Another way of doing text steganography that might not be as obvious would exploit fill-justified text written in fixed width fonts, the way some word processor

Another way of doing text steganography that might not be as obvious would exploit fill-justified text written in fixed width fonts, the way some word processors used to output documents in the old days before laser printers. If each line is filled to 80 characters and a given line has n words in it whose cumulative length is k, then it needs 80-k spaces distributed as equally as possible over n-1 gaps between the words. If 80-k is not a multiple of n-1, then there are t = (80-k) mod (n-1) gaps that need an extra space thrown in, and s = ((n-1) choose t) ways to assign them, hence the possibility of encoding floor(log_2(s)) bits in each line. I did this project for fun a long time ago and found pretty consistently a ratio of 137 bytes of text needed for each byte of concealed data. You might squeeze in a little more by considering multiple lines at a time.