Hi, I'm trying to find the equation for the volume of the toroidal section shown in the following image...

i.e. a torus that is sectioned by a plane that is tangential to the hole in the torus. I'm after the volume of the smaller piece.

The purpose of this problem is to derive the general equation in R & r for calculating the minimum volume of liquid required to create a seal between the two ends of a radial U-bend tube (like you have under your sink).

I've attempted calculus by taking annular slices down through the z-axis (above diagram shows x & y axes). Doing this yields the following equation that I need to integrate. Unfortunately, this is beyond me.

$\displaystyle 2\;\int\limits_0^r {T^2 } \cos ^{ - 1} \left( {{\raise0.7ex\hbox{$N$} \!\mathord{\left/

{\vphantom {N T}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$T$}}} \right) - N\sqrt {T^2 - N^2 } \;dx$ ...Equation 1

Firstly where $\displaystyle T = R + \sqrt {r^2 - x^2 }$

Secondly where $\displaystyle T = R - \sqrt {r^2 - x^2 }$

Where N is the radius of the torus hole $\displaystyle N = R - r$

All angles in radians

I should add that the integral doesn't yield the equation for the volume I'm after directly. The full equation for the volume will be the first integral minus the second integral both taken between the same limits.

Note: Equation 6 at the end is a much better candidate for an achievable integral.

I know the volume of a torus is given by the equation $\displaystyle 2\pi ^2 Rr^2$

So when the radius of the torus hole is zero (N=0 & R=r) the sectional plane will cut the torus in half and the volume I'm after will then be given by the equation:

$\displaystyle \pi ^2 r^3$

This can also be proved by substituting N=0 into the equation I need to integrate, which makes integration straightforward provided you know the following integral:

$\displaystyle \int {\sqrt {r^2 - x^2 } \;dx = \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} r^2 \left( {\sin ^{ - 1} \left( {{\raise0.7ex\hbox{$x$} \!\mathord{\left/

{\vphantom {x r}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$r$}}} \right) + x\frac{{\sqrt {r^2 - x^2 } }}

{{r^2 }}} \right)}$

So this is a pretty good indication that Equation 1 is correct.

So can anyone help me with this integral or think of a better way to derive the equation I'm after?

Addendum:

Okay, I've now written a program that calculates the integral numerically.

I've also derived a possible solution by reducing the problem down to two dimensions.

If we consider the two-dimensional torus; i.e. just a disc with a hole in the center as literally shown in the above image, then the total area of this 2D ring is given by:

$\displaystyle 4\pi Rr$

Compare this to the volume of the 3D torus:

$\displaystyle 2\pi ^2 Rr^2$

So moving from 2D to 3D has introduced the factor $\displaystyle \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \pi r$

Now look at the 2D equation for the area of the equivalent sector I'm after:

$\displaystyle \left[ {\left( {R + r} \right)^2 \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)} \right] - 2\left( {R - r} \right)\sqrt {Rr}$

If we multiply this by $\displaystyle \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \pi r$ in an attempt to change it into the 3D equation I'm after, we get:

$\displaystyle \left[ {\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \pi r\left( {R + r} \right)^2 \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)} \right] - \pi r\left( {R - r} \right)\sqrt {Rr}$ ...Equation 2

Now tantalizingly, this equation yields answers that are remarkable close to the numerically generated answers. In fact, when R=r this equation is correct, and then never deviates by more than 1.82% as R increases in relation to r.

As I see it, there are two possibilities:

Firstly, that Equation 2 is missing a subtle term, or secondly, that Equation 2 is correct and my computer program for generating the answer numerically is in error as R increases in relation to r, due to some form of compound rounding error.

Follow up:

I'm thinking of replacing $\displaystyle \cos ^{ - 1} \left( {{\raise0.7ex\hbox{$N$} \!\mathord{\left/

{\vphantom {N T}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$T$}}} \right)$ in Equation 1 with $\displaystyle \sec ^{ - 1} \left( {{\raise0.7ex\hbox{$T$} \!\mathord{\left/

{\vphantom {T N}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$N$}}} \right)$ which may be easier to integrate.

