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Lines of latitude are not geodesic except at the equator. So to follow perfect eastern course in the northern hemisphere in a vehicle a left turn is constantly required. We can express a turn as the radius of a circle that the turn would produce on the flat. So, at the pole the turning radius would be zero and at the equator the turning radius would be infinite. Near the pole the turning radius would be almost equal to the distance from the pole.

Does anyone know of an equation that would provide the turning radius at a given latitude required to follow that line of latitude?

I will try to clarify with some examples.

If I were to drive a car on an eastern course at a location of 100m south of the north pole, I could not drive in a straight line otherwise I would eventually be driving south. So, to maintain an eastern course, I must turn the car and circle the pole with an approximate turning radius of 100m.

If I were 200m south of the north pole the turning radius would be approximately 200m. However, the further south I am from the north pole the less turn is required and the more it differs from my distance to the pole.

At the equator no turn is required because I will be travelling a geodesic, the turning radius at this point is infinite (meaning no turn).

Is there a mathematical function (f) that given a latitude provides the turning radius required to preserve that latitude and maintain an eastern course?

r = f(l)

where: l - latitude, r – turning radius

I have performed extensive searches and cannot find an answer.

  • There are scores of duplicate posts on this topic. I suggest you review some of them. – Vince Aug 20 '17 at 2:22
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    I find your question very confusing. If I drive due East, I don't need to correct my course at all to follow the latitude line that I'm on. Do you instead mean how to follow a geodesic line? – mkennedy Aug 20 '17 at 3:18
  • I cannot find any posts that provide me with a method of determining the turning radius. – Gareth H Aug 20 '17 at 10:29
  • To keep a constant Eastern course you need to apply a turn correction. For example if I am in a car 100m south of the north pole, to travel east I would need to apply a turn radius of approximately 100m to circle the pole. – Gareth H Aug 20 '17 at 10:31
  • @gareth-h I believe that my answer provides you with the information you asked for. If you agree, please mark the answer accordingly. Thanks. – cffk Sep 4 '17 at 21:51
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Consider the circle of latitude φ; this has a radius R = a cosβ, where a is the equatorial radius and β is the parametric latitude. A cone with base radius R and apex semi-angle φ can be placed over the ellipsoid so that it is tangent to the ellipsoid at latitude φ. If the cone is unfolded and laid flat, then the base becomes a circle of radius R/sinφ.

The "turning radius" of a circle of latitude (the reciprocal of its geodesic curvature) is also R/sinφ. (The key here is that the geodesic curvature of a curve on a surface is an intrinsic property which does not depend on how the surface is embedded in 3-dimensional space.)

Two sanity checks: close to the pole sinφ → 1 and the radius of curvature is just R; on the equator sinφ → 0 and the radius of curvature is infinite.

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Uncharacteristically, I cut straight to a simplified version of cffk's answer (which I agree with). My answer, like that one, is in terms of a turning radius which is calculated equal to the radius of the earth (a near-constant 3,959 miles) divided by the tangent of the latitude. For example, at Latitude 30, the tangent is 0.577 which, divided into 3,959 miles, yields 6,857 miles turning radius (I picture a maypole over the near pole with a tether 6,857 miles long). At the pole, the turning radius calculates to zero and at the equator, infinity (straight ahead).

And my answer turns out to be cffk's, just folded into one equation through the trigonometric identity tan = cos/sin. That 30-degree maypole, by the way, would tower 3,959 miles above the pole. A POLAR maypole!

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