This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

Let a b and c d (Fig. 410) be given axes. Between one of the foci, f and f, and the centre e, mark any number of points, at random, as 1, 2, and 3; upon f and f, with b 1 for radius, describe arcs at g, g, g, and g; upon f and f, with a 1 for radius, describe arcs intersecting the others at g,g,g, and g; then these points of intersection will be in the curve of the ellipsis. The other points, h and i, are found in like manner, viz.: h is found by taking b2 for one radius, and a 2 for the other; i is found by taking b 3 for one radius, and a 3 for the other, always using the foci for centres. Then by tracing a curve through the points c, g, h, i, b, etc., the ellipse will be completed.

Fig. 410.

Fig. 411.

This problem is founded upon the same principle as that of the string. This is obvious, when we reflect that the length of the string is equal to the transverse axis, added to the distance between the foci. See Fig. 405, in which cf equals ae, the half of the transverse axis.

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