1
$\begingroup$

enter image description here To generate a solid ring torus around the cylinder, the circle (2) is revolved around the cylinder along a path $2\pi R$, where $R = r_{cylinder}+r_{circle}$. To generate a solid rectangular toroid, the rectangle (3) is revolved around the cylinder along a path $2\pi R$, where $R = r_{cylinder}+(h/2)$. In both of these cases, it seems that the radius of revolution to produce a toroid which has the cross-section of the original circle or rectangle must be the radius which extends to the center of area of the shape being revolved.

My question has two parts:

  1. Is it true that to generate a toroid with the cross section of the annular sector (left), I must first calculate the point which is the geometric center of the annular sector, and NOT the center of area of the annular sector? To be clear, I feel it would be some distance $r_3$ which is greater than $r_1$ + ($r_2$ - $r_1$)/2. My reasoning is that the center of mass of an annular sector, answered here could be a point outside of the sector itself. I believe I need to calculate the radius $r_3$ to a point in the center of the area of the annular ring, then worry about calculating the distance from that center of volume to the center of the cylinder axis.

  2. How is the radius $r_3$ to the geometric center of the annular sector in question calculated?

Thank you.

$\endgroup$
0
1
$\begingroup$

Let's think first why you need to rotate at that particular distance. It is done in such a way that when you take the cross section, in the radial direction, you get the desired figure. You also have the constraint that your cross section is touching the big cylinder.

It does not matter what your shape is. You inscribe it in a rectangle, with the height equal to the distance between the largest and the closest point to the cylinder. Rotate the rectangle as before, and your desired figure will be rotated accordingly.

$\endgroup$
2
  • $\begingroup$ I believe that this concise answer is what I needed. Would this same logic be true if the path that the shape is taking is not a circular path around the circumference of the cylinder, but instead a helical path along the surface of the cylinder? This is at the heart my actual question, but not sure if I should post another question. I know how to calculate the arc length of a helix given its radius, pitch, and number of revolutions. My true question is, should I pick a helical radius equal to the radius of the cylinder plus half of the height of the rectangle which inscribes my shape? $\endgroup$
    – kreeser1
    Dec 11 '18 at 23:08
  • 1
    $\begingroup$ Yes. The translation along the cylinder is irrelevant. You can decompose the motion into the circular motion above, and a translation. $\endgroup$
    – Andrei
    Dec 12 '18 at 0:03
0
$\begingroup$

If the substance is of uniform density, then the center of volume is the center of mass.

The center of volume of the torus is the point about which you are revolving.

Anyway, what I think this question is asking is answerd by Pappus' theorem.

https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem

We find the the area of the shape we are revolving ($A$), its centroid, the distance from the center of revolution to the centriod ($R+d$). And $V = 2\pi (R+d)A$

As for the centroid times the area of the anulus.

In Cartesian coordinates:

$2\int_0^{r_2} x\sqrt {r_2^2 -x^2}\ dx - 2\int_{0}^{r_1} x\sqrt {r_1^2 -x^2} \ dx$

In polar coordinates

$\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\int_{r_1}^{r_2} r^2\cos\theta \ dr\ d\theta$

Either way we get $\frac 23 (r_2^3 - r_1^3)$

$2\pi (R(\frac {\pi}{2}(r_2^2-r_1^2) + \frac 23(r_2^3 - r_1^3))$

$\endgroup$
0
$\begingroup$

I think your question is misphrased, because it wants us to work out the radius of revolution for the shape for generating the toroid, but the point upto which you measure the radius is subjective. In the case of the circle and rectangle, the radius extends till their geometric centres. Why not take it till the point where the circle and rectangle touch the cylinder $(R=r_{cylinder})$? A better question would be what do we conventionally take the radii of these tori as, or what is the average radii of the tori of different cross-sections.

Consider the case of a hollow ring torus. You take the radius till the centre of the cross-section (annulus of differential thickness), which is not on the cross section itself. So it is not necessary for the centre to always lie on the cross-section.

First note that the centre of mass of an object of constant mass-density coincides with its centre of volume. This is because the mass of the volume element is linked to its volume by a constant density:$$dM=\rho\ dV$$

The position vector of the centre of mass, given by $\displaystyle\bar r=\frac{\int\vec r\ dM}{\int dM}=\frac{\int\vec r\ dM}{\int dM}\cdot\frac{1/\rho}{1/\rho}=\frac{\int\vec r\ dV}{\int dV}$, which is the centre of volume.

I'm sure you realize that the answer in the post you linked is actually finding the centre of area (if that's a thing) of the cross-section. Calling it the centre of mass/volume means you are imparting an inconsequential differential thickness to the cross-section (in order for it to have mass/volume) only to ignore it later on by assuming that the mass/volume is proportional to the surface area, or that the mass density of the laminar cross-section is a constant.

In the case of your tori, the centre of mass/volume of elemental cross-sectional laminae is almost coincident with the centre of area of the cross-section. Almost, because the thickness of the element grows as we go radially outward. So the surface-mass/volume distribution is negligibly higher away from the centre of the torus, offsetting the centre of mass/volume of the element negligibly away from the centre of the torus. In most practical applications, this difference is too small to matter, and we can define the average radius of the torus either as the distance of the centre of mass/volume or the centre of area from the centre of the torus.

Torus

Do you notice how the elemental "slices" of the torus get thicker away from its centre?

$\endgroup$
1
  • $\begingroup$ Thank you for the detailed answer, and you're right, I need to make a few edits to my original question. This leaves the heart of my question unanswered, though it gets very close. In the case where I want to revolve the annular sector around the cylinder, with the constraint that the sector is touching the cylinder through its revolution, is my radius of revolution the radius which extends to the geometric center of the annular sector, or the radius which extends to half the height of a rectangle which fully compasses the annular sector, as Andrei suggested above? Thank you. $\endgroup$
    – kreeser1
    Dec 11 '18 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.