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What is the length of pair of wires after twisting them around each other?
Calulating length of cable running along exterior of an axleArea covered by a constant length segment rotating around the center of a square.Find depth of a half-filled parabolic cross-sectionWhat shape results from this deformation of a circle?Maximum tilt for a cylindrical glass without spillingIs this dress puzzle solvable?Cross Section of a Pancake Laying on a CylinderDimensions of paper needed to roll a cone (Updated with clarifications)
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$begingroup$
Given two wires each of diameter D and length L, what will be the new length after you twist the wires around each other 1 Turn?
I mean that when you start with the pair of straight wires and put them against a scale, the length is L. When you twist them around each other tightly, and place the straightened twisted wire against the scale, its new length will be reduced.
And what I mean by 1 Turn is that looking at the cross-section, if you begin with the Red wire on top, Black at bottom then after half Turn the Red wire is at bottom and after another half Turn (or Total 1 Turn) the Red is on top again. And both wires are twisting in same direction, say clockwise, when looking at the cross-section. Successive cross-sections will look like the number 8 rotating.
I suppose we should also give the pitch? But when you play with an actual mouse wire, twisting it around itself, you will realize that you can not reduce the pitch beyond a certain extent. So what is that minimum pitch when twisting the wires? What is causing that restriction? Can it be quantified?
Perhaps the question is not too clear, so please feel free to edit/clarify.
geometry
asked 8 hours ago
KumarKumar
1659 bronze badges
$endgroup$
add a comment |
$begingroup$
Given two wires each of diameter D and length L, what will be the new length after you twist the wires around each other 1 Turn?
I mean that when you start with the pair of straight wires and put them against a scale, the length is L. When you twist them around each other tightly, and place the straightened twisted wire against the scale, its new length will be reduced.
And what I mean by 1 Turn is that looking at the cross-section, if you begin with the Red wire on top, Black at bottom then after half Turn the Red wire is at bottom and after another half Turn (or Total 1 Turn) the Red is on top again. And both wires are twisting in same direction, say clockwise, when looking at the cross-section. Successive cross-sections will look like the number 8 rotating.
I suppose we should also give the pitch? But when you play with an actual mouse wire, twisting it around itself, you will realize that you can not reduce the pitch beyond a certain extent. So what is that minimum pitch when twisting the wires? What is causing that restriction? Can it be quantified?
Perhaps the question is not too clear, so please feel free to edit/clarify.
geometry
asked 8 hours ago
KumarKumar
1659 bronze badges
$endgroup$
add a comment |
$begingroup$
Given two wires each of diameter D and length L, what will be the new length after you twist the wires around each other 1 Turn?
I mean that when you start with the pair of straight wires and put them against a scale, the length is L. When you twist them around each other tightly, and place the straightened twisted wire against the scale, its new length will be reduced.
And what I mean by 1 Turn is that looking at the cross-section, if you begin with the Red wire on top, Black at bottom then after half Turn the Red wire is at bottom and after another half Turn (or Total 1 Turn) the Red is on top again. And both wires are twisting in same direction, say clockwise, when looking at the cross-section. Successive cross-sections will look like the number 8 rotating.
I suppose we should also give the pitch? But when you play with an actual mouse wire, twisting it around itself, you will realize that you can not reduce the pitch beyond a certain extent. So what is that minimum pitch when twisting the wires? What is causing that restriction? Can it be quantified?
Perhaps the question is not too clear, so please feel free to edit/clarify.
geometry
asked 8 hours ago
KumarKumar
1659 bronze badges
$endgroup$
Given two wires each of diameter D and length L, what will be the new length after you twist the wires around each other 1 Turn?
I mean that when you start with the pair of straight wires and put them against a scale, the length is L. When you twist them around each other tightly, and place the straightened twisted wire against the scale, its new length will be reduced.
And what I mean by 1 Turn is that looking at the cross-section, if you begin with the Red wire on top, Black at bottom then after half Turn the Red wire is at bottom and after another half Turn (or Total 1 Turn) the Red is on top again. And both wires are twisting in same direction, say clockwise, when looking at the cross-section. Successive cross-sections will look like the number 8 rotating.
I suppose we should also give the pitch? But when you play with an actual mouse wire, twisting it around itself, you will realize that you can not reduce the pitch beyond a certain extent. So what is that minimum pitch when twisting the wires? What is causing that restriction? Can it be quantified?
Perhaps the question is not too clear, so please feel free to edit/clarify.
geometry
geometry
asked 8 hours ago
KumarKumar
1659 bronze badges
asked 8 hours ago
KumarKumar
1659 bronze badges
edited 1 hour ago
Kumar
asked 8 hours ago
KumarKumar
1659 bronze badges
asked 8 hours ago
KumarKumar
1659 bronze badges
asked 8 hours ago
KumarKumar
1659 bronze badges
1659 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The axis of each wire will take the shape of a helix of radius $D/2$ and height $H$. The length of each helix is (for one turn)
$$
L=sqrtH^2+pi^2 D^2
$$
and we can invert this to obtain the desired new length:
$$
H=sqrtL^2-pi^2 D^2.
$$
(This is the vertical distance between the centers of the end sections of a wire, total length will be $H+D$).
