Solve the quartics

Quartic shapes require more maths than the lower degree polynomial shapes, but the result is so impressive that's its worth the effort! Here's the equation of the hollow cube shown here:

x^4 + y^4 + z^4 - 10*R^2*(x^2 + y^2 + z^2) + 40 * R^4

A quartic shape is defined by a fourth degree polynomial equation. In other terms, this equation is made of all the polynomial forms of the variables x, y and z, up to the 4th degree. These forms can be:

x^4, y^4, ...
x^3 * y, ...
x^2 * y^2, x^2* z^2, ...
x^3, y^3, ...
x^2 * y, x^2 * z, ...

There are 35 possible combinations, ranging from degree 0 to 4. As for the quadrics, the goal is to calculate the intersection between the ray and the shape. That leads to a fourth degree polynomial equation.

Solving such an equation requires a method such as Ferrari's method, based itself on the Cardano method that solves third degrees equations.

These two methods are included in TC-Ray.

The classical torus

The most popular quartic is the torus. It is quite easy to understand how it's made with parametric equations:

x = (R + l*Cos(Alpha)) * Cos(Theta)
y = (R + l*Cos(Alpha)) * Sin(Theta)
z = l*Sin(Alpha)

where R and l and constant lengths, Alpha and Theta ranging from 0 to 2*Pi. Let's study this: x = R*Cos(Theta) and y = R*Sin(Theta) define a circle in the horizontal plan, while u = l*Cos(Alpha) and v = l * Sin(Alpha) define another circle in a (u,v) plan. This new plan is mobile, that means we shall, for each point of the horizontal circle, draw a new vertical circle.

We need to transform this into a cartesian equation. In other terms, we need to get rid of Alpha and Theta.

1 and 2 give x^2 + y^2 = (R+l*Cos(Alpha))^2, i.e. l*Cos(Alpha)= SQR(x^2+y^2) - R

This new equation, merged with the third one, can be written:

z^2 + (SQR(x^2+y^2) - R)^2 = l^2

We can now develop this to get the cartesian equation of the torus, and solve it via the method of Ferrari:

x^4 + y^4 + z^4 + 2*(x^2*y^2 + x^2*z^2 + y^2*z^2) - 2*(R^2 + l^2)*(x^2 + y^2) + 2*(R^2 - l^2)*(1 + z^2) = 0

The bulbeous torus

Now that we have a first quartic cartesian equation, we can try other coefficients. On the basis of the classical torus, we can change just one parameter as follows:

x^4 + y^4 + z^4 + (x^2*y^2 + x^2*z^2 + y^2*z^2) - 2*(l^2 + R^2)*(x^2 + y^2) + 2*(R^2 - l^2)*(1 + z^2) = 0

See how the torus is dismorphed with this minor change!

The lemniscat

Let's go on playing around the torus... We can cut this shape with a vertical plan (y=a) and see what the cut looks like.

The most interesting case corresponds to y = R-l. The intersection is switching from separate shapes to a fully connected area. Replace y by (R-l) in the torus equation: this will provide the equation of a plan curve defined in y and z, that looks like z^2 = f(y). This curve is called a lemniscat.

Now, transform this plan curve into a 3D surface by rotating arounf the x axis. You will get a lemniscat of revolution. To obtain the corresponding equation, say that (y^2 + z^2) = f(y), f(y) being the above mentioned function.

The equation is finally:

x^4 + y^4 + z^4 + 2*(x^2*y^2 + x^2*z^2 + y^2*z^2) - 2*(R^2 + l^2)*x^2 + 4*(R^2 - l^2)*(y^2 + z^2) = 0

Note that the same rational can be followed with other f(y) functions...

Strange shapes appear...

The classical torus is based on the following equation:

z^2 + (SQR(x^2+y^2) - R)^2 = l^2

What happens if, instead of a constant radius in the right part of the equation, we use a function of x and y ? Well, strange things happen... You can work around formulas like F1 + F2*x*y + F3*(x^2-y^2), play with F1, F2 and F3, do a lot of tweaking. From time to time, you will get really surprising things. The best you can do to realise the possibilities is to visualise the video made with a version of TC-Ray.

This ultra-feminine shape was drawn like this!