Staying under Sun for 9 months straight

[exander]

TL;DR Summary: I am working on a story and would like helping figure out how a person can stay under sun for 9 months straight

Hello,

I am working on a story and I need to figure out how my protagonist can stay consecutively without sun setting over the horizon for at least 9 months.

My idea is to start at the North Pole:
1) Experience continuous daylight from March Equinox to September Equinox.
2) Before the September Equinox, arrive at the South Pole to experience continuous daylight from September Equinox to December Solstice.
3) Stay under sun during the whole time.

Most likely person can't stay exactly in the South or North Pole, so probably the closest city/station. So this would need to match.
I assume the transfer from the North Pole to the South Pole has to be quick, maybe spiraling around the planet to stay under the sun?

Can you offer ideas, calculation, how to best achieve this?

 

[DaveC426913]

Your protag doesn't need to stay directly at the poles. The sun stays above the horizon for long periods of time at latitudes much less than 90 degrees.

How is your protag getting around? Foot? Car? Boat? Plane? Each will have an effect on the path he can take. By plane, for example, he can theoretically travel to the antipodes in a half day of sunlight.

 

[exander]

Your protag doesn't need to stay directly at the poles. The sun stays above the horizon for long periods of time at latitudes much less than 90 degrees.

How is your protag getting around? Foot? Car? Boat? Plane? Each will have an effect on the path he can take. By plane, for example, he can theoretically travel to the antipodes in a half day of sunlight.

I know, but the further the poles, the shorted the times. He would most likely stayed the further north he could get and normally live there.

The question is what is the least technology he can get it, I will use it to set the Epoch. Currently, I don't have even hypothesis how to do it without a plane. And the question is how fast the plane needs to be. Or is there another way?

 

[Halc]

A train could get you from north to south, but it requires a track that nowhere exists.
So airplane it is.

Put your guy in high orbit. Problem solved.
Put him on the moon. He could then stay in the light even at the equator with a recreational vehicle, or if he's exceptionally athletic, a bicycle would do. But if he's close to a pole, walking is enough, and he doesn't even need to switch poles for the lunar winter.

 

[exander]

I actually thought about going to orbit, that pushes the story to be too futuristic. Lowest possible technology would be preferable. So it seems to be plane.

Could a ship theoretically do it as well?

 

[DaveC426913]

Put him on the moon. He could then stay in the light even at the equator with a recreational vehicle, or if he's

I think it was Joe Haldeman who wrote a story I've never forgotten. It was a about a crewpeson who got left behind on the Moon. The next ship was a month away. In order to stay alive she had to stay on the sunlit side of the Moon. So she had to hump it. That's 11,000km of hopping in 600 hours.

 

[DaveC426913]

Could a ship theoretically do it as well?

Probably not but it's hard to say.

Does the protag have control over when the ordeal occurs? eg. Can they choose optimial locations and time of year to their benefit? I'm sort of imagining it's essentially a bet - like Around the World in 80 Days.

The reason this is germane is because the protag will probably have to do the very same calculations we'e going to end up doing right here, before they can decide of it's even feasible enough to make the claim.

I'm starting with the assumption that we know they get from 90N to 90S in 12 hours - which requires a speed of 1,040mph. But can we do it any slower? (to make for better story).

If I could see this as an animation, that might give a clue:

Alternately, I'm toying with breaking the problem down to as simple a schematic as possible and using only geometry. But it's still a 3 dimensional problem: 2 space dimensions and one time dimension.

 

[exander]

Probably not but it's hard to say.

Does the protag have control over when the ordeal occurs? eg. Can they choose optimial locations and time of year to their benefit? I'm sort of imagining it's essentially a bet - like Around the World in 80 Days.

The reason this is germane is because the protag will probably have to do the very same calculations we'e going to end up doing right here, before they can decide of it's even feasible enough to make the claim.

I'm starting with the assumption that it is possible to thread a path through both space and time that is slower than the obvious 'get from 90N to 90S in 12 hours'.

If I could see this as an animation, that might give a clue:
View attachment 351901

Protag cannot change the start, but We can assume that he is already in the north during the polar day for at least a month. He can plan when he leaves to make it to the south.

The task can be simplified: protag is in polar day in the north and has to move into polar day in the south without the sun setting behind the horizon for him. How many months he spends in the north and in the south are actually not important. The question is how to make it between these two regions.

I could imagine sailing west from the north, We can take Norway as the point of origin for example. I would assume 18, maybe even 20 hours is there to make it (maybe even longer coming between north and South America), because he is moving west (like Around the World in 80 Days).

