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TomD.inPDX17-Peridot
17-Peridot
June 13, 2013
Solved

Help! Equation Solution Request.

  • June 13, 2013
  • 30 replies
  • 13330 views

How do I "define" a semi-circle by an equation when Y is linear?

I know the cartesian formula:

x = 4 * cos ( t * 180 )

y = 4 * sin ( t * 180 )

z = 0

But I need to compensate for the trajectory (Y) which has a linear value.

I need the formula applied only to X given Y is linear 0-1.

I want to sweep a VSS to make this shape. The trajpar is the given value (t as Y):

equation_question.PNG


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    Best answer by VladimirPalffy

    Dear Antonius

    I have tried to create SEMICIRCLE BY SWEEP feature - it is working (you can investigate my model)

    Note: I have used for sweep feature the Relation (for UPPER_HALF_OF_THE_SEMICIRCLE) sd4=(r^2 - (trajpar*r)^2) ^ (1/2)

    Regards,

    Vladimir

    30 replies

    VladimirPalffy14-Alexandrite
    14-Alexandrite
    June 14, 2013

    Dear Antonius,

    I think that you can combine Sweep feature with Graph feature - the values of graph will be inputs for sweep trajpar relations.

    Regards,

    Vladimir

    TomD.inPDX17-PeridotAuthor
    17-Peridot
    June 14, 2013

    ...or that The graph works nicely but there has to be a way to do the math.

      VladimirN24-Ruby IV
      24-Ruby IV
      June 15, 2013

      Hi Antonius,

      Maybe "Mathcad" community able to help you: http://communities.ptc.com/community/mathcad

      TomD.inPDX17-PeridotAuthor
      17-Peridot
      June 18, 2013

      I think the Mathcad community considers us an odd bunch and I rarely get replies... go figure

      Patriot_1776
      Patriot_177622-Sapphire II
      22-Sapphire II
      June 18, 2013

      I'd love to help, but I suck at math! I'd be graphing that baby!

        TomD.inPDX17-PeridotAuthor
        17-Peridot
        June 18, 2013

        I think that is why we are "visual" people... Geometry is so much simpler to "grasp" - literally

        Patriot_1776
        Patriot_177622-Sapphire II
        22-Sapphire II
        June 18, 2013

        Exactly! Same result, just visual!

          VladimirPalffy14-AlexandriteAnswer
          14-Alexandrite
          June 19, 2013

          Dear Antonius

          I have tried to create SEMICIRCLE BY SWEEP feature - it is working (you can investigate my model)

          Note: I have used for sweep feature the Relation (for UPPER_HALF_OF_THE_SEMICIRCLE) sd4=(r^2 - (trajpar*r)^2) ^ (1/2)

          Regards,

          Vladimir

            TomD.inPDX17-PeridotAuthor
            17-Peridot
            June 19, 2013

            That's the one! I suspected I would have to divide it into quarters.

            Here is the simplified "on the fly" version: sd4=sqrt((5^2 - (trajpar*5)^2))

            And as a bonus, it covers ellipses

            Thanks Vladimir!

            arc_traj_eq.PNG

              TomD.inPDX17-PeridotAuthor
              17-Peridot
              June 19, 2013

              Gopal, you didn't have to remove your reply. It is definitely worthy of this discussion:

              That's the formula of a circle x^2+y^2=radius^2

              with value of Y varying from with trajectory parameter.

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