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Henry Bushby Transcription pages

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265


Examination of I3.


[diagram][right of diagram] I3 2b[over]4c 1 [sigma] 2 not.
gives I2 2b[over]2c
[under diagram] 4c-2c = 2c. 1z+1z = 2z.
[line across page]
[diagram] I3 2b[over]4c. 1 [sigma] 2 Oro. gives
6c i. k. 2z.
4c+2c = 6c. 1z+1z=2z.

[diagram][to?] I3 2b[over]4c 1 [sigma] 3 not gives
I[2/z?] 3b[over]3c (overhand)
4c-1c = 3c. z=z.

[diagram scribbled out and diagram] [right of diagram] I3 2b[over]4c 1 [sigma] 3 Oro.
gives 5c. T.H.
[under diagrams]4c-1c [corrected with vertical pencil stroke to] 4c+1c = 3 [overwritten in pencil] 5c. z=z.


[diagram] = [diagram] [right of diagram] I3 3b[over]6c. 1 [sigma] 2 not.
gives 5c. i.k. zz.
[under diagrams] 6c-1c= 5c. 3z-1z = 2z.

[diagram] = [diagram] [right of diagram] I3 3b[over]6c 1 [sigma] 2 Oro.
gives 8c i.k. (^4c T.H. + ring)
[under diagrams] 6c+2c = 8c. 3z-1z = 2z.

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