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March 21st, 2017, 07:42 AM  #1 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  Curvilinear distance between two points on a cylinder (geodesic)
I would like to develop two formulas that would determine the curvilinear distance between two points on a cylinder. The radius R and the angle theta that a line between the two points makes with the cylinder's longitudinal axis are known in each case. In the first formula, the orthographic (vertical) distance L between the two points is known. Note: this is not the chord between the two points; it is the distance between the two points parallel to the circumferential axis. In the second formula the chord distance C between the two points along the inclined line in known. In both cases, the formulas would result in the geodesic G.
Last edited by skipjack; March 21st, 2017 at 07:56 AM. 
March 21st, 2017, 08:02 AM  #2 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 2,627 Thanks: 31 
$L=\sqrt{r^2 \theta^2 + x^2}$ where x is the distance along the axis between the two points. Essentially this is Pythagoras' theorem on the surface of a cylinder. The best way to visualize it is to assume that you can "unwrap" the surface of the cylinder and lay it flat. Last edited by Yooklid; March 21st, 2017 at 08:04 AM. 
March 21st, 2017, 09:07 AM  #3 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  I'm confused
To which of the two cases does this formula apply? The orthographic distance formula or the chord formula? or both? Is r in the formula equal to R in the figure? Is L in the formula equal to L in the figure? Is theta in radians? Is x in the formula equal to G in the figure?

March 21st, 2017, 09:33 AM  #4 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 2,627 Thanks: 31 
I found your terminology and the diagram to be very confusing. To me "orthographic" has a different meaning than what seemed to be implied on the diagram. So, it's possible that the formula I gave may not be what you're looking for. The formula that I gave is for the length of a helical arc (the geodesic) on the surface of the cylinder between two arbitrary points. It's not a chord length. r is the radius of the cylinder. Î¸ is the angular displacement between the two points (in radians). x is the axial displacement between the two points. Last edited by Yooklid; March 21st, 2017 at 09:40 AM. 
March 21st, 2017, 09:56 AM  #5 
Senior Member Joined: Jun 2015 From: England Posts: 566 Thanks: 147 
Isn't the orthographic distance the projection of a given distance onto the orthographic plane? So far as I can see the the projection in the diagram is parallel to the axis of the cylinder so all points on the surface will be projected onto the circumference of a circle on the orthographic plane and the orthographic distance will just be some arc of this circle. In effect all coils or part coils of a helix will collapse to the same circle in the projection. 
March 21st, 2017, 10:36 AM  #6 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 2,627 Thanks: 31 
Yes, but it was my understanding that the two points have some axial displacement from each other. So, the length would not be the same as the length of the projected arc.

March 21st, 2017, 12:24 PM  #7 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  Hopefully better diagrams
The dimension L is parallel with the circumferential axis. The line drawn through the two point is inclined from the circumferential axis. The desired geodesic is the surface distance between the two points that are inclined from the circumferential axis. If theta is 90 degrees then the geodesic would be simply the arc length along the circumferential axis. When theta is zero there is no arc length; the two points lie along the the longitudinal axis of the cylinder and the distance between them is purely linear. In the case of the desired chord formula the chord distance is equal to the chord along the inclined line drawn between the two points. In either case, what I seek is a formula in terms of G such that G = f(R,L,theta) and G = f(R,C,theta).

March 22nd, 2017, 01:45 PM  #8 
Senior Member Joined: Jun 2015 From: England Posts: 566 Thanks: 147 
I think I know what you are trying to achieve and why you can't find any formulae. The orthographic projection maps 3 dimensional space to a plane, which is two dimensional. This is why you require at least two views in an engineering drawing of a 3D object to fully detail all dimensions. An engineering drawing is an orthographic projection. In the sketch below of a cylinder it can readily be seen that all points on the surface along the straight line AB map to the one single point C on the projection. You are asking for an inverse of a many to one function which cannot be obtained. In other words the length L is not available at all in the projection plane. Strictly speaking an orthographic projection is made by lines parallel to a general viewing direction and the plane need not be perpendicular. However this would lead to the image of the cylinder being non circular and you have shown circular images in the projection views. So the projection direction is coaxial with the cylinder axis and the projection plane perpendicular to this direction. 
March 23rd, 2017, 02:53 AM  #9 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  New approach
Unfortunately I am having difficulty composing my query; consequently, I would like to attempt a new, and hopefully simpler approach. Forget any reference to orthographic planes and projections. The figures below describe the problem and the solution desired.

March 23rd, 2017, 03:15 AM  #10 
Senior Member Joined: Jun 2015 From: England Posts: 566 Thanks: 147 
OK that defines things better. In particular your distance X is not the orthographic distance. I am about to go out so I will offer this quickly, since you ae shown as online right now. I do not have time to complete the conversion to your coordinate system right now, but you can run with it as it is the solution to your problem. A curve in 3 D space is defined by two equations (in x, y and z), not one. That is it must satisfy both simultaneously. These equations may not be linear. We may project this curve onto the xy plane by eliminating z between the two equations to form a single one in x and y. This is known as the projecting cylinder. Similar projecting cylinders can be obtained by eliminating x or y instead. Any pair of these is enough to define the curve. Your curve is a very simple example of a space curve. 

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