How can I adjust a pantograph equation to account for an offset tip?












0












$begingroup$


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?



Offset pantograph






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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?



Offset pantograph






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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?



Offset pantograph






geometry







share|cite|improve this question











$endgroup$




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?



Offset pantograph






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 11:10







Daniele Procida

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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)} $.




  1. 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.


  2. The offset curve is given by the following equation
    $$begin{aligned}
    x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
    y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
    end{aligned}$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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)} $.




  1. 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.


  2. The offset curve is given by the following equation
    $$begin{aligned}
    x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
    y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
    end{aligned}$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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)} $.




  1. 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.


  2. The offset curve is given by the following equation
    $$begin{aligned}
    x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
    y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
    end{aligned}$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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)} $.




  1. 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.


  2. The offset curve is given by the following equation
    $$begin{aligned}
    x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
    y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
    end{aligned}$$







share|cite|improve this answer









$endgroup$



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)} $.




  1. 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.


  2. The offset curve is given by the following equation
    $$begin{aligned}
    x_C & = x + d ;frac{x cos( alpha) - y sin( alpha)}{sqrt{x^2+y^2}} \
    y_C & = y + d ;frac{x sin( alpha) + y cos( alpha)}{sqrt{x^2+y^2}}
    end{aligned}$$








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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