Spherical coordinate transformation

Question

We have the following picture ($r=1$. $AB$ is the prime meridian):

We can find the coordinates of $C$ using:

$$ x = \sin(b)\cos(a)$$

$$ y=\sin(b)\sin(a)$$

$$ z = \cos(b) $$

I understand this geometrically, but then they do this:

So basically, the z-axis gets shifted in a way that $B$ is now the north pole. They go on to say that the new coordinates of $C$ are:

$$ x' = \sin(a)\cos(180-b) = -\sin(a)\cos(b) $$

$$ y' = \sin(a) \sin(180-b) = \sin(a)\sin(b) $$

$$ z' = \cos(a) $$

I only understand that $$ z' = \cos(a)$$, but I can't geometrically visualize why $x'$ and $y'$ are as stated (why the "180- .." ?)

They go on to confuse me even more, saying that this gives us:

$$ - \sin(a)\cos(B) = \sin(b) \cos(A) \cos(c) - \cos(b) \sin(c) $$

$$ \sin(a)\sin(B) = \sin(b)\sin(A)$$

$$ \cos(a) = \sin(b)\cos(A)\sin(c) + \cos(b) \cos(c) $$

which gives us

$$ \cos(a) = \cos(b) \cos(c) + \sin(b)\sin(c)\cos(A)$$

I don't understand this part at all, I don't understand where they got this from and I don't understand the random capitalization of the letters. Can someone help me understand this badly-written booklet.

Answer 1

Yeah, that is confusing.

What I see is that you have a positive rotation about the $y$-axis through an angle $c$. Use your favorite version of the right-hand rule to convince yourself. Then:

$$\left(\begin{matrix} \cos c & 0 & \sin c \\ 0 & 1 & 0 \\ -\sin c & 0 & \cos c \end{matrix}\right) \left(\begin{matrix} \sin b \cos a \\ \sin b \sin a \\ \cos b \end{matrix}\right) = \left(\begin{matrix} \sin b \cos a \cos c + \sin c \cos b\\ \sin b \sin a \\ -\sin b \sin c \cos a + \cos c \cos b \end{matrix}\right)$$

That would be how I'd handle the rotation. But that's not what we see in your manual, so I'm at a bit of a loss.

The capitalization is random, though. All that appears to be done in the last two equations that you wrote is they rearranged the terms (ignoring capitalization).

I'd seek other ways of describing the rotations if you can.

Answer 2

The first part of your question: it is just a rotation about y-axis by angle c and the new coordinates of C is given by the following picture. The angle between OC and z-axis is a and the angle between the projection of OC in the second quad of xy-plane and negative y-axis is b. Note that OC are not in the first quad

For the second part, it is the relation between the two coordinates as the answer of John did.

Answer 3

In spherical trigonometry, authors sometimes use capital letters to denote the interior angle of a spherical triangle (in a tangent plane to the sphere), and small letters to denote the spherical length of the opposite side, a.k.a., the angle subtended by the arc at the origin.

FWIW, I can't see your first set of equations for $(x, y, z)$. However, $$ x = \sin(b) \cos(A),\qquad y = \sin(b) \sin(A),\qquad z = \cos(b) $$ looks fine, since $A$ is the longitude of $C$ from the arc $AB$. Assuming the proposed interpretation is correct, the transformed coordinates should read $$ x' = \sin(a) \cos(180 - B),\qquad y' = \sin(a) \sin(180 - B),\qquad z' = \cos(a); $$ the positive $x'$ axis has been placed at $B = 180$, so $180 - B$ is the longitude of $C$.

The equations with capitalization look plausibly like the spherical law of cosines, though I haven't checked carefully.