In the large diagram, the distance, A C, on the horizontal line
corresponds to the distance, N B, on the instrument. At A erect a
vertical line, and mark upon it a point B such that B C shall be
exactly eighteen times any convenient unit, B I. In the illustration B
C is 26 inches, and B I is 11/2 inches, so that B C is 27 inches in
length. About C as a center describe the arcs, B L, I K, and through I
draw a vertical line, cutting B L in D; draw the radius D C, cutting
the inner arc, I K, in J, through J draw another vertical, cutting B L
in E, and so on.
In the triangles, A B C, 1 D C, 2 E C, we have B I = D J = E F = 1/18
of the hypotenuse in each case, therefore the bases, A C, 1 C, 2 C,
are divided in the same proportion, as required, at the points 1, 2,
3. And we might extend the arcs, B L, I K, and repeat the above
operation until all the frets were located. But should that be done,
the diagram might become inconveniently large, and some of the
intersections might not be reliably determined. In order to avoid
this, the spacing of the outer arc may be stopped at any convenient
division, as L. The vertical by which that point is determined cuts B
C at B', and through B' a new arc, B' L', is described.
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