Conference rusure::math

    I imagine that the degree to which the previous observation is a
    surprise depends on whether or not one realizes that "A Thousand
    Days" is just about three years.  Eleven thousand days is about
    thirty years.
    
    If you already knew the relation, there is no surprise.  That reminds
    me of the definition of "information content" that I once read:
    If the signal contains a message that is a surprise to you, its
    information content is greater than if you expected the message,
    which is to say that when you answer the telephone and hear the
    word "hello" there is little information in that word.
    
    								matt
    

Check out the chapter on "Number Numbness" in Doug Hofstadter's book,
"Metamagical Themas". He has a number of examples of this kind, and some 
insights on how to become more facile at dealing with the realities of 
ultra-large numbers.

Reminds me of the story about the student who slept during a lesson at school,
when the teacher exaplained about the universe going to explode in about
one billion years from now.

"What? what?" the student suddenly woke up and screamed.

The teacher repeated the explanation about the universe going to explode in
one billion years from now.

"Oh, what a relief - I thought you said one MILLION years..." the student said.



Two more examples:

1. If you take 1 gram of Hydrogen gas, and replace every atom with a grain
   of sand, you can cover all the US with 3 feet of sand.

2. If you take a piece of paper and fold it about 35 times, it will be so
   thick, it will reach the moon.

/Nitsan

When you said "a gram of hydrogen", I first thought of a cubic centimeter.

But then I realized that a gram of WATER would be about a cubic centimeter.

A gram of hydrogen would be alot bigger at standard temperature and
pressure.

Anyone know about how large a gram of hydrogen is ?

Practical related question:

	Given an inflated child's balloon with helium, assuming
	spherical shape, measuring diameter with ruler, measuring
	pressure by measuring how much a kilogram weight sinks
	when placed on top, and measuring how much weiight the
	balloon is capable of lifting, can anyone calculate
	what mass of helium is in the balloon ? Oh yes, feel free
	to weigh empty balloon before inflating too.

/Eric

    re .4
    
    Doesn't that depend on temperature?
    

    I read somewhere that a trillion dollars (the amount Reagan wants to spend
    on SDI) is enough to buy every ant in the free world a Volvo.
    
    John

    RE: .6
    
    How much is a Volvo?  Assuming $20K, then only 50 million ants would
    be proud new owners.  I imagine there are many more ants than that.
    In fact, depending on location, there may be that many in the typical
    neighborhood.  Of course there would be a small problem parking all
    those cars in my neighborhood.

Re .3:

>2. If you take a piece of paper and fold it about 35 times, it will be so
>   thick, it will reach the moon.

I bet you will find it difficult to find a piece of paper you can fold in
half more than 8 times (with alternate folds perpendicular to each other).
Note that you problem is not true without a constraint like this - I
could easily put 35 folds in a piece of paper as long as most of them
are in the same direction.

Re .4:

A mole of an ideal gas occupies 22.4 liters at STP.  Hydrogen is a diatomic
molecule (H2) under those conditions.  Therefore a gram of hydrogen gas
would be half a mole of gas molecules, and would occupy about 11.2 liters
if it had all the properties of an ideal gas.

Re .5:

Well, you'll get different results depending on whether or not you wait for
the gas in the balloon to cool off when placing a weight on top, before
measuring the weight's deflection.

I suspect that the hardest part is accounting for the balloon's elastic
properties when attempting to measure the pressure inside by putting
weights on it.  I'm not sure whether other factors, such as ambient
atmospheric pressure, don't enter implicitly into the calculation, despite
the fact that they aren't given in the problem.  Then again, I recall
a college roommate discovering that a textbook "heating of water as it goes
over a waterfall" problem contained extra information, so perhaps not.
				/AHM

Even more restrictions are necessary.  Merely requiring folds to be
perpendicular to each other still allows for easy folding of paper
35 times.  We need to restrict that the paper must be entirely folded
IN HALF each time.

I'm beginning to think that 35 times might reach the moon.  I folded
my paper in half 15 times and it won't fit in my 7-ft high ceiling
anymore so I had to stop :-)

/Eric

Well, actually there may be a small error in my original calculations.
Considering a paper which is 0.1mm thick, and assuming a "tight" fold
(with no "air" in the middle), you need about 42 folds to reach the moon.

This is true, no matter which side you fold, as long as you fold in half
each time. Note that you can CUT in half and paste instead of folding.

Who said we need high tech to reach the moon?

/Nitsan

Consider actually FOLDING a .1mm thick paper in half 42 times to reach the
moon, instead of cutting it.

An interesting question is how WIDE the paper has to be if, say, we want
it at least 1cm wide after last fold.

The answer is *not* just 2**42 centimeters wide, if we consider that in
order to successfully fold a wad of paper in half, the outer layer must
now span around the rest of the thickness.

If you work on this puzzle, you could approximate the outer fold to always
being a rectangular configuration, so actually two folds are involved. 
Gee, I wish I could draw this for you on the screen.

Or, you could model it as a circular bend as the fold, which is more like
what paper does when you fold it.

/Eric

    re: .10
    
    How thick is each paste layer?