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113.

[[image:  drawing of a masonry wall cross section E A D O with a beam A B C D projecting perpendicularly out to the right with a weight L suspended from C. A dotted line I down the center-line of the beam, and marks G and H above and below the dotted line.]]

 Let us also apply these reasonings to ye solid A B C D fastened perpendicularly in ye wall E A D O, and let us suppose that if we should draw ye height into ye base perpendicularly there must be 600 pounds to break it, I say that if A D is divided into 3 equal parts by ye points G, H, and that C D be to D H as 60 to 1 ye weight L need be but 10 pound to break ye solid, when according to Galileus it should be 15 pound, since C D is to D I as 60 to 1 1/2 or 40 to a unite, and that 600 is ye product of 15 by 40.

  To prove this proportion let us suppose as hath been explained above, that ye fiber towards A ought to be extended 16 very small parts to be broken, and that there must be a like extension to break ye fibers towards G, I, & H, it is evident that these last do not resist with all their force to hinder ye breaking of ye fiber towards A, and that if they resist in proportion to their distance from ye point D, and if there must be 10 pounds in L to break ye fiber in A, there needs only 12 pounds to break ye fiber in G, 8 to break he fiber in I, and 4 to break ye fiber in H, but because that when ye fiber in A is broken, ye fiber in G shall be extended but 12 parts, that in I but 8, and that in H but 4, this makes another like proportion, and so instead of 12 pounds to break ye fiber towards G, there needs but 9 pounds ye 3/4 of 12, and 4 pounds to break ye fiber towards H.  Now 12 is ye mean proportional between 16 and 9, and 4 between 16 and 1, and by consequence these numbers 1, 4, 9, 16 being squares if we conceive ye length A D to be infinitely divided, ye resistances of all ye fibers would be in proportion as a series of squares from a unite, but if we take such a square numbers as we please in a series beginning at a unite 3 times their summ less by ye triangular number wch corresponds to ye half term of ye progression, shall be equal to ye product of ye greatest square by ye number of ye progression beginning with a cypher, and that triangular number exceeding shall be to ye half product according to this infinite progression 1/2 1/4 1/6 1/8 1/10 &c therefore this excess to infinity shall be a nothing, and by consequence all ye square to infinity shall be together but 1/3 of ye square equal to ye greatest adding 1 for ye first term 0 of ye progression, like as if we take ye following progression 0, 1, 2, 3, 4, 5, 6 &c ye sum of all.  These numbers is ye half of ye greatest by ye number of ye proportion:

  To prove by induction this property of a series of squares, let us take ye unite wch is ye first square, ye triple of ye unite is 3, ye unite multiplyed by ye number of terms of ye progression, 0, 1, is 2 wch is less than 3 by ye first triangular number 1 wch is 1/2 of ye number 2, 1 and 4 together are 8, 3 times 5 is 15, ye product by ye progression 0, 1, 2, is 12 less than 15 by 3 ye second triangular number and wch is 1/4 of 12 55 is ye summ of 5 first squares 3 times 55 is 165 ye greatest square 25 multiplyed by ye 6 terms of ye progression, 0, 1, 2, 3, 4,,5, 150 less than 165 by 15 wch is 1/10 of 150.

Transcription Notes:
mandc: Reviewed and amended the image description, changed J's to I's. Image: http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/QERNH1MN/pageimg&viewMode=images&mode=imagepath&pn=336&ww=0.2182&wh=0.1136&wx=0.7127&wy=0.0577