Math Is Fun Forum

I truly believe it is uncrackable because you cannot tell which letters are the message and which are fillers, which have been selected from a list that takes into account letter frequencies. It is a 26 letter wheel on top of a 26 letter wheel. But you don't know the wheel start  point, or how it rotates, or which letters in the given block are the message. Were I to send a second message, the letters forming the message would be differently located, of course. How can it be cracked? from a theoretical point of view? Since you don't even know how many letters comprise the message. The wheel start point and where the message starts, and the number of filler letters between message letters is all stated within the block so the recipient can decode it quickly and easily. Comments, please.

Here is my first substitution encrypted message. There are the same number of letters in each line
but the width of the letter spacing makes it appear as though there are not. I have used a randomly ordered alphabet on
an inner wheel and a different randomly ordered alphabet on the outer wheel. No more clues at this stage.

fhmxvwzwqcdicnxiyzwtyelazwczsyewdevyzmxhicuhvmhssl
qvvhwewjzzchvcsjcylenucxijsvlezicjzeyykplhvztjhzcz
uetvwzoecjhooujbxpysczgwtwfkwgsvfwvikxhxcecvlfxwne
iweujtjwywjtrczewcmrdidejeikourfzwjyiwfcocypwjishs
ecfehsdziyescqfwthwclwpgszvxwxjgvwzmzefykwgizhbijc
zeeiwsqlvyviwjxoxelejvhzjkwcvryehyewzjmewwhsajzjuz
ggmepidujviamzeavwwyzcwdyywwcfmufyzshmwwwvgipzwhbm
hnzcsclaledziwlehwswzzsizzjjdlveujvwszfsgjiaeimhvx
ejibyvtedycwfhwpcuiwczfhhfvavjdkzxsvfexsvjmwjwdumq
kzzhpzwiwxlfcvxyyczwhixwjwpzzrhfjwvwqsxwhiwzjhszee