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December 12

Most circular countries

Switzerland and Iceland are vaguely circular. Any better ideas? Probably some islands are pretty circular. I'd also be interested in sub-national divisions. LANTZY ^TALK 01:27, 12 December 2010 (UTC)[reply]

From a mathematical point of view, very few large objects are perfectly circular. This seems like more a geography question, to be honest. Unfortunately, I can't find a geography reference desk. May you might like to try Wikipedia:Reference desk/Miscellaneous. — Fly by Night (talk) 01:50, 12 December 2010 (UTC)[reply]

See Compactness measure of a shape. The "isoperimetric quotient" referred to in that article (I see the link doesn't go anywhere useful), also called the "circularity ratio," is the ratio of the area of a shape to the area of a circle with the same perimeter. It is given by

R={\frac {4\pi A}{P^{2}}}

, where

A

is the area of the shape and

P

is the shape's perimeter. This ratio will always be between 0 and 1. The areas and perimeters of countries and subnational divisions are easily available, so this is an easy measure to compute. I once compiled a list of countries sorted by circularity ratio, but it was deleted for being original research and not useful information. —Bkell (talk) 02:55, 12 December 2010 (UTC)[reply]

The problem with that approach is that many countries have infinite perimeter. This leads to infinite (or, if you use the truncated figures available in the sources you refer to, enormously high) circularity ratios, even for countries which are intuitively almost circular. To take an extreme example, a country which consisted of one large perfectly circular island and one tiny island the shape of a Koch snowflake would be infinitely far from circular by this measure, though in another sense it is very nearly a circle. The best idea I've had so far is to take an arbitrary circle, take the area of the symmetric difference between the country and the circle, express this as a percentage of the area of the circle, and take the minimum of this percentage over all possible choices of circle. This measure declares my circle+fractal example to be very nearly circular, as I think a good measure should, but I haven't thought about it enough to say if it's a good measure in general. Algebraist 03:12, 12 December 2010 (UTC)[reply]

Right, the circularity ratio has problems. Another problem with it is that for very large countries the ratio could be greater than 1 (if the country has a very smooth perimeter), because it ignores the curvature of the Earth. The advantage of the circularity ratio, of course, is that the required information is very easy to find. Another measure of compactness goes something like this: Find the area of the largest circle completely contained in the shape, and divide it by the area of the smallest circle that completely contains the shape. One problem with this measure is that a single narrow but deep "indentation" into the shape's perimeter (such as a fjord) will greatly decrease this ratio. —Bkell (talk) 06:56, 12 December 2010 (UTC)[reply]

Or a single tiny island a long way from the rest of the country. Algebraist 14:45, 12 December 2010 (UTC)[reply]

Ah, that's true—I hadn't thought about disconnected shapes. On the other hand, if you consider two countries, each with a mainland and an island, and each identically shaped except that one island is near its mainland while the other is far away, it seems reasonable to say that the country with the distant island is the less compact. The two-circles measure probably greatly overemphasizes that difference, but the other two measures we've discussed here ignore it completely (assuming that the nearby island is distant enough from the mainland to fall outside the optimizing circle in the symmetric-difference measure). —Bkell (talk) 19:28, 12 December 2010 (UTC)[reply]

True. Algebraist 20:36, 12 December 2010 (UTC)[reply]

You can fix the coastline paradox by approximating the perimeter to something polygonal (imagine measuring it using a ruler). As long as the sides are the same length for all the countries you are comparing, you should get a meaningful result (although it could vary slightly depending on where you start - a small side length should reduce the impact of that). --Tango (talk) 23:40, 12 December 2010 (UTC)[reply]

I don't know why you need to mess around with the boundary at all. I would suggest the following measure, that works for any non-pathological bounded set A in R²: First, compute the centroid of A, and also the area (2-d Lebesgue measure). Draw a disk (2-d circular ball) D, centered at the centroid of A, and having the same area. Now compute the area of the symmetric difference between D and A, and normalize by dividing by the area of A.

This gives a figure of merit equal to zero if A is a circle. Modifications for the non-flatness of the Earth are left as an exercise for the reader. --Trovatore (talk) 23:19, 13 December 2010 (UTC)[reply]

Oh, it looks like Algebraist made a very similar suggestion up above. --Trovatore (talk) 23:22, 13 December 2010 (UTC)[reply]

If I understand correctly, the problem comes when the coastlines of different countries have different fractal dimension (say, Norway vs. Hawaii, maybe). All of the coastlines' measured lengths increase as you decrease the length of your ruler, but those of higher fractal dimension will increase faster. Then the end result becomes very sensitive to your choice of ruler length, which we don't want. Eric. 82.139.80.124 (talk) 20:37, 14 December 2010 (UTC)[reply]

If we point out the most circular type of island, can we charge you atoll for this service ? :-) (See 3rd pic.) StuRat (talk) 01:02, 13 December 2010 (UTC)[reply]

For what it's worth, the deleted list had Mauritius followed by Niue as the most compact countries, and they are both reasonably circular, to the eye. Warofdreams talk 12:43, 14 December 2010 (UTC)[reply]

From the maps in this encyclopedia, Nauru seems more circular to me. – b_jonas 20:17, 18 December 2010 (UTC)[reply]

It'd be great to see the list Warofdreams, you should start the page. Otherwise, is there perhaps another way to see it? (Stpaul (talk) 11:08, 10 July 2012 (UTC))[reply]

Rocket to the Moon , HSC question.