I know that:

$\displaystyle \int {\sec ^{ - 1} \left( a \right)} = a\sec ^{ - 1} \left( a \right) - \ln \left( {a + \sqrt {a^2 - 1} } \right)$

Still beyond me

Okay, I've taken a different tac, and instead of calculating the integral by take sections down through the Z-axis, I've come up with an integral by taking circular cross-sections radially around the torus. That gives the integral...

$\displaystyle \int\limits_0^{\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)} {r^2 \cos ^{ - 1} \left( {{\raise0.7ex\hbox{$T$} \!\mathord{\left/

{\vphantom {T r}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$r$}}} \right)} - T\sqrt {r^2 - T^2 } \;d\theta$ ...Equation 3

Where $\displaystyle T = \left( {R - r} \right)\sec \left( \theta \right) - R$

Again the volume of the section I'm after equals twice this integral.

To check we can make R=r as we did before. This should produce an equation for the sectional volume of $\displaystyle \pi ^2 r^3$, however, it produces the equation $\displaystyle \pi ^2 r^2$, which is incorrect. I've lost an r somewhere. I'm guessing that it's because I've substituted the values of R & r before deriving the integral. I probably have to do it the other way around. I probably can't say that $\displaystyle \left( {R - r} \right)\sec \left( \theta \right) = 0$ when R=r, because at the bound when $\displaystyle \theta = \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$, $\displaystyle \sec \left( \theta \right) = \infty$, so $\displaystyle \left( {R - r} \right)\sec \left( \theta \right)$ is indeterminate.

To check, I've written a program to calculate this integral numerically, and the answers it yields are close enough to consider the integral correct. Interestingly, there is a small deviation between the numerical answers given by this program and the first program that uses the first integral (Equation 1). I suspect that there is a small compound error in both numerical solutions, which takes us back to the possibility that my guessed Equation 2 could be right.

Okay, I went back to the first method of integrating rings down through the z-axis, each cut by the same section, and looked at the ratio of the volume cut off by the section in relation to the volume of the whole ring for each ring section.

The first slice skims the torus such that the slice is almost a circle of radius R cut by the same section (R-r) from the center, the ratio of sectioned ring to whole ring is given by the equation:

$\displaystyle \frac{1}

{\pi }\cos ^{ - 1} \left( {1 - {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right)$

The last slice cuts the torus through the middle and creates a slice identical to the ring in the image above. Here the ratio of sectioned ring to whole ring is given by the equation:

$\displaystyle \left[ {\frac{{\left( {R + r} \right)^2 }}

{{4\pi Rr}}\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)} \right] - \frac{{\left( {R - r} \right)}}

{{2\pi \sqrt {Rr} }}$

Using this information I came up with an amended guess for the right equation:

Volume of section = $\displaystyle 2Rr^2 \cos ^{ - 1} \left( {1 - {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right)\; - \;r\left( {\pi - 1} \right)\left[ {\left( {R - r} \right)\sqrt {Rr} \; - \;\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \left( {R + r} \right)^2 \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)} \right]$ ...Equation 4

This yields results even closer to the first numerical solution than the previous guess (Equation 2) and is also exact when R=r. Unfortunately I have no way of knowing if this equation is correct.

I also believe that the following double integral will yield the equation for the volume I'm after:

$\displaystyle 2\;\int\limits_0^r {\;\int\limits_{R - r}^{R + r} {\;\sqrt {R^2 + r^2 - x^2 - z^2 - 2R\sqrt {r^2 - z^2 } } \;} } dx\;dz$ ...Equation 5

I'm not sure if this is easier to integrate than Equation 1 or Equation 3.

Also, the form of the first integral (Equation 1) was determined by working out the area of the cut section of each annular slice using trigonometry. However, this area can also be determined by integrating the equation of the annular ring from the cut section out to the radius of the ring. If this is done we end up with the following integral:

$\displaystyle 2\;\int\limits_0^r {\;\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \pi T^2 - T^2 \sin ^{ - 1} \left( {{\raise0.7ex\hbox{$N$} \!\mathord{\left/

{\vphantom {N T}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$T$}}} \right)} - N\sqrt {T^2 - N^2 } \;dx$

However, this is exactly the same as Equation 1, so gets us nowhere.

I've finally found a much simpler integral by taking vertical curved slices:

$\displaystyle 4\;\int\limits_{ - r}^{ + r} {\left( {R + x} \right)} \sqrt {r^2 - x^2 } \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right)\;dx$ ...Equation 6

This integral may be possible. So forget all previous integrals, concentrate on this one!

