Time Constants: much more than you wanted to know

vivianimbriotis | Aug. 12, 2025, 6:07 p.m.

Many authoritative physiology texts discuss the fabled respiratory time constant. This is the "amount of time it takes for an alveolus to inflate to 63% of its final volume". Without sufficient embarrassment, these books will then go on to tell us that the time constant is the product of resistance and compliance.

$$\tau = R \cdot C$$

What? Why?

Background: Compliance

Let's understand the products first. Compliance is the change in volume induced by a unit change in pressure:

$$C = \frac{\Delta V}{\Delta P}$$

Or, for the Leibniz enjoyers in the audience,

$$C = \frac{dV}{dP}$$

This is a horrible measurement, in that it is usually more mathematically elegant to refer to the elastance, which is the reciprocal:

$$Elastance = \frac{dP}{dV}$$

Compliance, for either the whole respiratory system or for a given alveolus, is not constant, and is also path dependent, i.e. it depends on whether inspiration or expiration is happening. This path dependence is called hysteresis.

At high volumes, compliance falls off. It also falls off at very low volumes, as alveoli collapse and require additional pressure to overcome the surface tension force which favors their collapse.

For our thought experiments today, we will assume that alveoli have constant compliance - that is, the compliance is the same for any volume.

Summary: Compliance is the change in volume that results from a unit change in pressure. A highly compliant alveolus balloons out with a small driving pressure, and recoils weakly. A poorly compliant alveolus expands a but a little with the same driving pressure and recoils strongly.

Background: Resistance

Resistance describes the amount of pressure needed to force a unit volume through a pipe in a unit time.

$$R = \frac{dP}{dQ}$$

Flow through a pipe can either be stokes flow - irrelevant to the human body - laminar flow (where the velocity of any given voxel is parallel to the direction of overall flow) or turbulent flow (where voxel velocity is going every which way). Which form of flow predominates is given by the Reynold's number:

$$Re = \frac{2 r \rho V}{\mu}$$

Where \(\rho\) is the fluid density and \(\mu\) is the fluid viscosity.

When Re < 1, stokes conditions predominate.

When Re < 2000, laminar conditions predominate.

When Re > 4000, turbulent conditions predominate.

Wait, why is the radius term in the numerator? Normally we associate turbulent flow and high resistance with the asthmatic or bronchitic patient, with threadlike bronchioles. But apparently they will have beautiful laminar flow?

The problem is the pesky velocity term in the numerator. Considering fluid in a pipe, we can wrangle this into a flow term because

$$Q = \pi r^2 * V$$

so

$$V = \frac{Q}{\pi r^2}$$

and therefore

$$Re = \frac{2 Q \rho V}{r \mu}$$

Now the radius is in the denominator.

This tells us that, even though, ceteris paribus, a smaller radius pipe reduces turbulent flow, for constant flow rate, reducing the radius of a pipe will increase the velocity of the gas, and the overall effect is to promote turbulent flow.

In laminar conditions, resistance is constant for any given pressure gradient, and given by the Hagen-Poiseuille equation:

$$R = \frac{8 l \mu}{\pi r^4}$$

$$Q = \frac{\Delta P}{R}$$

where l is the pipe length and r the pipe radius.

In turbulent conditions, flow varies with the square root of the pressure gradient, and resistance is no longer constant:

$$\Delta P \propto Q^2$$

Again, it would often be mathematically nicer to refer to the reciprocal of resistance, the hydraulic conductance, but centuries of physiology by malnourished britains has left us here.

As another aside, this is all a horrible oversimplification; the Navier-Stokes equations which describe fluid flow are famously not (generally) known to be solvable, and this is one of the millennium problems in mathematics.

Summary: Resistance is \(\frac{Pressure}{Flow}\). The main determinant is airway radius. It is also affected by gas density and airway length. In the context of a liquid, fluid viscosity also matters (but this is negligible in a gas).

