Equation of the cone about an arbitrary axis

[madhava.G...@gmail.com]

Hi ,
I am in an urgent need to determine the equation of the right
circular cone. The following are the details.

1. This cone is not oriented about any of the three co-ordinate axes
but is along an axis defined by an arbitrary straight line (known)
2. It opening angle of the cone is known ( the angle at the vertex of
the cone)
3. The vertex of the cone in XYZ co-ordinates is known.

Can any one help me with this. This equation needs to be in 3D. I need
this very urgently.

Thanks
Madhava

[hoff...@fho-emden.de]

madhava.Gadiche...@gmail.com schrieb:

M: vertex vector
No: unit axis vector
a: angle between axis vector and surface (half opening angle)
X: vector to point on the cone surface

[(X-M)^T*No]^2 = [(X-M)^T*(X-M)]*[cos(a)]^2

For better readability:

[(X-M)' *No]^2 = [(X-M)' *(X-M)] [cos(a)]^2

Best regards --Gernot Hoffmann

[madhava.G...@gmail.com]

hi Gernot,

Thanks for the early rep ly.
Can you please explain how you got there.
I also would like to know the meaning of T. you have not defined
it here. Also M being a point in XYZ co-ordinates, I dont understand
the (X-M) part. Please clarify

Thanks
Madhava

[hoff...@fho-emden.de]

> > M: vertex vector
> > No: unit axis vector
> > a: angle between axis vector and surface (half opening angle)
> > X: vector to point on the cone surface
> >
> > [(X-M)^T*No]^2 = [(X-M)^T*(X-M)]*[cos(a)]^2
> >
> > For better readability:
> >
> > [(X-M)' *No]^2 = [(X-M)' *(X-M)] [cos(a)]^2
> >
> > Best regards --Gernot Hoffmann
>
> hi Gernot,
>
> Thanks for the early rep ly.
> Can you please explain how you got there.
> I also would like to know the meaning of T. you have not defined
> it here. Also M being a point in XYZ co-ordinates, I dont understand
> the (X-M) part. Please clarify
>
> Thanks
> Madhava

Madhava,

A^T means 'transposed' - a row instead of a column.
A^T*B means ax*bx+ay*by+az*bz.
That's the same as the dot product (A.B) .

X-M is a vector which contains the known vector M
(which points to the vertex) and any arbitrary vector X
from the origin to the surface.
Sorry, without a drawing it's impossible to derive the
equation as mentioned in my recent post.

Best regards --Gernot Hoffmann

[Kaba]

> Can you please explain how you got there.

Recall the connection between a dot product of two vectors and their
angle:

dot(A, B) = cos(alpha) * ||A|| * ||B||

where alpha is the angle between A and B in the range [0, pi[.

Assume for a while ||A|| != 0 and ||B|| != 0. From this it is easy to
derive that the angle between two vectors A and B is given by:

alpha = arccos(dot(A, B) / (||A|| * ||B||))

Let M be the tip of the cone, N the axis vector of any length and alpha
the angle of the cone. Any point X on the cone is characterized by:
"vector (X - M) forms an angle alpha with N". In maths:

alpha = arccos(dot(X - M, N) / (||X - M|| * ||N||))

We assumed that ||X - M|| != 0. But clearly X = M is also part of the
cone: it is the tip of the cone. To include this point we need to
rearrange the equation so as to remove the singularity due to the
division:

cos(alpha) = dot(X - M, N) / (||X - M|| * ||N||)
=>
cos(alpha) * ||X - M|| * ||N|| = dot(X - M, N)

All the points X that satisfy this equation are on the cone. Note this
equation describes a "single-cone", not a double-cone. This is what
people usually mean by a cone.

For the double-cone that extends in both directions, square away both
sides to get rid of the sign in the dot product:

(Note ||A||^2 = dot(A, A))

cos^2(alpha) * dot(X - M, X - M) * dot(N, N) = dot^2(X - M, N)

If you know N is of unit length, then dot(N, N) = 1 and the equation
reduces to:

cos^2(alpha) * dot(X - M, X - M) = dot^2(X - M, N)

And this is the equation Gernot gave you.

--
Kalle Rutanen
http://kaba.hilvi.org