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User: eliasen

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  1. Re:A couple thoughts on Microsoft Patenting IM Translation? · · Score: 2, Informative
    I did this a long time ago as a learning experience. My Universal Translator applet translates between several languages.

    My favorite example was my friend Brian entering "Yo quiero un burrito" and the translation came out "I love a young donkey." Perfectly legitimate translation, too.

    This will never work for translating instant messages until people learn to spell correctly, though.

  2. Re:Basic Physics on NASA's Foam Test Offers Lesson in Kinetic Energy · · Score: 1
    There's a lot of bewailing about mixed/confused units in these postings. Computers can help! Modern calculating tools like Frink help you get the answer right even when you're mixing units. And it helps you convert them into units you're familiar with, so nobody has to complain.

    The kinetic energy involved is huge. I don't have a good intuitive grasp of "a basketball thrown at 500 mph" so I used Frink to put the kinetic energy into other terms.

    Using KE = 1/2 m v^2, the foam had the same energy as a 16-pound bowling ball travelling at 173 mph! (In Frink notation, which is usually just normal mathematical notation, this is:)

    KE = 1/2 1.7 lb 531 mph
    (KE / (1/2 16 lb))^(1/2) -> mph

    I still don't have a good intuition of a bowling ball being thrown that fast, (it's like dropping it from 300 feet!) but I have a very good feeling from dropping one on my foot from just a few inches that it could do some damage.

    Using a larger object, a 2000-lb car, it's like that car moving about 15 mph and crashing into the wing!

    (KE / (1/2 2000 lb))^(1/2) -> mph

    Put another way, this is like dropping that 2000-lb car from a height of 8 feet onto the wing! (Knowing potential energy is mass * gravity * height, this is, in Frink notation:)

    KE / (2000 lb gravity) -> feet

    The F=ma relation can give us a ballpark figure for the forces involved. From their statistics, the chunk of foam weighed 1.7 lb and had a volume of 1200 in^3. If it were a cube, it would be about 10 inches on a side, (in Frink, this is found by (1200 in^3)^(1/3)->in ) but we can see from the test photos that it was maybe more like 2 feet at its longest dimension.

    If the piece struck squarely, and compressed along its longest dimension, it would still have to decelerate from 531 mph to 0 within its length of 2 feet:

    2 feet/(531 mph)

    (If it didn't compress, the acceleration would be higher--think of running into a brick wall vs. running into foam.) That would take about 2.5 milliseconds. Going from 531 mph to 0 in that short of time would give an acceleration of about 9400 times the acceleration of gravity!

    531 mph / (2 feet/(531 mph)) -> gravity

    Taking F = m a, the force would be on the order of 16000 pounds-force on the wing!

    1.7 lb 531 mph / (2 feet/(531 mph)) -> lbf

    Even if the piece didn't strike squarely, and delivered only 1/10 of this force, that's still going to severely damage any wing that's light enough to get off the ground--much less one that has to be sealed so that superheated plasma can't get inside! Once there was an opening for the ultra-hot plasma to burn its way inside, it was probably already too late to save the shuttle.

    By the way, pound is a unit of mass.

  3. Cooling on PeltierBeer · · Score: 2, Interesting
    I was thinking of making my own Bar-Monkey style computerized drink mixer and thought of putting a Peltier junction and a heat-exchanger block on it to cool drinks on demand--until I ran the numbers too.

    I assumed that the heat capacity of beer was just about that of water (1 calorie/degC/g). So, using Frink, a calculating tool/programming language I've developed, the power needed to lower a lovely 12 floz beverage by a relatively scant 10 degrees F in a minute is given by:

    12 floz water (1 calorie/degC/gram) 10 degF/min -> W

    Which gives about 137 watts given perfect efficiency! You actually need to divide the left-hand side by the Coefficient of Performance of your Peltier junction which is probably--what--0.4? And then divide by all your other efficiency losses due to imperfect heat transfer and heat input from the environment...which, as the saying goes, "is left as an exercise for the reader."

    (You can use the web-based interface to Frink to plug in your own numbers and units like liters or degC, or K, or recalculate the numbers using the heat capacity of ultra-high-ethanol concoctions.)

    No wonder that Peltier-junction cooled ice chest I bought many years ago didn't work worth a lick. It kept things cool if you filled it with a big bag of ice. :)

    Wonderfully fun experiment, in any case. I'd sure like to see the thermometer placed in the liquid, though.