27.06.2018 Views

A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

MACEDONIA is GREECE and will always be GREECE- (if they are desperate to steal a name, Monkeydonkeys suits them just fine) ΚΑΤΩ Η ΣΥΓΚΥΒΕΡΝΗΣΗ ΤΩΝ ΠΡΟΔΟΤΩΝ!!! ΦΕΚ,ΚΚΕ,ΚΝΕ,ΚΟΜΜΟΥΝΙΣΜΟΣ,ΣΥΡΙΖΑ,ΠΑΣΟΚ,ΝΕΑ ΔΗΜΟΚΡΑΤΙΑ,ΕΓΚΛΗΜΑΤΑ,ΔΑΠ-ΝΔΦΚ, MACEDONIA,ΣΥΜΜΟΡΙΤΟΠΟΛΕΜΟΣ,ΠΡΟΣΦΟΡΕΣ,ΥΠΟΥΡΓΕΙΟ,ΕΝΟΠΛΕΣ ΔΥΝΑΜΕΙΣ,ΣΤΡΑΤΟΣ, ΑΕΡΟΠΟΡΙΑ,ΑΣΤΥΝΟΜΙΑ,ΔΗΜΑΡΧΕΙΟ,ΝΟΜΑΡΧΙΑ,ΠΑΝΕΠΙΣΤΗΜΙΟ,ΛΟΓΟΤΕΧΝΙΑ,ΔΗΜΟΣ,LIFO,ΛΑΡΙΣΑ, ΠΕΡΙΦΕΡΕΙΑ,ΕΚΚΛΗΣΙΑ,ΟΝΝΕΔ,ΜΟΝΗ,ΠΑΤΡΙΑΡΧΕΙΟ,ΜΕΣΗ ΕΚΠΑΙΔΕΥΣΗ,ΙΑΤΡΙΚΗ,ΟΛΜΕ,ΑΕΚ,ΠΑΟΚ,ΦΙΛΟΛΟΓΙΚΑ,ΝΟΜΟΘΕΣΙΑ,ΔΙΚΗΓΟΡΙΚΟΣ,ΕΠΙΠΛΟ, ΣΥΜΒΟΛΑΙΟΓΡΑΦΙΚΟΣ,ΕΛΛΗΝΙΚΑ,ΜΑΘΗΜΑΤΙΚΑ,ΝΕΟΛΑΙΑ,ΟΙΚΟΝΟΜΙΚΑ,ΙΣΤΟΡΙΑ,ΙΣΤΟΡΙΚΑ,ΑΥΓΗ,ΤΑ ΝΕΑ,ΕΘΝΟΣ,ΣΟΣΙΑΛΙΣΜΟΣ,LEFT,ΕΦΗΜΕΡΙΔΑ,ΚΟΚΚΙΝΟ,ATHENS VOICE,ΧΡΗΜΑ,ΟΙΚΟΝΟΜΙΑ,ΕΝΕΡΓΕΙΑ, ΡΑΤΣΙΣΜΟΣ,ΠΡΟΣΦΥΓΕΣ,GREECE,ΚΟΣΜΟΣ,ΜΑΓΕΙΡΙΚΗ,ΣΥΝΤΑΓΕΣ,ΕΛΛΗΝΙΣΜΟΣ,ΕΛΛΑΔΑ, ΕΜΦΥΛΙΟΣ,ΤΗΛΕΟΡΑΣΗ,ΕΓΚΥΚΛΙΟΣ,ΡΑΔΙΟΦΩΝΟ,ΓΥΜΝΑΣΤΙΚΗ,ΑΓΡΟΤΙΚΗ,ΟΛΥΜΠΙΑΚΟΣ, ΜΥΤΙΛΗΝΗ,ΧΙΟΣ,ΣΑΜΟΣ,ΠΑΤΡΙΔΑ,ΒΙΒΛΙΟ,ΕΡΕΥΝΑ,ΠΟΛΙΤΙΚΗ,ΚΥΝΗΓΕΤΙΚΑ,ΚΥΝΗΓΙ,ΘΡΙΛΕΡ, ΠΕΡΙΟΔΙΚΟ,ΤΕΥΧΟΣ,ΜΥΘΙΣΤΟΡΗΜΑ,ΑΔΩΝΙΣ ΓΕΩΡΓΙΑΔΗΣ,GEORGIADIS,ΦΑΝΤΑΣΤΙΚΕΣ ΙΣΤΟΡΙΕΣ, ΑΣΤΥΝΟΜΙΚΑ,ΦΙΛΟΣΟΦΙΚΗ,ΦΙΛΟΣΟΦΙΚΑ,ΙΚΕΑ,ΜΑΚΕΔΟΝΙΑ,ΑΤΤΙΚΗ,ΘΡΑΚΗ,ΘΕΣΣΑΛΟΝΙΚΗ,ΠΑΤΡΑ, ΙΟΝΙΟ,ΚΕΡΚΥΡΑ,ΚΩΣ,ΡΟΔΟΣ,ΚΑΒΑΛΑ,ΜΟΔΑ,ΔΡΑΜΑ,ΣΕΡΡΕΣ,ΕΥΡΥΤΑΝΙΑ,ΠΑΡΓΑ,ΚΕΦΑΛΟΝΙΑ, ΙΩΑΝΝΙΝΑ,ΛΕΥΚΑΔΑ,ΣΠΑΡΤΗ,ΠΑΞΟΙ

