hermanntrude Posted April 29, 2009 Posted April 29, 2009 because if they didnt, stars would get in the way when you were trying to read
Sisyphus Posted April 29, 2009 Posted April 29, 2009 What determines the apparent size of something? The angle it takes up out of your total field of vision. Simplify to one eye, and 2 dimensions. Hold 12 inch ruler 12 inches away. What angle, at your eye, is there between lines pointing at one end of the ruler and the other? Now move the ruler to 24 inches away. What angle is there now?
TimbaLanD Posted April 30, 2009 Author Posted April 30, 2009 The angle of vision is different from one person’s eye to the other, right? Are you saying not everyone will see something in the exact same way?
swansont Posted April 30, 2009 Posted April 30, 2009 It's the angle subtended by the object in your field of view. [math]\theta = \frac{s}{r}[/math] Something that's 1 meter tall and 100 meters away subtends 0.01 radians, so it looks the same size as something 1 cm tall and 1 meter away. Or, it's why the moon looks like it's the same size as the sun.
padren Posted April 30, 2009 Posted April 30, 2009 TimbaLanD, take a piece of graph paper, trace a quarter coin in the center, and put a dot in the center of that. Now, draw various circles on the paper, at different distances to the center circle - some pretty close, some pretty far - but all the same size. (maybe the size of the hole in a CD) Take any nearby circle, and draw lines from it's edges to the center dot, and lightly shade the space between those two lines. Do the same with any far off circle. Then, compare the amount that the edge of the inner 'coin sized center circle' is shaded as a result of the lines you drew to the two different objects. Each of those shaded sections take up a number of "degrees" out of 360, and when traced among the edge of that inner circle - it can become a distance. If you could graph this in three dimensions, it would be area instead of distance - and that's how the retina in the back of the eye works - light touches the surface over a certain area - but the geometry principles are identical.
Mr Skeptic Posted April 30, 2009 Posted April 30, 2009 Incidentally, we're talking about the same inverse square law as for gravity and electricity. In this case, it's size (area in your field of vision) instead of strength that decreases as the inverse square of the distance. Ie, something twice as far away looks half as tall and half as wide, for a total of 4 times smaller.
Sisyphus Posted April 30, 2009 Posted April 30, 2009 Incidentally, we're talking about the same inverse square law as for gravity and electricity. In this case, it's size (area in your field of vision) instead of strength that decreases as the inverse square of the distance. Ie, something twice as far away looks half as tall and half as wide, for a total of 4 times smaller. Which, to take it further, is a good way of intuitively understanding why those things work in an inverse square law. "I make up X% of that thing's 'field of view.' If I moved to half the distance, I make up 4X% of the field of view, so all omnidirectional radiating-like effects should be 4 times stronger."
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