Given a sphere, I am attempting to calculate the coordinates of a point on that sphere given some initial point, the radius of the sphere, and some arbitrary distance d the point will travel (along with the vector it is headed). This sphere is a planet and the problem is essentially getting the coordinate of point #2 after traveling some distance.
I've been reading up on Spherical Trigonometry but I'm having a hard time applying it here. I'm thinking applying Rodrigues' formula might be a solution..?
I do have the North Pole position if that's any help at all.
CONTEXT
Planet is a perfect sphere with radius r. The coordinates are in a Cartesian plane with the center of the planet being (0, 0, 0). The "height" of the object relative to the surface of the sphere should remain the same.
That's how I understand the problem in mathematical language: We have a sphere with radius $r$ and a Cartesian coordinate system with origin in the center of the sphere. Given is a point with coordinates $x_1,y_1,z_1$ on the sphere, that means $$x^2_1+y^2_1+z^2_1=r^2.$$ This point is to move for a distance $d$ on the sphere, starting in a direction given by a vector $\overrightarrow{v}=(v_x,v_y,v_z)$. Of course, it's crucial that this direction be tangential to the sphere (otherwise the point would take off into space, or bury into the ground), i.e. $$x_1v_x+y_1v_y+z_1v_z=0.$$ We'll assume that the length of this vector is one, i.e. $$v^2_x+v^2_y+v^2_z=1.$$ Then, the movement of the point is a rotation in the plane given by the origin, the point and the direction $\overrightarrow{v}$, the rotation angle defined by the arc, $$\alpha=\frac{d}r.$$ Since the normal vector $\overrightarrow{n}=(x_1/r,y_1/r,z_1/r)$ and $\overrightarrow{v}$ are orthogonal, this will transform $\overrightarrow{n}$ into $\cos\alpha\,\overrightarrow{n}+\sin\alpha\,\overrightarrow{v}$, so the new position is $$r\,(\cos\alpha\,\overrightarrow{n}+\sin\alpha\,\overrightarrow{v})=(x_1\,\cos\alpha+r\,v_x\,\sin\alpha,y_1\,\cos\alpha+r\,v_y\,\sin\alpha,z_1\,\cos\alpha+r\,v_z\,\sin\alpha).$$ For a very small distance $d$, we have $\sin\alpha\approx\alpha=d/r$ and $\cos\alpha\approx1$, so this becomes just $(x_1+d\,v_x,y_1+d\,v_y,z_1+d\,v_z)$, as one would expect. If, instead, $d=\pi\,r$, i.e. we're travelling half the circumference, we have $\cos\pi=-1$ and $\sin\pi=0$, and the new position will be $(-x_1,-y_1,-z_1)$ irrespective of the direction $\overrightarrow{v}$.