Harriot, Thomas, Mss. 6784

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18191
[Commentary:
The problem pursued in this and many other folios in Add MS 6784 is 'the cutting-off of an area', as set out in Pappus, Mathematicae collectiones (Pappus , Book 7. For a statement of the problem see Add MS 6784 f. . ]
1. problematis de spatij resectione 4a demonstratione et
[Translation: Problems of cutting off of an area, demonstration and most ]
Quatuor possunt esse
resectiones ab uno
puncto dato et duabus
solummodo lineis datis;
secundum quantitatem
spatij dati
aliquando tres:
semper
[Translation: There are four possible sections from one given point and just two given lines, according to the size of the given area; sometimes three, always two.
Exegesis.
producetur AC usque ad L et fiat
CL æqualis DE cui etiam sit æqualis
sit AM. Around M intervallo AL
agatur periferia FZO.
Ab E puncto, per O et F Ducatur rectæ
EF et EW.
Dico
[Translation: Exegesis.
Let AC be extended as far as L and make CL equal to DE to which AM is also equal. Around M with radius AL there is constructed the circumference FZO. From the point E, through O and F there are drawn the lines EF and EW.
I say ]
Poristice et inde
synthesis manifesta The poristic, and hence the synthesis, is obvious.
Si DE sit æqualis DC
tum:
CO=CG
Nam CLED erit quadratum
et punctum E [???]
[???].
Sit DE sit maior DC. tum
FC est maximum secta [???]
et CG
[Translation: If DE is equal to DC, then CO=CG.
For CLED will be a square, and the point E [???].
If DE is greater than DC, then FC is the maximum segment and CG the minimum.

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