colours; Besides that, here is oftentimes one colour generated without any
of the other appendant ones, which cannot be by the _Cartesian Hypothesis_.

There must be therefore some other propriety of refraction that causes
colour. And upon the examination of the thing, I cannot conceive any one
more general, inseparable, and sufficient, than that which I have before
assign'd. That we may therefore see how exactly our _Hypothesis_ agrees
also with the _Phaenomena_ of the refracting round body, whether _Globe_ or
_Cylinder_, we shall next subjoyn our _Calculation_ or _Examen_ of it.

And to this end, we will calculate any two Rays: as for instance;[10] let
EF be a Ray cutting the _Radius_ CD (divided into 20. parts) in G 16. parts
distant from C, and ef another Ray, which cuts the same _Radius_ in g 17.
parts distant, these will be refracted to K and k, and from thence
reflected to N and n, and from thence refracted toward P and p; therefore
the Arch Ff will be 5.d 5'. The Arch FK 106.d 30'. the Arch fk 101.d 2'.
The line FG 6000. and fg 5267. therefore hf. 733. therefore Fc 980, almost.
The line FK 16024. and fk 15436. therefore Nd 196. and no 147 almost, the
line Nn 1019 the Arch Nn 5.d 51'. therefore the Angle Nno is 34.d 43'.
therefore the Angle Non is 139.d 56'. which is almost 50.d more than a
right Angle.

It is evident therefore by this _Hypothesis_, that at the same time that ef
touches f. EF is arrived at c. And by that time efkn is got to n, EFKN is
got to d and when it touches N, the pulse of the other Ray is got to o. and
no farther, which is very short of the place it should have arriv'd to, to
make the Ray np to cut the _orbicular pulse_ No at right Angles: therefore
the Angle Nop is an acute Angle, but the quite contrary of this will
happen, if 17. and 18. be calculated in stead of 16. and 17. both which
does most exactly agree with the _Phaenomena_: For if the Sun, or a Candle
(which is better) be placed about Ee, and the eye about Pp, the Rays EFef
at 16. and 17. will paint the side of the luminous object toward np _Blue_,
and towards NP _Red_. But the quite contrary will happen when EF is 17. and
ef 18. for then towards NP shall be a _Blue_, and towards np a _Red_,
exactly according to the calculation. And there appears the _Blue_ of the
Rainbow, where the two _Blue_ sides of the two Images unite, and there the
_Red_ where the two _Red_ sides unite, that is, where the two Images are
just disappearing; w