Science!

[dvandom]

Well, more like science, not SCIENCE!. But still. Back in grad school I worked out an explanation of the Coriolis force that didn't require using all sorts of complicated principles. It had holes, and I found a better explanation. That has stood until recently, when a guy in Germany started emailing me about a problem with it. I'm still not totally sure what he's getting at, but clearly it's time for another revision. And here it is! The part I need to re-explain in the most depth isn't done yet, I'm still building the props.

Edit: I borrowed a basketball from the lab equipment room and used it and some micro-Gundam figures to finish up my explanation.

Comments

[dvandom.livejournal.com]

The great circle IS a deflection, that's the point. Something sent East won't keep going East, it'll follow the great circle instead.

[foomf.livejournal.com]

No. On a stationary sphere, there is no north, south, east, or west. These are imposed by the rotational movement and an arbitrary assignment of north/south.

Oh, wait, I see what you're trying to say here.
You're applying 2-d thinking by constraining the arc. You can define any arbitrary arc you like, the start and end points defined by a straight line passing through the surface of the sphere and intersecting at start and end. You can tip the center of the ellipsoid defined by that curve - nothing requires it to conform to the great circles of the sphere.

You stand at point A and lob an intercontinental ballistic spitwad at point B which is a distance of 1/4 of a diameter (of the sphere) away.

You define "up/down" in terms of the center of the sphere: a line between you and the center. This means that 'up' is a different direction for you and your target. If you construct a plane that intersects the two vectors, you will define a great circle on the sphere. Without rotation, this is as good an equator as any other. No deflection.
This plane will contain the shortest arc between you and the target, although you can bend the arc as much as you want to by pulling its 'top' away from the surface of the sphere.

More to come in a moment as I add points of view.

[dvandom.livejournal.com]

A stationary sphere can still have an arbitrarily defined north, actually. To define any coordinates on a sphere, you need to define an axis so that you know where to measure angles from. It just happens in this case that the pole is as arbitrary as the prime meridian.

[foomf.livejournal.com]

But there's no preferred north in this case - with a two-point problem, you never have to move off the equator no matter where it is.

And with regards to coriolis force, all this geometry serves only to identify the vectors involved. And the amusing thing about that: there's a right-hand-rule governing the expansion of a triangle to a sphere. The bulge will always be to the outside, which (standing at any vertex and facing out) will map the points to the right of where they would be on the planar view.

[dvandom.livejournal.com]

Part of defining an arbitrary north is that you're establishing a great circle upon which you're standing. If you say you're shooting something north on a non-spinning sphere, it will keep going north. If you say, "That's north, now I will shoot this east" and turn 90 degrees before firing, your shot will start to curve to the right as it follows a great circle rather than following the arbitrary line you draw on the ground.

I doubt this issue will bother the target audience, but I'll put in an aside anyway. :)

[foomf.livejournal.com]

The arbitrary line you draw on the ground is also a great circle segment. It's illusionary to think it's flat; on a sphere, there is no flat. You can demonstrate it by drawing all this on a balloon, then inflating it.

If you have a transparent bowl and the balloon is too opaque to see through, you can demonstrate the 'flat' part by showing the flat balloon under the bowl, then inflate the balloon and draw the line on the outside. The curve becomes immediately obvious when you look from the 'start.


Yeah, I figured all this out in third grade when the teacher was trying to explain parallel lines and that they don't ever meet, except of course in a curved space-time, and she didn't understand that. This is what happens when you teach new math before you teach arithmetic - the kids can grasp really interesting concepts but cannot perform simple calculations in reasonable time.

[foomf.livejournal.com]

AHA! I just spotted the problem.

No, if you shoot east and are facing east, the shot will not curve. The curve, as mapped to your planar view, is still entirely vertical and thus appears to be a line.

Now if you are facing North and shoot something to the northeast, and manage to keep the trajectory visible, you will see the rightwards bulge, of course, since the center of the translation is north. And it will arc back in again.

[foomf.livejournal.com]

And incidentally, you can see this effect in jet contrails.

[foomf.livejournal.com]

OK, so on your stationary sphere you add point C, at some arbitrary distance, say, 1/4 of a diameter from A and B, because otherwise the shape of the triangle adds confusion. There's still no rotation, so still no preferred directions.

The shortest distance between these points is defined by the plane intersecting them, and is a triangle; it can be projected onto the sphere and the lines will curve in the process.
If viewed from the 'overhead' defined by the line between the center of the triangle and the sphere, and projected down onto the common plane, the lines will indeed appear to be curved away from the center.

If C watches this intercontinental spitball fight, and if A were using an intercontinental mole-machine spitball, the straight-line distance from C to the spitball would decrease until the right-angle, then increase again.
The straight-line distance to the intercontinental ballistic spitball gets away from simple right-angles. There's an added component from the projection. Without deriving this component I can't tell if the distance would actually increase to the top of the arc or simply decrease less.

[foomf.livejournal.com]

And at this point my ability to visualize this has become fatigued so I'm going to stop and get ready for work.