Maths check again?

[ffutures]

Here's a passage from Weinbaum's "The Mad Moon" describing Io

"The whole little planet was mad—loonies, parcats, slinkers and Grant Calthorpe—all crazy. At least, anybody who ever ventured outside either of the two polar cities, Junopolis on the north and Herapolis on the south, was crazy. One could live there in safety from white fever, but anywhere below the twentieth parallel it was worse than the Cambodian jungles on Earth."

Now, if I'm reading this right it means that the danger zone extends to 20 degrees either side of the equator, and the rest of the moon is OK. Which means that they aren't actually short of space.

The trouble is that my geometry is so rusty that I can't remember how to work out what proportion of the surface is inside and outside these zones. I've worked out that since Io is 3660 Km in diameter, about half the size of Mars, the surface area of the whole moon is a little over 42 million square kilometers - about 42,083,519 to be precise. Anyone able to do the rest of the calculation?

Later I've got three different answers, all of which say that the safe zones are each bigger than the USA!

And I've just realised that this description doesn't work at all with a tidally locked world, despite Weinbaum actually giving the length of day as the same as the orbital period. Something really does not compute...

Later still Unless it rotates in the opposite direction to its orbit, of course.

Comments

[nelc.livejournal.com]

Instinctively, I'd guess that the proportion is (70/90)^2, in which case the answer is 25,457,931 sq.km in the safe zone. But I wouldn't swear to it until I've figured out the proof, or found an explanation elsewhere.

[ffutures.livejournal.com]

Looks like you're in the right ball park - I wasn't looking for a completely accurate answer, if I go with "roughly a third" for the equatorial zone and a third each for the N and S areas it's probably close enough.

[very-true-thing.livejournal.com]

http://www.mathlinks.ro/viewtopic.php?t=199613

So the area of the surface above 20° is S = 2πrh where h is r * (1 - sin(20°)). Double this to get both north and south, then subtract from the total area to find the area between 20° north and 20° south.

With a bit of canceling and combining the equation is 4πr²sin(20°). Which in hindsight is obvious.

So roughly 34% of the total area is inside the danger zone.

[ffutures.livejournal.com]

Not obvious to me, squire, I failed O-level maths three times before I passed.

Many thanks - so we get about a third of the planet as equatorial zone, and the other two-thirds as the safe polar zones, or about 14.3 million square km in the danger zone, 13.9 million square km in each of the polar zones. Really can't believe this is what Weinbaum intended - it's more area than the USA!

[pauldrye.livejournal.com]

The phrase you're looking at is "surface area of a zone", for which the formula is:

2 * Pi * radius * height

height has to be 670km (since the radius is 1830km and we have two angles, 20 degrees and 90 degrees). So one zone from equator to 20 degrees is 7,700,000 square kilometers (actually 7,703,813).

Twice that is 15,400,000 square kilometers.

[ffutures.livejournal.com]

Thanks - so the safe areas are pretty big.

[robertprior.livejournal.com]

I suppose he might have meant 20% from the pole. That's not how we use the term in geography, but it's a mistake my students sometimes make (they think the measurement is 20% away from whatever they are interested in).

More likely he just didn't do the math.

[ffutures.livejournal.com]

That was my idea too. Still a moderately large bit of territory, I think. Say Texas?

[armb.livejournal.com]

And if only the relatively small bit around the poles is habitable, it's perhaps more likely that people on Io would generally measure from them.
Though you would hope they didn't use the same word "parallel" or "latitude" if they were using it the other way around from Earth conventions.