 

[Baluncore]

And the question is how fast the plane needs to be.

Simple and uneconomic, without optimisation. The circumference of the arctic circle is; sin(23.5°) = 40% of the equator. You could travel east to west along the arctic circle, at 664 km/h in a plane.

 

[DaveC426913]

I could imagine sailing west from the north, We can take Norway as the point of origin for example. I would assume 18, maybe even 20 hours is there to make it (maybe even longer coming between north and South America), because he is moving west (like Around the World in 80 Days).

The problem is that any path that is not due north-south results in a greater distance to travel, which means they have to go that much faster - which pushes up your minimum allowable speed/technology - not much but a little.

 

[exander]

Simple and uneconomic, without optimisation. The circumference of the arctic circle is 40% of the equator. You could travel east to west along the arctic circle, at 664 km/h in a plane.

But, you would need to do the whole at least 3 months? Because once polar day is over there then you need to circle or sun sets behind the horizon. Moving to the South Pole seems the only option to me.

 

[DaveC426913]

Simple and uneconomic, without optimisation. The circumference of the arctic circle is; sin(23.5°) = 40% of the equator. You could travel east to west along the arctic circle, at 664 km/h in a plane.

Wait. Isn't it a requirement that they start near one pole and end up near the other?

I mean, if they can just stay in the northern hemisphere, they could do it pretty easily, no? They wouldn't even need a jet.

 

[exander]

Wait. Isn't it a requirement that they start near one pole and end up near the other?

I mean, if they can just stay in the northern hemisphere, they could do it pretty easily, no? They wouldn't even need a jet.

No, no requirement. I just don't have other idea. You have idea how to stay 90 days under sun only on north hemisphere?

 

[Baluncore]

Moving to the South Pole seems the only option to me.

If you must make one move only, it should be done pole to pole over the equinox.

Due N-S, in a 12 hour day, requires 20,000 km in 24 hours = 833.4 km/hour.

 

[DaveC426913]

Man,

Due N-S, in a 12 hour day, requires 20,000 km in 24 hours = 833.4 km/hour.

Doesn't it have to happen in 12 hours, not 24? That's how I got 1040mph(1675km/h).

 

[DaveC426913]

I'm trying to imagine a path that stays in the northern hemisphere, but does less than rotational speed of the Earth - per Baluncore's post #9.

Man it's hard to wrap my head around sunlight on a rotating body. Is there away to make an easier model?

 

[Baluncore]

Deosnt it have to happen in 12 hours, not 24? That's how I got 1040mph.

My error, 20 Mm / 12 hr = 1667 km/h.

 

[Baluncore]

I'm tring to imagine a path that stays in the northern hemisphere, but does less than rotational speed of the Earth.

That is a permanent flight along the arctic circle in winter.

 

[DaveC426913]

That is a permanent flight along the arctic circle in winter.

As you mentioned in post 9, yes. I just can't come to terms with that being the best (i.e. slowest ) he can do.

 

[Baluncore]

Start at the arctic circle, fly in a spiral for 45 days, at an ever decreasing speed, until you stop at the N-pole on the day of the vernal equinox. Wait 6 months for the autumnal equinox, then spiral out to the arctic circle in 45 days. That gives nine months of continuous sunlight in the Northern Hemisphere.

 

[DaveC426913]

Start at the arctic circle, fly in a spiral for 45 days, at an ever decreasing speed, until you stop at the N-pole on the day of the vernal equinox. Wait 6 months for the autumnal equinox, then spiral out to the arctic circle in 45 days. That gives nine months of continuous sunlight in the Northern Hemisphere.

OK, that still requires a speed of 664km/h for at least some portion of the journey, yes? So that still sets the same lower bound on the technology required.

A pity it makes for a too simple plot vehicle.

 

[Halc]

The problem is that any path that is not due north-south results in a greater distance to travel, which means they have to go that much faster - which pushes up your minimum allowable speed/technology - not much but a little.

Slower, not faster. Yes, the path is longer, but the diagonal (more like the shape of an integral sign) trip between poles buys many more hours to add to the 12, at a cost of only a little more distance.

We presume the guy isn't able to keep a plane in the air for 45 day, even though the tech to do so hypothetically exists. Yea, they can refuel in air, but can any engine really run that long?

No, finding the most efficient route between poles is what's needed. Can we keep the trip subsonic? The problem isn't one of calculus. It can be done with simple algebra.
I withdraw that. It's a calculus thing.

