Tracking the movement of the watch hand












5












$begingroup$


I love watches, and I had an idea for a weird kind of watch movement (all of the stuff that moves the hands). It is made up of a a central wheel, with one of the hands connected to it (in this case, it will be the hour hand). This hand goes through a pivot, and then displays the time. I attached a video of a 3d mock up here, because it is kinda hard to explain. My question is, is there any functions that would be able to graph the movement of the end of the hand? I don't want to make the real prototype just yet.






geometry functions trigonometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    That watch looks awesome but do you think it will be physically implementable?
    $endgroup$
    – Mohammad Zuhair Khan
    2 hours ago










  • $begingroup$
    I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird.
    $endgroup$
    – Aubrey Champagne
    2 hours ago










  • $begingroup$
    Which end of the hand are you talking about? The one that actually shows the hour, or the other end?
    $endgroup$
    – Arthur
    1 hour ago












  • $begingroup$
    The side of the hand that isn’t attached to the whee
    $endgroup$
    – Aubrey Champagne
    1 hour ago










  • $begingroup$
    I am also worried about the friction as it may slow down the transition. Use plenty of graphite.
    $endgroup$
    – Mohammad Zuhair Khan
    1 hour ago
















5












$begingroup$


I love watches, and I had an idea for a weird kind of watch movement (all of the stuff that moves the hands). It is made up of a a central wheel, with one of the hands connected to it (in this case, it will be the hour hand). This hand goes through a pivot, and then displays the time. I attached a video of a 3d mock up here, because it is kinda hard to explain. My question is, is there any functions that would be able to graph the movement of the end of the hand? I don't want to make the real prototype just yet.






geometry functions trigonometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    That watch looks awesome but do you think it will be physically implementable?
    $endgroup$
    – Mohammad Zuhair Khan
    2 hours ago










  • $begingroup$
    I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird.
    $endgroup$
    – Aubrey Champagne
    2 hours ago










  • $begingroup$
    Which end of the hand are you talking about? The one that actually shows the hour, or the other end?
    $endgroup$
    – Arthur
    1 hour ago












  • $begingroup$
    The side of the hand that isn’t attached to the whee
    $endgroup$
    – Aubrey Champagne
    1 hour ago










  • $begingroup$
    I am also worried about the friction as it may slow down the transition. Use plenty of graphite.
    $endgroup$
    – Mohammad Zuhair Khan
    1 hour ago














5












5








5


2



$begingroup$


I love watches, and I had an idea for a weird kind of watch movement (all of the stuff that moves the hands). It is made up of a a central wheel, with one of the hands connected to it (in this case, it will be the hour hand). This hand goes through a pivot, and then displays the time. I attached a video of a 3d mock up here, because it is kinda hard to explain. My question is, is there any functions that would be able to graph the movement of the end of the hand? I don't want to make the real prototype just yet.






geometry functions trigonometry







share|cite|improve this question









$endgroup$




I love watches, and I had an idea for a weird kind of watch movement (all of the stuff that moves the hands). It is made up of a a central wheel, with one of the hands connected to it (in this case, it will be the hour hand). This hand goes through a pivot, and then displays the time. I attached a video of a 3d mock up here, because it is kinda hard to explain. My question is, is there any functions that would be able to graph the movement of the end of the hand? I don't want to make the real prototype just yet.






geometry functions trigonometry




geometry functions trigonometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









Aubrey ChampagneAubrey Champagne

365




365












  • $begingroup$
    That watch looks awesome but do you think it will be physically implementable?
    $endgroup$
    – Mohammad Zuhair Khan
    2 hours ago










  • $begingroup$
    I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird.
    $endgroup$
    – Aubrey Champagne
    2 hours ago










  • $begingroup$
    Which end of the hand are you talking about? The one that actually shows the hour, or the other end?
    $endgroup$
    – Arthur
    1 hour ago












  • $begingroup$
    The side of the hand that isn’t attached to the whee
    $endgroup$
    – Aubrey Champagne
    1 hour ago










  • $begingroup$
    I am also worried about the friction as it may slow down the transition. Use plenty of graphite.
    $endgroup$
    – Mohammad Zuhair Khan
    1 hour ago


















  • $begingroup$
    That watch looks awesome but do you think it will be physically implementable?
    $endgroup$
    – Mohammad Zuhair Khan
    2 hours ago










  • $begingroup$
    I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird.
    $endgroup$
    – Aubrey Champagne
    2 hours ago










  • $begingroup$
    Which end of the hand are you talking about? The one that actually shows the hour, or the other end?
    $endgroup$
    – Arthur
    1 hour ago












  • $begingroup$
    The side of the hand that isn’t attached to the whee
    $endgroup$
    – Aubrey Champagne
    1 hour ago










  • $begingroup$
    I am also worried about the friction as it may slow down the transition. Use plenty of graphite.
    $endgroup$
    – Mohammad Zuhair Khan
    1 hour ago
















