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October 23

Two degrees off course

If I were two degrees off course, flying from Los Angeles to New York City, where would I land (north and or south) . In addition, if I were two degrees off course while flying to the moon, by how much would I miss my objective.

~~Thomas A. Coss (email removed) —Preceding unsigned comment added by 68.4.244.159 (talk • contribs)

This depends entirely of the distance you are going to travel and which direction the two degrees are relative to (example, two degrees CW or CCW from north). ☢ Ҡi∊ff⌇↯ 03:05, 23 October 2006 (UTC)[reply]

In the simplest aproximate way of presenting the answer it would be (sin 2º = 0.035) * 2500 miles = 87 miles for NY to LA. Just change the distance for the error for any other 2º deviation --hydnjo talk 05:20, 23 October 2006 (UTC)[reply]

Problem with this page

I just noticed that the first eight topics on this page have been duplicated. This seems to have been caused by RefDeskBot. --KSmrq^T 14:09, 23 October 2006 (UTC)[reply]

Eccentricity question

Is the eccentricity of a straight line 1 or infinity? I realize the article claimed it is 1 (I changed it recently, before realizing I should have asked for some sources), but I can't find another source via a brief Googling that corroborates or denies it. Based on my experience (mostly dealing with orbital mechanics, and thus a different interpretation from traditional mathematics), the defining equations, for any value of k, all give a value of 1 for an ellipse with a minor axis of 0 (hence a line). It may be possible it's just defined as such (presumably to differentiate it from a parabola), but again, I couldn't find a source that didn't apparently mirror Wikipedia.

I'd say you're right – it's ∞. I just replaced the first image in the article with a vector version (which I didn't make, it was already lying around) and the new one agrees with you. —Bromskloss 18:25, 23 October 2006 (UTC)[reply]

Oh, another thing. In what way would mathematics and orbital mechanics treat the subject differently? —Bromskloss 18:28, 23 October 2006 (UTC)[reply]

I'd say it is somewhat indeterminate, mathematically. On the one hand, considering the family of parabolas y = λx², all have e = 1 if λ ≠ 0. It would be reasonable then to extend this to the case λ = 0, a straight line. On the other hand, considering the family of hyperbolas μx²−y² = μ², as the parameter μ tends to 0 the eccentricity e = √(1+1/μ) tends to infinity, while the hyperbola does what it can to degenerate to a straight line. --Lambiam ^Talk 19:14, 23 October 2006 (UTC)[reply]

Sorry, I accidentally wrote that the article used to say 1, when I meant infinity. So everything Bromskloss wrote should say "dis"agree =O). In another interpretation I thought of, if one uses a conic section visualization, a line is really just the parabolic section passing through the (conic) origin, in which case the eccentricity would be one (assuming the eccentricity is proportional to the normalized slope of the plane). In this case an infinite eccentricity still refers to a hyperbola.

Re: orbital mechanics versus traditional interpretations, in OM we're mostly concerned with velocities and extrapolating positions, so the equations used to define the orbits (usually ellipses) tend to focus on time derivatives of position rather than position itself - also, the semi-minor axis is rarely mentioned. Ultimately everything is the same, but the biggest difference is the vocabulary and concepts used to describe the objects.

There is some confusion here. The article, at this moment, says a line is an ellipse with vanishing minor axis. Wrong; a line goes to infinity, an ellipse does not. Flattening a parabola by "spreading its legs" does produce a line, in the limit, but not a proper conic. A hyperbola loops through infinity twice, so flattens to a double line. The geometric answer is that a line is not a conic, and its eccentricity is not defined.

Take a circular cone around a vertical axis, with two halves meeting at a vertex. Slice it with a plane to get a conic section. Vertical planes produce hyperbolae. Sliding the plane to pass through the vertex produces a pair of lines intersecting at the vertex. Rotate the plane around the vertex until it is tangent to the cone; the two lines become one. Sliding the plane off the vertex produces a parabola. Hmm.

