tilting a disc in 3d space - need help












0












$begingroup$


Lets imagine you have a disc like a CD. Then you take that CD and rest it flat on a desk. Now you tilt the disc left to right and forward to back while touching the desk with 1 point on the edge of the disc.



-Assume you know the tilt angle forward to back as (y) and the tilt angle left to right as (x).
-You know the radius of the CD disc as R.

-You know the CD disc rotation angle between the radius of the point touching the desk and the point you want to figure the height for.



Edit//
Angle y is in the yz plane



Angle x is in the xz plane



The the angle between the radius of the 2 points is in the xy plane



//



Id like to know how to determine the height from the desk to any other point on that disc.






geometry trigonometry 3d







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Lets imagine you have a disc like a CD. Then you take that CD and rest it flat on a desk. Now you tilt the disc left to right and forward to back while touching the desk with 1 point on the edge of the disc.



    -Assume you know the tilt angle forward to back as (y) and the tilt angle left to right as (x).
    -You know the radius of the CD disc as R.

    -You know the CD disc rotation angle between the radius of the point touching the desk and the point you want to figure the height for.



    Edit//
    Angle y is in the yz plane



    Angle x is in the xz plane



    The the angle between the radius of the 2 points is in the xy plane



    //



    Id like to know how to determine the height from the desk to any other point on that disc.






    geometry trigonometry 3d







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Lets imagine you have a disc like a CD. Then you take that CD and rest it flat on a desk. Now you tilt the disc left to right and forward to back while touching the desk with 1 point on the edge of the disc.



      -Assume you know the tilt angle forward to back as (y) and the tilt angle left to right as (x).
      -You know the radius of the CD disc as R.

      -You know the CD disc rotation angle between the radius of the point touching the desk and the point you want to figure the height for.



      Edit//
      Angle y is in the yz plane



      Angle x is in the xz plane



      The the angle between the radius of the 2 points is in the xy plane



      //



      Id like to know how to determine the height from the desk to any other point on that disc.






      geometry trigonometry 3d







      share|cite|improve this question











      $endgroup$




      Lets imagine you have a disc like a CD. Then you take that CD and rest it flat on a desk. Now you tilt the disc left to right and forward to back while touching the desk with 1 point on the edge of the disc.



      -Assume you know the tilt angle forward to back as (y) and the tilt angle left to right as (x).
      -You know the radius of the CD disc as R.

      -You know the CD disc rotation angle between the radius of the point touching the desk and the point you want to figure the height for.



      Edit//
      Angle y is in the yz plane



      Angle x is in the xz plane



      The the angle between the radius of the 2 points is in the xy plane



      //



      Id like to know how to determine the height from the desk to any other point on that disc.






      geometry trigonometry 3d




      geometry trigonometry 3d






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 20 at 18:31







      Dan G.

















      asked Jan 20 at 14:55









      Dan G.Dan G.

      12




      12






















          1 Answer
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          0












          $begingroup$

          It is actually much simpler than that.



          You can set up a Cartesian 3D coordinate system where the disc radius is $R$ (diameter $2R$), the desk is at $z = 0$, the center of the disc is at $(0, 0, z_c)$, and the disc touches the desk at $(-x_t, 0, 0)$. This means the disc is always parallel to the $y$ axis (as if you were to hold the disc up in air using a thin wire at $x = 0$, $z = z_c$).



          Then, the vertical distance from any point on the disk is a function of the $x$ coordinate of that point,
          $$bbox[#ffffef]{z = z_c + x frac{z_c}{x_t} tag{1}label{NA1}}$$



          Note that even the disc radius is not a factor. This is because the same applies if you had a triangle, or indeed any planar shape that touches the desk at $(-x_t, 0, 0)$, and is angled in the $xz$ plane only (always parallel to the $y$ axis).



          This is easy to transform to any other coordinate system you prefer. However, the OP didn't specify any, so I suppose this should suffice.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
            $endgroup$
            – Dan G.
            Jan 20 at 18:17










          • $begingroup$
            @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
            $endgroup$
            – amd
            Jan 20 at 18:45












          • $begingroup$
            @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
            $endgroup$
            – Nominal Animal
            Jan 20 at 20:24











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          It is actually much simpler than that.



          You can set up a Cartesian 3D coordinate system where the disc radius is $R$ (diameter $2R$), the desk is at $z = 0$, the center of the disc is at $(0, 0, z_c)$, and the disc touches the desk at $(-x_t, 0, 0)$. This means the disc is always parallel to the $y$ axis (as if you were to hold the disc up in air using a thin wire at $x = 0$, $z = z_c$).



          Then, the vertical distance from any point on the disk is a function of the $x$ coordinate of that point,
          $$bbox[#ffffef]{z = z_c + x frac{z_c}{x_t} tag{1}label{NA1}}$$



          Note that even the disc radius is not a factor. This is because the same applies if you had a triangle, or indeed any planar shape that touches the desk at $(-x_t, 0, 0)$, and is angled in the $xz$ plane only (always parallel to the $y$ axis).



