Parametric equations for a helical pipe

  • Thread starter jaksweeney
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  • #1

Main Question or Discussion Point

I am trying to represent a helical pipe in x,y,z co-ordinants, would the x and y co-ordinants simply be multiplied by the equation of a circle if the growth of the helix is in the z direction?

Any help would be appreciated.
Thanks
 

Answers and Replies

  • #2
cristo
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The x and y equations would just be the parametric equations of a circle. If the helix grows in the z direction, then its equation would be z=ct, for some constant c.
 
  • #3
The basic line equation of a helix can be represented as x=acost, y=asint and z=ct, but I want to represent a coiled pipe so should there be another set of circular co-ordinates in the equation?
 
  • #4
HallsofIvy
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Then what are you looking for, exactly? x= acos(t), y= a sin(t), z= ct, are parametric equations for the center curve of the pipe but do you want parametric equations for the surface of the pipe? That would depend on two parameters, not 1. parametric equations for any point in the interior of the pipe would require 3 parameters.
 
  • #5
I am trying to find out how to determine a point of the surface of the pipe. How would that be represented?
 
  • #6
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All points on a coiled pipe may be represented by a two-parameter vector whose components are the Cartesian coordinates of the points. The derivation adopts the following strategy:

1) Compute a vector R(t) to a point on a helix (in the e3
direction) in terms of a parameter t.
2) Find the tangent to the helix at t and thereby the plane perpendicular
to the pipe at t, P(t).
3) Find the unit vector U(t) in plane P that lies along the projection of
e3 in this plane.
4) Find the unit vector in plane P that results from U by rotating through
an angle phi, S(t,phi)
5) The desired vector to any point on the surface of the pipe is then
R(t) + qS(t,phi), where q is the radius of the pipe.


1) Let e1, e2 and e3 be an orthonormal set
of vectors spanning 3-space. The centre of the pipe describes a helix
about the e3 axis. The helical curve is given by the vector R(t)
as a function of the parameter t:

R(t) = a cos(t) e1 + a sin(t) e2 + ct e3

2) The vector tangent to the helix at t is

dR/dt = -a sin(t) e1 + a cos(t) e2 + c e3.

The unit vector tangent to R(t) is

V(t) = (dR/dt)(a2+c2)-1/2

The plane perpendicular to the pipe is given by the unit bivector

P(t) = IV(t), where I = e1e2e3. i.e.
P(t) =(-a sin(t) e2e3 + a cos(t) e3e1
+ c e1e2)(a2+c2)-1/2.

3) We now wish to define a unit reference vector lying in the plane P(t). It
is convenient to take this as U(t) = W(t)/w, where W(t) is the projection
of vector e3 onto plane P(t) and w = ||W||:

W(t) = e3.P P-1 = -e3.P P since
P2 = -1. Substituting the expression for P(t) from above, we get
W(t) = (a2e3 + ac(sin(t) e1
- cos(t) e2))/(a2 + c2).
w = a (a2 + c2)-1/2.

The desired reference vector, U(t), is then W(t)/w:
U(t) = (e3 +(c/a)(sin(t) e1 - cos(t) e2))(1 + (c/a)2)-1/2.

4) This reference vector can now be rotated to produce all points on a unit
circle in the plane P(t), which is perpendicular to the pipe at t:

S(t,phi) = U(t) exp(P(t)phi) = U(t) (cos(phi) + P(t) sin(phi)); i.e. S(t,phi)

is a vector in the plane P(t) written in terms of its componets in the
reference direction U(t) and in a direction perpendicular to U,
U(t)P(t) = e3.P(t)/w. The latter vector should lie in the
e1e2 plane. Substituting the expression for P(t)
verifies this:

U(t)P(t) = cos(t) e1 + sin(t) e2.

5)Let the radius of the pipe be q. The points on the surface
of the coiled pipe are then given by
R(t) + q S(t,phi).

Please check my algebra before using this equation!
 

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