Objects at a distance....?

Question

If a one foot tall item appears to be one foot tall at one foot distance, how tall would the same item appear to be at 600 feet?

Is it simply 1/600th of a foot? Am I totally off the mark?

How would it compare to an identical 1 foot item still at 1 foot? I don't know how to phrase that differently.

If it's 1 foot high at 1 foot distance, how tall would it "appear" to be at 600 feet distance?

Is that better?

Raymond and Benoit, both of you answered that in terms that I can understand. Thanks, guys! Problem is, I can only pick one of you as the best and both of you helped just as much.

I'm going to flip a coin, so please understand.

Thanks to all of you!

Answer 1

This can depend on the position of the eye, but let's make a simple model... Draw a right triangle ABC where A is the position of your eye, B (right angle) is the foot of the object, and C is the top of the object. BC=1 foot, the height of the object.

You are saying that if AB=1, then angle A, which is the resulting angle taken by the image (45° or pi/4 rad) will make the object appear 1 foot tall.

We need to assume something else, that the apparent height of the object is proportional to angle A, the angle it uses in our field of vision.

If AB = x feet, then tanA = 1/x and so the object will appear to be A/45° = arctan(1/x) / 45° feet high. For x=600 this equals
arctan(1/600) / 45° = 0.002122 or about 1/471 ft.

EDIT: Answer below me is good too, it depends a lot on the position of the eye at a distance as close as 1 foot. But as you get further away it gets proportinal to the angle. That's because arctanx/arctany can be approximated by x/y for small angles.

This is a little too simplistic because our brain is good at adjusting for distance. This is why we don't perceive objects to be shrinking as they move away from us. But if we were unable to gauge distances and only estimated the size of objects based on how much of our angle of vision they took up, we'd get something like the above.

Answer 2

Apparent size is usually given as an angle.

A one foot object at a distance of one foot would sustend an angle of 45 degrees if the eye is level with the foot of the object or 60 degrees if the eye is level with the middle of the object.

Most problems of that nature usually avoid such close distances (because of the uncertainty, as shown above).

However, once you have picked a mode (let us say, eye level with the middle of the object), then angles vary as the inverse of the distance.

If an object subtends an angle of 60 degrees at a distance of 1 foot, then it will subtend 1 degree at a distance of 60 feet; it will subtend 0.1 degree at 600 feet.

The Sun (150 million km from us) is 400 times further away from us than is the Moon (0.375 million km from us near perigee, with observer on the Surface, not at the centre of Earth).

The Sun (diameter 1,392,530 km) is 400 times bigger than the Moon (diameter 3480 km).

That is why they have (approximately) the same apparent size (0.5 degree) when seen from Earth.

However, as the others imply, if you know that the object is 600 times further away and appears 600 times smaller, then you know that it still looks one foot tall.

Answer 3

The problems is how you are phrasing the question. Do you mean how tall will it appear compared to other objects at that same distance or compared to it's size at 1 foot distance?

Answer 4

It's still one foot :)