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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.
Or rather, having taken AS equal to 2/3 of AT, I observe that 3/2 of GD ought to be equal to 3/2 of SB, plus BQ; and, deducting both of them from FD or FQ, that FD less 3/2 of GD ought to be equal to FB less 3/2 of SB. And this last difference is a given length: and all that is required is to draw the straight line FD from the given point F to meet VG so that it may be thus.
If "the law" says that Tom still owns the boat, this makes me laugh: for how can Tom come to own two boats' good by the FD done on one only? Jack is ploughing the sea with a ship: just there he owns the sea, has taken, is moving, it for his good. He does not own the sea before, nor the sea behind, him: for the motions behind made by him have ceased to do good.
The two investigators had worked all day and hadn't come up with the slightest indication of an answer. This added the final section to my now voluminous report on the Lubbock affair. The next morning as I rode to the airport to catch an airliner back to Dayton I tried to put the whole puzzle together. It was hard to believe that all Fd heard was real.
Now it is easy to show that this wave will touch the arc BP at this point C. For since, by construction, FD + 3/2 DG + GL are equal to FB + 3/2 BA + AL; on deducting the equals LH, LA, there will remain FD + 3/2 DG + GH equal to FB + 3/2 BA. And, again, deducting from one side GH, and from the other side 3/2 of AS, which are equal, there will remain FD with 3/2 DG equal to FB with 3/2 of BS. But 3/2 of DG are equal to 3/2 of ES; then FD is equal to FB with 3/2 of BE. But DC was equal to 3/2 of EB; then deducting these equal lengths from one side and from the other, there will remain CF equal to FB. And thus it appears that the wave, the semi-diameter of which is DC, touches the arc BP at the moment when the light coming from the point L has arrived at B along the line LB. It can be demonstrated similarly that at this same moment the light that has come along any other ray, such as LM, MN, will have propagated the movement which is terminated at the arc BP. Whence it follows, as has been often said, that the propagation of the wave AH, after it has passed through the thickness of the glass, will be the spherical wave BP, all the pieces of which ought to advance along straight lines, which are the rays of light, to the centre F. Which was to be proved.
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.
I say to Jack "take a cigar"; he loudly says "yes!", but does not move it to his mouth, nor moves his mouth to it; instead, he moves a pen to his mouth; this makes me laugh: he has not taken a cigar. Jack is catching fish in a boat; Tom owns the boat: so Jack gives fish to Tom, until Jack's FD done on the fish is equivalent to Tom's FD done on the boat; and now Jack owns the boat.
Which is a problem quite similar to that which served for the first of these constructions, where FD plus 3/2 of GD had to be equal to a given length.
Animals and plants, by taking things from the planets presented to them, by moving things, raise Nature into wealth, and own things. To say "own" is to say "take"; to say "take" is to say "motion", i.e., the doing of work: "work done" being FD, i.e., Force used into Distance moved-over.
I cannot own the air: it is no good; I own the air in my lungs, having taken, moved, it, done FD on it: it is very good; and I own the air which, doing FD, moving to my face, I do not evade, but accept, take: it is very good.