As for orbits, it is one thing to restrict attention to satellites and the like; those orbits are ellipses. But suppose some perturbation sends an object in the Oort cloud inward to loop around the Sun. When we spot it with telescopes, we begin to accumulate data to construct an orbit. Perhaps it will reveal itself as a new periodic comet, whose orbit is elliptical. Perhaps it will be a one-time visitor to be flung back into space and lost forever, whose orbit is hyperbolic. Our methods for fitting Keplerian orbital elements must be prepared to cope. --KSmrq^T 12:26, 24 October 2006 (UTC)[reply]

I'm not really sure what you're getting at here - orbital mechanics also includes parabolas and hyperbolas, not just ellipses (though the last are obviously the most common). OM typically addresses the type and characteristics of an orbit through its "energy" (KE and PE, zeroed on a parabolic trajectory), which applies to all conics. Energy is calculated via Keplerian elements.

Consider a line L, a point P not on the line, and the locus of points Q in their plane such that |PQ| = b |LQ| (where |LQ| is the perpendicular distance). Amazingly enough, that locus is a conic with eccentricity b. As b increases toward infinity, the conic converges to L. —Tamfang 06:54, 24 October 2006 (UTC)[reply]

Duh, as the picture cited above illustrates. —Tamfang 06:55, 24 October 2006 (UTC)[reply]

Comment: This particular derivation also has the limit of eccentricity 0 (a circle) vanishing to a point focused at P, so I don't really know if it should also apply to the limit of an infinite eccentricity either (which, in my understanding, generates two lines, one at L and the other at infinity). 192.31.106.34 17:30, 27 October 2006 (UTC)[reply]

So ultimately the varied definitions of eccentricity (including the picture, which is just one user's interpretation) that seem to be used here imply (to me, anyways) that there is no set value of e for a line, unless someone has a definitive text that says so. —The preceding unsigned comment was added by 192.31.106.34 (talk • contribs) 15:23, 2006 October 24.

(In future, please, as requested at the top of this page:

Sign your question. Type --~~~~ at its end.

Actually, sign all your posts here and on talk pages. Among other things, it helps us know who is asking or stating what.) --KSmrq^T 19:06, 24 October 2006 (UTC)[reply]

In this specific case, it wouldn't, since I'm on a floating IP which seems to change every few hours =O) 192.35.35.34 19:09, 24 October 2006 (UTC)[reply]

System of DifEqs

It's been a couple years since I took difeq, and then I stumble across this in lab.

${\frac {dy_{1}}{dt}}=-Ay_{1}$

${\frac {dy_{2}}{dt}}=Ay_{1}-By_{2}$

${\frac {dy_{3}}{dt}}=By_{2}$

Any bright ideas? I thought I could solve the first eq and get y1 in terms of t, put that into the second to get an equation of just y2 and t, but that came out to be something like

${\frac {dy_{2}}{dt}}=ACe^{-At}-By_{2}$

which I'm not sure how to solve. --Registrar

Found it. If you're interested, Steady state (chemistry). --Registrar

Should you run into a situation with 17 reactants in your lab, a general way of solving this kind of linear ODEs can be found in our article Matrix exponential. --Lambiam ^Talk 19:27, 23 October 2006 (UTC)[reply]

Möbius Space Station

I was reading about the Möbius strip and I was wondering what would happen if a space station was designed like one. I was thinking a space station that is shaped like a ring (possible spinning to create artificial gravity) but had a twist in it. Would there be a part where the ceiling and floor met?Ed Dehm 22:18, 23 October 2006 (UTC)[reply]

You wouldn't be able to do that. A Möbius strip doesn't have an interior, the reason it works is that you're taking a 2 dimensional object and using the 3rd dimension to do the twist. (A space station has to have an interior in which people can live.) The nearest analogy in 3 dimensions is the Klein bottle (take a look at the Klein Bottle discussion above), which actually requires 4th dimension to pull off properly... and furthermore, a Klein Bottle has no inside or outside, so you wouldn't be able to protect astronauts from space. - Rainwarrior 23:27, 23 October 2006 (UTC)[reply]

