Construct a Cylinder from Points

[Hans J Wolters]

Hi,
I have a seemingly innocent problem. Given are points that I know lie
on the cylinder. I am trying to finding that cylinder from these
points, that means the axis and the radius.
If I had enough points I could certainly solve for the general quadric
equation and then find the axis by analyzing the symmetric matrix A.
But I am after somthing more constructive. It seems that one should be
able to do it from 5 points (that is my intution) but I do not know
how. Any ideas?

Thanks,

Hans

[Peter Glasson]

I would also like to know how to construct a cylinder from 5 points.
Have you found the answer at a different source? I can see that there
are degenerate problems, i.e. no unique cylinder for some sets of 5
points. Method should detect these cases.

[John Mason]

My colleagues and I have enjoyed toying with this problem. Given 5
points in general position in 3-space, is there a (right circular)
cylinder on which they all lie?

We note that if you take 3 points and consider the plane they
specify, for any direction there is an ellipse through the three
points in the plane with major axis in that direction. This enables
a cylinder of circular cross-section to pass through the three points
and have that ellipse as section.

Consider a fourth point not on that plane (because general position).

There is a family of cylinders based on ellipses at different
directions, and passing through the fourth point as well.

Now consider two points symmetrically placed with respect to that
plane (i.e one above and one below) and not on the cylinder specified
by the circumcircle of the first three points. Any cylinder through
the fourth point must be tilted, making it difficult to pass through
the fifth point.

JohnM

[Hans Wolters]

Hi,
Well I have not had time to think about it much more but I did not get
a satisfactory answer. I am pretty sure that in the general case
points are sufficient. A least squares set up of this problem might be
the most practical way to approach this by finding a line so that all
points have a distance of r to that line.
The very general setup by using ageneral representation for quadrics
swill probably cause trouble since one can not even be sure that the
quadric produced is a cylinder.
Did you find a way?

[Erik B]

Hi,
I'm not sure I've got the solution, but here's my reasoning:
You need 5 points on a plane to define an ellipse:

ellipse = Ax^2 + Bxy + Cy^2 + Dx + Ey + 1 = 0

so take five points in a plane, describe them with coords in the plane
and generate 5 linear equations using the above formula. This then is
obviously solvable.

It seems to me (haven't proved it or even tried) that the ratio
between long and short axis bears a 1-to-1 relationship with the
vector defining the cylinder's axis. Also, it seems to me that the
absolute value of the short (or long) axis bears a 1-1 relationship to
the radius of the cylinder (given the ratio long/short). That is to
say, I'm assuming that an ellipse can uniquely define a cylinder (is
this true?).

However, if you have 5 non-planar points you might have a problem:
1) 3 points define a plane, with an infinite number of ellipse passing
through them. This then means an infinite number of ellipses pass
through them.
2) As stated, 5 points define an ellipse. So we need 2 more points.
Now consider two points that lie on the cylinder. When projected along
a line parallel to the axis defined by (Rx,Ry,1) these two points
define two additional points on the mentioned ellipse, within two
unknowns Rx, Ry. We now have 5 equations with 7 unknowns. Two
additional points on the cylinder generate 2 additional equations
without adding more unknowns, resulting in 7 (non-linear) equations
with 7 unknowns.

Hence, it seems that you need 3+2+2 = 7 points to uniquely fix a
cylinder in space.....

What d'u think: do the arguments hold any water ?

cheers
Erik

[Walter Whiteley]

In general, it takes 9 points in 3-space to uniquely define a quadric surface.
[This is the analog of 5 points for a conic in the plane.]

So IF the 9 points lie on a cylinder, they will uniquely determine the answer.

The theory, much like below, is reflected in the equation:

A x^2 + B y^2 + C z^2 + Dxy + Eyz + Fxz + G x + Hy + Iz + J = 0

If someone gives you a point x_1, y_1, z_1 on he surface, this gives
one linear equation in the 10 variables above.
(Think of x^2, Y^2 xy etc as known numbers and see that A,B, ... J are
unknown.) In linear algebra, with 9 homogeneous equations in 10 variables, you
should expect a 1 paramter solution space. Indeed, any quadric can be
multiplied by a constant to give another equation of the same quadric.
What you did, below, was to set the last coordinate = 1 in order to take away
this one choice.

If you know the object is to be a cylinder, and you have decided which direction
the axis is to lie, this gives one point 'at infinity" which can be one of your
9 points. There are many options for which 9 points you need.
Having 3 of them collinear will also set up the axis direction, and guarantee
that the object being defined is not a sphere, ellipsoid, hyperboloid of two
sheets. It can still be something like a hyperbolic paraboloid (a saddle
surface) which contains lines. If you set up two parallel lines (3 points plus
three more) then you are close to a cylinder, but need a couple more points to
determine which conic section is the cross section. You now need three more
points to determine that cross section. In the extreme, of course, these points
might end up giving the answer of two parallel planes, and you will need to
include that as an extreme case of a cylinder!

Walter Whiteley