Consider a thin beam bent in 2 dimensions so that the torque at a point is proportional to its curvature, which is the commonest assumption.
We assume that the beam is at rest and in equilibrium, perhaps unstable.
In a plane, the torque is a linear function of the position and we assume here that the curvature at point <x, y> is Ty where T is a parameter.
We normalize by assuming that the curve passes thru <0, 1> parallel to the X axis there.
What is the shape of this curve?
It is easy to compute the curve it we take its arc length, τ, as the independent variable and consider x, y and α as dependent variables.
α is the angle, measured counter-clockwise from the positive X direction, in radians, of the tangent to the curve.
The initial conditions are
The differential equations are
- dα/dτ = –Ty
- dx/dτ = cos α
- dy/dτ = sin α
We describe a one parameter family of curves—T, for torque, is the parameter.
This too simple Java code, controlled by this makes the following pictures:
java Main 1 0 => 
java Main 2 0 => 
java Main 2.5 0 => 
java Main 3 0 => 
java Main 3.305 1.0 => 
Alas not a lemniscate!!
java Main 3.5 1.7 => 
java Main 3.9 1.7 => 
java Main 4.0 1.7 => 
Alas not a Folium
java Main 4.2 1.7 => 
java Main 4.4 1.7 => 
java Main 5 1.7 => 
In this family of curves there is a close mimic of the Folium of Descartes rotated by π/4 when T = 4, but it is not the same.
When T is about 3.305 it is nearly, but not quite a lemniscate.
For negative values of T we get a rescaled reprise of the curves from T > 4.
These curves bounded away from the X axis—indeed y ≥ 1.
Candidate RK code: xx