27784
Eæemp. V.
Sit DEB _Quadrans Circuli_, quem tangat recta BX;
tum linea
11Fig. 120,
121. AMO talis, ut in recta AV utcunque ſumptâ AP, quæ arcum BE
adæquet, erectáque PM ad AV normali, ſit PM æqualis arcûs BE
tangenti BG.
11Fig. 120,
121. AMO talis, ut in recta AV utcunque ſumptâ AP, quæ arcum BE
adæquet, erectáque PM ad AV normali, ſit PM æqualis arcûs BE
tangenti BG.
Sumpto arcu BF = AQ:
&
ductâ CFH;
demiſſis EK, FL
ad CB normalibus; nominentur CB = _r_. CK = _f_: KE = _g_.
Et quoniam eſt CE. EK: : arc. EF. LK; vel CE. EK: : QF.
LK; hoc eſt _r_. _g_: : _e_. {_ge_/_r_} = LK; erit CL = _f_ + {_ge_. /_r_} Et LF
= √ _rr_ - _ff_ - {2 _fge_/_r_} = √ _gg_ - {2 _fge_. /_r_}
ad CB normalibus; nominentur CB = _r_. CK = _f_: KE = _g_.
Et quoniam eſt CE. EK: : arc. EF. LK; vel CE. EK: : QF.
LK; hoc eſt _r_. _g_: : _e_. {_ge_/_r_} = LK; erit CL = _f_ + {_ge_. /_r_} Et LF
= √ _rr_ - _ff_ - {2 _fge_/_r_} = √ _gg_ - {2 _fge_. /_r_}
Eſt autem CL.
LF:
: (CB.
BH:
:) CB.
QN.
hoc eſt,
_f_ + {_ge_. /_r_} √ _gg_ - {2 _fge_/_r_}: : _r_. _m_ - _a_. vel (quadrando) _ff_ +
{2 _fge._ /_r_} _gg_ - {2 _fge_/_r_}: : _rr._ _mm_ - 2 _ma_. Unde (dimiſſis quæ
oportet) obtinetur æquatio, _rfma_ = _grre_ + _gmme_. unde
ſubſtituendo, eſt _rfmm_ = _grrt_ + _gmmt_. vel {_rfmm_/_grr_ + _gmm_} = _t_.
ſeu (quoniam eſt _m_ = {_rg_/_f_}) erit _t_ = {_rr_/_rr_ + _mm_} _m_ = {CB_q_/CG_q_} BG =
{CK_q_/CE_q_} BG.
_f_ + {_ge_. /_r_} √ _gg_ - {2 _fge_/_r_}: : _r_. _m_ - _a_. vel (quadrando) _ff_ +
{2 _fge._ /_r_} _gg_ - {2 _fge_/_r_}: : _rr._ _mm_ - 2 _ma_. Unde (dimiſſis quæ
oportet) obtinetur æquatio, _rfma_ = _grre_ + _gmme_. unde
ſubſtituendo, eſt _rfmm_ = _grrt_ + _gmmt_. vel {_rfmm_/_grr_ + _gmm_} = _t_.
ſeu (quoniam eſt _m_ = {_rg_/_f_}) erit _t_ = {_rr_/_rr_ + _mm_} _m_ = {CB_q_/CG_q_} BG =
{CK_q_/CE_q_} BG.
Hæc ſufficere videntur huic methodo elucidandæ.