Formula for Arclength of Geodesic Connecting Two Points in the Surface of a Cylinder












1












$begingroup$


Given two points laying on the surface of a cylinder, is there a simple equation for the arclength of the geodesic that connects those two points?



In my use case, the cylinder is oriented axially coincident with the x axis. I have two points for which I know their (x,y,z) locations, and I understand that I can convert these coordinates to cylindrical coordinates by the transformation x=x, y=rcos(theta), z=rsin(theta). Beyond that, I am not sure is there is a simple equation for calculating the geodesic length of between these two points without "unrolling" the cylinder into a plane and using the distance formula.



Can someone confirm for me if it is simply: L=SQRT(r^2θ^2+x^2), where x ix the axial distance which separates the points in my example? Is it this easy?



Thank you.






geometry







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$endgroup$












  • $begingroup$
    Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances.
    $endgroup$
    – Doug M
    Jan 11 at 21:38


















1












$begingroup$


Given two points laying on the surface of a cylinder, is there a simple equation for the arclength of the geodesic that connects those two points?



In my use case, the cylinder is oriented axially coincident with the x axis. I have two points for which I know their (x,y,z) locations, and I understand that I can convert these coordinates to cylindrical coordinates by the transformation x=x, y=rcos(theta), z=rsin(theta). Beyond that, I am not sure is there is a simple equation for calculating the geodesic length of between these two points without "unrolling" the cylinder into a plane and using the distance formula.



Can someone confirm for me if it is simply: L=SQRT(r^2θ^2+x^2), where x ix the axial distance which separates the points in my example? Is it this easy?



Thank you.






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances.
    $endgroup$
    – Doug M
    Jan 11 at 21:38
















1












1








1





$begingroup$


Given two points laying on the surface of a cylinder, is there a simple equation for the arclength of the geodesic that connects those two points?



In my use case, the cylinder is oriented axially coincident with the x axis. I have two points for which I know their (x,y,z) locations, and I understand that I can convert these coordinates to cylindrical coordinates by the transformation x=x, y=rcos(theta), z=rsin(theta). Beyond that, I am not sure is there is a simple equation for calculating the geodesic length of between these two points without "unrolling" the cylinder into a plane and using the distance formula.



Can someone confirm for me if it is simply: L=SQRT(r^2θ^2+x^2), where x ix the axial distance which separates the points in my example? Is it this easy?



Thank you.






geometry







share|cite|improve this question











$endgroup$




Given two points laying on the surface of a cylinder, is there a simple equation for the arclength of the geodesic that connects those two points?



In my use case, the cylinder is oriented axially coincident with the x axis. I have two points for which I know their (x,y,z) locations, and I understand that I can convert these coordinates to cylindrical coordinates by the transformation x=x, y=rcos(theta), z=rsin(theta). Beyond that, I am not sure is there is a simple equation for calculating the geodesic length of between these two points without "unrolling" the cylinder into a plane and using the distance formula.



Can someone confirm for me if it is simply: L=SQRT(r^2θ^2+x^2), where x ix the axial distance which separates the points in my example? Is it this easy?



Thank you.






geometry




geometry






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 21:01







kreeser1

















asked Jan 11 at 18:53









kreeser1kreeser1

10810




10810












  • $begingroup$
    Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances.
    $endgroup$
    – Doug M
    Jan 11 at 21:38




















  • $begingroup$
    Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances.
    $endgroup$
    – Doug M
    Jan 11 at 21:38


















$begingroup$
Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances.
$endgroup$
– Doug M
Jan 11 at 21:38






$begingroup$
Peel the skin off the the cylinder and unroll it. Now you have a rectangle. So yes, the square root of the sum of the squares of the shorter lateral distance and vertical distances.
$endgroup$
– Doug M
Jan 11 at 21:38












2 Answers
2






active

oldest

votes


















1












$begingroup$

Geodesic of cylinders are known to be




  • 1) either helixical arcs (the shortest helixical arc connecting the two points).


  • 2) or vertical segments



Let us consider case 1). When the cylinder, as a ruled surface is unrolled isometrically :




  • the ordinates of the points stay the same, whereas


  • abscissas are measured by the unrolling of arc lengthes $r theta$.



The geodesic (piece of an helix) is mapped isometrically onto the geodesic of the plane which is the line segment connecting points $(x_1=r theta_1,y_1)$ and $(x_2=r theta_2,y_2)$.



