Journal of the Society of Motion Picture Engineers (1930-1949)

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518 FKAYNE November The voltage developed across the track in the phototube is proportional to the length S uncovered in the clear area of the track. The shaded area is considered sufficiently opaque to contribute no voltage. Now S = Y sec « (27) where S = the length of the scanning line a = the azimuthal deviation of S, and u = S sin a (28) so that S = a0 sec a (l + sin ^ (x + S sin u)\ (29) Multiplying each side by sin a, and adding x to each side x + S sin a = x + a0 tan a + a0 tan a sin — (x + S sin a). (30) A Let y = -£ (x + S sin a) ? (x + a0 tan «) (31) A ft Then (30) becomes or y = z + ft sin ?/. This simplified form is identical with that previously developed7 for the exposure of a variable-density sound track with a two-ribbon light valve and the solution takes the same form. Thus, y — z can be expanded into a Fourier series of z having sine terms only. Thus, y — z = ^ an sin nz. (33) l