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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 (n,0) and radius n, where n is a natural number.

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Once that the generatrix circle is drawn, center the compass on the origin and draw another circle with radius r where 0r 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 y0 . The goniometric circle intersects the x-axis in two points, C3 in quadrant II and C1 in quadrant I, whereas it intersect the generatrix circle in C2 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 C2 and in C1 , 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 C2 and in C3 , 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 OC2Cn^ . 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 2θ . Therefore

[6] r2= ncos(2θ)

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Let us focus now on the isosceles triangle OPC1^ . That triangle has two sides equal to r and the base equal to ρ . It has also the angles opposite to the two sides of the same length equal to θ . Therefore

[7] ρcos(2θ)  = r(1  + cos(2θ))

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

[8] ρ = 2ncos(2θ)1 + cos(2θ)cos(θ)

However, since

[9] 1 + cos(2θ) = 2cos2(θ)

we get to

[10] ρ = 4 n cosθ cos2θ

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.

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