I enjoy playing Google's "a Google a day," which can help hone your searching skills and almost invariably features interesting morsels of food for thought! Today's features the expression "the twinkle of an eye" (as in "in a twinkle of an eye"), which comes originally from a Hellenistic Greek expression "en atomE" (the capital E represents the Greek letter Eta, thought to have been pronounced with a long "A" sound, as in "hey," for example), which literally means something like "in an atom [of time]," meaning a period of time so short that it cannot be cut up into any shorter time period (Dr. Sheldon Cooper of "The Big Bang Theory," a brilliant theoretical physicist who's a little more nerdy and OCD than I am, could tell you that today this would be understood as the Planck time, approximately 5.39106(+/-0.00032)x10^(-44) seconds, the time it takes light to travel one Planck length, about 1.616199(+/-0.000097)x10^(-35) meters, in a "vacuum"--until Sheldon or I or anyone else comes up with a viable quantum gravity theory, that's the smallest unit of time we can meaningfully imagine!)--apparently, in medieval times (Spoiler Alert! If you haven't already done today's a Google a day and you still want to, don't read any further!), a "twinkle of an eye" was taken to mean (1 minute)/376 = 0.1595744 seconds = 159.5744 millisec or so.
Today's question asks how many "twinkles of an eye" it takes you to get to the Renaissance festival if it takes you 20 minutes to drive there. I reasoned thusly: if there are 376 twinkles of an eye in 1 minute then surely there must be (376 twinkles of an eye/minute)x(20 minutes) = 7,520 twinkles of an eye in 20 minutes! Which, in fact, is the right answer in the real world! However, when I submitted "7,520 twinkles" as my answer, the Google of "Do No Evil" fame confidently informed me that this was not the correct answer!
Somewhat gobstruck, I went ahead and clicked to see what Google World thought the right answer was and this is their "answer": a "twinkle of an eye" is (1 minute)/376 = 160 milliseconds and there are (20 minutes[=1,200 seconds])/(1 millisecond[=10^(-3) seconds]) = 1,200,000 milliseconds in 20 minutes and (1,200,000 milliseconds)/(160 milliseconds/twinkle) = 7,500 twinkles!
Of course, Google's "answer" is certainly in the right ballpark and is pretty darn close to my correct (and exact!) answer of 7,520 twinkles. But, to assert that 7,520 twinkles, which is the right answer, is not the right answer and to go on to say that 7,500 twinkles is the right answer when it clearly is not is just plain wrong!
That happened earlier today when I first tried to submit the correct answer (7,520 twinkles). In the meantime, someone at Google must have discovered the problem, because when I resubmitted my correct answer just now as "7,520 twinkles" (I didn't use quotation marks, though, I'm just using them here to indicate exactly what I resubmitted!), this was now recognized as a correct answer! The marvels of modern technology! I guess Google didn't do evil after all!
Randy and Caroline
A lovely July in Seattle!
Saturday, February 18, 2012
Sunday, February 5, 2012
Random Ruminations and the Like!
Occasionally you run across something (on Wikipedia, for example) that is enchanting and interesting and coincidental and completely meaningless. A case in point happened today--while looking up data related to Earth's Moon on Wikipedia (what did we ever do before Wikipedia?!)--for no particular reason I noticed that the volume occupied by the Moon (V_Moon) is about 2.1958x10^10 km^3 or 21,958,000,000 cubic kilometers. Why did I find this intriguing? Because when I went to memorize that value, I realized with a shock that it will be ridiculously easy for me, in particular, to remember all five significant figures of this value, either because it encodes my birthdate (2/19/58) or my birthmonth (2/1958)!
Usually, it is standard operating procedure to memorize the radius R of a substantially spherical body and then calculate the volume V using the well-known formula for the volume of a sphere (known to the ancient Greek Archimedes long before Newton and Leibniz independently invented the infinitesimal calculus): V = (4/3)[pi]R^3, at least in 3 spatial Euclidean dimensions! Using this formula "in reverse," as it were, I can now always calculate the radius of the Moon (R_Moon) whenever I wish to, as long as I have access to a good enough calculator! Not to keep you all in undue suspense, the Moon's radius R_Moon is the delightfully palindromic 1.7371x10^3 km!
How does that compare to the Earth's radius R_Earth? Well, to answer this question, it helps to know (or remember) the very definition of the meter, originally! In the grips of the French Revolution, the good rational scientific-types in Paris wanted a standard unit of length that wasn't related to the random and variable lengths of some king's foot (foot) or thumb joint (inch) or forearm (cubit), but was somehow more grounded and less variable or changeable! What they came up with was that one meter would be equal to one ten-millionth of length of the meridian of longitude from the North Pole to the Equator that passes through Paris (where else?!), which means that nobody has any excuse anymore for not knowing that the circumference of our one and only home planet, Earth, is by definition equal to 4.0000000x10^7 meters, which is 2[pi]R_Earth, so that R_Earth is 6.3661977x10^6 meters or 6.3661977x10^3 km!
