§3.5#2ἐπεὶ οὖν ἡ Δ οὔτε πρὸς ἔλαττον τοῦ ΜΠ οὔτε πρὸς μείζω (ὁμοίως γὰρ δειχθήσεται), δῆλον ὅτι πρὸς αὐτὴν ἂν εἴη τὴν ἐφ᾿ ᾗ Μ Π. ὥστ᾿ ἔσται ὅπερ ἡ ΜΠ πρὸς ΠΚ, ἡ ΠΗ πρὸς τὴν ΜΠ.
Since, then, Δ is neither to a line less than MΠ nor to a greater (for it will be proved similarly), it is clear that it is to MΠ itself. So that as MΠ is to ΠK, so is ΠH to MΠ.
ἐὰν οὖν τῷ ἐφ᾿ ᾧ τὸ πόλῳ χρώμενος, διαστήματι δὲ τῷ ἐφ᾿ ᾧ Μ Π, κύκλος γραφῇ, ἁπασῶν ἐφάψεται τῶν γωνιῶν ἃς ἀνακλώμεναι ποιοῦσιν αἱ ἀπὸ τοῦ ΜΑ κύκλου· εἰ δὲ μή, ὁμοίως δειχθήσονται τὸν αὐτὸν ἔχουσαι λόγον αἱ ἄλλοθι καὶ ἄλλοθι τοῦ ἡμικυκλίου συνιστάμεναι, ὅπερ ἦν ἀδύνατον.
If, then, using Π as pole, and with MΠ as distance, a circle is described, it will touch all the angles which the lines reflected from the circle MA make; if not, they will be proved similarly to have the same ratio when constructed at other and other points of the semicircle, which was impossible.
ἐὰν οὖν περιαγάγῃς τὸ ἡμικύκλιον τὸ ἐφ᾿ ᾧ τὸ Α περὶ τὴν ἐφ᾿ ᾗ Η Κ Π διάμετρον, αἱ ἀπὸ τοῦ ΗΚ ἀνακλώμεναι πρὸς τὸ ἐφ᾿ ᾧ τὸ Μ ἐν πᾶσι τοῖς ἐπιπέδοις ὁμοίως ἕξουσι, καὶ ἴσην ποιήσουσι γωνίαν τὴν ΚΜ καὶ ἣν ποιήσουσι δὲ γωνίαν αἱ ΗΠ καὶ ΜΠ ἐπὶ τῆς ΗΠ, ἀεὶ ἴση ἔσται.
If, then, you rotate the semicircle at A about the diameter HKΠ, the lines reflected from HK to the point M will have the same relation in all the planes, and will make an equal angle; and the angle made with KM, and the angle which HΠ and MΠ make on HΠ, will always be equal.
τρίγωνα οὖν ἐπὶ τῆς ΗΠ καὶ ΚΠ ἴσα τῷ ΗΜΠ ΚΜΠ συνεστήκασι.
Triangles, then, equal to HMΠ and KMΠ are constructed on HΠ and KΠ.
τούτων δὲ αἱ κάθετοι ἐπὶ τὸ αὐτὸ σημεῖον πεσοῦνται τῆς ΗΠ καὶ ἴσαι ἔσονται.
And the perpendiculars of these will fall on the same point of HΠ and will be equal.
πιπτέτωσαν ἐπὶ τὸ Ο. κέντρον ἄρα τοῦ κύκλου τὸ Ο, ἡμικύκλιον δὲ τὸ περὶ τὴν ΜΝ ἀφῄρηται ἀπὸ τοῦ ὁρίζοντος· τῶν μὲν γὰρ ἄνω τὸν ἥλιον οὐ κρατεῖν, τῶν δὲ προσπτεριζομένων κρατεῖν, καὶ διαχεῖν τὸν ἀέρα· καὶ διὰ τοῦτο τὴν ἶριν οὐ συμβάλλειν τὸν κύκλον· γίγνεσθαι δὲ καὶ νύκτωρ ἀπὸ τῆς σελήνης ὀλιγάκις· οὔτε γὰρ ἀεὶ πλήρης, ἀσθενεστέρα τε τὴν φύσιν ὥστε κρατεῖν τοῦ ἀέρος· μάλιστα δ᾿ ἵστασθαι τὴν ἶριν, ὅπου μάλιστα κρατεῖται ὁ ἥλιος· πλείστη γὰρ ἐν αὐτῇ ἰκμὰς ἐνέμεινεν.
Let them fall on O. The center of the circle, then, is O, and the semicircle about MN is cut off by the horizon; for they say that the sun does not dominate the upper parts, but dominates the parts near the earth, and dissolves the air; and that for this reason the rainbow does not complete the circle; and that it also occurs at night from the moon, though rarely; for it is not always full, and is too weak by nature to dominate the air; and that the rainbow is most fixed where the sun is most dominated; for most moisture remains in it.
πάλιν ἔστω ὁρίζων μὲν ἐφ᾿ οὗ τὸ ΑΚΓ, ἐπανατεταλκέτω δὲ τὸ Η, ὁ δ᾿ ἄξων ἔστω νῦν ἐφ᾿ οὗ τὸ ΗΠ. τὰ μὲν οὖν ἄλλα πάντα ὁμοίως δειχθήσεται ὡς καὶ πρότερον, ὁ δὲ πόλος τοῦ κύκλου ὁ ἐφ᾿ Π κάτω ἔσται τοῦ ὁρίζοντος τοῦ ἐφ᾿ τὸ ΑΓ, ἀρθέντος τοῦ ἐφ᾿ ᾧ τὸ Η σημείου.
