So if you’re standing 3 miles from a 630 foot tall object. What is it’s apparent height? What is the general formula for this?
What do you mean by “apparent height”? It’ll look exactly as tall as a 630 foot object 3 miles away.
If you want the angular height, then it’s the arctangent of the ratio of the height to the distance. Or, if the distance is very large compared to the height, it can be very well approximated (in radians) as just that ratio.
Similar triangles say that a 630 foot tall object at 3 miles is the same as a 1.19 foot tall object 30 feet away.
So if you had no depth perception and you think the object is right in front of you, how tall would you think it was?
If you want it in linear measure you have to say it would appear to be x inches tall at y inches distant. So, you have to say what referrnce length you’d like (arm’s length, say). Which is why beowulf answered as he did.
What does “right in front of you” mean? An inch from your eye? A foot from your eye? Five feet from your eye? These will each give you a different answer.
–Mark
Ok, I get it. I can work from this. I’m trying to get a close object to appear to be a certain height relative to a distant object.
Also, considering the curvature of the Earth, you may want to more rigorously define what you mean by “how tall will this look at a distance”. You’re probably most concerned about the angular diameter of the object at that distance, but at sufficient distances your up-direction is not the same as the up-direction of the tall object. The tower will have a smaller angular diameter than a object one-tenth the height at one-tenth the distance away, as it will be pointed away from you more. The effect is probably negligible, but it depends on how precise of an answer you need.
It’ll also look much bluer - atmospheric perspective.
It this for your giant statue holding a croquet mallet? 
According to post #7, yes.
So maybe drive three miles from the St. Louis Arch and eyeball it.
Is it above you & skylined?
Is it below you and backed by water?
Salt flats at Bonneville?
Is it one foot wide or 30 feet wide?
Is it a lattice tower or a solid tank like tower?
This may help:
In all seriousness though, you just need to make sure the angle subtended at the eye is the same.
It looks about two degrees high.
angle = asin ( height/ distance) - Note: must be in same units for the two distances.
Technically arctan, not arcsin, unless by “distance” you mean from ground level to the top of the object. However, for small angles like this, the difference is much too small to worry about.
Degrees (or radians) isn’t really helpful though. Apparent height scales with distance for practical purposes so you can say a 630’ object 3 miles away is the height of something 210’ high a mile away, or a little over 10’ high a hundred yards away - about the height of the crossbar on a football (American) goal. However, that’s the apparent height above the ground; the crossbar’s only about four and a bit feet above your eye level. But if you’re reclining on the ground at one end of the football field, the crossbar at the other end is more or less in line with the top of your tall thing three miles away.
Here’s the St. Louis Arch from 3 miles away.
So it’s 0 feet tall at that point (for practical purposes). Looks like it’ll have to be close to the river to even see the arch.
Cost for the real estate just went up significantly.
Maybe it’s one of those Where’s Waldo things where I can see it because I know what I’m looking at. Let’s try a different angle.