which is on our chain also is under F, giving two values for one letter. Similar reasons reject most other positions: those left in are:
of these 2 contradicts the S + 3 = Y fit, 1 and 9 contradict the good tetra I + 3 = D: they are, therefore, unlikely to be right. 6 however is extremely interesting:
we scritched L R X C E and got the position:
In this alphabet F + 7 = Q and we have a good fit at this distance which was not used in our scritching. We have in fact 'picked up' this extra fit and thereby obtained a considerable factor in favour of this alphabet. We now fit in S in such a position that S + 3 = Y in accordance with our hexagram:
This makes S equal U and we notice at once that we pick up the tetragram D + 16 = U, not a very good tetra but valuable as a further contribution.
We now have to see for which wheels this alphabet is valid. The Turnovers on the wheels are in the positions marked on the alphabet above. From the fit R W C + 13 = RWL we know there is no turnover between C and L; if there were a turnover, the