Its arc length is thus :



$$sqrt{(r(theta_2-theta_1))^2+(y_2-y_1)^2} tag{1}$$



(almost as you, @user1998586, gave it ; why didn't you modify your answer instead of erasing it ?).



In the exceptional case where the geodesic is a vertical segment (corresponding to the case where the two points are on a same vertical line), happily, the isometrical mapping works the same : formula (1) is still valid with $theta_2=theta_1$ under the simplified form $$|y_1-y_2|$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
    $endgroup$
    – kreeser1
    Jan 11 at 21:30












  • $begingroup$
    Question 1) Yes, the r term has to be squared (I correct it at once).
    $endgroup$
    – Jean Marie
    Jan 11 at 21:33










  • $begingroup$
    Question 2) : what do you mean exactly by "oriented along the x-axis" ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:34










  • $begingroup$
    Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:39








  • 1




    $begingroup$
    @JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
    $endgroup$
    – user1998586
    Jan 12 at 7:39



















1












$begingroup$

You can "hide" the unrolling by using trig functions to convert chord-length to arc length.



A chord of length $l$ on a circle of radius $r$ gives an arc-length of $2 r arcsin frac{l}{2r}$



Given points on the cylinder with $Delta x = x_2 - x_1$, $Delta y = y_2 - y_1$, $Delta z = z_2 - z_1$ we then get the geodesic length:
$$
sqrt{(Delta x)^2 + 4r^2 arcsin^2 left(frac{1}{2r}sqrt{(Delta y)^2 + (Delta z)^2}right)}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
    $endgroup$
    – kreeser1
    Jan 11 at 20:15











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Geodesic of cylinders are known to be




  • 1) either helixical arcs (the shortest helixical arc connecting the two points).


  • 2) or vertical segments



Let us consider case 1). When the cylinder, as a ruled surface is unrolled isometrically :




  • the ordinates of the points stay the same, whereas


  • abscissas are measured by the unrolling of arc lengthes $r theta$.



The geodesic (piece of an helix) is mapped isometrically onto the geodesic of the plane which is the line segment connecting points $(x_1=r theta_1,y_1)$ and $(x_2=r theta_2,y_2)$.



Its arc length is thus :



$$sqrt{(r(theta_2-theta_1))^2+(y_2-y_1)^2} tag{1}$$



(almost as you, @user1998586, gave it ; why didn't you modify your answer instead of erasing it ?).



In the exceptional case where the geodesic is a vertical segment (corresponding to the case where the two points are on a same vertical line), happily, the isometrical mapping works the same : formula (1) is still valid with $theta_2=theta_1$ under the simplified form $$|y_1-y_2|$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
    $endgroup$
    – kreeser1
    Jan 11 at 21:30












  • $begingroup$
    Question 1) Yes, the r term has to be squared (I correct it at once).
    $endgroup$
    – Jean Marie
    Jan 11 at 21:33










  • $begingroup$
    Question 2) : what do you mean exactly by "oriented along the x-axis" ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:34










  • $begingroup$
    Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:39








  • 1




    $begingroup$
    @JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
    $endgroup$
    – user1998586
    Jan 12 at 7:39
















1












$begingroup$

Geodesic of cylinders are known to be




  • 1) either helixical arcs (the shortest helixical arc connecting the two points).


  • 2) or vertical segments



Let us consider case 1). When the cylinder, as a ruled surface is unrolled isometrically :




  • the ordinates of the points stay the same, whereas


  • abscissas are measured by the unrolling of arc lengthes $r theta$.



The geodesic (piece of an helix) is mapped isometrically onto the geodesic of the plane which is the line segment connecting points $(x_1=r theta_1,y_1)$ and $(x_2=r theta_2,y_2)$.



Its arc length is thus :



$$sqrt{(r(theta_2-theta_1))^2+(y_2-y_1)^2} tag{1}$$



(almost as you, @user1998586, gave it ; why didn't you modify your answer instead of erasing it ?).



