What exactly happens, when I do a bit rotation? I'm sitting in my office and having difficulties to get to know that exactly happens, when I do a bit rotation of a binary number. 
An example: 
I have the binary number 1110. By doing bit rotation, I get 1101, 1011, 0111 ... 
so I have 14 and get 7, 11, 13 and 14 again.
I can't get a rule out of it... can somebody help me?
Excuse my bad English, I'm just so excited about this problem.

If you have an $n$-bit binary number $m$, possibly with leading zeroes, a circular left shift gives you $$2\left(m-2^{n-1}\left\lfloor \frac{m}{2^{n-1}}\right\rfloor\right)+\left\lfloor \frac{m}{2^{n-1}}\right\rfloor=2m-(2^n-1)\left\lfloor\frac{m}{2^{n-1}}\right\rfloor,\tag{1}$$ and a circular right shift gives you $$\left\lfloor\frac{m}2\right\rfloor+2^{n-1}\left(m-2\left\lfloor\frac{m}2\right\rfloor\right)=2^{n-1}m-(2^n-1)\left\lfloor\frac{m}2\right\rfloor.\tag{2}$$
Changing the base from $2$ to $b$ merely requires changing $2$ to $b$ everywhere in $(1)$ and $(2)$.