I disagree. A paper Möbius strip has some thickness, and the thickness could be increased to be sufficient to accommodate a crew. However, it would be far more expensive to construct a space station shaped like that, and, if rotated about it's center, the apparent gravitational force would be toward the "floor", then one wall, then the "ceiling", then the other wall, then the "floor" again, as you ran around it. StuRat 23:55, 23 October 2006 (UTC)[reply]

Topologically speaking, the Möbius strip has no thickness, and if you think of it as a flattened-tube, it would be homeomorphic to just a regular tube, twisted or not (this has a lot to do with needing the extra dimension to accomplish the twist). So... you could have a Möbius strip painted on the inside of a toroidal tube, sure, but there'd be no real reason to call it the "floor", it would appear to go around the tube twice (crossing over once). - Rainwarrior 04:29, 24 October 2006 (UTC)[reply]

I don't see why you can't have a rectangular cross section with a twist in it (as it rotates around the center). Are you just saying that this isn't called a Möbius strip ? StuRat 04:47, 24 October 2006 (UTC)[reply]

A Möbius strip doesn't have a cross section. That negates its uniqueness. If you give it thickness it is homeomorphic to a torus, which is the regular space station shape we were starting with. A paper model of a Möbius strip is only a Möbius as much as you ignore its edges. If you include the edgies, then it has a continuous surface that goes all the way around (around its cross section), making it the same as any other ring. It is only with the 2-dimensional strip that the notion of "top" and "bottom" surfaces apply, and the peculiar property of the Möbius strip having only a single surface that is neither.

So, yes, you could construct a space station like you would a paper Möbius model, and ignore the "edges", the non-edge surface would be a Möbius strip (either the inside one or the outside one). But, since gravity is going to be radial, I don't think you can really say that it's the floor. If you built a space station with a twist in it, it's just going to be an unusual twisty hump in the floor, you won't continue onto the ceiling.

Maybe if you had a flexible station that continually moves the twisted portion along the length, then as long as you walked along it at the same speed that the twist is moving along, you could remain on the floor, which would eventually cover both the floor and ceiling. (Or, alternatively if you could generate gravity some other way that is local to the Möbius floor?)

The shape of the station though wouldn't have to have a rectangular cross section. It doesn't really matter what the cross section is shaped like, as long as the cross section is a continous curve that loops back on itself (homeomorphic to a circle), you can put a Möbius loop on its surface (inside or out). If you chose a circular cross section, the twist isn't going to look different from any other part in terms of cross section, but where you are assigning the "floor" would proceed in a twisted fasion. - Rainwarrior 05:12, 24 October 2006 (UTC)[reply]

So is there a name for a geometric figure like a Möbius strip, but with thickness added ? StuRat 05:23, 24 October 2006 (UTC)[reply]

A twisted prism would probably suffice. One can probably find such a shape (though not as flat) in a number of different failure modes related to buckling under torsion. I don't think there's an actual name for it, since it's not a very useful (or geometrically interesting) shape. Virogtheconq 06:04, 24 October 2006 (UTC)[reply]

Well, not really. As I said, with the thickness, topologically speaking you've just got some sort of deformed torus. You could call your shape the product of a Möbius loop and a short line segment, but I don't think there's a particular name for this. There's a Klein bottle, but that's a bit different (it's more like two Möbius loops grafted together). Also, the buckling suggestion is unlikely, as if a rectangular torus (or your rectangular torus) with no twist is going to be crushed and begin to buckle, under no circumstances would it be allowed to form a half twist in itself. The half twist has to be part of its construction, you can't add it later without tearing the object and reattaching it. - Rainwarrior 15:45, 24 October 2006 (UTC)[reply]

