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Updated: June 11, 2025
Having taken any point G in the line AK, and drawing the straight line IGL, its part GI will represent one of the incident rays, the refraction of which, GV, will then be found: and it is in this line that we must find the point D, one of those through which the curve DG ought to pass.
Let LG be a ray falling on the arc AK. Its refraction GV will be given by means of the tangent which will be drawn at the point G. Now in GV the point D must be found such that FD together with 3/2 of DG and the straight line GL, may be equal to FB together with 3/2 of BA and the straight line AL; which, as is clear, make up a given length.
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