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The tricky part is to find the nearest point on a meridian to a given point P. Let's fix the meridian M at φ=0, to keep the algebra simple.
Suppose P has spherical coordinates (θ, φ), with 0 ≤ θ ≤ π and -π < φ ≤ π. Let C be the great circle through P which is perpendicular to M; then we are looking for an intersection of C and M. (There are two such intersections; in the end we will simply choose the one that is on the same side of the equator as P).
The normal of C meets the sphere on the meridian M, at A = (ρ, 0), say. P is on C, so OP is perpendicular to OA, where O is the centre of the sphere. In cartesian coordinates, with r = 1 for simplicity:

A = (sin ρ, 0, cos ρ)
P = (cos φ sin θ, sin φ sin θ, cos θ)

The scalar product is

A.P = sin ρ cos φ sin θ + cos ρ cos θ

This must be zero, giving

tan ρ = -1/(cos φ tan θ)

The required point, Q say, is on M and perpendicular to A, so if Q = (σ, 0) in spherical coordinates, we have

tan σ = -1/tan ρ = cos φ tan θ

So

Q = (arctan (cos φ tan θ), 0)

If this Q is on the wrong side of the equator, take

Q = (-arctan π - arctan (cos φ tan θ), π)
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The tricky part is to find the nearest point on a meridian to a given point P. Let's fix the meridian M at φ=0, to keep the algebra simple.
Suppose P has spherical coordinates (θ, φ), with 0 <= ≤ θ <= ≤ π and -π < φ <= ≤ π. Let C be the great circle through P which is perpendicular to M; then we are looking for an intersection of C and M. (There are two such intersections; in the end we will simply choose the one that is on the same side of the equator as P).
The normal of C meets the sphere on the meridian M, at A = (ρ, 0), say. P is on C, so OP is perpendicular to OA, where O is the centre of the sphere. In cartesian coordinates, with r = 1 for simplicity:

A = (sin ρ, 0, cos ρ)
P = (cos φ sin θ, sin φ sin θ, cos θ)

The scalar product is

A.P = sin ρ cos φ sin θ + cos ρ cos θ

This must be zero, giving

tan ρ = -1/(cos φ tan θ)

The required point, Q say, is on M and perpendicular to A, so if Q = (σ, 0)0) in spherical coordinates, we have

tan σ = -1/tan ρ = cos φ tan θ

In spherical coordinates:

So

Q = (arctan (cos φ tan θ), 0)

If this Q is on the wrong side of the equator, take

Q = (-arctan (cos φ tan θ), π)
show/hide this revision's text 2 added 19 characters in body

The tricky part is to find the nearest point on a meridian to a given point P. Let's fix the meridian M at phi=0, φ=0, to keep the algebra simple.
Suppose P has spherical coordinates (theta, phi), θ, φ), with 0 <= theta θ <= pi π and -pi π < phi φ <= pi. π. Let C be the great circle through P which is perpendicular to M; then we are looking for an intersection of C and M. (There are two such intersections; in the end we will simply choose the one that is on the same side of the equator as P).
The normal of C meets the sphere on the meridian M, at A = (rho, ρ, 0), say. P is on C, so OP is perpendicular to OA, where O is the centre of the sphere. In cartesian coordinates, with r = 1 for simplicity:

A = (sin(rho), sin ρ, 0, cos(rho)) cos ρ)
P = (cos(phi) sin(theta), sin(phi) sin(theta), cos(theta)) cos φ sin θ, sin φ sin θ, cos θ)

The scalar product is

A.P = sin(rho) cos(phi) sin(theta) sin ρ cos φ sin θ + cos(rho) cos(theta) cos ρ cos θ

This must be zero, giving

tan(rho) 



tan ρ = -1/(cos(phi) tan(theta))
1/(cos φ tan θ)



The required point, Q say, is on M and perpendicular to A, so if Q = (sigma, σ, 0), we have



tan(sigma) 



tan σ = -1/tan(rho) 1/tan ρ = cos(phi) tan(theta)
cos φ tan θ



In spherical coordinates:



Q = (arctan (cos(phi) tan(theta)), cos φ tan θ), 0)



If Q is on the wrong side of the equator, take



Q = (-arctan (cos(phi) tan(theta)), pi)
cos φ tan θ), π)
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