22H Page 68
Therefore lim → Χ2 + x, and most of the bulge of 5 in ΔD comes against Χ2 DOTS. These two facts are shown by the following count of a Gurnard message.
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Fig. 22 (XI) |
(e) ΔD against B.M..
By an argument similar to that in (b) it can be seen that
P (ΔD = Θ | BM = x) = P (ΔD = Θ | TMx) | (H6) |
P (ΔD = Θ | BM = .) = { ½P (ΔP = Θ) + ½P (ΔD = Θ | TMx)} | (H7) |
(f) ΔD counts on 1 and 2 impulses.
From (E4) we get | δij ≡ Pβ[illegible] ( ΔDij = dot) = β'ij.Πij | ||
But | β'ij = β | ||
δij = Πij.β | (H8) |
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