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5
The rainbow can never be a circle nor a segment of a circle
greater than a semicircle. The consideration of the diagram will prove
this and the other properties of the rainbow. (See diagram.)
Let A be a hemisphere resting on the circle of the horizon, let
its centre be K and let H be another point appearing on the horizon.
Then, if the lines that fall in a cone from K have HK as their axis,
and, K and M being joined, the lines KM are reflected from the
hemisphere to H over the greater angle, the lines from K will fall
on the circumference of a circle. If the reflection takes place when
the luminous body is rising or setting the segment of the circle above
the earth which is cut off by the horizon will be a semi-circle; if
the luminous body is above the horizon it will always be less than a
semicircle, and it will be smallest when the luminous body culminates.
First let the luminous body be appearing on the horizon at the point
H, and let KM be reflected to H, and let the plane in which A is,
determined by the triangle HKM, be produced. Then the section of the
sphere will be a great circle. Let it be A (for it makes no difference
which of the planes passing through the line HK and determined by
the triangle KMH is produced). Now the lines drawn from H and K to a
point on the semicircle A are in a certain ratio to one another, and
no lines drawn from the same points to another point on that
semicircle can have the same ratio. For since both the points H and
K and the line KH are given, the line MH will be given too;
consequently the ratio of the line MH to the line MK will be given
too. So M will touch a given circumference. Let this be NM. Then the
intersection of the circumferences is given, and the same ratio cannot
hold between lines in the same plane drawn from the same points to any
other circumference but MN.
Draw a line DB outside of the figure and divide it so that
D:B=MH:MK. But MH is greater than MK since the reflection of the
cone is over the greater angle (for it subtends the greater angle of
the triangle KMH). Therefore D is greater than B. Then add to B a line
Z such that B+Z:D=D:B. Then make another line having the same ratio to
B as KH has to Z, and join MI.
Then I is the pole of the circle on which the lines from K fall. For
the ratio of D to IM is the same as that of Z to KH and of B to KI. If
not, let D be in the same ratio to a line indifferently lesser or
greater than IM, and let this line be IP. Then HK and KI and IP will
have the same ratios to one another as Z, B, and D. But the ratios
between Z, B, and D were such that Z+B:D=D: B. Therefore
IH:IP=IP:IK. Now, if the points K, H be joined with the point P by the
lines HP, KP, these lines will be to one another as IH is to IP, for
the sides of the triangles HIP, KPI about the angle I are
homologous. Therefore, HP too will be to KP as HI is to IP. But this
is also the ratio of MH to MK, for the ratio both of HI to IP and of
MH to MK is the same as that of D to B. Therefore, from the points
H, K there will have been drawn lines with the same ratio to one
another, not only to the circumference MN but to another point as
well, which is impossible. Since then D cannot bear that ratio to
any line either lesser or greater than IM (the proof being in either
case the same), it follows that it must stand in that ratio to MI
itself. Therefore as MI is to IK so IH will be to MI and finally MH to
MK.
If, then, a circle be described with I as pole at the distance MI it
will touch all the angles which the lines from H and K make by their
reflection. If not, it can be shown, as before, that lines drawn to
different points in the semicircle will have the same ratio to one
another, which was impossible. If, then, the semicircle A be
revolved about the diameter HKI, the lines reflected from the points
H, K at the point M will have the same ratio, and will make the
angle KMH equal, in every plane. Further, the angle which HM and MI
make with HI will always be the same. So there are a number of
triangles on HI and KI equal to the triangles HMI and KMI. Their
perpendiculars will fall on HI at the same point and will be equal.
Let O be the point on which they fall. Then O is the centre of the
circle, half of which, MN, is cut off by the horizon. (See diagram.)
Next let the horizon be ABG but let H have risen above the
horizon. Let the axis now be HI. The proof will be the same for the
rest as before, but the pole I of the circle will be below the horizon
AG since the point H has risen above the horizon. But the pole, and
the centre of the circle, and the centre of that circle (namely HI)
which now determines the position of the sun are on the same line. But
since KH lies above the diameter AG, the centre will be at O on the
line KI below the plane of the circle AG determined the position of
the sun before. So the segment YX which is above the horizon will be
less than a semicircle. For YXM was a semicircle and it has now been
cut off by the horizon AG. So part of it, YM, will be invisible when
the sun has risen above the horizon, and the segment visible will be
smallest when the sun is on the meridian; for the higher H is the
lower the pole and the centre of the circle will be.
