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Or rather, by deducting from each the length of LG, which is also given, it will merely be needful to adjust FD up to the straight line VG in such a way that FD together with 3/2 of DG is equal to a given straight line, which is a quite easy plane problem: and the point D will be one of those through which the curve BDK ought to pass.
From the point L let there be drawn to some point of the given line AK the straight line LG, which, being considered as a ray of light, its refraction GD will then be found.
To demonstrate the effect of the curve, let there be described about the centre L the circular arc AH, cutting LG at H; and about the centre F the arc BP; and in AB let AS be taken equal to 2/3 of HG; and SE equal to GD. Then considering AH as a wave of light emanating from the point L, it is certain that during the time in which its piece H arrives at G the piece A will have advanced within the transparent body only along AS; for I suppose, as above, the proportion of the refraction to be as 3 to 2.
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.
The medical faculty can have the benefit of my experience, and prescribe as follows for their rheumatic patients. "st. nt. o. lg. sl. S. r. = ther. z "Start at night on a long sleigh ride over a Siberian road with the thermometer below zero." A bouran arose in the afternoon of the second day, but was neither violent nor very cold.
Now we know that the piece of wave which is incident on G, advances thence along the line GD, since GV is the refraction of the ray LG. Then during the time that this piece of wave has taken from G to D, the other piece which was at S has reached E, since GD, SE are equal.
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