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Q: Would you rather land on water or marshmallow?

  1. Dec 16, 2008 #1
    Hello,

    Having a little chat in the pub last night, we got on to skydiving vs bungee jumping. I said I would rather try a HALO jump than a bungee jump.

    This took a slightly odd turn, and we were trying to figure out whether in the absence of a parachute, one would be better off landing on water or a cushion of marshmallow.

    I am aware that landing on water from a great height can kill you. Would the same be true of marshmallow? What are the contributing factors - surface tension and density?

    Any thoughts appreciated!

    Matt.
     
  2. jcsd
  3. Dec 16, 2008 #2

    Ranger Mike

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    having made over 65 jumps from helicopters, prop and jet cargo planes
    I'll take the marshmellow..water is " hard" ( surface tension}, if i reacll my notes from Jump Master School,
    you reach terminal velocity at about 120 MPH..reagrdless of 25000 ft AGL Halo or 600 ft static line jump
     
  4. Dec 16, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi Matt! Welcome to PF! :smile:

    I think the main factor is the incompressiblity of water …

    the sea has no "crumple zone" :biggrin:

    while marshmallow is a foam, and is highly compressible!

    Try googling compressible +marshmallow. :wink:

    (btw, good to know you aren't wasting your time in the pub discussing matters of no conceivable application to reality! :biggrin:)
     
  5. Dec 16, 2008 #4

    russ_watters

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    It would have to be a combination of both density and compressibility.
     
  6. Dec 16, 2008 #5
    100% marshmellows.

    Take a plunger and try to compress water - good luck.

    You can squeeze a marshmellow flat as a pancake with your fingers.

    (Im actually eating a bag of them as we speak! LOL) Yum.
     
  7. Dec 17, 2008 #6

    tiny-tim

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    skiing down marshmallow!

    oooh :!!) … gimme! gimme! :tongue2:​

    Can you ski down marshmallow :smile:, or would you just sink into it? :rolleyes:

    of course, you could survive by making a cave,
    and then eating your way out! :wink:
     
  8. Dec 17, 2008 #7

    Vanadium 50

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    After you land, it's a heck of a lot harder to drown in marshmallow.
     
  9. Dec 17, 2008 #8
    Lol :biggrin:

    Thanks all for responses, I think compressibility was the term / concept missing from our (very important and not at all time-wasting) discussion in the pub!

    And as for application to reality, well you just wait until I open the new Theme Park I've always dreamed of!

    Best - happy eating,
    Matt.
     
  10. Dec 17, 2008 #9
    just as a foot note ,, a 2hp motor attached to a Floating barge will make way as well as a speed boat with the same motor until as i recall the speed reaches about 3knots .. the water gets out of the way well enough upto that speed pretty much regardless .. at higher speeds a streamlined hull shape plays a greater role ..
     
  11. Dec 17, 2008 #10

    tiny-tim

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    eliminate global warming!

    Will this be a general Land of Chocolate theme park :tongue2:, or will it be specialising in food you can sink into, like marshmallow and candy floss and Baked Alaska? :smile:

    On a more serious note, why has nobody thought of eliminating global warming by covering the North Atlantic with marshmallow?

    This would vastly increase reflectivity, yet would not be an obstruction to shipping, since ships could just plough their way through it. :smile:
     
  12. Dec 17, 2008 #11

    FredGarvin

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    Hmmm...baked Alaska is served on fire...
     
  13. Dec 18, 2008 #12

    Dale

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    That depends, do you have a backpack full of graham crackers and chocolate instead of the parachute?
     
  14. Dec 19, 2008 #13
    It seems to me the marshmallow would be almost like quick sand. You would not be able to swim out, and then, and then the red ants come and.....well its too horrible to mention!
     
  15. Dec 20, 2008 #14
    Follow-up question: would you die hitting the marshmallow after a 30,000 foot fall?

    What depth - if any - of marshmallow would "save" you?
     
  16. Dec 20, 2008 #15
    I think graham crackers and chocolate is as good a reason as any for not having a parachute when jumping out of a plane.

    Mum packed the wrong thing again...
     
  17. Dec 20, 2008 #16

    Dale

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    :rofl:
     
  18. Dec 20, 2008 #17

    Dale

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    The height of the fall doesn't matter very much, only the conditions at impact. Not all falls are created equal, if you do a "spread eagle" fall you will hit much faster then if you did a "headfirst" fall. If we assume a hefty 100 kg skydiver, a respectable 60 m/s freefall, and a very squishy (barely able to keep him from sinking) 1000 N marshmallow cushion then the KE on impact is:
    1/2 100 kg (60 m/s)² = 180 kJ

    and the required stopping distance is
    180 kJ/1000 N = 180 m

    All of this changes for stiffer marshmallows, smaller skydivers, or different impact speeds, but I think that is a reasonable upper limit. Make it 400 m for an engineering safety factor of 2.2. I hope mom packed a LOT of graham crackers and chocolate.
     
  19. Dec 20, 2008 #18

    tiny-tim

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    terminal velocity

    ah … well, you would reach terminal velocity in air well before 30,000 feet …

    then the marshmallow would slow you down to terminal velocity in marshmallow …

    I think the two important questions are:

    i] what is terminal velocity for an average human through air?

    ii] what is the drag coefficient for an average human through marshmallow?

    From these (and http://en.wikipedia.org/wiki/Terminal_velocity), you can calculate the depth of marshmallow needed to bring you to a speed at which it would be safe to hit sugar rock. :smile:
     
  20. Dec 20, 2008 #19
    Well, assuming we model the marshmallows as massless springs with spring constant k, and you fall from rest at a height h, and assuming that the resistive force is approximately constant, ...

    mgh = 1/2 k x^2 => x = sqrt(2mgh/k) is the distance the spring compresses... so the height of the marshmallows

    The average force would simply be: F x = mgh => F = mgh/x = sqrt(mghk/2)

    So holding m, g, and h constant, we can find the maximum spring constant k if we know the maximum force F; in fact,

    k <= 2 (F_max)^2 / mgh

    If somebody wants to check the numbers, please do so. I'd be interested to see. F_max is probably online somewhere, and k can be found for a marshmallow by simple experiment. Then just put in your m, g = 10 m/s^2, and h = 10000 m and see how it turns out!
     
  21. Dec 20, 2008 #20
    My guess:

    F_max... a person will die, say, at 10 g's... so F_max = 10mg = 100m.
    Say m = 100 kg. So 100m = 10000 = 10^4.

    mgh = 100*10*1000 = 10^6.

    So k <= 100.

    So if the spring constant of a marshmallow is under 100 N/m, we may be in luck. Let's assume this is the case.

    Then the depth of marshmallows must be around sqrt(2*100*10*1000/100) = 141.

    So you'd need ~ 150 meters of marshmallows if my model is accurate (and we assume the worst-case spring constant; for more stretchy marshmallows, you could need many more, up to 10 times as deep).
     
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