## Thin-Walled Spherical Vessel

### Thin-Walled Sphere / The Law of Laplace

We'll start with an example familiar to all and derive the famous "Law of Laplace" for a thin-walled sphere with inner radius $$r_0$$ and wall thickness $$h$$; "thin walled" may be subject to interpretation and circumstances, but can also be defined mathematically in terms that will become apparent.  For the moment we will assume that $$r_0$$ is at least 10 times as large as $$h$$, a circumstance that is virtually never present in the heart. The "law" aspect of Laplace comes from the fact that we use Newton's laws and a free body diagram to determine a force equilibrium.  We can slice the sphere in half any way we choose such as that shown.

We'll assume that the internal pressure is $$p$$ and the external pressure is $$0$$.  The force due to pressure on the inside of the sphere turns out to be equal to the transmural pressure ($$\Delta p$$) multiplied by the cross-sectional area of the sphere (a force acting to push the sphere to the left).  This (and every) half of the sphere is stationary ( not accelerating ) so the pressure force must be balanced by the tension ($$T$$) in the wall suggested by the arrows pointing to the right.

$$\Large T = \Delta p \pi r_0^2$$

$$T$$ is in this equation is quite literally the force acting on the wall, as suggested by the arrows, to keep the sphere from distending/exploding.  This is more appropriately expressed as a ( mechanical ) stress which is the force divided by the surface area over which it acts.  For the thin-walled sphere, think of the wall area as a long, skinny rectangle of length $$2\pi r_0$$ and height $$h$$ ( the wall thickness ). The wall tension is equal to the wall stress multiplied by the area over which it acts.

$$\Large \sigma_{\theta} (2\pi r_0 h) = \Delta p \pi r_0^2$$

$$\Large \sigma_{\theta} =\Delta p \frac{r_0}{2\, h}$$

This is the so-called law of Laplace for a thin-walled sphere. I've seen the equation with the 2 missing in the denominator and that would be incorrect.  We'll see shortly that this part of the shape factor is related to the geometry and it's 2 for a sphere.  Invariably people invoke Laplace's law to support the claim that both wall tension and wall stress are increased for a larger heart, i.e. greater radius $$r_0$$.  However this notion skirts the important aspects of the relationship.  First, the wall tension per se is of no interest - certainly not to the heart.  Yes a horse heart has more tension in it than a mouse heart as does a dilated heart.  However there's no indication that tension per se can be perceived in the heart or in any load bearing material.  The mechanical behavior of materials is determined by mechanical stress – force per unit area ( end of story ).  The equation above shows that wall stress increases with pressure and also with the ratio ( shape factor ) $$r_0/(2h)$$.  The wall stress doesn't depend on radius alone!  It depends on this ratio so the thickness of the wall always comes into play  regardless of how thin the wall may be.  In clinical terminology it's not the size of the ventricle that determines wall stress, it's the relative wall thickness (and the transmural pressure).  While relative wall thickness (e.g. $$RWT_d = (IVS_d+LVW_d)/LVID_d$$) is one of the clinical parameters used to quantify the relationship between ventricular diameter and wall thickness, several or many are possible.

Before proceeding, let's agree that we are on shaky ground in the first place with this thin-walled sphere approximation.  Nevertheless this is a good place to start before working towards more complex models of how vessels behave.  For our next trick, let's try to determine the compliance of the thin-walled sphere.  To accomplish this, we actually need some information about how the material behaves in response to stress.  So for the next shaky assumption in this process, we'll assume that the sphere is made of a linear elastic material that behaves according to Hooke's law:

$$\Large \sigma = E \epsilon$$

$$E$$ in this case is a property of the material itself called Young's modulus of elasticity. The equation describes a linear relation between mechanical  stress ($$\sigma$$) and strain ($$\epsilon$$).  Biological materials do not behave according to this relationship although they may appear to do so over a small range of strain. There are also different ways to define strain, but we'll go with a small strain definition (also not necessarily appropriate).

$$\Large \epsilon = \frac{l-l_0}{l_0} = \frac{\Delta l}{l_0}$$

$$l_0$$ is the initial length.  For a sphere, we have the change in circumference divided by the original circumference as the tangential strain $$\epsilon_{\theta}$$:

$$\Large \epsilon_{\theta} = \frac{2\pi \Delta r}{2\pi r_0} = \frac{\Delta r}{r_0}$$

(The latter should remind you of the fractional shortening.)

$$\Large \sigma_{\theta} = E \epsilon_{\theta} =\Delta p \frac{r_0}{2\, h}$$

$$\Large \sigma_{\theta} = E \frac{\Delta r}{r_0} =\Delta p \frac{r_0}{2\, h}$$

Rearranging

$$\Large \Delta r =\Delta p \frac{r_0^2}{2\,E h}$$

$$\Large \frac{\Delta r}{\Delta p} = \frac{r_0^2}{2\,E h}$$

The volume of a sphere:

$$\Large V_s = \frac{4}{3}\pi r^3$$

$$\Large \frac{dV_s}{dr} = 4\pi r^2$$

Consequently:

$$\Large \frac{dV}{dp} = \frac{dV_s}{dr} \frac{\Delta r}{\Delta p} = (4\pi r^2) \frac{r_0^2}{2\,E h}$$

$$\Large \frac{dV}{dp} = 2 \pi \frac{r_0^4}{E h}$$

That would be the compliance of a thin-walled sphere if I've worked it out correctly.  $$r_0^4/h$$ has units of volume and $$E$$ has units of stress (pressure), so at least we've got the physical units straight.  Now it appears that the magnitude of the compliance depends rather dramatically on the size of the vessel.

$$\Large \frac{dV}{dp} = 2 \pi r_0^3 \frac{r_0}{E h}$$

Here the compliance has been rewritten in such a way as to separate the main geometric factor, $$r_0/h$$, From the rest of the formula.  Now we're going to divide the whole formula by the volume of the sphere:

$$\Large \frac{dV}{dp}/V = \frac{3}{2} \frac{r_0}{E h}$$

This quantity, volume distensibility, is not dependent on the size of the vessel.  For the specific geometry (sphere), it depends only on Young's modulus and a ratio, $$3r_0/ (2h)$$, having to do with the geometry. It does not depend on the size of the sphere; it depends on a ratio.  This would be a good thing to have on hand in case we run across any other thin-walled spheres.