          This is easy to transform to any other coordinate system you prefer. However, the OP didn't specify any, so I suppose this should suffice.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
            $endgroup$
            – Dan G.
            Jan 20 at 18:17










          • $begingroup$
            @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
            $endgroup$
            – amd
            Jan 20 at 18:45












          • $begingroup$
            @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
            $endgroup$
            – Nominal Animal
            Jan 20 at 20:24
















          0












          $begingroup$

          It is actually much simpler than that.



          You can set up a Cartesian 3D coordinate system where the disc radius is $R$ (diameter $2R$), the desk is at $z = 0$, the center of the disc is at $(0, 0, z_c)$, and the disc touches the desk at $(-x_t, 0, 0)$. This means the disc is always parallel to the $y$ axis (as if you were to hold the disc up in air using a thin wire at $x = 0$, $z = z_c$).



          Then, the vertical distance from any point on the disk is a function of the $x$ coordinate of that point,
          $$bbox[#ffffef]{z = z_c + x frac{z_c}{x_t} tag{1}label{NA1}}$$



          Note that even the disc radius is not a factor. This is because the same applies if you had a triangle, or indeed any planar shape that touches the desk at $(-x_t, 0, 0)$, and is angled in the $xz$ plane only (always parallel to the $y$ axis).



          This is easy to transform to any other coordinate system you prefer. However, the OP didn't specify any, so I suppose this should suffice.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
            $endgroup$
            – Dan G.
            Jan 20 at 18:17










          • $begingroup$
            @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
            $endgroup$
            – amd
            Jan 20 at 18:45












          • $begingroup$
            @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
            $endgroup$
            – Nominal Animal
            Jan 20 at 20:24














          0












          0








          0





          $begingroup$

          It is actually much simpler than that.



          You can set up a Cartesian 3D coordinate system where the disc radius is $R$ (diameter $2R$), the desk is at $z = 0$, the center of the disc is at $(0, 0, z_c)$, and the disc touches the desk at $(-x_t, 0, 0)$. This means the disc is always parallel to the $y$ axis (as if you were to hold the disc up in air using a thin wire at $x = 0$, $z = z_c$).



          Then, the vertical distance from any point on the disk is a function of the $x$ coordinate of that point,
          $$bbox[#ffffef]{z = z_c + x frac{z_c}{x_t} tag{1}label{NA1}}$$



          Note that even the disc radius is not a factor. This is because the same applies if you had a triangle, or indeed any planar shape that touches the desk at $(-x_t, 0, 0)$, and is angled in the $xz$ plane only (always parallel to the $y$ axis).



          This is easy to transform to any other coordinate system you prefer. However, the OP didn't specify any, so I suppose this should suffice.






          share|cite|improve this answer









          $endgroup$



          It is actually much simpler than that.



          You can set up a Cartesian 3D coordinate system where the disc radius is $R$ (diameter $2R$), the desk is at $z = 0$, the center of the disc is at $(0, 0, z_c)$, and the disc touches the desk at $(-x_t, 0, 0)$. This means the disc is always parallel to the $y$ axis (as if you were to hold the disc up in air using a thin wire at $x = 0$, $z = z_c$).



          Then, the vertical distance from any point on the disk is a function of the $x$ coordinate of that point,
          $$bbox[#ffffef]{z = z_c + x frac{z_c}{x_t} tag{1}label{NA1}}$$



          Note that even the disc radius is not a factor. This is because the same applies if you had a triangle, or indeed any planar shape that touches the desk at $(-x_t, 0, 0)$, and is angled in the $xz$ plane only (always parallel to the $y$ axis).



          This is easy to transform to any other coordinate system you prefer. However, the OP didn't specify any, so I suppose this should suffice.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 16:52









          Nominal AnimalNominal Animal

          7,0432517




          7,0432517












          • $begingroup$
            I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
            $endgroup$
            – Dan G.
            Jan 20 at 18:17










          • $begingroup$
            @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
            $endgroup$
            – amd
            Jan 20 at 18:45












          • $begingroup$
            @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
            $endgroup$
            – Nominal Animal
            Jan 20 at 20:24


















          • $begingroup$
            I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
            $endgroup$
            – Dan G.
            Jan 20 at 18:17










          • $begingroup$
            @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
            $endgroup$
            – amd
            Jan 20 at 18:45












          • $begingroup$
            @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
            $endgroup$
            – Nominal Animal
            Jan 20 at 20:24
















          $begingroup$
          I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
          $endgroup$
          – Dan G.
          Jan 20 at 18:17




          $begingroup$
          I can’t seem to grasp how it is parallel to the y axis if your tilting the disc along 2 axis. I need a bit of an explanation please.
          $endgroup$
          – Dan G.
          Jan 20 at 18:17












          $begingroup$
          @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
          $endgroup$
          – amd
          Jan 20 at 18:45






          $begingroup$
          @DanG. The coordinate system has been chosen so that this is the case. It’s not the same coordinate system that you’re using in your question, though.
          $endgroup$
          – amd
          Jan 20 at 18:45














          $begingroup$
          @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
          $endgroup$
          – Nominal Animal
          Jan 20 at 20:24




          $begingroup$
          @DanG.: There is no need to use two tilt angles. Instead, rotate the disc so that the point where the disc touches the desk and the center of the disc are both on the $x$ axis.
          $endgroup$
          – Nominal Animal
          Jan 20 at 20:24


















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