 

[Baluncore]

OK, that still requires a speed of 664km/h for at least some portion of the journey, yes?

Now I see that, there is no need to start at the arctic circle, since only 1.5 months = 45 days appears at each end of the 6 month sun-up.
The declination of the Sun will then be -16.5°, not -23.4°. So the first and last circle latitude will be at 90° - 16.5° = 73.5°.

... finding the most efficient route between poles is what's needed.

True, and the flight must be centred on the day of the equinox.
The impossible ideal transfer flight would be the minimum distance at infinite speed.

The Sun is up, on the equinox, for only one minute, at noon, at the poles. That restricts both the time of departure, and the time of arrival to noon, which for a 12-hour flight would be on the opposite side of the Earth. Departure from the North Pole would be towards the Sun, due south obviously, with arrival at the South Pole, away from the Sun, from due north.

The problem comes down to the 'S' shaped path that connects the poles and crosses the equator. That flight will be continuously under the Sun, fixed by the noon departure and arrival times, on the same day. The flight duration is freely variable, but must be at a speed greater than the equatorial velocity of 40 Mm / 24 h = 1667 km/h, since it must cross the equator, at noon, under the Sun.

 

[Orodruin]

TL;DR Summary: I am working on a story and would like helping figure out how a person can stay under sun for 9 months straight

Most likely person can't stay exactly in the South or North Pole, so probably the closest city/station.

https://en.m.wikipedia.org/wiki/Amundsen–Scott_South_Pole_Station

 

[exander]

That was established November 1956, so that also puts some significant bounds on the story.

 

[DaveC426913]

The Sun is up, on the equinox, for only one minute, at noon, at the poles.

There is a little wiggle room in a real world that the protag may be able to leverage.

The sun is not a point, so the time it is visible is a hair longer than exactly 12 hours.
This can be expanded by taking advantage of altitude, which makes the sun set later and rise earlier.

In my opinion these are the bits that make the story. The threading of needle, as it were. The tweaking of 'It's a preposterous claim!' onto the 'And yet here I stand having done it.'.

 

[exander]

There is a little wiggle room in a real world that the protag may be able to leverage.

The sun is not a point, so the time it is visible is a hair longer than exactly 12 hours.
This can be expanded by taking advantage of altitude, which makes the sun set later and rise earlier.

In my opinion these are the bits that make the story. The threading of needle, as it were. The tweaking of 'It's a preposterous claim!' onto the 'And yet here I stand having done it.'.

Yes, I was actually thinking if balloon or something on the poles can give you longer daylight. But I am unsure how to incorporate it. I would like the way to be interesting and scientifically accurate on this planet.

 

[Vanadium 50]

Do it slower? Take a helical path.

Note that if you have a conveyance that will take you from the N to the S pole in under 12 hours, you also have one that will take you around the world in under 24. You don't need to mess with the poles.

Is this your first story? Lots of first timers come here thinking The Only Really Important Thing Is Setting. The market disagrees.

 

[Baluncore]

Do it slower? Take a helical path.

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).

There is a little wiggle room in a real world that the protag may be able to leverage.
The sun is not a point, so the time it is visible is a hair longer than exactly 12 hours.

I agree there is some wiggle room, but not much.

During the flight, there must be an average longitude rate of 15° west per hour. At the poles the sunlight will be very short, so direction will be critical.

When it comes to crossing the equator, there will be 12 hours of daylight. The lowest fixed-speed flight appears to be one which departs the North Pole at noon, before the equinox. Then as the day lengthens, it gets ahead by following the line of dawn, before making a dash across the equator to the line of dusk, and then moving in to arrive at the South Pole, as the line of dusk merges with the noon meridian.

A cardinal point, in the optimum flight, would be to cross the Equator at the instant of the equinox.

 

[Vanadium 50]

That was established November 1956, so that also puts some significant bounds on the story.

I hope this isn't "guess my constraints". That is a game with no winners,

No, it can't be done in 1956. It can't be done today.

There is one (western) aircraft capable of polar operations, the LC-130. Even it cannot operate dawn-to-dusk. So that ends that.

Even if it could, it's slow. It takes about 4 hours minimum to make it to McMurdo, leaving you ~8 hours to go 12000 miles to the North pole. That's 1500 mph for the whole trip and there is no aircraft that can do this. The SR71 was close, but it will need to refiel 6 or 7 times.

And of course, it can't land at the pole. Your hero will need to bail out.

 

[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.