The Kansas University Quarterly, Vol. I, No. 2, October 1892 — John Shaqi
The Kansas University Quarterly, Vol. I, No. 2, October 1892Various
History
The Kansas University Quarterly, Vol. I, No. 2, October 1892
Various
Natural history -- Periodicals; Science -- Periodicals
e 1
r = ——— cos_x_ + ——— ;
p p
where e and p are constants, being respectively the eccentricity and
semi-latus rectum of the conic.
From the above equation it is readily seen that the curve may be traced
by drawing from a fixed point O on a circle any number of chords
and laying off a constant length on each of these lines, measured
from the circumference of the circle. The point O is the node of the
limaçon; and the fixed circle, which I shall call the base circle, is
the inverse of the directrix of the conic. This is readily shown as
follows:—the polar equation of the directrix is r = p/(e cos_x_). Hence
the equation of its inverse is r = (e cos_x_)/p, which is the equation of
the base circle of the limaçon.
If the conic which we invert be an ellipse, the point O will be an
acnode on the Limaçon; if the conic be a hyperbola, the point O is a
crunode. If the conic be a parabola, O is then a cusp and the inverse
curve is called the Cardioid.
The limaçon may also be traced as a roulette.
Let the circle A C have a diameter just twice that of the circle A B.
Then a given diameter of A C will always pass through a fixed point Q
on the circle A B, (Williamson’s Diff. Cal. Art. 286) and will have its
middle point on the circle A B. Now any point P on the diameter of A C
will always be at a fixed distance from C and will therefore describe a
limaçon of which A B will be the base circle.
The pedal of a circle with respect to any point is a limaçon. This may
be inferred from the general theorem that the pedal of a curve is the
inverse of its polar reciprocal, (Salmon’s H. P. C. Art. 122). For the
polar reciprocal of a conic from its focus is a circle and hence its
pedal is a limaçon.
The base circle is the locus of the instantaneous centre for all points
on the limaçon. Let B O P be a line cutting a circle in B and Q. Let
the line revolve about B, Q following the circle; the point P will
trace a limaçon.
Now, for any instant, the instantaneous center will be the same whether
Q be following the circle or the tangent at the point where the line
cuts the circle. Therefore the instantaneous center for the point P
is found by erecting a perpendicular to the line P B, through B, and
a normal to the circle at Q. (Williamson’s Diff. Cal. Art. 294). The
intersection (C) of these two lines is the instantaneous center for the
curve at the point P. But by elementary geometry C is on the circle.
Now as the line P B revolves through 360° around B, the line B C which
is always perpendicular to it also makes a complete revolution and the
instantaneous center C moves once round the base circle.
Public-domain text, read in full here on John Shaqi.
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