Moving a point on a sphere

Question

I have a sphere and a point on it. Now I want to move this point in the direction of a randomly choosen angle "a" for for the arc length "l", over the surface of the sphere.

What are the mathematical equations for that?

I thought about maybe using great-circles somehow, but I don't know how to do that practically.

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I also found this, but it don't applies to my case, because that case is reduced to the use of Vectors:

Moving a point around a Sphere

How is your point on the sphere represented? Latitude and longitude? Please edit the answer into your question. — TonyK, Oct 17 '18 at 13:47

user · Answer 1 · 2018-10-17T13:55:19.543

1

Given the radius $r$ and the arch length $l$, the angle $\theta$ is given by

$$\theta=\frac l r$$

to define the new position we need to define the rotation axis, expressed by the unit vector $\vec u=(u_x,u_y,u_z)$, and use the following Rotation matrix from axis and angle

$$R = \begin{bmatrix} \cos \theta +u_x^2 \left(1-\cos \theta\right) & u_x u_y \left(1-\cos \theta\right) - u_z \sin \theta & u_x u_z \left(1-\cos \theta\right) + u_y \sin \theta \\ u_y u_x \left(1-\cos \theta\right) + u_z \sin \theta & \cos \theta + u_y^2\left(1-\cos \theta\right) & u_y u_z \left(1-\cos \theta\right) - u_x \sin \theta \\ u_z u_x \left(1-\cos \theta\right) - u_y \sin \theta & u_z u_y \left(1-\cos \theta\right) + u_x \sin \theta & \cos \theta + u_z^2\left(1-\cos \theta\right) \end{bmatrix}$$

edited Oct 17 '18 at 13:55

answered Oct 17 '18 at 13:42

user

154,566

okay, thank you, but how can I move my point now? – jackjuni Oct 17 '18 at 13:44
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This answer doesn't look likely to help the OP. – TonyK Oct 17 '18 at 13:44
so, I have my point (r,θ,ϕ) and I want to move it by l into direction a. – jackjuni Oct 17 '18 at 13:48
@TonyK Yes I'll try to understand better what the asker is looking for. – user Oct 17 '18 at 13:48
@jackjuni How do you define the direction $\vec a$? – user Oct 17 '18 at 13:49
@gimusi Which unit is θ in this case, radians or degrees or smth else? – jackjuni Oct 17 '18 at 14:00
@gimusi So far I define a just as degrees. – jackjuni Oct 17 '18 at 14:01
@jackjuni Radians of course! To summarize, given $r$ and $l$ and a point on a sphere $P(x,y,z)$ with $x^2+y^2+z^2=r^2$ we can define a random $\vec u$ with $|\vec u|=1$ and find the new position $P'$ by $R\cdot (x,y,z)^T$. – user Oct 17 '18 at 14:03
@jackjuni It is meaningless define a direction for rotation by a vector $\vec a$ in degrees!The angle is defined once we have $r$ and $l$. To determine the new position we need to define a rotation axis or something else which is equivalent. – user Oct 17 '18 at 14:04
Let us continue this discussion in chat. – jackjuni Oct 17 '18 at 14:07

Moving a point on a sphere

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