Arc length between two points on a cylindrical shell

rww88

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I searched for a week trying to find a formula to determine the arc length between two points on a cylindrical shell. I finally derived my own formulas through trial and error. Can someone please verify that these are correct?. Thanks.
 

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What are the equations of this specific arc on the cylinder? Where are they used in your formula? Or are you calculating the length of the shortest path between two points on the cylinder? What if A and B lie on a vertical line? Then your "\(\displaystyle \gamma\)" is 0 and you are dividing by 0.
 
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Hopefull a better explanation

Actually I am attempting to find the length of the geodesic, or the shortest possible line between two points along the surface of a sphere or other curved surface. I probably need to qualify the equations to be viable, if in fact they are, for angles of gamma greater than zero degrees?
 
The shortest distance between two points on a sphere is on

the intersection of the sphere and the plane formed by the two points and the center of the sphere. The shortest distance between two points on a cylinder can be found by cutting the cylinder through one of the two points vertically and flatten the cylinder to make it a rectangle. On the rectangle, just find the distance of the two points. For any curve in space, if you know the function of the curve, the length can be found by the integral formula.
 
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Thank you Yma16 for your assistance

Thank you Yma16 for your assistance. If in my original drawing Rm = 22, L =30, x = 11, and gamma = 50 degrees, would you demonstrate your suggested method for finding the length of the geodesic from point A to point B? Thanks so much for your help.
 
is L the axial displacement between the two points.

Thank you Yma16 for your assistance. If in my original drawing Rm = 22, L =30, x = 11, and gamma = 50 degrees, would you demonstrate your suggested method for finding the length of the geodesic from point A to point B? Thanks so much for your help.

You need to tell me what L is.
 
Sorry for the long delay in answering

The variable L is the linear distance between the two points orthographically projected to the plan view of the tank.
 
that is what i thought. the result is

The variable L is the linear distance between the two points orthographically projected to the plan view of the tank.

about 35.61723. x and gamma are duplicates. One can determine the other. Just imagine when you project the two points vertically on the bottom of the plane. The two points are on the circle. The x determine the gamma and vise versa. But your x and gamma do not match. If I use x and ignore gamma, the result is 37.82546.
 
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Thank you Yma16 for your assistance. If in my original drawing Rm = 22, L =30, x = 11, and gamma = 50 degrees, would you demonstrate your suggested method for finding the length of the geodesic from point A to point B? Thanks so much for your help.

Here's how I would do this.

Your points A and B, as I understand it, are on the cylindrical surface where the two vertical lines intersect the circle in the top view. If we assign coordinates to the horizontal diameter shown (with 0 at the center), the two points on it are at coordinates x (positive as shown) and x-L (negative as shown because L>x).

You have not stated the vertical distance between A and B in the side view, but it can be calculated as z = L tan(gamma).

"Unwrapping" the cylindrical surface to make a rectangle, the horizontal distance between A and B is the arc length on the circle. The angles to A and B are, respectively, arccos(x/r) and arccos((x-L)/R), so the arc length between them is y = R[arccos((x-L)/R) - arccos(x/r)].

Therefore, by the Pythagorean theorem, the distance from A to B on the surface is sqrt(z^2 + y^2), that is,

sqrt(L^2 tan^2(gamma) + R^2 [arccos((x-L)/R) - arccos(x/r)]^2)

For your numbers, I get z = 35.75, y = 34.45, so the distance is 49.65.
 
Geodesic distance between two point on a cylinder

Thank you Dr. Peterson. Your explanation is logically straightforward and will be very useful. Looking at the expressions in your solution, it appears that my originally derived formulas were at least headed in the right direction.
 
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