Note: the root in this integral should be $\displaystyle \pm$ and the multiplier at the start should be 2. I've simply removed the $\displaystyle \pm$ to make it easier to understand and made the multiplier 4.

I've now taken a significant step toward deriving this integral. I've worked out that...

$\displaystyle \int\limits_{ - r}^{ + r} {\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right)} \;dx\quad =$

$\displaystyle \left[ {\left( {R + x} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right) - \left( {R - r} \right)\ln \left( {\frac{{R + x + \sqrt {\left( {R + x} \right)^2 - \left( {R - r} \right)^2 } }}

{{R - r}}} \right)} \right]_{ - r}^{ + r}$

or

$\displaystyle \;\left[ {\left( {R + x} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right) - \left( {R - r} \right)\cosh ^{ - 1} \left( {\frac{{R + x}}

{{R - r}}} \right)} \right]_{ - r}^{ + r}$

Taken between limits yields...

$\displaystyle \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right) - \left( {R - r} \right)\ln \left( {\frac{{R + r + 2\sqrt {Rr} }}

{{R - r}}} \right)$

or

$\displaystyle \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right) - \left( {R - r} \right)\cosh ^{ - 1} \left( {\frac{{R + r}}

{{R - r}}} \right)$

$\displaystyle \int\limits_{ - r}^{ + r} { \pm \sqrt {r^2 - x^2 } } \;dx\; = \;\pi r^2$

I'm getting there slowly. I've now calculated the following integral that represents two-thirds of Equation 6:

$\displaystyle \int\limits_{ - r}^{ + r} {\left( {R + x} \right)} \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right)\;dx$

$\displaystyle = \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \left[ {\left( {R + x} \right)^2 \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right) - \left( {R - r} \right)\sqrt {\left( {R + x} \right)^2 - \left( {R - r} \right)^2 } } \right]_{ - r}^{ + r}$

Taken between limits yields:

$\displaystyle \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \left( {R + r} \right)^2 \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right) - \left( {R - r} \right)\sqrt {Rr}$

Interestingly, if we multiply this equation by $\displaystyle \pi r$ we get back to Equation 2 my first guess.

We can also differentiate the above equation:

$\displaystyle y = \left( {R + x} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right)$

$\displaystyle \frac{{dy}}

{{dx}} = \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + x}}} \right) + \frac{{R - r}}

{{\sqrt {\left( {R + x} \right)^2 - \left( {R - r} \right)^2 } }}$

Using this we can attempt to integrate Equation 6 by parts setting U equal to the above equation and dV equal to the root term. Taken between limits, this yields a first term for the sectional volume we are looking for as:

$\displaystyle \pi r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

The remaining integral in the parts approach is too difficult, but this at least provides a start for another guess as to the desired equation.

Okay, I've now determined the following:

$\displaystyle {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 1\quad V \to \pi r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

$\displaystyle {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0\quad V \to \frac{{64}}

{{15}}r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

I think the general form of the equation is:

$\displaystyle V = f\left( {R,r} \right)r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

Where f(R,r) has yet to be determined.

Actually the function "f" doesn't have variables of R & r, but just one variable r/R. So we can rewrite the general equation:

$\displaystyle V = f\left( {{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right)r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

Because I can work out the value of each volume (V) numerically from the original integral (Equation 6), it is possible to work out what f(r/R) should be for every possible r/R. So I plotted r/R along the x-axis from 0 to 1, and f(r/R) up the y-axis from $\displaystyle \pi$ to 64/15 and scaled so that each axis was the same length. This produced a curve that I was convinced was an arc from a greater circle. This made perfect sense, as a torus is made by two circular rotations. So I worked out the radius of this larger circle and superimposed that on the graph. It seemed to match perfectly. So all I needed to do was to use the equation of this mystery circle as the base for the equation for the function f(r/R). Here it is:

$\displaystyle f\left( {{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right) = \left( {x - y} \right)\left( {\frac{{64}}

{{15}} - \pi } \right) + \pi$

$\displaystyle x = \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} + \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \sqrt {2m^2 - 1}$

$\displaystyle y = \sqrt {m^2 - \left( {x - \frac{r}

{R}} \right)^2 }$

Where m=2.2762118 and is the radius of our mystery circle of which the function f(r/R) appears to be part. This makes m & x unchanging constants for all tori.

