Tracking the movement of the watch hand
$begingroup$
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
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
$endgroup$
add a comment |
$begingroup$
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
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
$endgroup$
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
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
asked 2 hours ago
Aubrey ChampagneAubrey Champagne
365
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
OldboyOldboy
7,4581834
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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 $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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 $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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 $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
$endgroup$
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 $left(frac 1{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft}),frac m{sqrt{1+m^2}}(L-sqrt{2R^2-2Rcos ft})right)$
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
answered 1 hour ago
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
$begingroup$
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.
answered 1 hour ago
OldboyOldboy
7,4581834
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
OldboyOldboy
7,4581834
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
OldboyOldboy
7,4581834
$endgroup$
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.
answered 1 hour ago
OldboyOldboy
7,4581834
answered 1 hour ago
OldboyOldboy
7,4581834
answered 1 hour ago
OldboyOldboy
7,4581834
answered 1 hour ago
OldboyOldboy
7,4581834
7,4581834
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
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