Drawing a Topographical Map

Question

The gradient is a mathematical operator that may help you.

If you can turn your interpolation into a differentiable function, the gradient of the height will always point in the direction of steepest ascent. All curves of equal height are perpendicular to the gradient of height evaluated at that point.

Your idea about starting from the highest point is sensible, but might miss features if there is more than one local maximum.

I'd suggest

1. pick height values at which you will draw lines
2. create a bunch of points on a fine, regularly spaced grid, then walk each point in small steps in the gradient direction towards the nearest height at which you want to draw a line
3. create curves by stepping each point perpendicular to the gradient; eliminate excess points by killing a point when another curve comes too close to it-- but to avoid destroying the center of hourglass like figures, you might need to check the angle between the oriented vector perpendicular to the gradient for both of the points. (When I say oriented, I mean make sure that the angle between the gradient and the perpendicular value you calculate is always 90 degrees in the same direction.)

Alternately, there is the marching squares algorithm which seems appropriate to your problem, although you may want to smooth the results if you use a coarse grid.

The topo curves you want to draw are isosurfaces of a scalar field over 2 dimensions. For isosurfaces in 3 dimensions, there is the marching cubes algorithm.

In response to your comment to @erickson and to answer the point about calculating the gradient of your function. Instead of calculating the derivatives of your 300 term function you could do a numeric differentiation as follows.

Given a point [x,y] in your image you could calculate the gradient (direction of steepest decent)

g={  ( f(x+dx,y)-f(x-dx,y) )/(2*dx), 
  {  ( f(x,y+dy)-f(x,y-dy) )/(2*dy) 

where dx and dy could be the spacing in your grid. The contour line will run perpendicular to the gradient. So, to get the contour direction, c, we can multiply g=[v,w] by matrix, A=[0 -1, 1 0] giving

c = [-w,v]

I've wanted something like this myself, but haven't found a vector-based solution.

A raster-based solution isn't that bad, though, especially if your data is raster-based. If your data is vector-based too (in other words, you have a 3D model of your surface), you should be able to do some real math to find the intersection curves with horizontal planes at varying elevations.

For a raster-based approach, I look at each pair of neighboring pixels. If one is above a contour level, and one is below, obviously a contour line runs between them. The trick I used to anti-alias the contour line is to mix the contour line color into both pixels, proportional to their closeness to the idealized contour line.

Maybe some examples will help. Suppose that the current pixel is at an "elevation" of 12 ft, a neighbor is at an elevation of 8 ft, and contour lines are every 10 ft. Then, there is a contour line half way between; paint the current pixel with the contour line color at 50% opacity. Another pixel is at 11 feet and has a neighbor at 6 feet. Color the current pixel at 80% opacity.

alpha = (contour - neighbor) / (current - neighbor)

Unfortunately, I don't have the code handy, and there might have been a bit more to it (I vaguely recall looking at diagonal neighbors too, and adjusting by sqrt(2) / 2). I hope this enough to give you the gist.