Background: the general time constant

Say you have a quantity \(y\), such that the rate of change of y depends on the value of y \(\dot{y} \propto y\). Let's call our constant of proportionality \(k\) the rate constant, and write:

$$\dot{y} = k y$$

Alternatively, we can rewrite the rate constant as a time constant \(\tau = \frac{1}{k}\) so

$$\dot{y} = \frac{y}{\tau}$$

(note: \(\tau\) is the ratio of the value to its own derivative, \(\tau = frac{y}{{\dot{y}}\))

This is the simplest ordinary differential equation.

$$\frac{dy}{dt} = \frac{y}{\tau}$$

$$\int \frac{1}{y} \frac{dy}{dt} dt = \int \frac{1}{\tau} dt$$

And by the substitution rule for integrals:

$$\int \frac{1}{y} dy = \int \frac{1}{\tau} dt$$

Which ends up giving us

$$y = y_0 e^{\frac{t}{\tau}}$$

where \(y_0\) is the initial value for y. This is the case for anything where the rate of change depends linearly on the value - for heat transfer, for one-compartment renal excretion of drugs, for current through an R-C circuit...

There are two cases. If \(\tau\)>0, then the function starts at \(y_0\) and blows up to infinity. If \(\tau\)<0, then the function starts at \(y_0\) and decays to zero. 

Let's consider a decaying function and 

$$y = y_0 e^{\frac{-t}{\tau}}$$

After N time-constants worth of time has passed, the remaining y, as a proportion of \(y_0\), is 

$$\frac{y}{y_0} = \frac{1}{e^N}$$

Summary: a quantity y that behaves like \(\dot{y} = \frac{-y}{\tau}\) will decay exponentially. After \(t=N\tau\), then \(\frac{y}{y_0} = \frac{1}{e^N}\)

Respiratory time constants: intuitively

So the claim here goes like this:

Respiratory units have roughly linear compliance. As they deflate (or inflate) with a constant airway pressure, their volume follows an exponentially decaying curve. The time constant of this curve is the product of their resistance and their compliance.

Let's try to get at this intuitively first. Longer time constants mean a longer time until inflation is complete.

If resistance is low, then for a given driving pressure, flow will be high. This means the rate of change in volume will be high. The alveolus will fill up quickly.

Units with a low compliance, on the other hand, simply do not need much flow to fill up, since their change in volume (for a given driving pressure) will be very small. They only need a tiny bit of gas, and they're finished.

So a low resistance OR a low compliance will result in rapid filling, and a short (small) time constant.

Time constants: like an adult

Definitions:

$$C = \frac{\Delta V}{\Delta P}$$

$$R = -\frac{\Delta P}{Q}$$

so the product

$$C \cdot R = - \frac{\Delta V}{Q}$$

But the flow Q is just the rate of change of the volume V, so immediately we can recognize this whole thing as \(\tau\)!

Not satisfied? Let's keep going:

$$C \cdot R = -\frac{\Delta V}{Q}$$

$$Q = -\frac{\Delta V}{C \cdot R}$$

$$\frac{dV}{dt} = -\frac{\Delta V}{C \cdot R}$$

$$V = V_0 e^{-\frac{t}{C \cdot R}}$$

Implications of different time constants in the lung

Imagine two alveoli, one with half the compliance of the other.

As the lung is filled during inspiration, the alveolus with the lower compliance will fill first; it has a shorter time constant. No worries though, eventually both will fill up, the higher compliance one to twice the volume of the low compliance one.

But if we exhale before the highly compliant alveolus can fill, then most of the air ends up in the poorly compliance alveolus. This makes the overall system appear less compliant!

If inspiration is halted, but the breath is held, then at end-inspiration, pressure will be higher in the poorly compliant alveolus (it will be closer to 100% filled, at which time it's internal pressure will be the same as the airway pressure). This will drive gas from the poorly compliant alveolus into the more compliant alveolus, which was mostly unfilled and has low internal pressure. This redistribution of gas results in improved compliance as the breath is held, and contributes to hysteresis.

This makes dynamic compliance (the apparently compliance measured during normal breathing) worse the higher the respiratory rate is, because short inspiratory times mean that the most non compliant alveoli will be filled. This frequency-dependance is best exemplified by ARDS.