1

You may simplify Earth as a sphere with 40 Mm circumference.

I thought about it when I read the Mercator projection has straight lines for constant bearing courses. You'd spiral around the north pole infinite times, and the line on the Mercator projection is infinite, but I suspect the actual distance is finite. I haven't tried to calculate that at all.

darsie
  • 113
  • What are your thoughts? What have you tried? Where does this question come from? – Christoph Feb 16 '19 at 13:28
  • @Christoph I thought about it when I read the Mercator projection has straight lines for constant bearing courses. You'd spiral around the north pole infinite times, and the line on the Mercator projection is infinite, but I suspect the actual distance is finite. I haven't tried to calculate that at all. – darsie Feb 16 '19 at 13:32
  • Please add this information (your response to Christoph) to the body of the question, by clicking the tiny edit. Otherwise this post is likely to be closed due to lack of context via community votes. – Lee David Chung Lin Feb 16 '19 at 15:33

1 Answers1

2

For a sphere of radius $1$ centered at the origin in $\mathbb R^3$, you get a curve \begin{align*} x(t) &= \frac{\cos t}{\sqrt{\sinh^2 t + 1}}, \\ y(t) &= \frac{\sin t}{\sqrt{\sinh^2 t + 1}}, \\ z(t) &= \frac{\sinh t}{\sqrt{\sinh^2 t + 1}}, \\ \end{align*} with time $t\in[0,\infty)$.

This curve has finite length $$ L = \int_0^\infty \sqrt{\dot x(t)^2+\dot y(t)^2+\dot z(t)^2}\,\mathrm dt = \frac{\pi}{\sqrt 2}. $$

Multiply by the radius of the earth to obtain the real length of the trip.

Christoph
  • 24,912
  • Ok, so that's 14,153 km instead of the direct 10,000 km. thanks. A general answer (any angle, any starting point) would trump this simple and elegant answer. – darsie Feb 16 '19 at 15:13