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116.

  A cane of glass of 5/3 of a line thick is broken by its proper weights at 6 foot (de Saillis)

  A cylinder of black marble of 5 lines diameter sustained horizontally 190 pounds that is to 10 1/2 pounds at ye distance of 48 lines, ye square of 5/3 is 25/9 its product by a foot long or 144 lines is 1600 or 400 lines therefore ye feet weigh 2400 cubic lines, as 14 is to 11 so is 2400 to 1886 lines, and because [[strikethrough]] and [[/strikethrough]] a cubic inch or 1728 lines weighs 2 ounces 1 dram, 1886 lines shall weigh about 2 ounces 3 drams.

  The half of ye length of 6 feet is 36 inches or 432 lines as the third of 5/3 of a line viz 5/9 is to 432 so is 2 ounces 1/3 to 1814 wch divided by 16 ounces gives 113 pounds 6 ounces wch ought to be ye weights to support perpendicularly that cylinder of glass of 5/3 of a line.

  A rod of glass of 1 3/4 of a line thickness and 12 inches long being placed upon 2 rules 9 inches distant one from ye other and of an inch large and thick; and being charged at ye middle with 1 3/4 pound suspended by a cord is broken in ye middle: a like rod placed after ye same manner but fastened by its ends between ye two rules and 2 small flat pieces of wood of ye same largenesss as ye rules, is broken by 3 pound 1 ounce hung at its middle, ye rupture is made at ye two ends 1 joyned ye rules and likewise one of ye ends was broken 3 lines within futher than ye point of support;  so that we may take it for a rule that ye two extremitys next ye support are broken in this last case and by consequence there is need of twice as much force as when ye extreems are free and it is broken in ye midle.

  A like rod placed with its midle upon an edge of a knife (we put Spanish wax towards ye ends to hinder ye slipping of ye cords wch should sustain ye weight and to make their distance wch was 9 inches) there needed but 1 1/2 pound and about 3 ounces to break it, that is to say that having put at each end 2 weights each of a pound less by 2 ounces and [[?9/2]], it was broken 3 lines from ye edge, there was a white mark to mark ye middle of ye rod.

  A sword blade placed by its end in a hole oblique from below upwards supported 68 pound and a small plate of from 80:

[[image:  Drawing of two walls G and F supporting a beam A B between them and a weight L hanging from the mid point of the beam E.]]

  It is manifest that if a solid A B is broken by a weight L suspended at its midle E;  being supported by ye ends upon ye two rules J and F, that it ought to break also if ye support is in E and ye two powers in A and B equal to to another and both together to ye force of ye weight L since it is always ye same effort wch is made in E.  Galileus hath demonstrated that ye same weight wch breaks it in E will break ye solid of ye same thickness fastened in a wall to ye point A if its length be equall to A E, whence it follows what I have often found by experience viz that a glass plate A B of 12 inches long placed and supported by its ends being broken by one pounds 10 ounces and 5 drams is also broken by 3 pounds 5 ounces 4 drams when ye ends are inserted in ye supporters because they then ought to break in A and B joyning to ye supports and because these 2 resist by their 2 ends twice as much as E A alone by its end A, there must be a double weight in L.

  The same author also demonstrated that if ye supports are in a double distance ye half of ye weight wch was in E will suffice to break ye solid, whereas ye reason is that ye lever becomes 2 times as long, and ye weight by consequence hath 2 times as much force ye counter lever not changing;  but if ye solid is twice as thick there must be 4 times ye weight because one side there is twice as many parts to disjoyn and so ye force of ye lever is half diminished, wch makes that ye weight ought to be quadruple, and generally ye weights ought to be in a double proportion of ye thickness.

Transcription Notes:
mandc: Reviewed and amended image description. By the math (432 lines/36 inches = 12), one ofMariotte's "line" equals 1/12 of an inch.