Again, let the horizon be the one on which is AKΓ, and let H have risen, and let the axis now be the one on which is HΠ. All other things, then, will be proved in the same way as before, but the pole of the circle, which is at Π, will be below the horizon at AΓ, since the point at H has risen.
ἐπὶ δὲ τῆς αὐτῆς ὅ τε πόλος καὶ τὸ κέντρον τοῦ κύκλου καὶ τὸ τοῦ ὁρίζοντος νῦν τὴν ἀνατολήν· ἔστι γὰρ οὗτος ἐφ᾿ ᾧ τὸ HΠ. ἐπεὶ δὲ τῆς διαμέτρου τῆς ΑΓ τὸ ΚΗ ἐπάνω, τὸ κέντρον εἴη ἂν ὑποκάτω τοῦ ὁρίζοντος πρότερον τοῦ ἐφ᾿ ᾧ τὸ ΑΓ, ἐπὶ τῆς ΚΠ γραμμῆς, ἐφ᾿ οὗ τὸ Β. ὥστ᾿ ἔλαττον ἔσται τὸ ἐπάνω τμῆμα ἡμικυκλίου τὸ ἐφ᾿ ᾧ Ψ Υ· τὸ γὰρ ΨΥΟ ἡμικύκλιον ἦν, νῦν δὲ ἀποτέτμηται ἀπὸ τοῦ ΑΓ ὁρίζοντος.
And on the same line are the pole, the center of the circle, and that of the horizon, now towards the rising; for this is the line HΠ. And since KH is above the diameter AΓ, the center would be below the former horizon at AΓ, on the line KΠ, at B. So that the segment above, on which is ΨY, will be less than a semicircle; for ΨYO was a semicircle, but now it has been cut off by the horizon AΓ.
τὸ δὴ ΟΥ ἀφανὲς ἔσται αὐτοῦ, ἐπαρθέντος τοῦ ἡλίου· ἐλάχιστον δ᾿, ὅταν ἐπὶ μεσημβρίας· ὅσον γὰρ ἀνώτερον τὸ Η, κατώτερον ὅ τε πόλος καὶ τὸ κέντρον τοῦ κύκλου ἔσται.
Therefore, the part OY of it will be invisible, as the sun rises; and it is least when the sun is on the meridian; for the higher H is, the lower the pole and the center of the circle will be.
ὅτι δ᾿ ἐν μὲν ταῖς ἐλάττοσιν ἡμέραις ταῖς μετ᾿ ἰσημερίαν τὴν μετοπωρινὴν ἐνδέχεται ἀεὶ γίγνεσθαι ἶριν, ἐν δὲ ταῖς μακροτέραις ἡμέραις ταῖς ἀπὸ ἰσημερίας τῆς ἑτέρας ἐπὶ τὴν ἰσημερίαν τὴν ἑτέραν περὶ μεσημβρίαν οὐ γίγνεται ἶρις, αἴτιον ὅτι τὰ μὲν πρὸς ἄρκτον τμήματα πάντα μείζω ἡμικυκλίου καὶ ἀεὶ ἐπὶ μείζω ἡμικυκλίου, τὸ δ᾿ ἀφανὲς μικρόν, τὰ δὲ πρὸς μεσημβρίαν τμήματα τοῦ ἰσημερινοῦ, τὸ μὲν ἄνω τμῆμα μικρόν, τὸ δ᾿ ὑπὸ γῆν μέγα, καὶ ἀεὶ δὴ μείζω τὰ πορρώτερα· ὥστ᾿ ἐν μὲν ταῖς πρὸς θερινὰς τροπὰς ἡμέραις διὰ τὸ μέγεθος τοῦ τμήματος, πρὶν ἐπὶ τὸ μέσον ἐλθεῖν τοῦ τμήματος καὶ ἐπὶ τὸν μεσημβρινὸν τὴν τὸ Η, κάτω ἤδη τελέως γίγνεται ἡ τὸ Π, διὰ τὸ πόρρω ἀφεστάναι τῆς γῆς τὴν μεσημβρίαν διὰ τὸ μέγεθος τοῦ τμήματος.
The reason why in the shorter days after the autumn equinox a rainbow can always occur, but in the longer days from the other equinox to the other, no rainbow occurs about noon, is that all the northern segments are greater than a semicircle and always go on a segment greater than a semicircle, and the invisible part is small, while of the segments of the equinoctial towards the south, the upper segment is small, and the part under the earth is large, and those further away are always larger; so that in the days near the summer solstice, because of the size of the segment, before H reaches the middle of the segment and the meridian, Π is already completely below, because the meridian is far removed from the earth owing to the size of the segment.
ἐν δὲ ταῖς πρὸς τὰς χειμερινὰς τροπὰς ἡμέραις, διὰ τὸ μὴ πολὺ ὑπὲρ γῆς εἶναι τὰ τμήματα τῶν κύκλων, τοὐναντίον ἀναγκαῖον γίγνεσθαι· βραχὺ γὰρ ἀρθείσης τῆς ἐφ᾿ ᾧ τὸ Η, ἐπὶ τῆς μεσημβρίας γίγνεται ὁ ἥλιος.
But in the days near the winter solstice, because the segments of the circles are not far above the earth, the contrary must happen; for when the point H has risen but a little, the sun is on the meridian.