$begingroup$
That watch looks awesome but do you think it will be physically implementable?
$endgroup$
– Mohammad Zuhair Khan
2 hours ago




$begingroup$
That watch looks awesome but do you think it will be physically implementable?
$endgroup$
– Mohammad Zuhair Khan
2 hours ago












$begingroup$
I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird.
$endgroup$
– Aubrey Champagne
2 hours ago




$begingroup$
I think it would be. The only thing I am kinda afraid of would be friction at the pivot, and telling the hour from 5-7 might be tough, but who knows. I love weird watches, and its certainly weird.
$endgroup$
– Aubrey Champagne
2 hours ago












$begingroup$
Which end of the hand are you talking about? The one that actually shows the hour, or the other end?
$endgroup$
– Arthur
1 hour ago






$begingroup$
Which end of the hand are you talking about? The one that actually shows the hour, or the other end?
$endgroup$
– Arthur
1 hour ago














$begingroup$
The side of the hand that isn’t attached to the whee
$endgroup$
– Aubrey Champagne
1 hour ago




$begingroup$
The side of the hand that isn’t attached to the whee
$endgroup$
– Aubrey Champagne
1 hour ago












$begingroup$
I am also worried about the friction as it may slow down the transition. Use plenty of graphite.
$endgroup$
– Mohammad Zuhair Khan
1 hour ago




$begingroup$
I am also worried about the friction as it may slow down the transition. Use plenty of graphite.
$endgroup$
– Mohammad Zuhair Khan
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Sure, pick your coordinate system. I will take the origin to be the place the hand slides through, $y$ vertical positive up, $x$ horizontal positive right. Let the hand have length $L$ and the circle radius $R$. It appears $L$ is a little greater than $2R$, so it sticks out of the pivot even when the left end is at the farthest left point.



The position of the left end is $(Rcos ft-R,Rsin ft)$ where $f=frac {2 pi}{ 12 hours}$

The distance from the left end to the pivot is $sqrt{(Rcos ft-R)^2+(Rsin ft)^2}=sqrt{2R^2-2Rcos ft}$

The slope of the hand is $frac {R sin ft}{Rcos ft-R}=m$

The length of the hand to the right of the pivot is $L-sqrt{2R^2-2Rcos ft}$

The position of the right end of the hand is $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    enter image description here



    Denote with $l$ the length of the hand and with $R$ the radius of the circle.



    Parametric coordinates of point $M$ (as a function of $alphain[0,2pi)$) are:



    $$x_M=(l-2Rcosfracalpha2)cosfracalpha2=lcosfracalpha2-R(1+cosalpha)$$



    $$y_M=-(l-2Rcosfracalpha2)sinfracalpha2=-lsinfracalpha2+Rsinalpha$$



    As an exercise you can eliminate angle $alpha$ and obtain an implicit relation between coordinates of point $M$, but there is not much that you can do with it. It is better to work with parametric equations. Select $l,R$ and calculate coordinates for a range of $alpha$ angles.






    share|cite|improve this answer









    $endgroup$













      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%2f3082779%2ftracking-the-movement-of-the-watch-hand%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Sure, pick your coordinate system. I will take the origin to be the place the hand slides through, $y$ vertical positive up, $x$ horizontal positive right. Let the hand have length $L$ and the circle radius $R$. It appears $L$ is a little greater than $2R$, so it sticks out of the pivot even when the left end is at the farthest left point.



      The position of the left end is $(Rcos ft-R,Rsin ft)$ where $f=frac {2 pi}{ 12 hours}$

      The distance from the left end to the pivot is $sqrt{(Rcos ft-R)^2+(Rsin ft)^2}=sqrt{2R^2-2Rcos ft}$

      The slope of the hand is $frac {R sin ft}{Rcos ft-R}=m$

      The length of the hand to the right of the pivot is $L-sqrt{2R^2-2Rcos ft}$

      The position of the right end of the hand is $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Sure, pick your coordinate system. I will take the origin to be the place the hand slides through, $y$ vertical positive up, $x$ horizontal positive right. Let the hand have length $L$ and the circle radius $R$. It appears $L$ is a little greater than $2R$, so it sticks out of the pivot even when the left end is at the farthest left point.



        The position of the left end is $(Rcos ft-R,Rsin ft)$ where $f=frac {2 pi}{ 12 hours}$

        The distance from the left end to the pivot is $sqrt{(Rcos ft-R)^2+(Rsin ft)^2}=sqrt{2R^2-2Rcos ft}$

        The slope of the hand is $frac {R sin ft}{Rcos ft-R}=m$

        The length of the hand to the right of the pivot is $L-sqrt{2R^2-2Rcos ft}$

        The position of the right end of the hand is $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Sure, pick your coordinate system. I will take the origin to be the place the hand slides through, $y$ vertical positive up, $x$ horizontal positive right. Let the hand have length $L$ and the circle radius $R$. It appears $L$ is a little greater than $2R$, so it sticks out of the pivot even when the left end is at the farthest left point.