MACEDONIA is GREECE and will always be GREECE- (if they are desperate to steal a name, Monkeydonkeys suits them just fine)

ΚΑΤΩ Η ΣΥΓΚΥΒΕΡΝΗΣΗ ΤΩΝ ΠΡΟΔΟΤΩΝ!!!

ΦΕΚ,ΚΚΕ,ΚΝΕ,ΚΟΜΜΟΥΝΙΣΜΟΣ,ΣΥΡΙΖΑ,ΠΑΣΟΚ,ΝΕΑ ΔΗΜΟΚΡΑΤΙΑ,ΕΓΚΛΗΜΑΤΑ,ΔΑΠ-ΝΔΦΚ, MACEDONIA,ΣΥΜΜΟΡΙΤΟΠΟΛΕΜΟΣ,ΠΡΟΣΦΟΡΕΣ,ΥΠΟΥΡΓΕΙΟ,ΕΝΟΠΛΕΣ ΔΥΝΑΜΕΙΣ,ΣΤΡΑΤΟΣ, ΑΕΡΟΠΟΡΙΑ,ΑΣΤΥΝΟΜΙΑ,ΔΗΜΑΡΧΕΙΟ,ΝΟΜΑΡΧΙΑ,ΠΑΝΕΠΙΣΤΗΜΙΟ,ΛΟΓΟΤΕΧΝΙΑ,ΔΗΜΟΣ,LIFO,ΛΑΡΙΣΑ, ΠΕΡΙΦΕΡΕΙΑ,ΕΚΚΛΗΣΙΑ,ΟΝΝΕΔ,ΜΟΝΗ,ΠΑΤΡΙΑΡΧΕΙΟ,ΜΕΣΗ ΕΚΠΑΙΔΕΥΣΗ,ΙΑΤΡΙΚΗ,ΟΛΜΕ,ΑΕΚ,ΠΑΟΚ,ΦΙΛΟΛΟΓΙΚΑ,ΝΟΜΟΘΕΣΙΑ,ΔΙΚΗΓΟΡΙΚΟΣ,ΕΠΙΠΛΟ, ΣΥΜΒΟΛΑΙΟΓΡΑΦΙΚΟΣ,ΕΛΛΗΝΙΚΑ,ΜΑΘΗΜΑΤΙΚΑ,ΝΕΟΛΑΙΑ,ΟΙΚΟΝΟΜΙΚΑ,ΙΣΤΟΡΙΑ,ΙΣΤΟΡΙΚΑ,ΑΥΓΗ,ΤΑ ΝΕΑ,ΕΘΝΟΣ,ΣΟΣΙΑΛΙΣΜΟΣ,LEFT,ΕΦΗΜΕΡΙΔΑ,ΚΟΚΚΙΝΟ,ATHENS VOICE,ΧΡΗΜΑ,ΟΙΚΟΝΟΜΙΑ,ΕΝΕΡΓΕΙΑ, ΡΑΤΣΙΣΜΟΣ,ΠΡΟΣΦΥΓΕΣ,GREECE,ΚΟΣΜΟΣ,ΜΑΓΕΙΡΙΚΗ,ΣΥΝΤΑΓΕΣ,ΕΛΛΗΝΙΣΜΟΣ,ΕΛΛΑΔΑ, ΕΜΦΥΛΙΟΣ,ΤΗΛΕΟΡΑΣΗ,ΕΓΚΥΚΛΙΟΣ,ΡΑΔΙΟΦΩΝΟ,ΓΥΜΝΑΣΤΙΚΗ,ΑΓΡΟΤΙΚΗ,ΟΛΥΜΠΙΑΚΟΣ, ΜΥΤΙΛΗΝΗ,ΧΙΟΣ,ΣΑΜΟΣ,ΠΑΤΡΙΔΑ,ΒΙΒΛΙΟ,ΕΡΕΥΝΑ,ΠΟΛΙΤΙΚΗ,ΚΥΝΗΓΕΤΙΚΑ,ΚΥΝΗΓΙ,ΘΡΙΛΕΡ, ΠΕΡΙΟΔΙΚΟ,ΤΕΥΧΟΣ,ΜΥΘΙΣΤΟΡΗΜΑ,ΑΔΩΝΙΣ ΓΕΩΡΓΙΑΔΗΣ,GEORGIADIS,ΦΑΝΤΑΣΤΙΚΕΣ ΙΣΤΟΡΙΕΣ, ΑΣΤΥΝΟΜΙΚΑ,ΦΙΛΟΣΟΦΙΚΗ,ΦΙΛΟΣΟΦΙΚΑ,ΙΚΕΑ,ΜΑΚΕΔΟΝΙΑ,ΑΤΤΙΚΗ,ΘΡΑΚΗ,ΘΕΣΣΑΛΟΝΙΚΗ,ΠΑΤΡΑ, ΙΟΝΙΟ,ΚΕΡΚΥΡΑ,ΚΩΣ,ΡΟΔΟΣ,ΚΑΒΑΛΑ,ΜΟΔΑ,ΔΡΑΜΑ,ΣΕΡΡΕΣ,ΕΥΡΥΤΑΝΙΑ,ΠΑΡΓΑ,ΚΕΦΑΛΟΝΙΑ, ΙΩΑΝΝΙΝΑ,ΛΕΥΚΑΔΑ,ΣΠΑΡΤΗ,ΠΑΞΟΙ

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

THE COLLECTION. BOOK IV 381<br />

Hence / lies on a certain curve.<br />

Therefore E, its projection<br />

on the plane ABO, also lies on a curve.<br />

In the particular case where the given ratio <strong>of</strong> EF <strong>to</strong> the<br />

arc CD is equal <strong>to</strong> the ratio <strong>of</strong> BA <strong>to</strong> the arc CA, the locus <strong>of</strong><br />

E is a quadratrix.<br />

[The surface described <strong>by</strong> the straight line LH is a plec<strong>to</strong>id.<br />

