Staying under Sun for 9 months straight

[Halc]

the problem remains, how do you cross the equator while maintaining 15° of longitude per hour.

Don't have to. You have 12 hours of free time. You can cross the equator at minimal longitude per hour. Get it over with ASAP.

At the poles the sunlight will be very short, so direction will be critical.

On the day of the equinox, the day is 12 hours long everywhere. The day is not shorter near the poles, except exactly at the poles, where the sun is at the horizon at all times.

So you start north and spiral westward at say 600 mph, arriving at latitude 57 at dawn. This is as far south as you can go and keep up with dawn. Optimal path from there is curved, but sub-optimally, you can go diagonal southwest for 18 hours. That's 114 degrees south and 107 west.
The south component gets you close enough to the south pole that you can outrun dusk. The 107 west buys 7 hours on top of the 12 of daylight, for 19 total hours of daylight, one hour to spare.

The above calculation is crude and wrong. The real path is actually more efficient than that since the westward progress is greater at both ends. I assumed a cylindrical world where a straight path would be optimal.

It can probably be done at about 500 mph, pole to pole, always in the sun. I'm not paid enough to write a program to simulate the optimal path at the slowest speed.

 

[DaveC426913]

It can probably be done at about 500 mph, pole to pole, always in the sun.

I guess the upshot is this: if any point of the journey requires a speed of 500mph, then that is the slowest mode of transport you'll need.

It can't be done by train or boat alone.

Still, it might be a cooler story if the modes of transport are meted out as parsimoniously as possible - a little like AtWi80D, but that's not my call.

I'm not paid enough to write a program to simulate the optimal path at the slowest speed.

You know what? I'm feeling generous. Let's double your usual rate for writing programs for PF!

 

[Baluncore]

Don't have to. You have 12 hours of free time. You can cross the equator at minimal longitude per hour. Get it over with ASAP.

You have misquoted me. My actual statement was:

That would be nice, but the problem remains, how do you cross the equator while maintaining 15° of longitude per hour. (A possible solution follows).

You have then repeated a crude version of the solution I gave later in that same post.

... by following the line of dawn, before making a dash across the equator to the line of dusk,...

On the day of the equinox, the day is 12 hours long everywhere. The day is not shorter near the poles, except exactly at the poles, where the sun is at the horizon at all times.

The day is shorter in "kilometres of longitude" near the pole. So the speedy flight must depart the pole directly towards the Sun. Anything else fails immediately, in the mathematical computer model, and puts you on the wrong part of the face of the Earth to reach and follow the line of dawn.

It can probably be done at about 500 mph, pole to pole, always in the sun. I'm not paid enough to write a program to simulate the optimal path at the slowest speed.

There is a "lowest steady speed" at which it is possible. I want to know that exact numerical value, modelled on the WGS 84 ellipsoid, not a crude guess.
Have you even considered what type of algorithm you would employ?

 

[Halc]

You have then repeated a crude version of the solution I gave later in that same post.

OK, we're mostly on the same page. I did it in mph (mostly because Earth spins at not much over 1000 mph), but I actually prefer metric. It would be funny if the minimum speed turned out to be exactly half that.

So the speedy flight must depart the pole directly towards the Sun.

Why? Why not follow the line of dawn? Much shorter that way. You can depart in any direction towards the light, which is 180 deg of choice, but following the dusk line is just silly.
The computation can stop at the equator with the sun directly overhead. Symmetry gets you the rest of the way.

Have you even considered what type of algorithm you would employ?

Guess at the speed required. Follow the dawn line until you can't, then follow a trajectory that maximizes the ratio of drop in latitude vs change in local sun position. This is dangerous right when you leave the dawn line since the optimal path is a circle (ratio 0/0), getting you nowhere. You have to initially angle down at some minimal angle else the path goes forever.

Compute where the sun is when the equator is crossed, and adjust speed guess accordingly, until it's narrowed down.

The path obviously curves, compared to my crude straight geodesic, but that was just a ballpark figure.

My method assumes an aircraft up to the task, refueled in-flight if necessary.

 

[Baluncore]

I think we agree that from the North Pole, the flight must get onto the moving dawn line early, then somehow transition into the equator crossing run, before finishing on the moving dusk line at the South Pole.

Ending the flight on the dusk line is inherent in the symmetry of the problem, as is crossing the equator at the instant of the equinox.

All possible optimum geodesics share only one known point, that is fixed in space and time. The model should begin the computation at that point, on the equator, at the instant of the equinox, at the longitude where the Sun is on the meridian for that year's equinox.

The route could be called crepuscular, if it wasn't for (the mad dogs and Englishmen) crossing the equator under the midday sun.
You can trust us all day at Crepusculair. We will fly you from dawn to dusk.

It does appear to be an optimum route to avoid vampire activity. Maybe the vampires could time-share the polar accommodation, by employing the antipodean, reciprocal, "red-eye" overnight route. Maybe they already do.