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Compass-and-straightedge

Every point of a solenoid can be constructed using compass alone. The process is very simple and requires a generatrix circle with centre coordinates $\left(n,0\right)$ and radius $n$, where $n$ is a natural number.

Once that the generatrix circle is drawn, center the compass on the origin and draw another circle with radius $r$ where $0\le r\in \mathrm{ℝ}\le 2n$. This circle is called goniometric circle.

Let us focus on the first and second quadrant of the Cartesian plane, that is the two quadrands where $y\ge 0$. The goniometric circle intersects the x-axis in two points, ${C}_{3}$ in quadrant II and ${C}_{1}$ in quadrant I, whereas it intersect the generatrix circle in ${C}_{2}$ in quadrant I. Of course it intersect the generatrix circle also in quadrant IV, but since the siluroid is symmetric with respect to the x-axis, we need only to build it on the first two quadrants and mirror it in the third and fourth one.

Now, let us center the compass in ${C}_{2}$ and in ${C}_{1}$, always with radius $r$, and draw the corresponding circles. They intersect each other in two points: the origin and $P$.

$P$ is a point of the main lobe of the siluroid.

You can likely obtain the secondary lobes by centering the compass in ${C}_{2}$ and in ${C}_{3}$, always with radius $r$. The two points of intersection are now $Q$ and the origin. $Q$ is a point of the secondary lobe lying in quadrant II.

Demonstration

To demonstrate that $P$ is a point of a siluroid, let us focus on the isosceles triangle $\stackrel{^}{O{C}_{2}{C}_{n}}$. That triangle has two sides equal to $n$ and the base equal to $r$. It has also the angles opposite to the two sides of the same length equal to $2\theta$. Therefore

[6] $\frac{r}{2}= n \mathrm{cos}\left(2\theta \right)$

Let us focus now on the isosceles triangle $\stackrel{^}{OP{C}_{1}}$. That triangle has two sides equal to $r$ and the base equal to $\rho$. It has also the angles opposite to the two sides of the same length equal to $\theta$. Therefore

[7]

Thence, by substituting the $r$ in [7] by using [6] we obtain

[8]

However, since

[9]

we get to

[10] $\rho =4n\mathrm{cos}\left(\theta \right)\mathrm{cos}\left(2\theta \right)$

that is, the polar equation of the siluroid.

Q.E.D.

A similar demonstration can be used to proof that $Q$ belongs to the siluroid too.

My Profiles

The Siluroid

The Siluroid Curve and Me!