Yes, that was the question, what if the space station was originally designed with a twist. StuRat 06:39, 25 October 2006 (UTC)[reply]

Perhaps I should have said such a twisted shape was then bent into a circle - I realize there's no way to get a failure quite like that in reality (aside from a sudden cold weld, perhaps?). And I withdraw my comment on not being a geometrically interesting shape, because it's a three dimensional object that only has four sides, which I think is the minimum required for any 3d shape. That being said, I don't think there's a name for it, and it's still not a good shape for a space station, since there's no inherent gain on volume or surface area in the design over a traditional torus (within orders of magnitude, anyway), and it's technically more challenging (though not impossible) to build. Virogtheconq 14:46, 26 October 2006 (UTC)[reply]

I'd never heard of cold welding before. That's pretty interesting. - Rainwarrior 16:04, 26 October 2006 (UTC)[reply]

At a much higher level, for those who are interested, we're saying that the real projective line bundle over the circle (that is, the Möbius strip) has no nonzero section and thus is a nontrivial vector bundle (and is nonorientable as a manifold and so on). Conversely, you can get a rank-2 vector bundle by taking the trivial rank-2 real bundle on the interval [0,1], then identifying the endpoints 0 and 1 of the base space while identifying their fibers by the antipodal map (which is the same as a half-twist). This vector bundle is equivalent to a "thickened" Möbius strip, but thanks to the added dimension this bundle has a nonzero section. In fact, it has 2 orthogonal nonzero sections and thus is the trivial bundle

S^{1}\times \mathbb {R} ^{2}

. Tesseran 07:13, 26 October 2006 (UTC)[reply]

Think outside the box. How about a Möbius Ringworld? :-D --KSmrq^T 12:32, 24 October 2006 (UTC)[reply]

Terry Pratchett says: "I'll start thinking outside the box when I see any evidence that there's any thought going on inside the box." :) JackofOz 10:20, 25 October 2006 (UTC)[reply]

I thought that was what he was talking about - a ring-like world/station/whatever, with a half-twist in it somewhere along its length. The problem is, as they said (at some point, if you wade back through the rest of it) if you base the gravity off rotation, you've essentially just made a normal ringworld with an inconvenient sort of hump thing partway along, where coming along the inside, the land turns over and exposes you (fatally) to space. Imagine hiking down into a valley with no floor. If the twist were drawn out farther along the ring, more of the world would be sideways and it would be pretty well impossible to live on that section, unless you like living on the side of a vertical mountain face. Sadly, this marvel of topology rather loses its luster when you try to actually use it for something. One thing that might be interesting, though, going with the half-turn idea, would be to build a donut-shaped station, made up of a bunch of donut-shaped tubes, with a half-twist that attaches each tube to a different one. You could be walking down that hallway for a very long time, just going round and round the station, without realizing how many loops you'd completed. Not that you'd have any way of judging that in the first place, though. Black Carrot 22:30, 27 October 2006 (UTC)[reply]

bases

teacher says that [101(base b)]^5 = (b^2 + 1)^5

but didn't explain where this came from or what the general case was. anyone seen this strange identity before? some kind of binomial theorem?

It's true that 101(base b) = b^2 + 1. I wouldn't worry about the ^5. Melchoir 00:20, 24 October 2006 (UTC)[reply]

Yeah, indeed it's true.

  1    0    1
× b^2  b^1  b^0
= b^2 + 1

The 5 is completely irrelevant here. ☢ Ҡi∊ff⌇↯ 03:30, 24 October 2006 (UTC)[reply]

It might help to think about this in the case of base 10 -- in that case, all you're saying is that 101 is 10*10 + 1. The biggest digit is the 100's column, which is the 10-squared column. The middle digit is the 10's column, and the smallest digit is the 1's column (10-to-the-zero-power). -- Creidieki 21:13, 27 October 2006 (UTC)