Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving a 'skew' quadrilateral for vertex position.

  1. Jul 12, 2012 #1
    I've been scratching my head over this one. I'm trying to find a system of equations to solve for points R and S. The known parameters are: Point Q, tangent vector t and the axis vector a.

    The following vectors are perpendicular to each other:
    a,d
    a,b
    b,c
    c,t
    d,t

    The other known parameter is that vector c has a magnitude of 0.0625. This creates a quadrilateral that is skew, with vector c and a being at an angle to each other and pulling the sides out of plane.

    I'm trying to find the mathematical way of solving this across a number of points. SolidWorks can find one valid answer for the sample (and fully defines the topology) but I cannot get a system of equations to solve to match in Mathcad.

    Any method would be appreciated? I'm trying to get it so that I don't have to rely on the SolidWorks sketch engine to reduce computation times.
     

    Attached Files:

  2. jcsd
  3. Jul 12, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey athuss and welcome to the forums.

    So looking at your diagram, here is what we know:

    Q + c = R
    R + b = S
    T + a = S
    Q + d = T

    tunit x cunit = dunit

    The approach I am going to take is to use the tangent vector to get the plane that contains the points Q, R, and T as well as the vectors c and d.

    We know the plane equation since we have a point on the plane Q and our normal vector t. If we use n to be the normalized vector for t we get our plane equation to be n . (r - Q) = 0.

    Now from this point we need to establish how the direction of a affects the direction of either c or d directly.

    Now we know that a and c are parallel since a,b orthogonal and b,c orthogonal which means ahat = chat. But because we know this, it means that we can calculate dhat by calculating dhat = that x chat where the hat vectors are normalized.

    Now we have the vector c since we have its magnitude so we get c = chat*||c||.

    From this we have R by using R = Q + c. So that's R down.

    Please tell if I've screwed up anywhere with regards to assumptions.
     
    Last edited: Jul 12, 2012
  4. Jul 13, 2012 #3
    Hello chiro, thanks for the help.

    I worked through your reply and I think that there is one assumption that isn't true. That is:

    tunit x cunit = dunit

    tunit and cunit are perpendicular, as are tunit and dunit, but cunit and dunit are not.

    This means that a and c are not parallel and the vector c is still unknown.

    I've attached another view to show the skew of the quadrilateral. I think there is enough information to solve this but I haven't yet found it.
     

    Attached Files:

  5. Jul 13, 2012 #4

    chiro

    User Avatar
    Science Advisor

    Thanks for that athuss! I'll take a look later though it's getting a little late here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Solving a 'skew' quadrilateral for vertex position.
  1. Four vertex theorem (Replies: 6)

Loading...