To test out the full equation (V) for the volume I'm after (now including the actual equation for f(r/R)) I plugged in values of r & R and compared the results to those obtained from the numerical integration.

Bugger!

Amazingly the results are fractionally out. Surely the function f(r/R) can't be so close to a circular arc without actually being a circular arc? So I plotted the function f(r/R) in green and a perfect circular arc in red at a much higher resolution to show up any minor deviation. Here is that graph:

Graph1

Note: If the crossed lines appear broken, then you have auto image scaling enabled in your browser. To see the graph at the correct scale hover your mouse pointer over the image and click the icon that appears.

As can be seen, f(r/R) in green does deviate slightly from the perfect arc in red.

This is totally amazing! How can an equation that relates to all tori be so close to a circular arc without actually being a circular arc! The great God of mathematics is surely pulling my chain!

I'm starting to believe that this planar section through a torus is related to the impossible problem of squaring the circle. This could actually be the case. But I'll persevere.

I thought that maybe my numerical method for calculating the integral was responsible for the deviation, but I'm quite confident that this is not the case.

As can be seen, the curve deviates the most on the left-hand side of the graph between r/R=0 and r/R=0.5. Actually, the middle of the curve is reached when r/R=0.42 (0.4203507?).

Because the deviation occurs when R is large in relation to r, I'm guessing that there is a missing constant somewhere. This missing term may actually be related to (R-r)/(R+r) which is central to this problem.

Let's go back a bit to:

$\displaystyle {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 1\quad V \to \pi r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

$\displaystyle {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0\quad V \to \frac{{64}}

{{15}}r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

The general form for the equation might be:

$\displaystyle V = \pi r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right) + f\left( {R,r} \right)$

Where:

$\displaystyle f\left( {R,r} \right) \to 0\quad as\quad{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 1$

$\displaystyle f\left( {R,r} \right) \to\left( {\frac{{64}}

{{15}}-\pi} \right)r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)\quad as\quad{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0$

We can simplify this last limit because...

$\displaystyle r^2 \left( {R + r} \right)\cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right) \to 2r^2 \sqrt {Rr}\quad as\quad{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0$

so,

$\displaystyle f\left( {R,r} \right) \to 2r^2 \sqrt {Rr} \left( {\frac{{64}}

{{15}}-\pi} \right)\quad as\quad{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0$

We already know that:

$\displaystyle V \to \pi ^2 r^3 \quad as\quad {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 1$

But we now also know:

$\displaystyle V \to \frac{{128}}

{{15}}r^2 \sqrt {Rr} \quad as\quad {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0$

Again we can plot r/R against f(r/R), but this time we need to fix r=1. So r/R tends to zero as R tends to infinity. Our volume (V) and also function f(r/R) also tend to infinity as r/R tends to zero. So f(r/R) will go from zero to infinity as r/R goes from 1 to zero. So we can say immediately that this curve will not be an arc from a circle.

Before we do that, let's have a look at our equation for V at those limits. We could write:

$\displaystyle V \to \pi ^2 r^2 \sqrt {Rr} \quad as\quad {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 1$

$\displaystyle V \to \frac{{128}}

{{15}}r^2 \sqrt {Rr} \quad as\quad {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0$

So we could write the equation for our volume (V) as:

$\displaystyle V = f\left( {{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right)r^2 \sqrt {Rr}$

Where:

$\displaystyle f\left( {{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right) \to \pi ^2 \quad as\quad {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 1$

$\displaystyle f\left( {{\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}} \right) \to \frac{{128}}

{{15}}\quad as\quad {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}} \to 0$

Numerical results suggest that this function has one variable r/R instead of two separate variables R & r, just like our previous function.

I think this is a much more direct function to try and plot. I suspect this function will involve our old friend:

$\displaystyle \cos ^{ - 1} \left( {\frac{{R - r}}

{{R + r}}} \right)$

Which of course can be written:

$\displaystyle \cos ^{ - 1} \left( {\frac{{1 - {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}}}

{{1 + {\raise0.7ex\hbox{$r$} \!\mathord{\left/

{\vphantom {r R}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$R$}}}}} \right)$

This new function f(r/R) turns out to be almost a straight line, but as we've come to expect from this problem, not exactly a straight line.