In the exceptional case where the geodesic is a vertical segment (corresponding to the case where the two points are on a same vertical line), happily, the isometrical mapping works the same : formula (1) is still valid with $theta_2=theta_1$ under the simplified form $$|y_1-y_2|$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
    $endgroup$
    – kreeser1
    Jan 11 at 21:30












  • $begingroup$
    Question 1) Yes, the r term has to be squared (I correct it at once).
    $endgroup$
    – Jean Marie
    Jan 11 at 21:33










  • $begingroup$
    Question 2) : what do you mean exactly by "oriented along the x-axis" ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:34










  • $begingroup$
    Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:39








  • 1




    $begingroup$
    @JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
    $endgroup$
    – user1998586
    Jan 12 at 7:39














1












1








1





$begingroup$

Geodesic of cylinders are known to be




  • 1) either helixical arcs (the shortest helixical arc connecting the two points).


  • 2) or vertical segments



Let us consider case 1). When the cylinder, as a ruled surface is unrolled isometrically :




  • the ordinates of the points stay the same, whereas


  • abscissas are measured by the unrolling of arc lengthes $r theta$.



The geodesic (piece of an helix) is mapped isometrically onto the geodesic of the plane which is the line segment connecting points $(x_1=r theta_1,y_1)$ and $(x_2=r theta_2,y_2)$.



Its arc length is thus :



$$sqrt{(r(theta_2-theta_1))^2+(y_2-y_1)^2} tag{1}$$



(almost as you, @user1998586, gave it ; why didn't you modify your answer instead of erasing it ?).



In the exceptional case where the geodesic is a vertical segment (corresponding to the case where the two points are on a same vertical line), happily, the isometrical mapping works the same : formula (1) is still valid with $theta_2=theta_1$ under the simplified form $$|y_1-y_2|$$






share|cite|improve this answer











$endgroup$



Geodesic of cylinders are known to be




  • 1) either helixical arcs (the shortest helixical arc connecting the two points).


  • 2) or vertical segments



Let us consider case 1). When the cylinder, as a ruled surface is unrolled isometrically :




  • the ordinates of the points stay the same, whereas


  • abscissas are measured by the unrolling of arc lengthes $r theta$.



The geodesic (piece of an helix) is mapped isometrically onto the geodesic of the plane which is the line segment connecting points $(x_1=r theta_1,y_1)$ and $(x_2=r theta_2,y_2)$.



Its arc length is thus :



$$sqrt{(r(theta_2-theta_1))^2+(y_2-y_1)^2} tag{1}$$



(almost as you, @user1998586, gave it ; why didn't you modify your answer instead of erasing it ?).



In the exceptional case where the geodesic is a vertical segment (corresponding to the case where the two points are on a same vertical line), happily, the isometrical mapping works the same : formula (1) is still valid with $theta_2=theta_1$ under the simplified form $$|y_1-y_2|$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 21:33

























answered Jan 11 at 21:19









Jean MarieJean Marie

30.6k42154




30.6k42154












  • $begingroup$
    Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
    $endgroup$
    – kreeser1
    Jan 11 at 21:30












  • $begingroup$
    Question 1) Yes, the r term has to be squared (I correct it at once).
    $endgroup$
    – Jean Marie
    Jan 11 at 21:33










  • $begingroup$
    Question 2) : what do you mean exactly by "oriented along the x-axis" ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:34










  • $begingroup$
    Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:39








  • 1




    $begingroup$
    @JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
    $endgroup$
    – user1998586
    Jan 12 at 7:39


















  • $begingroup$
    Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
    $endgroup$
    – kreeser1
    Jan 11 at 21:30












  • $begingroup$
    Question 1) Yes, the r term has to be squared (I correct it at once).
    $endgroup$
    – Jean Marie
    Jan 11 at 21:33










  • $begingroup$
    Question 2) : what do you mean exactly by "oriented along the x-axis" ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:34










  • $begingroup$
    Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
    $endgroup$
    – Jean Marie
    Jan 11 at 21:39








  • 1




    $begingroup$
    @JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
    $endgroup$
    – user1998586
    Jan 12 at 7:39
















$begingroup$
Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
$endgroup$
– kreeser1
Jan 11 at 21:30






$begingroup$
Thank you @Jean Marie, this looks like exactly what I need. Two questions: in your equation (1), should the "r" term be squared? Additionally, how do I modify this equation to fit my stipulation that the cylinder is oriented along the x-axis?
$endgroup$
– kreeser1
Jan 11 at 21:30














$begingroup$
Question 1) Yes, the r term has to be squared (I correct it at once).
$endgroup$
– Jean Marie
Jan 11 at 21:33




$begingroup$
Question 1) Yes, the r term has to be squared (I correct it at once).
$endgroup$
– Jean Marie
Jan 11 at 21:33












