How can I adjust a pantograph equation to account for an offset tip?
$begingroup$
In How can I calculate the position of pantograph's tip from the angles at its base?, the answer gives an equation for the relationship between the angles of arms a and b from the 0˚ vertical and the base, and the resulting position of the free-moving tip of the rhombus.
In the version below, everything is just the same, but the arrangement is complicated by an offset.
How can the relationship be calculated, if we know the length of the offset vertex from the arm, and the distance from the tip at its base?
That is: given a position in X/Y co-ordinates of the offset tip, what will the angles of the two arms be from the vertical?
geometry
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
$endgroup$
add a comment |
$begingroup$
In How can I calculate the position of pantograph's tip from the angles at its base?, the answer gives an equation for the relationship between the angles of arms a and b from the 0˚ vertical and the base, and the resulting position of the free-moving tip of the rhombus.
In the version below, everything is just the same, but the arrangement is complicated by an offset.
How can the relationship be calculated, if we know the length of the offset vertex from the arm, and the distance from the tip at its base?
That is: given a position in X/Y co-ordinates of the offset tip, what will the angles of the two arms be from the vertical?
geometry
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
$endgroup$
$begingroup$
Is the offset tip the free-moving tip? By offset, do you mean the pantograph's sides aren't equal in length, or that it's rotated so the free-moving tip isn't vertically above the fixed base?
$endgroup$
– J.G.
Dec 21 '18 at 10:43
$begingroup$
@J.G. It's exactly the same arrangement as the one in the question I referred to, with the addition of an offset tip, which is in a fixed relationship with the arm it's joined to. The relationship between the free-moving tip and the angles is given in math.stackexchange.com/questions/3047602/…, but now it's a question of the relationship between the offset tip and those angles.
$endgroup$
– Daniele Procida
Dec 21 '18 at 11:08
add a comment |
$begingroup$
In How can I calculate the position of pantograph's tip from the angles at its base?, the answer gives an equation for the relationship between the angles of arms a and b from the 0˚ vertical and the base, and the resulting position of the free-moving tip of the rhombus.
In the version below, everything is just the same, but the arrangement is complicated by an offset.
How can the relationship be calculated, if we know the length of the offset vertex from the arm, and the distance from the tip at its base?
That is: given a position in X/Y co-ordinates of the offset tip, what will the angles of the two arms be from the vertical?
geometry
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
$endgroup$
In How can I calculate the position of pantograph's tip from the angles at its base?, the answer gives an equation for the relationship between the angles of arms a and b from the 0˚ vertical and the base, and the resulting position of the free-moving tip of the rhombus.
In the version below, everything is just the same, but the arrangement is complicated by an offset.
How can the relationship be calculated, if we know the length of the offset vertex from the arm, and the distance from the tip at its base?
That is: given a position in X/Y co-ordinates of the offset tip, what will the angles of the two arms be from the vertical?
geometry
geometry
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
edited Dec 21 '18 at 11:10
Daniele Procida
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
asked Dec 21 '18 at 10:36
Daniele ProcidaDaniele Procida
1185
1185
$begingroup$
Is the offset tip the free-moving tip? By offset, do you mean the pantograph's sides aren't equal in length, or that it's rotated so the free-moving tip isn't vertically above the fixed base?
$endgroup$
– J.G.
Dec 21 '18 at 10:43
$begingroup$
@J.G. It's exactly the same arrangement as the one in the question I referred to, with the addition of an offset tip, which is in a fixed relationship with the arm it's joined to. The relationship between the free-moving tip and the angles is given in math.stackexchange.com/questions/3047602/…, but now it's a question of the relationship between the offset tip and those angles.
$endgroup$
– Daniele Procida
Dec 21 '18 at 11:08
add a comment |
$begingroup$
Is the offset tip the free-moving tip? By offset, do you mean the pantograph's sides aren't equal in length, or that it's rotated so the free-moving tip isn't vertically above the fixed base?
$endgroup$
– J.G.
Dec 21 '18 at 10:43
$begingroup$
@J.G. It's exactly the same arrangement as the one in the question I referred to, with the addition of an offset tip, which is in a fixed relationship with the arm it's joined to. The relationship between the free-moving tip and the angles is given in math.stackexchange.com/questions/3047602/…, but now it's a question of the relationship between the offset tip and those angles.
$endgroup$
– Daniele Procida
Dec 21 '18 at 11:08
$begingroup$
Is the offset tip the free-moving tip? By offset, do you mean the pantograph's sides aren't equal in length, or that it's rotated so the free-moving tip isn't vertically above the fixed base?
$endgroup$
– J.G.
Dec 21 '18 at 10:43
$begingroup$
Is the offset tip the free-moving tip? By offset, do you mean the pantograph's sides aren't equal in length, or that it's rotated so the free-moving tip isn't vertically above the fixed base?
$endgroup$
– J.G.
Dec 21 '18 at 10:43
$begingroup$
@J.G. It's exactly the same arrangement as the one in the question I referred to, with the addition of an offset tip, which is in a fixed relationship with the arm it's joined to. The relationship between the free-moving tip and the angles is given in math.stackexchange.com/questions/3047602/…, but now it's a question of the relationship between the offset tip and those angles.
$endgroup$
– Daniele Procida
Dec 21 '18 at 11:08
$begingroup$
@J.G. It's exactly the same arrangement as the one in the question I referred to, with the addition of an offset tip, which is in a fixed relationship with the arm it's joined to. The relationship between the free-moving tip and the angles is given in math.stackexchange.com/questions/3047602/…, but now it's a question of the relationship between the offset tip and those angles.
