NavList:

A couple of quick thoughts.

If you assume a perfect sphere, then the radius of the Earth is 60·180/pi nautical miles. That's my favorite magic number: 3438. 

For your specific problem, the most important issue here is that the longitudes are nearly the same. This implies that you can already see the great circle path on a common globe. It's a line of longitude. And distance measured anywhere along that great circle is just difference in latitude. And this will also be nearly identical to the rhumbline distance (exactly identical if the difference in longitude is exactly zero). Even if we did not have this special case of nearly identical longitudes, the distance is short enough to guarantee that the great circle and the rhumbline path (and distance) are insignificantly different. As a rule of thumb (or a rhumb of thule in colder climes), if the distance is less than 1000 nautical miles, you can ignore the difference between great circles and rhumblines except at high latitudes.

Suppose you could bore a tunnel that's a true straight line from Auckland to Wellington. It would not be practical, but just for a super-tech game, let's go with it. Let's make the train float on friction-less wheels or a magnetic field with no resistance. And --now getting really impractical-- let's evacuate the tunnel so that it is a hard vacuum ensuring no air resistance. The train sits at the station in Wellington. It is not level, of course, and is angled slightly downhill toward Auckland, held at the station by a hook. Now un-hook. It will fall, very slowly at first, down that shallow incline towards its destination. At the mid-point, it will gliding along at a pretty good clip, but after that it will start to rise uphill and the train will decelerate. And suddenly this is a classic physics 101 puzzler: will this unpowered train reach its destination? How long will it take to get there? And how long would it take if we dug the same straight tunnel to Honolulu instead? Spoiler: it arrives at the end point just as its speed drops to zero and the trip takes 45 minutes (which for deep and fascinating reasons) is half the orbital period of a satellite in LEO (Low Earth Orbit).

Frank Reed
PS: This train biz is off-topic (and we should wind that down within 24 hours) but the parts about great circles and rhumblines and 3438 and all that are reasonably on-topic, I think. :)