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Thread: Calculate angle between lines projected on a plane

  1. #1
    Apr 2009

    Calculate angle between lines projected on a plane

    Hello All

    This may be an easy answer, but I've been racking my brain about it for a couple hours and just cant figure it out. What I have is an inclined plane (30 degrees from horizontal). On this plane there are 2 lines, 20 degrees apart. What I need to figure out is how to calculate the angle between the two lines when they are projected to the horizontal plane, as shown in the picture (I am looking for the "In Plane Angle").


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  2. #2
    Apr 2009
    Hi Jt,

    I don't if this is the simpler solution, but it's a solution

    We need to find the coordinates of the 2 points lying on the inclined plane. I don't know if you have those available, but I'm assuming the worst case, which is not.

    Let's throw some names:

    $\displaystyle \theta$: is half the angle between the lines (10deg in your example)
    $\displaystyle \alpha$: is the angle between the planes (30deg in your example)
    $\displaystyle P0$: is the point where the 2 lines intercept. I'm assuming this will always be the origin, like in your picture
    $\displaystyle P1$ and [tex]P2[\math]: are the points that define the $\displaystyle \theta$ angle

    I'll take these points first lying on the horizontal plane, then rotate them by $\displaystyle \alpha$ degrees to find the new coordinates.

    $\displaystyle P0=(0, 0, 0)$
    $\displaystyle P1=(sin(\theta), cos(\theta), 0)$
    $\displaystyle P2=(-sin(\theta), cos(\theta), 0)$

    They will rotate about the X axis. The rotation matrix for this axis is:

    $\displaystyle Rx=\left(\begin{array}{ccc}1&0&0\\0&cos(\alpha)&-sin(\alpha)\\0&sin(\alpha)&cos(\alpha)\end{array}\ right)$

    Multiply the point by the matrix and you have the coordinates of the rotated point:

    $\displaystyle P1r=Rx*P1$
    $\displaystyle P2r=Rx*P2$

    This 2 points are now projected on the horizontal plane. In this particular case x and y coordinates are the same, and z = 0. So, the projected points are:

    $\displaystyle P1p=(sin(\theta), cos(\theta)*cos(\alpha), 0)$
    $\displaystyle P2p=(-sin(\theta), cos(\theta)*cos(\alpha), 0)$

    These 2 points make 2 vector with the origin:

    $\displaystyle u=P1-P0$
    $\displaystyle v=P2-P0$

    These 2 vector lie on the horizontal plane, now just find the angle between them using the definition of dot product:

    $\displaystyle cos(\theta projected)=\frac{u.v}{|u|*|v|}$

    Solving this leads to:

    $\displaystyle \theta projected=arccos \frac{cos(\theta)^2*cos(\alpha)^2-sin(\theta)^2}{cos(\theta)^2*cos(\alpha)^2+sin(\th eta)^2}$

    Finally, for your example we have $\displaystyle \theta projected=23.01678deg$

    I tried to give you all the steps so you could understand where the answer came from. I hope that was it you needed =)

    Best regards,
    Rodrigo Basniak
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  3. #3
    Apr 2009
    Thank you Rodrigo.

    I am going to be using this with different cases (the angle between planes will change). My end goal is to find the X and Y coordinates of the end points of the vectors. I drew the example up in AutoCAD, and can therefore check the angle measurement.... In this example CAD is giving me 22.7959deg, which is different than the math from your example gives. Any thoughts?
    Last edited by jontaylor; Apr 30th 2009 at 06:07 AM.
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