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Re: physics/pedagogy of coffee-mixing



Here is the output of my little simulation.
It confirms that each time the output is
slightly different but that the fraction of
tea in coffee is always the same as the
fraction of coffee in tea. The listing of
the program is also shown. Each cup has the
capacity of 1000 balls, the capacity of
a spoon was chosen to be 150 (for this
output). The initial conditions can be
changed to create the asymmetry. I will
do this after posting this message and
will display the output very soon.

Cup A has 9.2 % tea, Cup B has 9.2 % coffee
Cup A has 9.1 % tea, Cup B has 9.1 % coffee
Cup A has 9.1 % tea, Cup B has 9.1 % coffee
Cup A has 9.4 % tea, Cup B has 9.4 % coffee
Cup A has 8.7 % tea, Cup B has 8.7 % coffee
Cup A has 8.9 % tea, Cup B has 8.9 % coffee
Cup A has 9.2 % tea, Cup B has 9.2 % coffee

program balls
! cup A initially black, cup B initially red
! 1st transfer from cup A to cup B.
! 2nd transfer from cup B to cup A.
DO ! infinite number of repetitions
let sp=100 ! spoon size (such as 10 or 52)
if sp<1 or sp>999 then
print "illegal spoon size (must be 1 to 999)"
stop
end if
let sp=int(sp) ! make it an integer
let BLK_A=1000-sp ! Black in cup A after transfer #1
let BLK_B=sp ! Black in cup B after trasfer #1
let RED_B=1000 ! Red in cup B after transfer #1
let pB=sp/(1000+sp)! initial prob. of black in cup B
let BLK_S=0 ! to count blacks in 2nd spoon
let RED_S=0 ! to count reds in 2nd spoon
! rnd (a random between 0 and 1)
for i=1 to sp ! compose the second spoon
if rnd<=pB then
let BLK_S=BLK_S+1 ! into the spoon -->A
let BLK_B=BLK_B-1 ! from cup B
else
let RED_S=RED_S+1 ! into the spoon -->A
let RED_B=RED_B-1 ! from cup B
end if
let pB=BLK_B/(RED_B+BLK_B) ! FOR NEXT PICK
next i
! next empty 2nd spoon to A
let BLK_A=BLK_A+BLK_S
let RED_A=RED_S
print "Cup A has ";100*RED_A/(RED_A+BLK_A);" % tea, ";
print "Cup B has ";100*BLK_B/(RED_B+BLK_B);" % coffee"
LOOP
end

brian whatcott wrote:

At 16:00 6/25/00 -0400, you wrote:
...
....John is providing an amusing illustration of the dangers of using
simplifying assumptions, another favorite device of physicists....

And that is how I show that his incisive qualititative analysis is wrong.
...
my solution holds to an arbitrarily high degree of approximation, and the
more carefully you follow the directions the more accurately it holds.

Let's try again.

For any partition of quantities into two vessels other than exact halves,
John's solution is wrong.

A numerical example again:
first cup 501 red balls, second cup 499 red balls, 1000 black balls.
This represents two different mix ratios.

This is not exactly hard to see. How could it happen?
If the 'spoonful' quantities into the first cup turn out to contain fewer
balls than the spoonfuls into the second cup. (This is inevitable in
physical partitioning at the molecular level)


In Ludwik's model of discrete balls, my solution holds exactly.

I used Ludwik's ball model again to show the error in John's assertion.

Sincerely