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March 24th, 2017, 02:22 AM  #11 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,323 Thanks: 39 
Okay, from your latest diagram, your value of x is the same as I used, and which I called the axial displacement between points. In other words, it's the distance down the cylinder from the first point to the second. As for the angle Î¸, it appears that you've defined it differently than what I had assumed. It's too late for me to think about it tonight, but I'll have a look at it tomorrow. It shouldn't be difficult to translate it from one to the other. 
March 24th, 2017, 08:06 AM  #12 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  Maybe a solution?
I think I may have solved my own problem. Regarding the diagram with a series of views of the same cylinder, if the cylinder is "rolledout" into a flat development it appears as if the arc length between points A and B is independent of the radius and is simply equal to x/sin theta. Does anyone agree with this logic? Now, what if the distance y is given. Does anyone know the equation for the arc length between points A and B in terms of y, r, and theta? 
March 24th, 2017, 12:06 PM  #13 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,323 Thanks: 39 
Yes. That's what I said in my first post. Roll it out flat and you get a straight line between A and B.

March 27th, 2017, 06:13 AM  #14 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  Yooklid and studiot please verify these solutions
These solutions are based on trig and do not involve the Pythagorean theorem. As a reminder, x is the distance between points A and B measured on the cylinder's longitudinal axis and y is the distance between points A and B measured along the cylinder's circumferential axis.

April 3rd, 2017, 09:02 PM  #15 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,323 Thanks: 39 
Your first formula, looks correct. The second formula will only work if your "circumferential axis" is parallel to the end projection of chord AB, which in your diagram, it is not. Also note that you can simplify your denominator, because cos(90Î¸)=sin(Î¸) Do you have a practical way to measure the distance y and angle Î¸? It might be a lot easier if you tell us which angles and distances you can measure most easily. Then perhaps we can give you a more straightforward solution. Last edited by Yooklid; April 3rd, 2017 at 09:04 PM. 
April 5th, 2017, 06:04 AM  #16 
Newbie Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0  Thanks everybody; here is my workedout solution
Thanks everybody for your input. Here is my workedout solution to the problem. This original graphic will be included in my soontobe published engineering text "Design and Analysis of Openings in Tanks and Pressure Vessels". Please, if anyone sees a flaw in my discovered solution, alert me. Thanks again.


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bewteen, curvalinear, curvilinear, cylinder, distance, geodesic, points 
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