Harriot, Thomas, Mss. 6784

List of thumbnails

< >
411
411 (206)
412
412 (206v)
413
413 (207)
414
414 (207v)
415
415 (208)
416
416 (208v)
417
417 (209)
418
418 (209v)
419
419 (210)
420
420 (210v)
< >
page |< < (269) of 862 > >|
537269
[Commentary:
The problem pursued in this and many other folios in Add MS 6784 is that of 'inclination' or 'neusis', as set out in Pappus, Mathematicae collectiones (Pappus , Book 7. For a statement of the problem see Add MS 6784 f. . ]
De
[Translation: On ]
prop.
[Translation: Proposition ]
Sit semicirculus ACB. et contingens BD: oportet puncto A ducere rectam
lineam ad contingentem BD ita ut intersecta CD sit æqualis datæ lineæ H
[Translation: Let there be a semicircle ACB and a tangent BD. It is required to draw a line from a point A to the tangent BD so that the intercepted line CD is equal to the given line H.
Ducatur FA perpendicularis ad AB. et producatur versus E.
Fiat FA Dimidium lineæ H. Agatur lineam FB* cui fiat æqualis FE.
centro A intervallo AE describatur periferia EC quæ secabit perferiam
ACB in puncto C. Per A et C Ducatur lineam usque ad D.
Dico quod CD est æqualis lineaæ H
[Translation: Draw FA perpendicular to AB and extend it towards E.
Make FA half of the line H. Construct the line FB* and make FE equal to it.
With centre A and radius AE draw the circumference EC which will cut the circumference ACB at the point C. Through A and C draw a line as far as D.
I say that CD is equal to the line H.
Aliter:
* a qua abscindatur FG æqualis FA.
Inscribatur AC æqualis GB et producetur ad D.
Dico quod CD est æqualis
[Translation: Another way:
* from which there must cut off FG equal to FA.
Inscribe AC equal to GB and extend it to D.
I say that CD is equal to the given line.
Aliter: et optime.
Producatur DB versus I. et fiat BI dimidium H.
Agatur AI et sit OK æqualis IB. centro A et intervallo
AK describatur periferia KC, quæ secabit periferiam ACB
in puncto G. Per A et C Ducatur linea usque ad D.
Dico quod CD est æqualis
[Translation: Anotoher way, and the best.
Extend DB towards I and make BI half of H.
Construct AI and let OK be equal to IB. With centre A and radius AK draw the circumference KC, which will cut the circumference ACB at the point G. Through A and C draw a line as far as D.
I say that CD is equal to the given line.
Nota quod Triangulum
ABI continet totum
effectionum huius
[Translation: Note that triangle ABI contains all the construction of this equation.

Text layer

  • Dictionary

Text normalization

  • Original

Search


  • Exact
  • All forms
  • Fulltext index
  • Morphological index