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
          1






          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080678%2ftilting-a-disc-in-3d-space-need-help%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080678%2ftilting-a-disc-in-3d-space-need-help%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
          Read more