What is the Value of a Snowflake?
Let’s sketch out an approach that snow-ologists might refine, by digitizing a value for a snowflake.
Thangham notes that snowflakes have “six-fold symmetry.” Let’s concentrate on one-sixth of a snowflake, isolating the area between two major arms. That should be an equilateral triangle, since the angle at the centre is 360/6=60 degrees and the sides from that vertex have the same length.
The argument is that any one triangular plate, or lace pattern, is repeated for the whole snowflake. Therefore we only need to examine one triangle per snowflake.
Charles Q. Choi states that the largest snowflake’s diameter is about 50.8mm, or two inches. So the length of one side of this equilateral triangle is 25.4mm. The area of any triangle is one-half the base times the height, where the height is the distance of a line from the opposite vertex that meets the base at a right angle.
However, the area of any equilateral triangle is (3^(1/2)) * (s^2) /4, after using the Pythagorean theorem to determine the height from the side ‘s’.
By rounding down the length of the side to 25mm and approximating the square root of three as 1.73, the area of this triangle is approximately 270.3 square millimetres.
Resolving a Snowflake’s Value
Orange Coast College introduces microscopes by saying that the “resolution limit of the human eye is about 100 micrometers” or 0.1mm. As well, the “compound light microscope has a resolution limit of about 0.2 micrometers under ideal conditions.” So a microscope can provide up to 500 times greater resolution.
Let’s use the naked eye to resolve each snowflake. If our limit is 0.1mm, then there are about 250 pixels along the side of the snowflake triangle, and 270.3*10*10 = 27030 pixels inside the area of that triangle.
We can digitize the snowflake’s triangle by saying that each pixel is one bit of information: either it has a grain of ice, or not. Let ‘one’ represent the grain of ice, and ‘zero’ its absence.
We can create a unique number for the snowflake by weaving from the central point, whose value is almost certainly ‘1’. The next bit is on the left arm beside the centre; then use the next bits heading toward the other arm. On either arm, step one bit away from the centre and then record the values of the bits heading to the other arm.
The total information is 2^27030 = 10^821.7846 approximately. (See How to Convert the Base of an Exponent with Logarithms for this useful conversion technique). Yes, we need an 822-digit base-10 number to describe a perfectly flat snowflake’s lacey pattern. Let’s round up to 10^822 different possible snowflakes; that’s the ‘d’ in Weisstein’s birthday problem.
How Many Snowflakes Fall per Year?
Choi quoted expert Jon Nelson’s estimates for the mass of one snowflake (“1 millionth of a gram” = 10^-9 Kg) and the total snowfall per year (“1 million billion” = 10^15 Kg). So 10^24 snowflakes fall per year. Let’s invite them all to Weisstein’s party.
The Probability of Identical Digitized Snowflakes
Now we have ‘n’ = 10^24 snowflakes per year, to be compared on ‘d’ = 10^822 values. The equation for the probability that at least two do match is “P = 1 – ( 1 – ( n/(2*d) ) )^(n-1)”.
By substitution, “P = 1 – ( 1 – ( (10^24)/(2*(10^822)) ) )^((10^24)-1)”.
Let’s examine some of the terms.
The “n/(2*d)” term, by itself, is “(10^24)/(2*(10^822))” = “1/(2*(10^(792)” = 0.00…005 with 791 “zeroes” between the decimal point and the ‘5’. Subtracting that from ‘1’ leaves 0.99…95 with 791 “nines”.
To a reasonable approximation, taking “0.99….99^((10^24 )- 1)” removes the 24 least-important “nines”. We now have “0.99…99” with only about 767 “nines”.
Finally, we have the leading “1 – 0.99…99” = 0.0…01 with over seven hundred “zeroes” after the decimal point.
Assigning these values to this mathematical model makes it extremely improbable that any two snowflakes are identical.
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