Shortest Path Between Two Points On A Sphere

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Күн бұрын

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@betaorionis2164 Ай бұрын

That was an excellent exercise of calculus and of course I gave you a thumbs up. However, you can calculate the distance between two points on a sphere in a a waaaay shorter mode with straightforward geometrical calculations.

@Freeball99 Ай бұрын

I'd like to hear more about this. Note that I wasn't trying to calculate the actual distance between two points, but rather was trying to find the geometry of the shortest path.

@betaorionis2164 Ай бұрын

@@Freeball99 What do you understand by "the geometry of the shortest path"? ¿The equation of the great circle between those two points? I was under the impression that you were looking for the shortest distance, which can be obtained through the Haversine formula, itself derived from the properties of the spherical triangles, which can be obtained from geometry.

@Freeball99 Ай бұрын

@@betaorionis2164 The haversine formula can indeed be used to calculate the shortest distance between two points on a sphere, but it presupposes that the shortest path lies along a great circle. It does not derive the fact that the shortest path is a great circle; rather, it uses this assumption to compute the arc length directly.

@betaorionis2164 Ай бұрын

@@Freeball99 Ok, I understand now.

@Freeball99 Ай бұрын

@@betaorionis2164 👍 I'll pin this to the top of the comments section because I suspect that others will have a similar concern.

@sonyaraman Ай бұрын

Clear and enjoyable, I believe this is the best explanation of calculus of variations in regards to this problem

@IanHarris-fp6dq Ай бұрын

dumb old me derived the trigonometric great circle equation from first principles decades ago. But am latterly enthralled by this demonstrated sophistication. Wow! Thankyou

@MrPoornakumar Ай бұрын

Brilliant! You took me through my under-grad course. Thanks.

@notsodope834 Ай бұрын

Nice, using a constraint that all those paths lie on the sphere. Can see how really difficult the integral becomes for questions of shortest path between two points in an irregular 3d shape . Lucky that in case of sphere its doable.

@pseudolullus 14 күн бұрын

Pretty nice how the functional is very similar to the plane one except for the extra sine squared

@5eurosenelsuelo Ай бұрын

Thank you very much for the explanation. It'd be interesting to see the problem applied to other surfaces such as a cone or a cylinder. I know they're flat so it'd be possible to develop them and draw the straight line but I'm not familiar enough with the Lagrangian to do it this way. Well, the cylinder maybe I could because of its similarities to a circle. Maybe even the cone if I follow this video slowly. But I'm still missing the foundation. I need to rewatch your video on the Lagrangian.

@Freeball99 Ай бұрын

Try apply this exact technique to a cylinder. The kinematics are slightly different, but the integral is far simpler to carry out - hint: you should find that z' = dz / dθ = const (which is trivial to integrate). Use x = r cos θ, y = r sin θ, z = z and substitute into the distance formula. You should arrive at the conclusion that the shortest path between 2 points on a cylinder is a helix.

@papkenhartunian186 Ай бұрын

I loved your material on this video. I would like to know how you can prove a straight line is a circle and Vis versa.

@Freeball99 Ай бұрын

I'm not sure I am clear on your question. Are you asking how to prove that the intersection of a plane with a sphere is a circle?

@bobh6728 Ай бұрын

I think that is first time I’ve seen a mathematical proof that the great circle is the shortest distance. I’ve seen intuitive explanations before.

@mathunt1130 Ай бұрын

I think you've missed a trick: You can rotate your sphere in such that the P and Q can be put on the axis where either \theta or \phi is constant. That should reduce the complexity dramatically.

@Freeball99 Ай бұрын

You make a good point, however, rotating to make one angle constant actually presupposes a great circle solution. I am treating the problem from the point-of-view of not knowing a priori what the form of the solution is. The Euler-Lagrange method derives this result from first principles, without assuming the answer. It demonstrates how calculus of variations reveals geometric truths purely from mathematics.

@MehrdadMohajer-p1m 2 күн бұрын

Thx. & as usual well done. If you plan to come to Hamburg any time soon …., Be happy to stay with me for couple of days, working on an tangible Project together.

@r2k314 Ай бұрын

That was really fun. I quess I never saw this application of the E.L. equation before because it's usually demonstrated with the geodesic equation, correct?

@Freeball99 Ай бұрын

To be clear, the geodesic equation does not, in fact, prove that the shortest path between two points on a sphere is the arc of a great circle, but rather it presupposes this fact; it the uses this to derive a formula for calculating the distance. This is analogous to presupposing that the shortest path between two points on a plane is a straight line (as opposed to proving that) and then using the distance formula to calculate the path length.

@r2k314 Ай бұрын

@@Freeball99 oh wow. Thanks. I'm going to go back and look into the derivation of the geodesic equation again. Appreciate your clear presentation s very much.

@vishalmishra3046 Ай бұрын

*Shortest length between 2 points on a sphere* is simply the direct arc length between them = radius x angle (in radians) between the 2 points subtended at the center of the sphere. Let Center C = (Cx,Cy, Cz) and the points on the sphere are P = (Px, Py, Pz) and Q = (Qx,Qy,Qz), with angle T between them (calculated using arc cos of dot product below). Then, radius of the sphere, R = |P-C| = |Q-C| and related unit vectors are p = (P-C) / |P-C| and q = (Q-C) / |Q-C| with cos T = p . q (dot product). So, shortest arc length = R x T = Radius of the Sphere x Angle in Radians between the 2 points at the center of the Sphere. *Simple, right* ?

@Freeball99 Ай бұрын

Thank you for sharing this. You're right that it's a simple and effective way to calculate the shortest distance between two points on a sphere - I find it far more intuitive than the haversine formula which performs a similar calculation. HOWEVER, it assumes that the shortest path is along a great circle arc. The Euler-Lagrange approach I used actually proves why this is the case, deriving the result from first principles without assuming the nature of the solution beforehand.