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Essentially, I've been wrapping the pictured object tightly with string to exert a confinement pressure on its exterior. It's been difficult however to make a good estimate of how much pressure is exerted on the object. Theoretical models I've found of string wrapped around a cylinder/disk seem to be too simple for making precise measurements. Right now I'm using a pressure sensor to take measurements from the center of the object where the two interior pillars meet (the sensor doesn't work well with large surface areas). I can make precise measurements of the pressure within the pillars using the sensor, the pressure between the pillars clearly doesn't give the value of the pressure exerted on the exterior, as it doesn't match other setups I've tried or the models.

I've tried to work out how the pressure between the interior pillars relates to the confinement pressure. However, I don't have much confidence in my results.

Just in case this must be solved numerically, the dimensions are as follows:

Radius of the disk: 17.5 mm Radius of the hollow region: 11 mm Width/Height of disk: 14 mm Radius of central pillar contact surface: 4.5 mm Depth of lipped edge: 4.5 mm

Pictured on the left is the object of discussion from a front-face view. On the right is an image looking vertically upward at the object's cross section.

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The string will exert some pressure $P$ over the outer area of the cylinder $A$. If the height of the object/cylinder is $h$, and has outer radius $R$, then the area of the outer curved surface (against which the string is wound) of the top half-cylinder is $\mathbf{A}=\pi Rh\hat{\mathbf{r}}=A\hat{\mathbf{r}}$. By symmetry we can tell that the net force from the string on the top hemisphere must be directed downward. We can find that force by integrating the vertical component of the directed area vector multiplied by the pressure from the string:

$$\begin{align*}\mathbf{F}&=-RhP\hat{\mathbf{y}}\int_0^\pi\mathrm{d}\theta\,\sin\theta\\&=-2RhP\hat{\mathbf{y}}\end{align*}$$

You can also see this by just noticing that the "projected area" (the cross-section you see by looking from above) is just a rectangle of size $2R\times h$, and thus the total force must be $F=2RhP$.

This force is then transmitted across a region of area $a$, which corresponds to a measured pressure $p$ at the center of the device of $p = F/a = 2RhP/a$. Therefore, we can write the "confinement pressure" $P$ in terms of the measured central pressure $p$, the height of the cylinder $h$, the area of the central spot $a$, and the outer radius of the cylinder $R$ as:

$$P=\frac{pa}{2Rh}$$

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  • $\begingroup$ Thank you! It's actually not too dissimilar from what I derived, excepting that I included a factor of pi in the expression for F. I do have one question though. The two hemispheres are not a single object, and are only bound together by the confinement pressure. Would that not mean that the force is distributed across a region that doesn't just include the central pillar where the sensor is located, but also the two cross-sections of the outer rim? In other words, call the area of the central spot 'a' and the area of the two cross-sections 'b'. Then, p=F/(a+b), no? $\endgroup$
    – Modestas Botha
    Commented Jun 13 at 16:19
  • $\begingroup$ @ModestasBotha Indeed that would change things, my analysis is assuming there is no contact through the outer rim. I agree with yours in the case of contact! $\endgroup$
    – Riley Scott Jacob
    Commented Jun 13 at 16:21
  • $\begingroup$ @ModestasBotha To be clear as well, it depends on the exact nature and geometry of the pressure sensor. As a real physical object, it will have some thickness no doubt, and probably not precisely match the area of the inner contact surface either (unless they were designed for this purpose). A better area to use in this case would be the cross sectional area of the sensor. Another assumption implicit in the above analysis is that the string-wrap covers the entire outer area of the cylinder $\endgroup$
    – Riley Scott Jacob
    Commented Jun 13 at 21:21
  • $\begingroup$ No worries about the pressure sensor. It's as thin as it can be (0.2 mm) and the area of the sensor is approximately equal to the area of the central contact surface. As for the wrapping, the string is wrapped uniformly along the outer rim inside the spool lips (the lipped edges are visible on the second figure, they stick out from the "rectangle"). I know that they have a moderate effect on the results, at the moment I'm just going with the assumption that I could use a proportionally smaller 'h' to make up for it. I plan to take a measurement soon without them to check it. $\endgroup$
    – Modestas Botha
    Commented Jun 14 at 0:00

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