§3.5#15 ὅτι δ᾿ οὔτε κύκλον σἷόν τε γενέσθαι τῆς ἴριδος οὔτε μεῖζον ἡμικυκλίου τμῆμα, καὶ περὶ τῶν ἄλλων τῶν συμβαινόντων περὶ αὐτήν, ἐκ τοῦ διαγράμματος ἔσται θεωροῦσι δῆλον.
5 That it is impossible for a circle of the rainbow to be formed, or a segment greater than a semicircle, and concerning the other things that happen about it, will be clear to those who study the diagram.
ἡμισφαιρίου γὰρ ὄντος ἐπὶ τοῦ ὁρίζοντος κύκλου τοῦ ἐφ᾿ ᾧ τὸ Α, κέντρου δὲ τοῦ Κ, ἄλλου δέ τινος ἀνατέλλοντος σημείου ἐφ᾿ ᾧ τὸ Η, ἐὰν ἀπὸ τοῦ Κ γραμμαὶ κατὰ κῶνον ἐκπίπτουσαι ποιῶσιν ὡσπερεὶ ἄξονα τὴν ἐφ᾿ ᾗ ΗΚ, καὶ ἀπὸ τοῦ Κ ἐπὶ τὸ ἐπιζευχθεῖσαι ἀνακλασθῶσιν ἀπὸ τοῦ ἡμισφαιρίου ἐπὶ τὸ Η ἐπὶ τὴν μείζω γωνίαν, πρὸς κύκλου περιφέρειαν προσπεσοῦνται αἱ ἀπὸ τοῦ Κ·
For, there being a hemisphere over the horizon-circle at A, and its center being K, and another rising point being at H, if the lines falling from K in a cone make the line HK as if it were an axis, and being joined from K to [M], are reflected from the hemisphere to H at a greater angle, the lines from K will fall upon the circumference of a circle.
καὶ ἐὰν μὲν ἐπ᾿ ἀνατολῆς ἢ ἐπὶ δύσεως τοῦ ἄστρου ἡ ἀνάκλασις γένηται, ἡμικύκλιον ἀποληφθήσεται τοῦ κύκλου ὑπὸ τοῦ ὁρίζοντος τὸ ὑπὲρ γῆν γιγνόμενον, ἐὰν δ᾿ ἐπάνω, ἀεὶ ἔλαττον ἡμικυκλίου· ἐλάχιστον δέ, ὅταν ἐπὶ τοῦ μεσημβρινοῦ γένηται τὸ ἄστρον.
And if the reflection occurs at the rising or setting of the star, the part of the circle which is above the earth will be cut off by the horizon as a semicircle; but if the star is above, it is always less than a semicircle; and it is least when the star is on the meridian.
ἔστω γὰρ ἐπ᾿ ἀνατολῆς πρῶτον, οὗ τὸ Η, καὶ ἀνακεκλάσθω ἡ ΚM ἐπὶ τὸ Η, καὶ τὸ ἐπίπεδον ἐκβεβλήσθω ἐν ᾧ ἡ Α, τὸ ἀπὸ τοῦ τριγώνου ἐν ᾧ τὸ ΗΚΜ. κύκλος οὖν ἡ τομὴ ἔσται τῆς σφαίρας ὁ μέγιστος.
For let it first be at the rising, where H is, and let KM be reflected to H, and let the plane in which A is, namely that from the triangle in which is HKM, be extended. The intersection of the sphere, then, will be a great circle.
ἔστω ὁ ἐφ᾿ ᾧ Α· διοίσει γὰρ οὐδὲν ἂν ὁποιονοῦν τῶν ἐπὶ τῆς ΗΚ κατὰ τὸ τρίγωνον τὸ ΚΜΗ ἐκβληθῇ τὸ ἐπίπεδον.
Let it be the one at A; for it will make no difference if the plane is extended along the triangle KMH through any point on HK.
αἱ οὖν ἀπὸ τῶν ἀναγόμεναι γραμμαὶ ἐν τούτῳ τῷ λόγῳ οὐ συσταθήσονται τοῦ ἐφ᾿ ᾧ Α ἡμικυκλίου πρὸς ἄλλο καὶ ἄλλο σημεῖον· ἐπεὶ γὰρ τὰ τε Κ Η σημεῖα δέδοται καὶ ἡ HΚ, δεδομένη ἂν εἴη καὶ ἡ ΜΗ, ὥστε καὶ λόγος τῆς ΜΗ πρὸς ΜΚ. δεδομένης οὖν περιφερείας ἐφάψεται τὸ Μ. ἔστω δὴ αὕτη ἐφ᾿ ἧς τὰ ΝΜ· ὥστε ἡ τομὴ τῶν περιφερειῶν δέδοται.
Therefore, the lines drawn up from [H, K] in this ratio will not be constructed to different points on the semicircle at A; for since the points K and H and the line HK are given, MH would also be given, so that the ratio of MH to MK is also given. M, then, will touch a given circumference. Let this be that on which NM is; so that the intersection of the circumferences is given.
πρὸς ἄλλη δέ γε ἢ τῇ ΜΝ περιφερείᾳ ἀπὸ τῶν αὐτῶν σημείων ὁ αὐτὸς λόγος ἐν τῷ αὐτῷ ἐπιπέδῳ οὐ συνίσταται.