Graph2

This is so close to a straight line that we can't really see its shape. The next graph multiplies the displacement between the function and the diagonal by a factor of 30. At the top-right the function actually undershoots the diagonal. The insert image shows this section magnified and with the displacement to the diagonal increased by a factor of 100.

Graph3

The first half of the function from the origin approximates to an arc from a circle like our previous function. The top half obviously doesn't. Because this difference is quite subtle in the un-magnified graph, I suspect that this explains why our first function deviated slightly from a perfect arc, despite only being related to this function.

So what exactly is this new function? It obviously oscillates between the end points, but many curves do this. Weird that the amplitude is so tiny and the second half-cycle is so small compared to the first.

There are three points of interest along this curve: the maximum deviation (L) above the diagonal, the point the curve crosses the diagonal and the maximum deviation below the diagonal. Numerical analysis has determined these points to be:

Max above:r/R = 0.3772161Crossing point:

f(r/R) = 9.05215854

L = 0.0088448351r/R = 0.9634874Max below:

f(r/R) = 9.82081382

L = 0r/R = 0.9829642

f(r/R) = 9.84674677

L = 0.000055870519

Note: L is measured at right-angles to the diagonal and with the vertical axis unscaled; i.e. with a diagonal of 53.19 degs to the horizontal as opposed to the 45 degs in the previous graphs due to the scaling of the vertical axis to form a square with the horizontal. In fact this scaling was only done because the previous f(r/R) looked like an arc under these conditions. As the new function is nothing like an arc, there really is no reason to scale the length of the vertical axis. This axis can be left at 1.3362711 units long in comparison to the horizontal axis that is 1.0 units in length.

Well I’ve been all around this problem now. I’ve found half a dozen different integrals, but each has the same problem. What I need to do is to determine whether the section of the torus I’m after, namely the section that is tangential to the central hole, is a special case. I know that a torus with a zero diameter hole (r/R=1) is a special case because here the section runs through the centre and so the volume of the section I’m after equals half the volume of the torus. But what about my section?

To determine if my tangential section is special like the central section I have plotted the volume of all tori and all sections on a single graph.

Graph4

In this graph the horizontal axis represents the distance from the center to the outside edge of a torus with unit overall radius (R+r=1). The vertical blue lines represent the radius of the inner hole for eleven different tori with holes ranging in size form zero radius (left-most blue line) to the entire radius of the torus (right-most blue line) in steps of 0.1. The red lines represent the volume of the smaller piece cut off by a straight sectional plane passing down through the torus, starting off passing through the center and then moving to the outside edge of the torus. For each torus the volume will start off as half the volume of the torus (left-hand side) and end up as zero (right hand side). The starting volume has been scaled to the same height for each of the eleven tori. The lowest red line plots the volume of the sweeping section for the torus with a hole of zero size (left-most blue line r/R=1), the red line up from that plots the volume of the sweeping section for the torus with a hole 0.1 in size (2nd blue line), and so on until the upper red line plots the volume of the sweeping section for the torus with a hole equal to the overall diameter of the torus (right-most blue line r/R=0). The intersection of the vertical blue lines with their corresponding red lines represents the volume for the section that I’m after, where the sweeping section is tangential to the hole. The green line plots this for all tori from r/R=1 (left-hand side) to r/R=0 (right-hand side).

There doesn’t appear to be anything special about our tangential section represented by the green line as it weaves its way through all tori from r/R=1 (lower red line) to r/R=0 (upper red line). However, if we flip the green line about its diagonal axis, it ends up looking quite similar to the some of the lower red curves. There’s a very good reason for this. This green line represents our tangential section, but that too sweeps from the center to the outside edge of the torus as the hole increases in size. The only difference is that the green line section sweeps out as the torus itself changes in hole size, whereas the red line section sweeps out across a torus of fixed hole diameter. It will be interesting to see if the green line can be made to match exactly with a particular red line.

This won’t necessarily get us anywhere, because we started out looking for the green line to be a special case that would help us derive the equation for the tangential volume. We’ve ended up showing that far from being a special case, the tangential case may represent all sectional volumes. So if we can’t solve the case for all sectional volumes, it’s doubtful we’ll solve the problem for our tangential section. Ho hum.