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  • one year ago

Playing with clay question!

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  1. Empty
    • one year ago
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    Take some clay, and make it into a nice star like this with a cookie-cutter: |dw:1436272462408:dw| Now smash it under a plate and it looks like this! |dw:1436272485317:dw| Is it possible that if we're allowed to make any measurements on the smashed star that we can exactly find the area of the unsmashed star?

  2. SolomonZelman
    • one year ago
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    Finding the volume of unsmashed star based on the volume of the smashed star wouldn't be a hard task (because the volume remains the same). A harder task would be the area, but I think you still can find the area of unsmashed, based on area of the smashed, if you know: 1) Area of smashed star 2) How thin is the smashed star 3) How many times thinner is the smashed star {in proportion/ when compared to} the unsmashed star. `~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~` At least I would imagine that it is so....

  3. Empty
    • one year ago
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    I am going to say we aren't allowed to know the thinness of the original star. Particularly what I'm interested in is if we can "unmash" the curve itself since it appears to be roughly star shaped and knowing how clay distorts when it's mashed is sort of the idea I'm considering. I just sorta made up this question for fun, so we can go ahead and we can sorta just add or take away whatever we want, if you want to consider solving this problem somehow by adding your 3rd assumption (My problem already allows you to know your first and second ones) then go for it, since we could have smashed another shape altogether and not known it was originally a star, it sort of makes this a weird or possibly unsolvable problem. Maybe we can only assign a "probable area" to a range of possibilities that all deform into this one single shape when smashed? Or does each smash correspond to only a single shape?

  4. SolomonZelman
    • one year ago
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    Well, the "smasher" (or whatever this equipement is called) can smash, and then do the exact same movement back. `My assumption:` The smasher will smash till a certain thinnes (if that makes sense) regardless of the object being smashed (not talking about smashing metal, but any food, dough or something of this sort). With this assumption we can find out the original thickness (of the unsmashed star) if we were to take a mass, put it in a (for example) plasit bad. Using glue we can stik it to the bottom and the top, and unsmash it. (that is the first that comes to my mind)

  5. SolomonZelman
    • one year ago
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    This arguement is faulty, tho' because the smasher can go up (backwards) as much as we want, so it could be a tiny but super thick star as well.....

  6. Empty
    • one year ago
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    Hmmm I like this, luckily we're just mashing mathematical dough and so I think we can sorta allow weird things like this. My idea is that the masher will always mash objects that are 1 inch high to 1/2 inch high. I was thinking that it seems like any unique shape we put in will mash uniquely to a different shape, so it should be a bijective function, meaning it's an invertible transformation.

  7. Empty
    • one year ago
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    I guess in my mind the focus is on the area, not the actual volume. Another way to think about it is to imagine that there is a hole in the roof creating a puddle on the floor, can we determine the shape of the hole in the roof from this weird, idealized situation? (picture shows water dripping from star-shaped hole to create similar looking puddle. |dw:1436275212368:dw|

  8. SolomonZelman
    • one year ago
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    if you know the thickness of the current star, and the volume of the previous (unsmashed) and current (smashed) star (without taking hard conditions like lose of water in this case or perhaps some other conditions) then you can perhaps find the area of the initial star.

  9. SolomonZelman
    • one year ago
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    I like the whole proposal tho':) Very creative! We would need to recall some physics to be precise...but, right now I have to depart for a second (apologize). Tnx for sharing!

  10. Empty
    • one year ago
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    Yeah no problem, this is just sorta for fun, I'm sorta thinking of the idealized situation where the actual physics aren't so much important. Flattening out clay will also realistically tend to crack as it spreads out. The problem with the water dripping through the ceiling is that it tends to make the problem seem a little more vague and further from the mathematical conception I have in my mind. Maybe it would be better to assume that the floor is a carpet where the water doesn't disperse very quickly so that it holds the rough shape better.

  11. SolomonZelman
    • one year ago
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    :D

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