A history of Greek mathematics - Wilbourhall.org

THE COLLECTION. BOOK IV 381
Hence / lies on a certain curve.
on the plane ABO, also lies on a curve.
Therefore E, its projection
In the particular case where the given ratio of EF to the
arc CD is equal to the ratio of BA to the arc CA, the locus of
E is a quadratrix.
[The surface described by the straight line LH is a plectoid.
The shape of it is perhaps best realized as a continuous spiral
staircase, i.e. a spiral staircase with infinitely small steps.
The quadratrix is thus produced as the orthogonal projection
of the curve in which the plectoid is intersected by a plane
through BC inclined at a given angle to the plane ABC. It is
not difficult to verify the result analytically.]
(2) The second method uses a right cylinder the base of which
is an Archimedean spiral.
Let ABC be a quadrant of a circle, as before, and EF, perpendicular
at F to BC, a straight
line of such length that EF is
to the arc DC as AB is to the
arc ADC.«
Let a point on AB move uniformly
from A to B while, in the
same time, AB itself revolves
uniformly about B from the position BA to the position BC.
The point thus describes the spiral AGB. If the spiral cuts
BD in G,
or BG :
BA:BG = (arc ADC) :
(arc DC) = BA :
(arc ADC).
(arc DC),
Therefore BG = EF.
Draw GK at right angles to the plane ABC and equal to BG.
Then GK, and therefore K, lies on a right cylinder with the
spiral as base.
But BK also lies on a conical surface with vertex B such that
its generators all make an angle of \tt with the plane ABC.
Consequently K lies on the intersection of two surfaces,
and therefore on a curve.
Through K draw LK1 parallel to BD, and let BL, EI be at
right angles to the plane ABC.
Then LKI, moving always parallel to the plane ABC, with
one extremity on BL and passing through K on a certain