Convert a map's scale to z height Assume the following example: 


*

*Given a device with a 10cm x 10cm screen (just an example)

*Assume the screen shows a fully zoomed in map of Los Angeles

*Zoom out the map, until you reach a map scale of 1:2000

*Let's assume a bird has the same field of vision (10cm x 10cm). Of course he doesn't, but just to bring the "height" into context. As you zoom out the map, the bird flies higher.


Question: what is the math to calculate the bird's distance to the ground?
NB: This is about a 3D map style definition, to convert a 2D map style definition defined as scale factor into a 3D object distance between a camera and the ground. But I prefer the bird for the example =)
 A: Here is how perspective works:

The brownish far right vertical line is the ground.
The green vertical line in the middle is the projection plane. 
The blue lines, and the angle $\beta$, denote the field of view, the part of the ground that is visible in the projection. $\beta$ is also called the angle of view. 
In the diagram, $Y$ is the distance from ground to the projection plane, and $y$ is the distance from the projection plane to the observer. $d$ is the distance from observer to ground, i.e. $d = Y + y$.
In the diagram, $X$ is drawn as the size of the ground that fits in the view, and $x$ as the size of the projection. However, you can just as well think of $X$ as the size of some detail on the ground, and $x$ as its projected size.
If you have a map in $1:2000$ scale, it just means that $x:X = 2000$, i.e. $x = \frac{X}{2000}$. It does not tell us anything about $Y$, $y$, or $\beta$.
If we look at the diagram, we can see that the blue triangle is actually two right triangles $\beta/2$, halving the blue triangle vertically. Trigonometry tells us that
$$\tan\left(\frac{\beta}{2}\right) = \frac{x}{2 y} = \frac{X}{2 d} = \frac{X}{2 y + 2 Y}$$
We can use any two equal parts, say
$$\frac{x}{2 y} = \frac{X}{2 y + 2 Y} \; \Leftrightarrow \; Y = \frac{y}{x}(X - x) \; \Leftrightarrow \; y = \frac{x Y}{X - x}$$
In order to find $d$, $y$, or $Y$, you need at least $X$, $x$, and one of $\beta$, $y$, or $Y$.
Note that it is easy to calculate your own preferred angle of view, if you have a 3D picture of something interesting. Move closer or further away from your display, until the depth seems "real", and there is no distortion. Measure the distance from your eye to the center of the 3D picture $y$ (using a ruler or maybe a piece of string), and the size of the picture $x$ on the display (using a physical ruler on top of the display, not in pixels!). Then,
$$\beta = 2 \arctan\left(\frac{x}{2 y}\right)$$
For most binocular animals with eyes placed horizontally, the angle of view is a lot wider horizontally than vertically.
In 3D games, a wider field of view means you see more of what is happening around you, and is therefore preferable. If you perceive distortion, try moving closer or further out from your display device. (If you move closer to your display device, you'll also want to reduce the brightness of your display device if the room you are in is not brightly lit.)
The "surprise" view sometimes used in movies, where the person or object in the focus stays the same size, but things behind zoom out, is done by manipulating the angle of view, using a zoom lens. The camera starts close to the focus, but zoomed out, which causes the angle of view to be larger than normal. The camera moves away, but zooms in, which causes the angle of view to decrease. If the zooming and moving the camera is done at just the right rates, the effect feels very weird. (Actually, I don't remember which way -- moving in and zooming out, or moving out and zooming in -- the effect is better, and more often done.)

what is the math to calculate the bird's distance to the ground?

You find $d$ in the above diagram. For example:
$$d = \frac{y}{x} X = \frac{X}{2\tan\left(\frac{\beta}{2}\right)}$$
