In geometry, a sectrix of Maclaurin is defined as the curve swept out by the point of intersection of two lines which are each revolving at constant rates about different points called poles. Equivalently, a sectrix of Maclaurin can be defined as a curve whose equation in biangular coordinates is linear. The name is derived from the trisectrix of Maclaurin (named for Colin Maclaurin), which is a prominent member of the family, and their sectrix property, which means they can be used to divide an angle into a given number of equal parts. There are special cases known as arachnida or araneidans because of their spider-like shape, and Plateau curves after Joseph Plateau who studied them.
We are given two lines rotating about two poles
P
P1
P=(0,0)
P1=(a,0)
t
P
\theta=\kappat+\alpha
P1
\theta1=\kappa1t+\alpha1
\kappa
\alpha
\kappa1
\alpha1
t
\theta1=q\theta+\theta0
q=\kappa1/\kappa
\theta0=\alpha1-q\alpha
q
Q
\psi
Q
\psi=\theta1-\theta
r
P
Q
{r\over\sin\theta1}={a\over\sin\psi}
r=a
\sin\theta1 | |
\sin\psi |
=a
\sin[q\theta+\theta0] | |
\sin[(q-1)\theta+\theta0] |
The case
\theta0=0
q=n
n
r=a
\sinn\theta | |
\sin(n-1)\theta |
The case
\theta0=0
q=-n
n
r=a
\sinn\theta | |
\sin(n+1)\theta |
A similar derivation to that above gives
r1=(-a)
\sin[(1/q)\theta1-\theta0/q] | |
\sin[(1/q-1)\theta1-\theta0/q] |
r1
\theta1
a
Let
q=m/n
m
n
\theta1=q\theta+\theta0
n\theta1=m\theta+n\theta0
z=x+iy
\theta=\arg(z), \theta1=\arg(z-a)
n \arg(z-a)=m \arg(z)+n \theta0
m \arg(z)-n \arg(z-a)=\arg(zm(z-a)-n)=const
\operatorname{Re | |
(z |
m(z-a)-n)}{\operatorname{Im}(zm(z-a)-n)}=const.
w=zm(z-a)-n
arg(w)=const.
|w|=const
|z|m | |
|z-a|n |
=const.
Let
q=m/n
m
n
\theta=np
p
x=a
\sin[mp+\theta0]\cosnp | |
\sin[(m-n)p+\theta0] |
, y=a
\sin[mp+\theta0]\sinnp | |
\sin[(m-n)p+\theta0] |
x=a
\sin[mp+\theta0]\cosnp | |
\sin[(m-n)p+\theta0] |
=a+a
\cos[mp+\theta0]\sinnp | |
\sin[(m-n)p+\theta0] |
={a\over2}+{a\over2}
\sin[(m+n)p+\theta0] | |
\sin[(m-n)p+\theta0] |
x={a\over2} ⋅
\sin[(m+n)p+\theta0] | |
\sin[(m-n)p+\theta0] |
,y=a
\sin[mp+\theta0]\sinnp | |
\sin[(m-n)p+\theta0] |
\theta0=0
x={a\over2}
\sin(m+n)p | |
\sin(m-n)p |
,y=a
\sinmp\sinnp | |
\sin(m-n)p |
The inverse with respect to the circle with radius a and center at the origin of
r=a
\sin[q\theta+\theta0] | |
\sin[(q-1)\theta+\theta0] |
r=a
\sin[(q-1)\theta+\theta0] | |
\sin[q\theta+\theta0] |
=a
\sin[(1-q)\theta-\theta0] | |
\sin[((1-q)-1)\theta-\theta0] |
q, | 1 |
q |
, 1-q,
1 | , | |
1-q |
q-1 | , | |
q |
q | |
q-1 |
Let
q=m/n
m
n
\theta0
\theta0
\varphi
P
P1
P1
\varphi+\theta0
Q
PQ
\theta
\varphi+\theta0=\theta1=q\theta+\theta0
\theta=
n\varphi | |
m |
\theta
\varphi
\varphi/m
Now draw a ray with angle
\varphi
P
Q'
P'Q'
\theta1=q\theta+\theta0=q\varphi+\theta0
\theta0
q\varphi=
m\varphi | |
n |
\varphi/n
Finally, draw a ray from
P
\pi/2-\varphi-\theta0
P'
\pi/2+\varphi+\theta0
C
PP'
C
P
P'
\anglePCP'=2(\varphi+\theta0)
\varphi+\theta0
P
P'
Q''
\varphi+\theta0=\anglePQ''P'=\psi=\theta1-\theta=(q-1)\theta+\theta0
\varphi=
(m-n)\theta | , \theta= | |
n |
n\varphi | |
m-n |
\varphi/(m-n)
This is the curve
r=a
\sin\theta0 | |
\sin(-\theta+\theta0) |
(a,0)
This is a circle containing the origin and
(a,0)
r=a
\sin(\theta+\theta0) | |
\sin\theta0 |
|z-a| | |
|z| |
=const.
(0,0)
(a,0)
These curves have polar equation
r=a
\sin(-\theta+\theta0) | |
\sin(-2\theta+\theta0) |
arg(z(z-a))=const.
x2-y2-x=c(2xy-y)
\theta=\theta0/2
\theta0/2+\pi/2
(a/2,0)
|z||z-a|=c
(0,0)
(a,0)
In the case where
q=3
q=1/3
\theta0=0
r=a
\sin3\theta | |
\sin2\theta |
={a\over2}
4\cos2\theta-1 | |
\cos\theta |
={a\over2}(4\cos\theta-\sec\theta)
In the case where
q=3/2
q=2/3
\theta0=0
r=a
\sin\tfrac{3 | |
2 |
\theta}{\sin\tfrac{1}{2}\theta}=a(3\cos2\tfrac{1}{2}\theta-\sin2\tfrac{1}{2}\theta)=a(1+2\cos\theta)
The equation with the origin take to be the other pole is the rose curve that has the same shape
r=-a
\sin\tfrac{2 | |
3 |
\theta}{\sin-\tfrac{1}{3}\theta}=2a\cos\tfrac{1}{3}\theta