But the same ratio from the same points in the same plane is not constructed to any other circumference than the circumference MN.
ἐκκείσθω οὖν τις γραμμὴ ἡ ΔΒ, καὶ τετμήσθω ὡς ἡ ΜΗ πρὸς ΜΚ ἡ Δ πρὸς B. μείζων δὲ ἡ ΜΗ τῆς ΚΜ, ἐπείπερ ἐπὶ τὴν μείζω γωνίαν ἡ ἀνάκλασις τοῦ κώνου· ὑπὸ γὰρ τὴν μείζω γωνίαν ὑποτείνει τοῦ τριγώνου.
Let, then, a line ΔB be set out, and let it be cut so that Δ is to B as MH is to MK. And MH is greater than KM, since the reflection of the cone is at a greater angle; for it subtends the greater angle of the triangle.
προσπεπορίσθω οὖν πρὸς τὴν B, ἐφ᾿ ἧς τὸ Ζ·
Let there be added, then, to B, a line on which is Z; so that as Δ is to B, so BZ is to Δ.
ὥστ᾿ εἶναι ὅπερ τὴν Δ πρὸς τὴν B, τὴν ΒΖ πρὸς τὴν Δ. εἶτα ὅπερ ἡ Ζ πρὸς τὴν ΚΗ, ἡ τὸ B πρὸς ἄλλην πεποιήσθω τὴν ΚΠ, καὶ ἀπὸ τοῦ ἐπὶ τὸ Μ ἐπεζεύχθω ἡ τὸ ΜΠ. ἔσται οὖν τὸ Π πόλος τοῦ κύκλου, πρὸς ὃν αἱ ἀπὸ τοῦ Κ γραμμαὶ προσπίπτουσιν·
Then let B be made to another line KΠ as Z is to KH, and from Π to M let MΠ be joined. Π, then, will be the pole of the circle to which the lines from K fall; for as Z is to KH, so will B be to KΠ, and Δ to ΠM.
ἔσται γὰρ ὅπερ ἡ Ζ πρὸς ΚΗ, καὶ ἡ Β πρὸς ΚΠ, καὶ ἡ Δ πρὸς ΠΜ. μὴ γὰρ ἔστω, ἀλλ᾿ ἢ πρὸς ἐλάττω ἢ πρὸς μείζω τῆς ΠΜ·
For if it is not, let it be to a line either less or greater than ΠM; for it will make no difference.
οὐδὲν γὰρ διοίσει.
Let it be to ΠP.
ἔστω πρὸς ΠΡ. τὸν αὐτὸν ἄρα λόγον αἱ καὶ ΚΠ καὶ ἡ ΠΡ πρὸς ἀλλήλας ἕξουσιν ὃνπερ αἱ ΔΒ Ζ. αἱ δὲ Δ Β Ζ ἀνὰ λόγον ἦσαν, ὅνπερ ἡ Δ πρὸς Β, ἡ ΖΒ πρὸς Δ· ὥστε ὅπερ ἡ ΠΗ πρὸς τὴν ΠΡ, ἡ τὸ ΠΡ πρὸς τὴν ΠΚ. ἂν οὖν ἀπὸ τῶν Κ αἱ ΗΡ καὶ ΚΡ ἐπὶ τὸ Ρ ἐπιζευχθῶσιν, αἱ ἐπιζευχθεῖσαι αὗται τὸν αὐτὸν ἕξουσι λόγον ὅνπερ ἡ ΗΠ πρὸς τὴν ΠΡ·
The lines KΠ and ΠP, therefore, will have the same ratio to one another as Δ, B, Z. But Δ, B, Z were in proportion, as Δ is to B, so ZB is to Δ; so that as ΠH is to ΠP, so is ΠP to ΠK. If, then, from [H, ] K the lines HP and KP are joined to P, these joined lines will have the same ratio as HP to ΠP; for the sides of the triangle HPΠ and KRP are proportional about the same angle.
περὶ γὰρ τὴν αὐτὴν γωνίαν τὴν ἀνάλογον αἵ τε τοῦ ΗΠΡ τριγώνου καὶ τοῦ ΚΡΠ. ὥστε καὶ ἡ ΠΡ πρὸς τὴν ΚΡ τὸν αὐτὸν ἕξει λόγον, καὶ ἡ τὸ ΗΠ πρὸς τὴν ΠΡ. ἔχει δὲ καὶ ἡ ΜΗ πρὸς ΚΜ τοῦτον τὸν λόγον·
Therefore, ΠP to KP will also have the same ratio, and HΠ to ΠP.
ὅνπερ γὰρ ἡ τὸ Δ πρὸς τὴν Β ἀμφότεραι.
But MH also has this ratio to KM; for both are as Δ to B.
ὥστε ἀπὸ τῶν ΗΚ σημείων οὐ μόνον πρὸς τὴν Ν περιφέρειαν συσταθήσονται τὸν αὐτὸν ἔχουσαι λόγον, ἀλλὰ καὶ ἄλλοθι· ὅπερ ἀδύνατον.
So that from the points H, K they will be constructed to have the same ratio not only to the circumference NM, but also elsewhere; which is impossible.