Welcome to today's science lesson!

I chose to base my code on the Periodic Table of the Elements. When Isaac posted his example in his "Crypto tricks" post and used "Helium" as his key, it was such a marvelous coincidence that I couldn't resist adding in "a key such as Isaac referred to in his last post." The key that I posted in that followup post was also Helium.

"#" = "Helium"

If you look at a table of ASCII codes, # is number 35. 33 is the first number on the aforementioned table of ASCII codes that gives a non-blank character. So, let us subtract 33 from 35. That leaves 2, and 2 is the Atomic Number of Helium. And that is also a marvelous coincidence, because that is not what my code is based on. The Atomic Weight of Helium is approximately 4.0026. Let us round that to the nearest integer, and subtract the number of protons in a Helium atom which is equal to the Atomic Number, in this case 2. That gives the number of neutrons in the most common isotope of Helium, which is also 2. Now replace the "#" with the symbol for Helium, "He", and that is the basis of my code.

I used the words "positive" and "neutral" in that second post to refer to the sub-atomic particles used arrive at the correct numbers.

Now, looking at my actual code: k ? ? ) é + k é ½ ; ) ( A 1

"k" is 107 on the ASCII table. Subtract 33 and you get 74. I had anticipated an extra layer of protection here, in that even if people knew what the method was, they would be unfamiliar enough with the elements that this would be a lengthy process of trial and error as they added various Atomic Numbers until they arrived at a suitable Atomic Weight. This particular aspect of my plan was foiled by Google, who will be happy to tell you all about Iodine if you enter a search for "74 neutrons." The Atomic Number for Iodine is 53. 53 + 74 = 127 which is the Atomic Weight of that element's most common isotope. The symbol of Iodine is "I" so "k" = "I".

Moving on, "?" is 181 on the ASCII table. Subtracting 33 will give 148. If someone was still under the mistaken impression that we were dealing with Atomic Numbers at this point, this would convince them otherwise, as the Periodic Table currently only goes up to 118. Solving for neutrons, we can find that we are looking at good old Americium, symbol "Am".

Continue this for each character in the code and

k ? ? ) é + k é ½ ; ) ( A 1

becomes

I Am Am O Dy F I Dy Ac Ti O N Co P

Clean up a little spacing

I Am A mODyFIDy AcTiON CoP

and you have my role claim. I know, "modified" is incorrectly spelled as "mODyFIDy" but that was unavoidable given the limitations of the character set I was working with. I trust that it is still recognizable.