For this reason, the results may be believed. The British doubted one weak
character in Delta X5 shown on the flag sheet, tried it as a dot (so there
would be 12 crosses) and tried to integrate to a X5 wheel. This failed.
(See flag sheet, Page 85.) So they tried the doubted character as a cross,
as the score actually indicated it to be, and reversed the delta X5
pattern to get 12 crosses. This integrated.
The delta X5 pattern was then pushed through each rectangle.
It was pushed through in inversed form, because they hadn't versed Delta-X5,
as yet (Pages 86-89.) This gave embryonic Delta-X wheels on all impulses.
These embryonic wheels were then scribbled down on the wheels sheet shown, in
true versed form. See wheels sheet, Page 90. In doing this they doubted
any character which shoved less than 3 pips, and did not record it. Since
lOLog10 of (1 + Beta)/(1 - Beta) equals 3 decibans if b = 2/3, the average
rectangle pip is worth three decibans, and they therefore were doubting
any character of less than 9 decibans.
The embryonic wheels of the wheel sheet were then held up to
the Turingery sheet, and the delta key on every impulse excepting the second
wan de-delta-chied. (The known X2oneback limitation would give a good X2 by other
means, so it was not used) Thus the resultant was deltaPSI' 1,3,4,5.
The second impulse of delta psi prime was then "counted out."
This was done by examining the other delta psi prime impulses at each character
and if they were dots, to assume the motor key was a dot and therefore the
unknown delta-psi-prime impulse had to be a dot. (See the count sheets and
also the Turingery sheet.) Scoring was done as follows:
If other deltaPSI' impulses gave: Then the deltaPSI' impulse in question was
scored:
4 dots 0 crosses 16 db., favor dot.
3 dots 0 crosses 12 db. in favor of a dot
2 dots 0 crosses 7 db. in favor of a dot.
1 dot 0 crosses 3 db. in favor of a dot.