MATLAB: Set boundaries to a plane equation

boundariesplane equation

I have a surface with four edges, namely (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), (x4, y4, z4). Those four points are always coplanar and they form a rectangle. I can find a plane equation with any of 3 coordinates above, however, that equation can be applied infinitely along x, y and z. How to limit the equation so that it is valid within the surface only? In other words, how to set boundaries to a plane equation in matlab coding

i tried searching but to no avail. Thank you.

Best Answer

In THREE dimensions, FOUR points do NOT define a plane. (Well, you might get lucky, and all 4 points lie in the same plane. Luck never seems to happen for me.)

As you state, you can pick any three of those points and define a plane from them. Think of it as if 4 points define a tetrahedron. In general, they do. There are 4 planes that define a tetrahedron, the 4 facets of the tetrahedron. As well, there are 4 ways to choose 3 points from the set of 4. That is not a coincidence.

So you might decide to bound a volume as the intersection of 4 half-spaces, thus the set of points that lie entirely on one side of a plane. That would create a tetrahedron if the resulting volume was closed and bounded.

But I also sincerely don't think that is what you are asking. You talk about 4 edges, defined by 4 points, so essentially a quadrilateral region.

I think you are looking for some way to define a surface by 4 points in three dimensions. Most of the time, that surface will not be a planar one, unless as I said before, you got lucky. Good luck on that ever happening. In fact though, there are infinitely many ways to define a general surface through 4 points in 3-dimensions. This is a general problem of interpolation, one that has plagued users of MATLAB (and mathematics) forever and will do so forever, because as I said, there is no unique, simple solution.

You will typically need to choose an interpolation method, then accept what and where it brings you. But maybe what you really want is that tetrahedron. In that case, it is just 4 planes, and you already know how to do that.

Related Solutions

MATLAB: Robust version of convexhulln

You can want convhulln to work as you wish, but wishing won't make it so. Code does as it is designed to do. In this case, it is clear that convhulln is not designed to work with degenerate cases. That is a choice made which is entirely valid.

Anyway, I'm not really sure that I agree with all of your "definitions".

For example, to call a line segment the convex hull of two points in a 1-d space seems wrong. In general a convex hull will be an object of manifold dimension one less than the set representing the volume described. So the convex hull of a triangle in 2-dimensions will be composed of line segments. We might call it a 1-manifold.

So in one dimension, the convex hull of two points will at most be a 0-manifold, composed of the endpoints of the included line segment.

As far as the convex hull of a triangle in 3-dimensions goes, it is apparently a decision made by TMW to fail. For example:

>> convhulln(rand(2,3))
Error using cgprechecks (line 40)
Not enough unique points specified.
Error in convhulln (line 41)
cgprechecks(x, nargin, cg_opt);

In fact, I completely agree with that decision. The presumption is that the convex hull of a list of points in n dimensions will be an (n-1)-manifold. If that must fail to be true, then the code will return an error.

So the convex hull of two points in R^3 will also fail, for the same reason. Such a convex hull would be computationally useless, even were you able to define it as a convex hull. Testing if a point lies "inside" that convex hull would virtually always fail due to floating point arithmetic. To know that a point lies EXACTLY on any given line is difficult. So to know that a point lies inside a such a degenerate convex hull will be a test of tolerance.

Similarly, this will fail:

convhulln([1 1;2 2;3 3])

It generates a rather messy looking message. The convex hull is degenerate. Yes, you can fix that using a joggle, convincing the convex hull to think the set is non-degenerate. Still, any test for inside that set will be problematic.

Are there things you can do? Yes, of course. You can put a wrapper around convhulln for your own usage, that tests for degeneracy in these forms and deals with any degeneracies as you desire. That is a virtue of MATLAB. You can always make it do as you like. Of course, testing for a point "inside" a degenerate convex hull can be nasty. You will want to work with projections into a lower dimensional subspace, working with (user supplied) tolerances on the result.

MATLAB: The length of the intersection curve between a sphere and a cylinder

I won't do your homework. I'll get you off to a good start, in fact, probably more than you deserve, since you appear to have made no serious effort.

The sphere is a sphere centered at the origin, of radius 4.

The cylinder is a circle in the (x,y) plane, so the same for all z. Completing the square, it can be written as:

(x^2 - 2*x +1) + y^2 = 1

(x - 1)^2 + y^2 = 1

So the cylinder has center (1,0,z), for all z.

That tells me the two objects do intersect, as well, they will intersect in TWO disjoint curves, one for positive z, the second for negative z.

In fact, if you view the intersection curve from the top, it will look perfectly circular. And this should be the clue you need.

Suppose you thought about this in terms of polar coordinates? To make it simple, translate the problem so the cylinder is centered at (0,0) in (x,y). Then the equation of the sphere is just:

(x+1)^2 + y^2 + z^2 == 16

The cylinder is now

x^2 + y^2 == 1

Nothing changes about the length of the curve. But perhaps now you can visualize that happens when we re-write this all in polar form. Actually, cylindrical coordiantes, thus (r,theta,z)

In cylindrical coordinates, the cylinder has equation

r == 1

Now, substitute into the sphere equation. Remember, since we now know that r==1,

x = r*cos(theta) = cos(theta)

y = r*sin(theta) = sin(theta)

now, lets look at what happened to the sphere.

(I've deleted the stuff that gives you a formula for z as a function of theta. In fact there is a remarkably simple form for z that will result.)

z = stufffffffffffff

Can you now find the arc length of that function, as theta varies from 0 to 2*pi? (I hope so.) A quick numerical computation of my own had the arc length as roughly 6.3864. My guess is only the first few digits of that are accurate, because of the approach I took for that first pass. But it should be close to 2*pi, but a bit larger, just as a visual assessment from a plot.

How do you find the arc length of a curve? I might be lazy, and use my arclength utility, as posted on the file exchange.

theta = linspace(0,2*pi,10000);
x = cos(theta);y = sin(theta);z = ????????;
arclength(x,y,z,'spline')
ans =
          6.39448891369661

You should be able to do the same however, using an integration. A google search for the phrase

"arc length of a parametric curve"

might give you some ideas. Don't forget that this curve is in the form (x(theta),y(theta),z(theta)). x(theta) is just cos(theta), etc. In fact, when I do that computation, integral tells me the arc length should be

ans =
          6.39448891369662

int did not find a symbolic solution however. Que sera, sera. But vpaintegral gives me:

ans =
6.3944889136966193144

I would note that the answer is indeed just a wee bit more than 2*pi, as I predicted it should be.

So, now all you need to do is do the algebra, then do some numerical methods work to replicate my results. IF you do decide to make a serious effort, then show what you did as a comment here. If you come at least close, showing that you have thought about the problem, then I might raise some of the veil on my answer.

Best Answer

Related Solutions

MATLAB: Robust version of convexhulln

MATLAB: The length of the intersection curve between a sphere and a cylinder

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