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44.

[[image: three dimensional drawing of a long, square "J" shaped tube, the long left branch is labeled A at the open top end, B across the bottom branch and C at the top of the short right branch. Halfway down the left branch is labeled H. Then there are five dotted lines across both branches labeled as follows: P O, N M, G F, L I, D E]]

of ye air E C is to ye extent I C. But if we would reduce ye same air into ye space M C of 3 inches, wch is ye fourth of E C; we must put 84 inches of mercury into ye branch D A above ye Horizontal Line M N, and we shall find that proportion by he following calculation: as M C 3 inches is to M E 9 inches, so 28 inches ye weight of ye atmosphere, is to 84 for by changing 84 shall be to 25 as 9 is to 3, and compounding 84 more 28 ^[[that is]] [[strikethrough]]one shall[[/strikethrough]] one [[strikethrough]]?[[/strikethrough]] 12 shall be to 28 as 9 more 3  ^[[that is]] [[strikethrough]]one shall[[/strikethrough]] E C to 3, and if we would know what height of ye pipe it is necessary to reduce that air into ye space O C at one inch, we must say as O C one inch is to O E 11 inches, so 28 inches of mercury ye weight of ye Atmosphere to 308 and 308 shall be there vertical height wch must be given to ye mercury above ye point O or P by wch is known that ye branch D A must be heigher than 308 inches, that is that it must be above 320 to ye end that there may remain a place above ye mercury to hinder its running out.

Transcription Notes:
mandc: Read "9 more 3" as "9 + 3." http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/QERNH1MN/pageimg&start=131&viewMode=images&mode=imagepath&pn=174&ww=0.0598&wh=0.2416&wx=0.8671&wy=0.2198