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Olympic medals and their rankings March 3, 2010
As the winter olympics in Vancouver drew to a close last Sunday, the media went through the usual exercise of tabulating national medal tallies. Although the International Olympic committee doesn’t recognise any of these arrangements, they make for interesting analysis. While some of the medal tallies are arranged in descending order of total medals won, some others are ordered based on the number of gold medals won, with silver and bronze medals used only to break ties. The latter scheme apparently enjoys wider appeal and is used in the wikipedia entries of the medal tables too but the total medal count is favoured by most American newspapers. Each scheme has advantages and disadvantages.
Ranking performances based on total medal count obviously treats all medals the same and hence fails to appropriately credit a Gold medal over silver or bronze. Also, this scheme falls prey to one handicap of the system where the third best performance in a discipline is awarded a medal alsongside the first and second while the fourth is not. So, a country which had one 3rd place performance is ranked above another with many 4th place performances.
Rankings based primarily on the number of Gold medals won and using silver and bronze medals progressively to split ties, while failing to allay the injustice to the 4th place completely is at least not as glaring. Any number of third places now don’t make up for a second or first place. But, it instinctively seems unfair to rank 1 gold medal over 10 silver medals either.
A fair system of assigning weights ot Gold, Silver and Bronze medals is required to compare them against each other in compiling the final medals table. Wikipedia lists such schemes that have been previously used by newspapers like 5:2:1, 3:2:1 and 4:2:1 as the ratio for Gold, Silver and Bronze medal weights respectively. Perhaps a poll among past olympians in a field can be used to set the right ratio for olympic medals in that field. It seems to be a matter of opinion but that shouldn’t stop us from exploring the possibilities!
Applying these schemes to some of the top nations in the 2008 olympics, we have the following rank list-
|United States||China||United States||United States||United States|
|Great Britain||Great Britain||Great Britain||Great Britain||Great Britain|
|South Korea||France||South Korea||France||South Korea|
|France||South Korea||France||South Korea||France|
The three weighted schemes used are obviously attempts at compromising the colour-blind medals list and gold first list. Considering this list, it isn’t surprising why the American media insisted on using the total medals are their ranking criterion!
Trailblazing December 20, 2009
The Royal Society of The UK is celebrating its 350th anniversary this year. In this regard, they have made available some of the trailblazing leaps in Science, as seen in the proceedings of the Royal Society over the years, freely on the internet. From Isaac Newton’s theory of Light and Colours in 1672 to Benjamin Franklin flying a kite in an electric storm in 1752, Bayes’ essay on chance in 1763, Maxwell’s theory of the Electromagnetic field in 1865, Dirac’s theory of the electron in 1928 and Watson and Crick’s DNA structure in 1954.
It was interesting to me how the only example from the 21st century they chose to highlight was a paper on Geoengineering. I wonder if this amounts to an endorsement for research into the field by the Royal Society. With the current international impasse on emissions reductions, it is very likely that Geoengineering will become increasingly prominent in the near future.
Partly due to my background in Electrical Engineering, my favourite paper among all the highlights is Maxwell’s masterpiece tying up Electricity and Magnetism into the unified Electromagnetic theory. Although the paper itself makes difficult reading today, even a cursory look betrays the rigour of the analysis and genius of Maxwell. Even today, it is perhaps the most elegant unification in history of forces and fields that were previously thought to be separate. For the sake of reductionist Physics and the intellectual clarity that goes with it, I hope it doesn’t remain so much longer!
Mixes made equal November 9, 2009
Today, I shall explore a simple puzzle which illustrates the advantages of extreme conditions in getting pointers towards the answer. Consider this scenario found at the Times’ Tierneylab blog a few weeks ago. Essentially, the question is-
If we have two vessels, X and Y, X containing a volume “a” of liquid A and Y containing a volume “b” of liquid B which mixes with liquid A uniformly, and we pour a volume x of liquid A from X to Y first followed by pouring a volume x of the mixture of liquids in Y back to X, will there now be more of liquid A in Y than of liquid B in X? Assume that b is greater than a.
Before solving this algebraically, let’s think about the problem. Since the volume b is more than a, there is more of liquid B than liquid A. When x of b is poured into X, X will now have a+x amount of a mixture of A and B whereas Y will have b-x amount of just liquid B. When x amount of the mixture in X is poured back to Y, X will go back to having just a amount, but of the mixture of A and B, and Y will go back to having b amount, again of the mixture of A and B. There doesn’t seem to be any immediately compelling reason why the amount of B in X and A in Y should be the same or for the percentages to be the same either. So, let’s consider an extreme case for some help with the arguments.
Let us consider the case where x=0. Then, we can see that no liquid is exchanged between X and Y and so at the end the amount of B in X and A in Y are the same, which is zero. Now consider the case where x=a. After the first exchange, we’ll have 2a amount of liquid in X, which is made of a amount of A and a amount of b, i.e. 50% of each. When x=a amount is poured back from X, this would involve a/2 amount each of A and B. Adding this to the b-a amount of B that was left in Y, we’ll have b-(a/2) amount of B and a/2 amount of A there. At the same time, X will have a/2 amount each of A and B. So, the amount of B in X and A in Y are the same- a/2.
In both the special cases we considered, the amount of B in X and amount of A in Y are the same. It is interesting to see that the concentration of B in X and of A in Y are not the same because the volumes a and b are not the same. Given these two special cases, which give the same result for very different values of x, it seems likely that the result might hold for all x. These examples do not prove so, but it does motivate us as to the answer to look for. So, while we solve it algebraically, not only will we have a higher confidence in the result if we prove it, it is likely that we are more likely to get to it since we know what we are looking for!
Getting to the algebra now, in the general case, after the first exchange of liquids, X will now have a-x amount of A and Y will have a total amount of b+x, of which there is b of B and x of A. When x amount of the liquid from Y is now poured back to X, since the liquids in Y mixed well, the amount poured back will involve [b/(b+x)]*x of B and [x/(b+x)]*x of A. In the proportion of their volumes, this now adds up to x amount poured back from Y to X. So, the amount of A left in Y is x-[(x*x)/(b+x)] which is bx/(b+x). The amount of B now in X would just be the amount of B poured from Y to X, which again is bx/(b+x). Voila, we have the same number!
This means that the concentration of A in Y is much less than the concentration of B in X. So, if I gulp in a little sea water and spit out the same amount, there will be the same amount of sea water left in my mouth as the amount of my spit in the sea, but my mouth will taste much saltier than the sea will of my spit!
Featured on scientia pro publica! September 22, 2009
Thank you for the honour Lab Rat. It sure motivates me to keep blogging on my observations and thoughts on science.