$endgroup$
– Daniele Procida
Dec 21 '18 at 11:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't know what the pantograph equation is, but you are asking about an offset to a curve let me offer the following procedure used on cam followers (and other paths of rolling bodies).
Consider a shape given in cartesian coordinates $pmatrix{x(t) & y(t)} $ in terms of some parameter $t$. We want to offset this curve by a fixed distance $d$, to find the curve $pmatrix{x_C(t) & y_C(t)} $.
For each $t$, calculate the lead angle $alpha(t)$ $$ tan(alpha) = - frac{x' cos(t)+y' sin(t)}{sqrt{x^2+y^2}} $$ here $x'$ and $y'$ are the derivatives of $x$ and $y$ in terms of $t$. This is why you need an analytic curve to do this.
The offset curve is given by the following equation
$$begin{aligned}
x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
end{aligned}$$
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
$endgroup$
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
add a comment |
Your Answer
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%2f3048379%2fhow-can-i-adjust-a-pantograph-equation-to-account-for-an-offset-tip%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
$begingroup$
I don't know what the pantograph equation is, but you are asking about an offset to a curve let me offer the following procedure used on cam followers (and other paths of rolling bodies).
Consider a shape given in cartesian coordinates $pmatrix{x(t) & y(t)} $ in terms of some parameter $t$. We want to offset this curve by a fixed distance $d$, to find the curve $pmatrix{x_C(t) & y_C(t)} $.
For each $t$, calculate the lead angle $alpha(t)$ $$ tan(alpha) = - frac{x' cos(t)+y' sin(t)}{sqrt{x^2+y^2}} $$ here $x'$ and $y'$ are the derivatives of $x$ and $y$ in terms of $t$. This is why you need an analytic curve to do this.
The offset curve is given by the following equation
$$begin{aligned}
x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
end{aligned}$$
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
$endgroup$
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
add a comment |
$begingroup$
I don't know what the pantograph equation is, but you are asking about an offset to a curve let me offer the following procedure used on cam followers (and other paths of rolling bodies).
Consider a shape given in cartesian coordinates $pmatrix{x(t) & y(t)} $ in terms of some parameter $t$. We want to offset this curve by a fixed distance $d$, to find the curve $pmatrix{x_C(t) & y_C(t)} $.
For each $t$, calculate the lead angle $alpha(t)$ $$ tan(alpha) = - frac{x' cos(t)+y' sin(t)}{sqrt{x^2+y^2}} $$ here $x'$ and $y'$ are the derivatives of $x$ and $y$ in terms of $t$. This is why you need an analytic curve to do this.
The offset curve is given by the following equation
$$begin{aligned}
x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
end{aligned}$$
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
$endgroup$
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
add a comment |
$begingroup$
I don't know what the pantograph equation is, but you are asking about an offset to a curve let me offer the following procedure used on cam followers (and other paths of rolling bodies).
Consider a shape given in cartesian coordinates $pmatrix{x(t) & y(t)} $ in terms of some parameter $t$. We want to offset this curve by a fixed distance $d$, to find the curve $pmatrix{x_C(t) & y_C(t)} $.
For each $t$, calculate the lead angle $alpha(t)$ $$ tan(alpha) = - frac{x' cos(t)+y' sin(t)}{sqrt{x^2+y^2}} $$ here $x'$ and $y'$ are the derivatives of $x$ and $y$ in terms of $t$. This is why you need an analytic curve to do this.
The offset curve is given by the following equation
$$begin{aligned}
x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
end{aligned}$$
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
$endgroup$
I don't know what the pantograph equation is, but you are asking about an offset to a curve let me offer the following procedure used on cam followers (and other paths of rolling bodies).
Consider a shape given in cartesian coordinates $pmatrix{x(t) & y(t)} $ in terms of some parameter $t$. We want to offset this curve by a fixed distance $d$, to find the curve $pmatrix{x_C(t) & y_C(t)} $.
For each $t$, calculate the lead angle $alpha(t)$ $$ tan(alpha) = - frac{x' cos(t)+y' sin(t)}{sqrt{x^2+y^2}} $$ here $x'$ and $y'$ are the derivatives of $x$ and $y$ in terms of $t$. This is why you need an analytic curve to do this.
The offset curve is given by the following equation
$$begin{aligned}
x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
end{aligned}$$
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
answered Dec 21 '18 at 15:54
ja72ja72
7,57212044
7,57212044
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
add a comment |
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
$begingroup$
Thank you for the detailed information, though I am sorry to say I don't have the knowledge to apply it to my case. I don't understand where for example the length of the arms (all are the same length) apply in these equations.
$endgroup$
– Daniele Procida
Dec 21 '18 at 19:07
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%2f3048379%2fhow-can-i-adjust-a-pantograph-equation-to-account-for-an-offset-tip%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
$begingroup$
Is the offset tip the free-moving tip? By offset, do you mean the pantograph's sides aren't equal in length, or that it's rotated so the free-moving tip isn't vertically above the fixed base?
$endgroup$
– J.G.
Dec 21 '18 at 10:43
$begingroup$
@J.G. It's exactly the same arrangement as the one in the question I referred to, with the addition of an offset tip, which is in a fixed relationship with the arm it's joined to. The relationship between the free-moving tip and the angles is given in math.stackexchange.com/questions/3047602/…, but now it's a question of the relationship between the offset tip and those angles.
$endgroup$
– Daniele Procida
Dec 21 '18 at 11:08