The slope $theta$ of a wire is given by $tantheta=Hoverpi D$. Height $H$ cannot be less than $Dovercostheta$ (see diagram), because a wire cannot penetrate into itself. Hence the minimum slope of the helix must satisfy
$$
tantheta=D /costhetaoverpi D=1overpicostheta.
$$
This equation can be solved to get the minimum slope:
$$
sintheta=1overpi.
$$
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
$endgroup$
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The axis of each wire will take the shape of a helix of radius $D/2$ and height $H$. The length of each helix is (for one turn)
$$
L=sqrtH^2+pi^2 D^2
$$
and we can invert this to obtain the desired new length:
$$
H=sqrtL^2-pi^2 D^2.
$$
(This is the vertical distance between the centers of the end sections of a wire, total length will be $H+D$).
The slope $theta$ of a wire is given by $tantheta=Hoverpi D$. Height $H$ cannot be less than $Dovercostheta$ (see diagram), because a wire cannot penetrate into itself. Hence the minimum slope of the helix must satisfy
$$
tantheta=D /costhetaoverpi D=1overpicostheta.
$$
This equation can be solved to get the minimum slope:
$$
sintheta=1overpi.
$$
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
$endgroup$
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
add a comment |
$begingroup$
The axis of each wire will take the shape of a helix of radius $D/2$ and height $H$. The length of each helix is (for one turn)
$$
L=sqrtH^2+pi^2 D^2
$$
and we can invert this to obtain the desired new length:
$$
H=sqrtL^2-pi^2 D^2.
$$
(This is the vertical distance between the centers of the end sections of a wire, total length will be $H+D$).
The slope $theta$ of a wire is given by $tantheta=Hoverpi D$. Height $H$ cannot be less than $Dovercostheta$ (see diagram), because a wire cannot penetrate into itself. Hence the minimum slope of the helix must satisfy
$$
tantheta=D /costhetaoverpi D=1overpicostheta.
$$
This equation can be solved to get the minimum slope:
$$
sintheta=1overpi.
$$
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
$endgroup$
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
add a comment |
$begingroup$
The axis of each wire will take the shape of a helix of radius $D/2$ and height $H$. The length of each helix is (for one turn)
$$
L=sqrtH^2+pi^2 D^2
$$
and we can invert this to obtain the desired new length:
$$
H=sqrtL^2-pi^2 D^2.
$$
(This is the vertical distance between the centers of the end sections of a wire, total length will be $H+D$).
The slope $theta$ of a wire is given by $tantheta=Hoverpi D$. Height $H$ cannot be less than $Dovercostheta$ (see diagram), because a wire cannot penetrate into itself. Hence the minimum slope of the helix must satisfy
$$
tantheta=D /costhetaoverpi D=1overpicostheta.
$$
This equation can be solved to get the minimum slope:
$$
sintheta=1overpi.
$$
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
$endgroup$
The axis of each wire will take the shape of a helix of radius $D/2$ and height $H$. The length of each helix is (for one turn)
$$
L=sqrtH^2+pi^2 D^2
$$
and we can invert this to obtain the desired new length:
$$
H=sqrtL^2-pi^2 D^2.
$$
(This is the vertical distance between the centers of the end sections of a wire, total length will be $H+D$).
The slope $theta$ of a wire is given by $tantheta=Hoverpi D$. Height $H$ cannot be less than $Dovercostheta$ (see diagram), because a wire cannot penetrate into itself. Hence the minimum slope of the helix must satisfy
$$
tantheta=D /costhetaoverpi D=1overpicostheta.
$$
This equation can be solved to get the minimum slope:
$$
sintheta=1overpi.
$$
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
edited 2 hours ago
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
answered 6 hours ago
AretinoAretino
27.1k3 gold badges19 silver badges47 bronze badges
27.1k3 gold badges19 silver badges47 bronze badges
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
add a comment |
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
Thank you for answering. Given that the wires are tightly wound and can not seem to be twisted below a certain threshold H, is there a relation between D & L that would quantify that threshold?
$endgroup$
– Kumar
5 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
This is unclear. why is H = D Cosθ at minimum? Or should it be H/2 = D Cosθ ??
$endgroup$
– Kumar
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
I was wrong: the correct relation is $H=D/costheta$, see my edit.
$endgroup$
– Aretino
3 hours ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
$begingroup$
Still can't wrap my head around as to why the length will be H+D and not just H the length of the axis. The figure makes it even more confusing. I was under the impression that the cross section is the figure 8 and distance between the 2 centers is always D. Thanks anyway.
$endgroup$
– Kumar
1 hour ago
add a comment |
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