          The position of the left end is $(Rcos ft-R,Rsin ft)$ where $f=frac {2 pi}{ 12 hours}$

          The distance from the left end to the pivot is $sqrt{(Rcos ft-R)^2+(Rsin ft)^2}=sqrt{2R^2-2Rcos ft}$

          The slope of the hand is $frac {R sin ft}{Rcos ft-R}=m$

          The length of the hand to the right of the pivot is $L-sqrt{2R^2-2Rcos ft}$

          The position of the right end of the hand is $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$






          share|cite|improve this answer









          $endgroup$



          Sure, pick your coordinate system. I will take the origin to be the place the hand slides through, $y$ vertical positive up, $x$ horizontal positive right. Let the hand have length $L$ and the circle radius $R$. It appears $L$ is a little greater than $2R$, so it sticks out of the pivot even when the left end is at the farthest left point.



          The position of the left end is $(Rcos ft-R,Rsin ft)$ where $f=frac {2 pi}{ 12 hours}$

          The distance from the left end to the pivot is $sqrt{(Rcos ft-R)^2+(Rsin ft)^2}=sqrt{2R^2-2Rcos ft}$

          The slope of the hand is $frac {R sin ft}{Rcos ft-R}=m$

          The length of the hand to the right of the pivot is $L-sqrt{2R^2-2Rcos ft}$

          The position of the right end of the hand is $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Ross MillikanRoss Millikan

          293k23197371




          293k23197371























              2












              $begingroup$

              enter image description here



              Denote with $l$ the length of the hand and with $R$ the radius of the circle.



              Parametric coordinates of point $M$ (as a function of $alphain[0,2pi)$) are:



              $$x_M=(l-2Rcosfracalpha2)cosfracalpha2=lcosfracalpha2-R(1+cosalpha)$$



              $$y_M=-(l-2Rcosfracalpha2)sinfracalpha2=-lsinfracalpha2+Rsinalpha$$



              As an exercise you can eliminate angle $alpha$ and obtain an implicit relation between coordinates of point $M$, but there is not much that you can do with it. It is better to work with parametric equations. Select $l,R$ and calculate coordinates for a range of $alpha$ angles.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                enter image description here



                Denote with $l$ the length of the hand and with $R$ the radius of the circle.



                Parametric coordinates of point $M$ (as a function of $alphain[0,2pi)$) are:



                $$x_M=(l-2Rcosfracalpha2)cosfracalpha2=lcosfracalpha2-R(1+cosalpha)$$



                $$y_M=-(l-2Rcosfracalpha2)sinfracalpha2=-lsinfracalpha2+Rsinalpha$$



                As an exercise you can eliminate angle $alpha$ and obtain an implicit relation between coordinates of point $M$, but there is not much that you can do with it. It is better to work with parametric equations. Select $l,R$ and calculate coordinates for a range of $alpha$ angles.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  enter image description here



                  Denote with $l$ the length of the hand and with $R$ the radius of the circle.



                  Parametric coordinates of point $M$ (as a function of $alphain[0,2pi)$) are:



                  $$x_M=(l-2Rcosfracalpha2)cosfracalpha2=lcosfracalpha2-R(1+cosalpha)$$



                  $$y_M=-(l-2Rcosfracalpha2)sinfracalpha2=-lsinfracalpha2+Rsinalpha$$



                  As an exercise you can eliminate angle $alpha$ and obtain an implicit relation between coordinates of point $M$, but there is not much that you can do with it. It is better to work with parametric equations. Select $l,R$ and calculate coordinates for a range of $alpha$ angles.






                  share|cite|improve this answer









                  $endgroup$



                  enter image description here



                  Denote with $l$ the length of the hand and with $R$ the radius of the circle.



                  Parametric coordinates of point $M$ (as a function of $alphain[0,2pi)$) are:



                  $$x_M=(l-2Rcosfracalpha2)cosfracalpha2=lcosfracalpha2-R(1+cosalpha)$$



                  $$y_M=-(l-2Rcosfracalpha2)sinfracalpha2=-lsinfracalpha2+Rsinalpha$$



                  As an exercise you can eliminate angle $alpha$ and obtain an implicit relation between coordinates of point $M$, but there is not much that you can do with it. It is better to work with parametric equations. Select $l,R$ and calculate coordinates for a range of $alpha$ angles.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  OldboyOldboy

                  7,4581834




                  7,4581834






























                      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%2f3082779%2ftracking-the-movement-of-the-watch-hand%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

                      Liste der Baudenkmale in Friedland (Mecklenburg)

                      Image
                      In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
                      Read more

                      Single-Malt-Whisky

                      Image
                      Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
                      Read more

                      Czorneboh

                      Image
                      Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
                      Read more