. Carnegie Institution of Washington publication. METHODS OF PERIODIC ANALYSIS. 91 Now, taking analyzing lines aa1 and bbl in figure 31 as horizontal, and letting the sweep be inclined as a small angle 5 with the analyzing lines, the number of lines required to cross the sweep in the direction ab perpendicular to analyzing lines will be increased and hence the value in years between two analyzing lines will be decreased; hence — cos 5 = years per line from a to 6. If the fringe is perpendicular to the analyzing lines, its period is the distance ab in years and we have for this special case: ys
. Carnegie Institution of Washington publication. METHODS OF PERIODIC ANALYSIS. 91 Now, taking analyzing lines aa1 and bbl in figure 31 as horizontal, and letting the sweep be inclined as a small angle 5 with the analyzing lines, the number of lines required to cross the sweep in the direction ab perpendicular to analyzing lines will be increased and hence the value in years between two analyzing lines will be decreased; hence — cos 5 = years per line from a to 6. If the fringe is perpendicular to the analyzing lines, its period is the distance ab in years and we have for this special case: ys ,, p\=— cos o. i. FIG. 31.—Diagram of theory of differential pattern in periodograph analysis. If, however, the fringe takes some other slant, as the direction ac, making the angle 6 with the analyzing lines, then the period desired is the time in years between a and c. That equals the time between a and 6 less the time from b to c. Now be in years would equal ab cot 6 except for the fact that the horizontal scale along be is greater than the vertical scale along ab in the ratio - '— and therefore a definite space sin o interval along it means fewer years in the ratio of !!E_. Hence we have: cos d be (in years) = ab (in years) tan 5 cot 6 or P = pi(i — tan 5 cot 6) which is the period required. The separation of the fringes needs to be known at times in order to find whether one or more actual cycles are appearing in the period under test. In figure 31 ab = ad = sin 5 — s sin (0—5) ae = — sin d which is the width Please note that these images are extracted from scanned page images that may have been digitally enhanced for readability - coloration and appearance of these illustrations may not perfectly resemble the original Carnegie Institution of Washington. Washington, Carnegie Institution of Washington
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