Harriot, Thomas, Mss. 6784

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[Commentary:
There is a reference on this page to Proposition II.4 from Conicorum libri quattuor (Apollonius . ]
Alhazen. pag.
Ducere rectam (ad) a puncto (a)
ut (ed) sit æqualis data (hz
[Translation: Draw a line (ad) from the point (a) so that (ed) is equal to a given (hz).
Ap: 4,2i. fiat per punctum t et hxz
hyperbola. tp. Continuetur tx ad o
et fiat xo=xt.
fiat per punctum o et nxk
hyperbola contraposita ou
Tum a puncto t intelligatur duci tr
quæ sit minima omnium quæ duci
possunt ad sectionem rocu.
Si tr esset sit æqualis diametro bg et
t fiat centrum et periferia transeat
per r, non secabit sectionem sed
tanget in puncto r.
Seu cum bg sit >tr. centro tintevallo
bg describatur periferia cγ quæ secabit
oppositum sectionem in duabus punctis c et \gamme.
Agantur tc, tγ: utræque æqualis bg
[Translation: Apollonius II.4. Through the point t and hxz make a hyperbola tp.
Continue tx to o and make x0=xt.
Through the point o and nxk make the opposite hyperbola.
Then from the point t there is understood to be ocnstructed tr, which is the minimum of all that can be constructed to the segment rocu.
If tr is equal to the diameter bg and t is the centre, and the circumference passes through r, it will not cut the segment but touch at the point r.
Or when bg>tr, with centre t and distance bg, draw a circumference cγ which will cut the opposite segment at the points c adn γ.
Construct tc and tγ: both are equal to bg.
tc, secabit zn in q. et hl in f.
[…]
Dico quod: ed=hz
[Translation: tc will cut zn at q and hl at f.

I say that ed=hz
(per construct.
[Translation: (by the construction

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