 3m long object from 10m away and that's that.
closed account (G309216C)
also see the degree you are seeing it from. That way I suggest you look at it from a straight line then find the formula then add new things to it such as integrating angles to it. I doubt it that you are the first trying it, there must be a formula already available.
Okay, now assume you take a picture of that 3m object and you measure it, what would the relationship be?
EDIT:
@Catfish meh.
EDIT:
@Catfish meh.
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None. It depends on how large you print the picture.
The problem with your question is that it assumes that the image of something has a size, rather than being just a large number of photons that happen to converge towards your retina. All that can be said about an object at the point where it is being seen is that it has an apparent size measured as a solid angle. For example, one unit of solid angle is the surface area of the projection of the object in question onto the unit sphere. So the apparent size of an object that occupies the entire night sky would be 2*pi. The apparent size of a d=3 m circle held 10 m away from your eye's focal point would be
a = atan(1.5/10)
b = 1 - cos(a)
A = 2 * pi * b ~= 695 cm^2
The object's true surface area is about 70685.83 cm^2, so this is a linear magnification of 1/10(relative to holding it 1 m away from the focal point, I imagine).
Actually relative to having an infinitely large planar eye at infinity, parallel to the circle.
The problem with your question is that it assumes that the image of something has a size, rather than being just a large number of photons that happen to converge towards your retina. All that can be said about an object at the point where it is being seen is that it has an apparent size measured as a solid angle. For example, one unit of solid angle is the surface area of the projection of the object in question onto the unit sphere. So the apparent size of an object that occupies the entire night sky would be 2*pi. The apparent size of a d=3 m circle held 10 m away from your eye's focal point would be
a = atan(1.5/10)
b = 1 - cos(a)
A = 2 * pi * b ~= 695 cm^2
The object's true surface area is about 70685.83 cm^2, so this is a linear magnification of 1/10
Actually relative to having an infinitely large planar eye at infinity, parallel to the circle.
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@helios While I thank you for your answer I must say I did not understand some parts, would you mind explaining the following please:
What is a solid angle?
r=1 or d=1?
How did you get that?
Where did these formulas come from?
d=300cm
r=150cm
SA=4*pi*r^2
=4*pi*22500
=90 000*pi
~=282743.34
| measured as a solid angle |
What is a solid angle?
| unit sphere |
r=1 or d=1?
| So the apparent size of an object that occupies the entire night sky would be 2*pi |
How did you get that?
| a = atan(1.5/10) b = 1 - cos(a) A = 2 * pi * b ~= 695 cm^2 |
Where did these formulas come from?
| The object's true surface area is about 70685.83 cm^2 |
d=300cm
r=150cm
SA=4*pi*r^2
=4*pi*22500
=90 000*pi
~=282743.34
Solid angles are generalizations of regular angles to three-dimensional space.
The unit circle and unit spheres have a radius of 1.
The imaginary night sky projected onto the unit sphere is half of it. The surface are of the unit sphere is 4*pi.
A point on the edge of the circle is at (10, 1.5) (eliminating one dimension to simplify calculations is possible because the circle has rotational symmetry), which corresponds to (cos(a), sin(a)). sin(a)/cos(a) = tan(a). a = atan(tan(a)), since we're on the first quadrant.
b is h for the formula to get the surface of a spherical cap. A is the surface of a unit spherical cap (http://en.wikipedia.org/wiki/Spherical_cap ).
I stated the object was a circle of diameter 3. Even if it was a sphere, its total surface area would be of no use, since not all of it is projected onto your eye when you look at it. Obtaining the viewable surface area of a sphere is more complicated because you have to figure the size of the minimal spherical cap every point of whose edge is tangent to your line of vision.
The unit circle and unit spheres have a radius of 1.
The imaginary night sky projected onto the unit sphere is half of it. The surface are of the unit sphere is 4*pi.
A point on the edge of the circle is at (10, 1.5) (eliminating one dimension to simplify calculations is possible because the circle has rotational symmetry), which corresponds to (cos(a), sin(a)). sin(a)/cos(a) = tan(a). a = atan(tan(a)), since we're on the first quadrant.
b is h for the formula to get the surface of a spherical cap. A is the surface of a unit spherical cap (http://en.wikipedia.org/wiki/Spherical_cap ).
I stated the object was a circle of diameter 3. Even if it was a sphere, its total surface area would be of no use, since not all of it is projected onto your eye when you look at it. Obtaining the viewable surface area of a sphere is more complicated because you have to figure the size of the minimal spherical cap every point of whose edge is tangent to your line of vision.
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