And of course resolves the problem of implementing that moon armor idea from yesterday.
I dunno, ladders don’t have a good record for space exploration (SFW):
(page 1) Ladder 1 (page 2)Ladder 1
The rare xkcd that I only understood after reading the alt-text. And then a sensible chuckle.
Now that’s funny. Your comment I mean.
I got the idea of the dialog instantly, before I even got to the last word, but had to puzzle for awhile to get the point of the hover-text.
Brains are weird. And evidently idiosyncratic.
Yep. For me, the idea of map projection problems being about the ratio of area scaling was just too ingrained for the original comic to hit immediately. But somehow the idea in the alt-text of “well yes, that’s the only place where you get a true 1:1 scale” explained the original comic retroactively.
Must resist urge to compute the width of the ring…
Well, it’s not too hard. If the map is a meter across, then the scale ring is a meter in circumference. The thickness of the ring depends on what you consider “drastically,” but if you want 10% error, then the ring would have an inner circumference of 0.9 m and an outer of 1.1, which works out to a width of 3.2 cm.
I don’t think that’s correct… Remember, we’re in a hugely nonlinear regime, here. The y coordinate of a Mercator projection map is proportional to the tangent of the latitude. Even a micron-wide band inside the perfect-fit ring would result in an area on the map much greater than the area of the entire circular cap on the real Earth.
Oh, and speaking of which, a Mercator projection isn’t actually rectangular. It’s infinitely tall.
Locally, it’s still quite linear. That’s one of the biggest advantages to the Mercator projection: you can zoom in anywhere, and squares always look like squares (the aspect ratio stays 1:1). So relatively small amounts of error will have the same magnitude in the projection vs. not.
It gets more dicey if “drastically” means like 1000% vs. 110%. Which is why I didn’t assume that 
.
Yes, that’s the difficulty. To compute the width accurately, we need to take into account the distance on the map of the ring from the equator.
Any sufficiently-small square still looks like a square. But up that close to the pole, “sufficiently small” is very small indeed.
And 3.2 cm is very small indeed compared to the circumference of Earth.
The equation for the y coordinate is y=R\ln[\tan(\frac{\pi}{4} + \frac{\phi}{2})], where \phi is latitude. R is the radius of the globe, so for a 1 m circumference it comes to 0.16. If the ring’s centerline radius is at 0.16 m, then I claim the outer radius for 10% error is at 0.176 m. The latitudes for those are 1.57079632279 and 1.57079632239 respectively. That comes to y-values of 3.0937 and 3.0784 meters. The difference is 0.0153 m, which is very close to the 0.016 m I added in physical space.
I’m actually mildly surprised we only get a 6 meter tall map here, but that ln is doing a lot of work.
I’m reminded of Steven Wright’s deadpan line:
I bought a map of the world. Full-size, one inch equals one inch.
Here, I’ll unfold it…
Stop pandering to those nutso flat earthers with this projection map stuff. We’re on a globe, an oblate sphereoid dammit.
/unhinged rant off
I would imagine that the biggest problem isn’t how big the circle is, but rather exactly where it is.
The funny thing is that if we ignore the spherical / flat conversion issue, it would only take 39 folds in half to reduce the area of the folded map to 1m square. So something a human could easily carry.
If we ignore its immense weight and thickness. And the difficulty near impossibility of folding anything in half that many times.
I get 49 folds. The surface area of the earth is 510 trillion m2, so 39 folds will get you down to 928 m2, and 49 folds to 0.9 m2. The Mercator projection is about 3x the area, so you’d need 50-51 folds to get to 1 m2.
A quick google says the thinnest paper is 0.02 mm thick and 70 g per m2, so at 49 “folds” it would be 11.3 million km thick – 30x the distance to the moon – and weigh 35.7 trillion kg.
Exponential growth is a beast.
I get log (510*10^9)/log(2)\approx 39
Brian