$begingroup$
Question 2) : what do you mean exactly by "oriented along the x-axis" ?
$endgroup$
– Jean Marie
Jan 11 at 21:34




$begingroup$
Question 2) : what do you mean exactly by "oriented along the x-axis" ?
$endgroup$
– Jean Marie
Jan 11 at 21:34












$begingroup$
Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
$endgroup$
– Jean Marie
Jan 11 at 21:39






$begingroup$
Maybe you mean that the first point has coordinates $(x=0,y=r)$ meaning you start with a $theta=pi/2$ angle, then you proceed in a clockwise manner to point $(x=r,y=0)$ reaching $theta=0$ ?
$endgroup$
– Jean Marie
Jan 11 at 21:39






1




1




$begingroup$
@JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
$endgroup$
– user1998586
Jan 12 at 7:39




$begingroup$
@JeanMarie the reason I deleted my original answer was because on re-reading the question, I interpreted the wording 'without "unrolling" the cylinder' in the question to be asking for a formula using Euclidean not polar co-ordinates.
$endgroup$
– user1998586
Jan 12 at 7:39











1












$begingroup$

You can "hide" the unrolling by using trig functions to convert chord-length to arc length.



A chord of length $l$ on a circle of radius $r$ gives an arc-length of $2 r arcsin frac{l}{2r}$



Given points on the cylinder with $Delta x = x_2 - x_1$, $Delta y = y_2 - y_1$, $Delta z = z_2 - z_1$ we then get the geodesic length:
$$
sqrt{(Delta x)^2 + 4r^2 arcsin^2 left(frac{1}{2r}sqrt{(Delta y)^2 + (Delta z)^2}right)}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
    $endgroup$
    – kreeser1
    Jan 11 at 20:15
















1












$begingroup$

You can "hide" the unrolling by using trig functions to convert chord-length to arc length.



A chord of length $l$ on a circle of radius $r$ gives an arc-length of $2 r arcsin frac{l}{2r}$



Given points on the cylinder with $Delta x = x_2 - x_1$, $Delta y = y_2 - y_1$, $Delta z = z_2 - z_1$ we then get the geodesic length:
$$
sqrt{(Delta x)^2 + 4r^2 arcsin^2 left(frac{1}{2r}sqrt{(Delta y)^2 + (Delta z)^2}right)}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
    $endgroup$
    – kreeser1
    Jan 11 at 20:15














1












1








1





$begingroup$

You can "hide" the unrolling by using trig functions to convert chord-length to arc length.



A chord of length $l$ on a circle of radius $r$ gives an arc-length of $2 r arcsin frac{l}{2r}$



Given points on the cylinder with $Delta x = x_2 - x_1$, $Delta y = y_2 - y_1$, $Delta z = z_2 - z_1$ we then get the geodesic length:
$$
sqrt{(Delta x)^2 + 4r^2 arcsin^2 left(frac{1}{2r}sqrt{(Delta y)^2 + (Delta z)^2}right)}
$$






share|cite|improve this answer











$endgroup$



You can "hide" the unrolling by using trig functions to convert chord-length to arc length.



A chord of length $l$ on a circle of radius $r$ gives an arc-length of $2 r arcsin frac{l}{2r}$



Given points on the cylinder with $Delta x = x_2 - x_1$, $Delta y = y_2 - y_1$, $Delta z = z_2 - z_1$ we then get the geodesic length:
$$
sqrt{(Delta x)^2 + 4r^2 arcsin^2 left(frac{1}{2r}sqrt{(Delta y)^2 + (Delta z)^2}right)}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 19:30

























answered Jan 11 at 19:21









user1998586user1998586

22114




22114












  • $begingroup$
    Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
    $endgroup$
    – kreeser1
    Jan 11 at 20:15


















  • $begingroup$
    Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
    $endgroup$
    – kreeser1
    Jan 11 at 20:15
















$begingroup$
Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
$endgroup$
– kreeser1
Jan 11 at 20:15




$begingroup$
Thank you for your input. It seems like there must be a more concise equation if we were to convert the Cartesian coordinates to cylindrical coordinates, and fix r to a constant such that the two points lie on the coordinate surface r=C, and the two points are then only defined by their theta and x coordinate. Is this logic reasonable?
$endgroup$
– kreeser1
Jan 11 at 20:15


















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