Heya I have been doing HSC past papers in order to study for an upcoming physics exam and this one question is bothering me as I cannot seem to get a rational / logical answer.

In his science fiction novel From the Earth to the Moon, Jules Verne describes how to launch a capsule from a cannon to land on the moon. To reach the moon, the capsule must leave the cannon with a speed of 1.06 × 10^4 m s−1. The cannon has a length of 215 m, over which the capsule can be assumed to accelerate constantly. (a) Calculate the magnitude of the acceleration required to achieve this speed using this cannon

I have tried dividing the end speed needed by the length of the cannon , but this gives a seemingly tiny result. —Preceding unsigned comment added by 175.39.1.6 (talk) 02:13, 12 December 2010 (UTC)[reply]

Should that be 1.06*10⁴ m/s? Taemyr (talk) 02:54, 12 December 2010 (UTC)[reply]

Also constant acceleration means that the accelaration pr meter will decrease. This because the velocity is increasing, so you need to factor that in. Call the acceleration a, can you define the time spent traveling 215m in terms of a, given a 0 starting velocity? What is the speed at the end? What does a need to be for this speed to be the velocity you are looking for? Taemyr (talk) 03:02, 12 December 2010 (UTC)[reply]

Indeed, simply dividing the end speed by the length of the cannon gives a result in s^-1; you need something in m/s^2. The article Equations of motion might have the relevant equation. -- Bk314159 (Talk to me and find out what I've done) 03:51, 12 December 2010 (UTC)[reply]

You are close. Acceleration is change in speed per unit time, not per unit distance. --Tango (talk) 23:44, 12 December 2010 (UTC)[reply]

Time t, length s, speed v, and acceleration a satisfy v = at and s = at²/2. Eliminate t and solve for a. Bo Jacoby (talk) 10:37, 13 December 2010 (UTC).[reply]

A Logic Question

Recently, I am being asked a question which I can't solve it, I need wiki's help! The question is that:

There is three bags that are enclosed on the desk. One of them contains a red apple and two of them contain green apples. Now, I choose one of the bags and keep it(not knowing what is inside). Then, someone gets one of the remaining two bags and opens it. The bag contains a red apple. So, the question comes, is it neccessary to exchange my bag with the remaining one bag on the desk to increase the possibility of getting the green apple?

What I am thinking is that the initial possibility is 33%, when the person takes away that bag, the possibility increases to 50%. I am just not sure if I am correct, so please answer my question in details. Thanks for attention! —Preceding unsigned comment added by 218.102.101.36 (talk) 10:43, 12 December 2010 (UTC)[reply]

I think you mean two red apples and one green apple. Then it's just the Monty Hall problem. Feezo (Talk) 11:27, 12 December 2010 (UTC)[reply]

It is usually emphasized the importance that Monty knows what's behind the doors, though I'm not entirely sure why; it seems to me that the conclusion is the seem even if the other person "just happened" to open a door with a goat/pick a bag with a red apple. --COVIZAPIBETEFOKY (talk) 15:07, 12 December 2010 (UTC)[reply]

Nvm, it's explained in the article. The op's problem doesn't really specify whether "someone" knows what's in the bags, so it's not certain which way this particular problem goes. However, the probability of the op's selected bag containing the green apple never exceeds 1/2, so switching will certainly never hurt your chances of getting the green apple.--COVIZAPIBETEFOKY (talk) 15:54, 12 December 2010 (UTC)[reply]

As soon as the other person opens his bag and shows a red apple you know that the both the bag on the table, and the bag behind your back, contain green apples. There's a probability of 1 that your bag has a green apple in it, and there's a probability of 1 that the bag on the table has a green apple in it. So you don't change your chances at all. You're certain to have a green apple if you swap or not. I think the Monty Hall problem would apply if the other person revealed a green apple, and not the only red one.— Fly by Night (talk) 18:13, 12 December 2010 (UTC)[reply]

If the other person knows what is in all three bags, has a preference for green apples, and deliberately picks a bag which he knows contains a green apple, then we have the Monty Hall problem. If the other person just picks any bag, and chance would have it that the apple in it is green, then the Monty Hall problem does not turn up.

I suspect the original question is incorrectly stated, because if there are two greens and one red, and the other person picked up the red apple, then it is trivial that the two remaining are green. Sjakkalle (Check!) 16:17, 13 December 2010 (UTC)[reply]

Me too. — Fly by Night (talk) 22:39, 13 December 2010 (UTC)[reply]