Hills with a specific grade

Bug Reports
1

I am trying to figure this out myself, but I am having difficulty with it… I searched as well, but couldn’t find an answer.

Is there a way with WM to create a hill with a specific grade?

Say I want a hill with a 5% grade (rise 1meter per run of 20 meters), what would I need to do this?

Thanks.

2

There’s no automated way to do it, but You could figure this out with a little math.

For example, create a Radial Gradient using one of the shapes with linear sides (say, the pyramid shaped one).

If you want a 5% grade and the radius is 8km, then the hill should be 400m tall (8km * 1/20).

3

What about running a Scalar Generator into the Radial Gradient (with diamond or cone setting) and then you would be able to essentially go from a vertical slope to a flat horizontal slope in the scalar range of 0-1.
Maybe could convert the slope to want to radians and then get the 0-1 scalar value you need to plug into the Scalar Generator node.

Hummm, am I thinking about this correctly? For a 5% grade:

grade / 100 * 90 = degrees
5 / 100 * 90 = 4.5 degrees
and also
grade / 100 * Pi / 2 = radians
5 / 100 * 3.14159 / 2 = .0785 radians

Take the cosine of the radian angle to get the scalar value needed to put into the Scalar Generator node:
cos(radians) = scalar value
cos(.0785) = .99692

Been too long since I tried to do this stuff. :roll:

4

AFAIK slope percentages are not related to the angle but to the amount of height variation w.r.t the horizontal position. A 5% slope means 1 meter every 20 meters, not 5% of 90 degrees (tan(4.5) = .079, which is much larger than .05).

Bye!!!

5

Ah, I see. Thanks latego.

6

Sorry I haven’t been back to this topic. I was expecting to get an email when it was responded to I guess.

Yes, latego, that is correct. A 5% slope is approximately 2.86 degrees.

tan(A) = rise/run

So, in this case:
tan(a) = 1/20
tan(a) = .05 (.05 is the 5% grade).
a = 2.86 degrees

Thanks marmil, I’ll try that out with the formula changes noted above.

7

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Cheers,