The shape <strong>of</strong> it is perhaps best realized as a continuous spiral<br />

staircase, i.e. a spiral staircase with infinitely small steps.<br />

The quadratrix is thus produced as the orthogonal projection<br />

<strong>of</strong> the curve in which the plec<strong>to</strong>id is intersected <strong>by</strong> a plane<br />

through BC inclined at a given angle <strong>to</strong> the plane ABC. It is<br />

not difficult <strong>to</strong> verify the result analytically.]<br />

(2) The second method uses a right cylinder the base <strong>of</strong> which<br />

is an Archimedean spiral.<br />

Let ABC be a quadrant <strong>of</strong> a circle, as before, and EF, perpendicular<br />

at F <strong>to</strong> BC, a straight<br />

line <strong>of</strong> such length that EF is<br />

<strong>to</strong> the arc DC as AB is <strong>to</strong> the<br />

arc ADC.«<br />

Let a point on AB move uniformly<br />

<strong>from</strong> A <strong>to</strong> B while, in the<br />

same time, AB itself revolves<br />

uniformly about B <strong>from</strong> the position BA <strong>to</strong> the position BC.<br />

The point thus describes the spiral AGB. If the spiral cuts<br />

BD in G,<br />

or BG :<br />

BA:BG = (arc ADC) :<br />

(arc DC) = BA :<br />

(arc ADC).<br />

(arc DC),<br />

Therefore BG = EF.<br />

Draw GK at right angles <strong>to</strong> the plane ABC and equal <strong>to</strong> BG.<br />

Then GK, and therefore K, lies on a right cylinder with the<br />

spiral as base.<br />

But BK also lies on a conical surface with vertex B such that<br />

its genera<strong>to</strong>rs all make an angle <strong>of</strong> \tt with the plane ABC.<br />

Consequently K lies on the intersection <strong>of</strong> two surfaces,<br />

and therefore on a curve.<br />

Through K draw LK1 parallel <strong>to</strong> BD, and let BL, EI be at<br />

right angles <strong>to</strong> the plane ABC.<br />

Then LKI, moving always parallel <strong>to</strong> the plane ABC, with<br />

one extremity on BL and passing through K on a certain<br />

382 PAPPUS OF ALEXANDRIA<br />

curve, describes a certain plec<strong>to</strong>id, which therefore contains the<br />

point /.<br />

Also IE = EF, IF is perpendicular <strong>to</strong> BG, and hence IF, and<br />

therefore 7, lies on a fixed plane through BG inclined <strong>to</strong> ABG<br />

at an angle <strong>of</strong> ^w.<br />

Therefore I, lying on the intersection <strong>of</strong> the plec<strong>to</strong>id and the<br />

said plane, lies on a certain curve. So therefore does the<br />

projection <strong>of</strong> I on ABG, i.e. the point E.<br />

The locus <strong>of</strong> E is clearly the quadratrix.<br />

[This result can also be verified analytically.]<br />

(S) Digression: a spiral on a sphere.<br />

Prop. 30 (chap. 35) is a digression on the subject <strong>of</strong> a certain<br />

spiral described on a sphere, suggested <strong>by</strong> the discussion <strong>of</strong><br />

a spiral in a plane.<br />

Take a hemisphere bounded <strong>by</strong> the great circle KLM,<br />

with H as pole. Suppose that the quadrant <strong>of</strong> a great circle<br />

HNK revolves uniformly about the radius HO so that K<br />

describes the circle KLM and returns <strong>to</strong> its original position<br />

at K, and suppose that a point moves uniformly at the same<br />

time <strong>from</strong> H <strong>to</strong><br />

K at such speed that the point arrives at K<br />

at the same time that HK resumes its original position. The<br />

point will thus describe a spiral on the surface <strong>of</strong> the sphere<br />

between the points H and K as shown in the figure.<br />

Pappus then sets himself <strong>to</strong> prove that the portion <strong>of</strong> the<br />

surface <strong>of</strong> the sphere cut <strong>of</strong>f <strong>to</strong>wards the pole between the<br />

spiral and the arc HNK is <strong>to</strong> the surface <strong>of</strong> the hemisphere in

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!