Tracking the movement of the watch hand












6












$begingroup$


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
    1 hour 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
    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
    56 mins ago










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
















6












$begingroup$


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
    1 hour 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
    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
    56 mins ago










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














6












6








6


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 1 hour ago









Aubrey ChampagneAubrey Champagne

385




385












  • $begingroup$
    That watch looks awesome but do you think it will be physically implementable?
    $endgroup$
    – Mohammad Zuhair Khan
    1 hour 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
    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
    56 mins ago










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


















  • $begingroup$
    That watch looks awesome but do you think it will be physically implementable?
    $endgroup$
    – Mohammad Zuhair Khan
    1 hour 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
    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
    56 mins ago










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
















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




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






$begingroup$
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
56 mins ago




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












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




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









$endgroup$





















    1












    $begingroup$

    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









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      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 57 mins ago









          Ross MillikanRoss Millikan

          293k23197371




          293k23197371























              1












              $begingroup$

              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









              $endgroup$


















                1












                $begingroup$

                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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 44 mins ago









                  OldboyOldboy

                  7,4481834




                  7,4481834






























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