# A Torus and Its Dual, Part II

After I published the last post, which I did not originally intend to have two parts, this comment was left by one of my blog’s followers. My answer is also shown.

A torus can be viewed as a flexible rectangle rolled into a donut shape, and I had used 24 small rectangles by 24 small rectangles as the settings for Stella 4 for the torus, and its dual, in the last post — which, due to the nature of that program, are actually rendered as toroidal polyhedra. To investigate my new question, I increased 24×24 to 90×90, and these three images show the results. The first shows a 90×90 torus, the second shows its dual, and the third shows the compound of the two.

When I compare these images to those in the previous post, it is clear that these figures are approaching a limit as n, in the expression “nxn rectangle,” increases. What’s more, I recognize the dual now, of the true torus, at the limit, as n approaches infinity — it’s a cone. It’s not a finite-volume cone, but the infinite-volume cone one obtains by rotating a line around an axis which intersects that line. This figure, not a finite-volume cone, is the cone used to define the conic sections: the circle, ellipse, parabola, and hyperbola.

What’s more, I smell calculus afoot here. I do not yet know enough calculus.

“Learn a lot more about calculus” is definitely on my agenda for the coming Summer, for several reasons, not the least of which is that I plainly need it to make more headway in my understanding of geometry.

[Note: Stella 4d, the program used to make these images, may be found at http://www.software3d.com/Stella.php.]

# The Cone Problem (The Easier Sequel to the Hemisphere Problem)

That hemisphere problem (see previous two posts) was quite difficult. I’m going to unwind a  bit with the much easier cone version of the same problem: at what height x above the ground, expressed as a fraction of h, must a cone of height h and radius r be cut, in order for the two pieces produced by the cut to have equal volume? The fact that a path down the lateral surface of a cone is a straight line, not a curve, should make this much easier than the hemisphere problem.

Since the volume of a cone is (1/3)πr²h, and the smaller cone created above the cut would be half that volume, it follows that

(1/3)πr²h = (2/3)π(r′)²h′                [equation 1]

By cancellation of (1/3)π, this equation becomes

r²h = 2(r’)²h’               [equation 2]

Also, based on divisions of the cone’s altitude, we know that

h = h′ + x                [equation 3]

Furthermore, since the problem asks that the height x be expressed as a fraction of h, we can let that fraction (a decimal between zero and one) be represented by f, so that

x = fh               [equation 4]

Also, by using similar right triangles’ corresponding legs, we know that

r/h = r′/h′                [equation 5]

which rearranges to

rh′ = r′h                  [equation 6]

There is a proportionality constant in play here, p, defined as the fraction of the length of one part of the larger cone which equals the length of the corresponding part of the smaller cone. As equations, then,

r′ = pr         and          h′ = ph              [equations 7a and 7b]

Also, because p is the fraction of h which is h′, and f is the fraction of h which is x, and h = h′ + x, it follows that

p + f = 1                  [equation 8].

Next, by substituting equations 7a and 7b into equation 2 for r′ and h′, we know that

r²h = 2(pr)²ph               [equation 9]

Which reduces to

1 = 2p³               [equation 10]

When equation 10 is solved for p, it becomes

p = (1/2)^(1/3)                [equation 11]

And, since equation 8 states that p + f = 1, it follows that f = 1 – p, and f is the fraction we seek. By substituting equation 11 for p in f = 1 – p, the following value for f can be determined:

f = 1 – (1/2)^(1/3)               [equation 12]

This leads to the following cleaned-up solution to the problem, shown in standard exact form, and with a decimal approximation as well.

The cut, therefore, should be made approximately 20.6% of the way from the bottom to the top of the full cone.

To check this answer, I need only find the volume of the smaller cone, times two, and show that it equals the value of the larger cone.

2(volume of smaller cone) = (2/3)π(r′)²h′ = (2/3)π(pr)²ph =

(2/3)πp³r²h = (2/3)π(cube root of ½)³r²h = (2/3)π(1/2)r²h = (1/3)πr²h,

which is the volume of the full cone, as it should be. The problem has now been solved, and the solution f (by way of p, which equals 1 – f, by a rearrangement of equation 8) has been checked.