1
$\begingroup$

Imagine that I am standing at a place on Earth, using coordinates of say N41 W74. Now the Earth's axis rolls 90 degrees, causing the N/S axis to become the equator, and rotation resumes as before. Points that formerly had been on the equator become the new north pole and south pole.

New lines of latitude and longitude would now be drawn to this new Earth.

What formulas do I need to determine the new coordinates of the place I was standing before? Is is possible to use decimal degrees for this formula?

thank you

$\endgroup$
  • $\begingroup$ "causing the N/S axis to become the equator": The north/south axis is a line segment, whereas the equator is a circle. It would take a significant deformation, not just a rotation, to turn a line segment into a circle. $\endgroup$ – joriki May 2 '13 at 3:44
0
$\begingroup$

It sounds like you take the earth as it is and start rotating about an axis that has two endpoints on the equator. I am assuming you are considering the earth to be spherical. You have not chosen what those endpoints are. So let us assume the new North pole is what was $0^\circ$ longitude (and zero latitude) and the new South pole was at $180^\circ$ longitude. I am measuring angles in degrees, as you asked. Now we have defined the location of the new North and South pole, which gives every point a latitude. We still have not defined the zero of longitude. If we put it through Greenwich again, the old North pole has longitude $\phi=0^\circ$ and the old South pole has $\phi=180^\circ$

This is a rotation of $90^\circ$ of the coordinates around the $y$ axis, or $-90^\circ$ rotation of the points, so the rotation matrix is $\begin {bmatrix} 0&0&1\\0&1&0\\-1&0&0 \end {bmatrix}$, so you take your point, which in Cartesian coordinates is $(R\cos \lambda \cos \phi,R\cos \lambda \sin \phi,R\sin \lambda)^T$ and it becomes $(-R \sin \lambda, R \cos \lambda \sin \phi, R \cos \lambda \cos \phi)^T$. The new latitude is $\arcsin(\cos \lambda \cos \phi)$ and the new longitude is $\arctan(\sin \phi \cot \lambda)$ where you have to worry about the quadrant of the longitude-you may have to add $180^\circ$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.