So R_Earth is 3.6648 times bigger than R_Moon or just a little under 11/3 times bigger or almost 4 times bigger! In American-friendlier units (as the 1988 Physics Nobel Laureate Leon Lederman said "We're inching toward the metric system!"), since there are just about 5 furlongs in 1 kilometer (as I've shown elsewhere on this blog, exhaustively--see below!), 6,400 km = 32,000 furlongs! Since there are 8 furlongs in 1 mile, we see instantly that the radius of the Earth R_Earth is very close to 4,000 miles, a nice roundish number! Similarly, the Earth's "belt size" is 40,000 km, which is 200,000 furlongs, which is 25,000 miles, which is the distance every point on Earth's Equator must travel in a day, which is 24 hours, so that every point on Earth's Equator must travel at a speed of more than 1,000 miles per hour just to stay in the same place (relative to the Earth, of course)! And that's not even counting the speed of the Earth about the Sun or the speed of the Solar System about the Milky Way Galaxy or the speed of the Milky Way Galaxy in our Local Group of Galaxies or the speed of our Local Group of Galaxies toward the Shapley Supercluster (or Shapley Concentration), etc., etc.!
On an entirely different subject altogether, lest I forget to mention it, my most recent CT scan at MD Anderson Cancer Center was on Wednesday, January 11, 2012, and we are very pleased to report that there continues to be No Evidence of Disease (NED)! Thanks so much for all your many prayers and positive thoughts and best wishes--they are all deeply appreciated and are definitely working, so please keep them coming our way! I'm scheduled for a colonoscopy just in time for the one (1!) year anniversary of my 12 hour operation last April 12, 2011. The preparation for the colonoscopy will be jolly good fun (as ever!), but the whole procedure should take about half as long as before since I now only have a semicolon! I'm also scheduled for another CT scan in April, before I meet again with my wonderful surgeon, Dr. Paul Mansfield at MDACC! Assuming there continues to be NED, I may even get put on a 6-month CT scan regimen rather than my current 3-month one. The CT scans are painless, but each CT scan gives a dose of x-rays equivalent to about 300 chest x-rays or about 1,200 x-ray scans at an airport! The worst part about a CT scan is waiting to get the results! And the best part about a CT scan is getting the result that there's NED!
Usually, it is standard operating procedure to memorize the radius R of a substantially spherical body and then calculate the volume V using the well-known formula for the volume of a sphere (known to the ancient Greek Archimedes long before Newton and Leibniz independently invented the infinitesimal calculus): V = (4/3)[pi]R^3, at least in 3 spatial Euclidean dimensions! Using this formula "in reverse," as it were, I can now always calculate the radius of the Moon (R_Moon) whenever I wish to, as long as I have access to a good enough calculator! Not to keep you all in undue suspense, the Moon's radius R_Moon is the delightfully palindromic 1.7371x10^3 km!
How does that compare to the Earth's radius R_Earth? Well, to answer this question, it helps to know (or remember) the very definition of the meter, originally! In the grips of the French Revolution, the good rational scientific-types in Paris wanted a standard unit of length that wasn't related to the random and variable lengths of some king's foot (foot) or thumb joint (inch) or forearm (cubit), but was somehow more grounded and less variable or changeable! What they came up with was that one meter would be equal to one ten-millionth of length of the meridian of longitude from the North Pole to the Equator that passes through Paris (where else?!), which means that nobody has any excuse anymore for not knowing that the circumference of our one and only home planet, Earth, is by definition equal to 4.0000000x10^7 meters, which is 2[pi]R_Earth, so that R_Earth is 6.3661977x10^6 meters or 6.3661977x10^3 km!
So R_Earth is 3.6648 times bigger than R_Moon or just a little under 11/3 times bigger or almost 4 times bigger! In American-friendlier units (as the 1988 Physics Nobel Laureate Leon Lederman said "We're inching toward the metric system!"), since there are just about 5 furlongs in 1 kilometer (as I've shown elsewhere on this blog, exhaustively--see below!), 6,400 km = 32,000 furlongs! Since there are 8 furlongs in 1 mile, we see instantly that the radius of the Earth R_Earth is very close to 4,000 miles, a nice roundish number! Similarly, the Earth's "belt size" is 40,000 km, which is 200,000 furlongs, which is 25,000 miles, which is the distance every point on Earth's Equator must travel in a day, which is 24 hours, so that every point on Earth's Equator must travel at a speed of more than 1,000 miles per hour just to stay in the same place (relative to the Earth, of course)! And that's not even counting the speed of the Earth about the Sun or the speed of the Solar System about the Milky Way Galaxy or the speed of the Milky Way Galaxy in our Local Group of Galaxies or the speed of our Local Group of Galaxies toward the Shapley Supercluster (or Shapley Concentration), etc., etc.!
On an entirely different subject altogether, lest I forget to mention it, my most recent CT scan at MD Anderson Cancer Center was on Wednesday, January 11, 2012, and we are very pleased to report that there continues to be No Evidence of Disease (NED)! Thanks so much for all your many prayers and positive thoughts and best wishes--they are all deeply appreciated and are definitely working, so please keep them coming our way! I'm scheduled for a colonoscopy just in time for the one (1!) year anniversary of my 12 hour operation last April 12, 2011. The preparation for the colonoscopy will be jolly good fun (as ever!), but the whole procedure should take about half as long as before since I now only have a semicolon! I'm also scheduled for another CT scan in April, before I meet again with my wonderful surgeon, Dr. Paul Mansfield at MDACC! Assuming there continues to be NED, I may even get put on a 6-month CT scan regimen rather than my current 3-month one. The CT scans are painless, but each CT scan gives a dose of x-rays equivalent to about 300 chest x-rays or about 1,200 x-ray scans at an airport! The worst part about a CT scan is waiting to get the results! And the best part about a CT scan is getting the result that there's NED!
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