[Self] The horizontal curvature of the line of latitude 85° on Earth (perfect sphere)

This was something I did in another reddit thread, and I thought was pretty cool (not linked to avoid potential anti-brigading rules). I just used the second definition of curvature on wikipedia https://en.wikipedia.org/wiki/Curvature . The calculation is easily adaptable for other latitudes too (just replace pi/36 with whatever you want). Notably, it is infinite at the north and south poles and zero at the equator, just as you'd expect.

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If we let the radius of the earth be R, and assume for the moment that it's a perfect sphere (ie: ignoring the flatening at the poles and any mountains), then we have earth is the sphere of radius R centered at the origin.

The latitude 85° line is given by the equation in terms of x:

(sin(x)sin(pi/36) ; cos(x)sin(pi/36) ; cos(pi/36) ) * R which we will simplify to L(x) * R

With this parametrisation, x is moving at speed R * sin(pi/36) everywhere on the curve

Now given some y, the tangeant plane at the point L(y) * R is given by T(y) = {z: L(y).z = R} where '.' refers to scalar product.

Since we are only interested in the horizontal curvature, we project the latitude 85° line into T(y) to eliminate the downwards curvature, which gives us:

 L(x) * R + (R - (L(x) * R).L(y))L(y) 

= [ L(x) + (1-L(x).L(y))L(y) ] * R

For simplicity I'll take y=0. Although note that the curvature remains the same at all points on this curve due to symmetry so we should get the same result for all values of y; the only difference is y=0 is easier to work with.

 L(0) = ( 0 ; sin(pi/36) ; cos(pi/36) ) 

 L(0).L(x) = cos(x) * sin(pi/36)^2 + cos(pi/36)^2 

so the projection of the curve onto the T(0) is:

[ L(x) + (1-L(x).L(0))L(0) ] * R
= [ L(x) + (1 - cos(x)sin(pi/36)^2 - cos(pi/36)^2)  ( 0 ; sin(pi/36) ; cos(pi/36) ) ] * R
= [ (sin(x)sin(pi/36) ; cos(x)sin(pi/36) + sin(pi/36) - cos(x) * sin(pi/36)^3 - cos(pi/36)^2 * sin(pi/36) ; cos(pi/36) + cos(pi/36)-cos(x) * sin(pi/36)^2 * cos(pi/36) - cos(pi/36)^3 ) ] * R
= L(0) * R +  [ (sin(x) * sin(pi/36) ; (cos(x) - 1) * sin(pi/36) * cos(pi/36)^2 ; (1 - cos(x)) * sin(pi/36)^2 * cos(pi/36) ) ] * R
= L(0) * R + [ (sin(x) ; (cos(x) - 1) * cos(pi/36)^2 ; (1 - cos(x)) * sin(pi/36) * cos(pi/36) ) ] * R * sin(pi/36)

Now we differentiate by x to find the tangeant of this new curve, which means removing all terms that do not depend on x, replacing all instances of cos(x) with -sin(x) and replacing all instances of sin(x) with cos(x):

[ (cos(x) ; -sin(x) * cos(pi/36)^2 ; sin(x) * sin(pi/36) * cos(pi/36) ) ] * R * sin(pi/36)

We need to normalise by the speed which was R * sin(pi/36) in the sphere, and hence it is also that in the projection when close to the x=0. This gives us the unit tangeant at x when x is close to 0:

 (cos(x) ; -sin(x) * cos(pi/36)^2 ; sin(x) * sin(pi/36) * cos(pi/36) ) 

The curvature is found by differentiating by x at x=0 (and dividing by the speed again) and taking absolute value:

|| (-sin(0) ; -cos(0) * cos(pi/36)^2 ; cos(0) * sin(pi/36) * cos(pi/36) ) || / (R * sin(pi/36))
= || (0 ; - cos(pi/36)^2 ; sin(pi/36) * cos(pi/36) ) || / (R * sin(pi/36))
= ( cos(pi/36)^4 + sin(pi/36)^2 * cos(pi/36)^2 ) / (R * sin(pi/36))
= cos(pi/36) ^2 / (R * sin(pi/36))
=~ 11.39/R
=~ 0.000001788 m^-1

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Therefore the curvature of this line is 0.000001788 m^-1. For comparison, the same path on a flat earth (meaning the path at the same distance from the north pole) would have curvature 0.000001799 m^-1