In the shorter days after the autumn equinox there may be a
rainbow at any time of the day, but in the longer days from the spring
to the autumn equinox there cannot be a rainbow about midday. The
reason for this is that when the sun is north of the equator the
visible arcs of its course are all greater than a semicircle, and go
on increasing, while the invisible arc is small, but when the sun is
south of the equator the visible arc is small and the invisible arc
great, and the farther the sun moves south of the equator the
greater is the invisible arc. Consequently, in the days near the
summer solstice, the size of the visible arc is such that before the
point H reaches the middle of that arc, that is its point of
culmination, the point is well below the horizon; the reason for
this being the great size of the visible arc, and the consequent
distance of the point of culmination from the earth. But in the days
near the
winter solstice the visible arcs are small, and the
contrary is necessarily the case: for the sun is on the meridian
before the point H has risen far.
6
Mock suns, and rods too, are due to the causes we have described.
A mock sun is caused by the reflection of sight to the sun. Rods are
seen when sight reaches the sun under circumstances like those which
we described, when there are clouds near the sun and sight is
reflected from some liquid surface to the cloud. Here the clouds
themselves are colourless when you look at them directly, but in the
water they are full of rods. The only difference is that in this
latter case the colour of the cloud seems to reside in the water,
but in the case of rods on the cloud itself. Rods appear when the
composition of the cloud is uneven, dense in part and in part rare,
and more and less watery in different parts. Then the sight is
reflected to the sun: the mirrors are too small for the shape of the
sun to appear, but, the bright white light of the sun, to which the
sight is reflected, being seen on the uneven mirror, its colour
appears partly red, partly green or yellow. It makes no difference
whether sight passes through or is reflected from a medium of that
kind; the colour is the same in both cases; if it is red in the
first case it must be the same in the other.
Rods then are occasioned by the unevenness of the mirror-as
regards colour, not form. The mock sun, on the contrary, appears
when the air is very uniform, and of the same density throughout. This
is why it is white: the uniform character of the mirror gives the
reflection in it a single colour, while the fact that the sight is
reflected in a body and is thrown on the sun all together by the mist,
which is dense and watery though not yet quite water, causes the sun's
true colour to appear just as it does when the reflection is from
the dense, smooth surface of copper. So the sun's colour being
white, the mock sun is white too. This, too, is the reason why the
mock sun is a surer sign of rain than the rods; it indicates, more
than they do, that the air is ripe for the production of water.
Further a mock sun to the south is a surer sign of rain than one to
the north, for the air in the south is readier to turn into water than
that in the north.
Mock suns and rods are found, as we stated, about sunset and
sunrise, not above the sun nor below it, but beside it. They are not
found very close to the sun, nor very far from it, for the sun
dissolves the cloud if it is near, but if it is far off the reflection
cannot take place, since sight weakens when it is reflected from a
small mirror to a very distant object. (This is why a halo is never
found opposite to the sun.) If the cloud is above the sun and close to
it the sun will dissolve it; if it is above the sun but at a
distance the sight is too weak for the reflection to take place, and
so it will not reach the sun. But at the side of the sun, it is
possible for the mirror to be at such an interval that the sun does
not dissolve the cloud, and yet sight reaches it undiminished
because it moves close to the earth and is not dissipated in the
immensity of space. It cannot subsist below the sun because close to
the earth the sun's rays would dissolve it, but if it were high up and
the sun in the middle of the
heavens, sight would be dissipated.
Indeed, even by the side of the sun, it is not found when the sun is
in the middle of the sky, for then the line of vision is not close
to the earth, and so but little sight reaches the mirror and the
reflection from it is altogether feeble.
Some account has now been given of the effects of the secretion
above the surface of the earth; we must go on to describe its
operations below, when it is shut up in the parts of the earth.
Just as its twofold nature gives rise to various effects in the
upper region, so here it causes two varieties of bodies. We maintain
that there are two exhalations, one vaporous the other smoky, and
there correspond two kinds of bodies that originate in the earth,
'fossiles' and metals. The heat of the dry exhalation is the cause
of all 'fossiles'. Such are the kinds of stones that cannot be melted,
and
realgar, and
ochre, and
ruddle, and
sulphur, and the other
things of that kind, most 'fossiles' being either coloured
lye or,
like
cinnabar, a stone compounded of it. The vaporous exhalation is
the cause of all metals, those bodies which are either fusible or
malleable such as iron, copper, gold. All these originate from the
imprisonment of the vaporous exhalation in the earth, and especially
in stones. Their dryness compresses it, and it congeals just as
dew or
hoar-frost does when it has been separated off, though in the
present case the metals are generated before that segregation
occurs. Hence, they are water in a sense, and in a sense not. Their
matter was that which might have become water, but it can no longer do
so: nor are they, like savours, due to a qualitative change in
actual water. Copper and gold are not formed like that, but in every
case the evaporation congealed before water was formed. Hence, they
all (except gold) are affected by fire, and they possess an
admixture of earth; for they still contain the dry exhalation.
This is the general theory of all these bodies, but we must take
up each kind of them and discuss it separately.
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