3d volume of concave and convex hull of open surface

Hi,

I have a large set of 3d (x,y,z) data. I have "before" and "after" data sets for a surface. I used them to create 3d surface maps. The "before" looks like a hilly surface(ridges) with one giant hill in the center. The "after" looks like a hilly surface (ridges) with a large dip (missing a large chunk) in the center. I need to find the combined volume of the hill and large dip in the center, in order to determine the amount of material loss.

Does anyone have any suggestions how I can go about solving this? Essentially, I need to find the volume within a certain region of an open surface. I tried using convhulln but the numbers didn't look right, and I can't use it for the concave dip.

Thanks in advance

Answer 1

[Sean de Wolski]

Is the dip convex?

If it is then you could use the second output of convhulln and subtract the convex 'dip' from the convex hill. Else, look at some of Luigi Giacarra's files on the FEX.

[Joan]

I suppose the dip is convex in the sense that if you rotate it by 180, then it would be a convex.

By second output, do you mean K?

[Sean de Wolski]

Nope. I mean 'v'

doc convhulln

[K, v] = ...

v is the volume...

[Sean de Wolski]

Also, something that is convex, will be convex regardless of its rotation.

[Joan]

Ok, so say the surface looks like a series of ridges (looking at a cross section, the outline looks like it's a sin wave), mountains and valleys. Is the volume calculated going to be the positive or negative volume? How can I make sure it only does one or the other?

I ask this because I had used convhulln for both the hill and dip, but the numbers didn't look right. I have a surface map for the part before any wear testing, after 1 million cycles, and after 10 million cycles. The volume for the piece after 10 million cycles should be greater than for after 1 million cycles (because the surface is worn away more). However, this wasn't the case. Can you explain to me what convhulln is actually calculating?