**3. Non-dimensional ventilatory index**

For disease detection, it is more convenient to formulate and employ a non-dimensional number to serve as a ventilatory performance index *LVI* (to characterize ventilatory function), as:

$$LVI\_1 = \left[ \left( R\_a \mathbb{C}\_a \right) \left( \text{Ventilation rate in s}^{-1} \right) 60 \right]^2 = \pi\_a^2 \left( BR \right)^2 60^2 \tag{10}$$

where *BR* is the breathing rate.

836 Biomedical Science, Engineering and Technology

Lun

Lun

g Volume (L)

Point of inflection where *V* = 0

*V* = 0.29 L

*ti* = 1.18 s

g Pressure

*PN*(*t*) (cmH

O2 ) *Pm*

*P*0 *Pk*

(a)

*tv* = 2.02 s

Fig. 3. (a)The pressure curve represented by Equation (3) matched against the pressure data

(b)

*Ca* = 0.2132 L(cmH2O)-1 and *Ra* = 2.275 (cmH2O)sL-1, matched against the volume data represented by dots. In Figure 3(a), the terms *P0*, *Pm* and *Pk* refer to Equation (11). At *t* = *tv*, *V*

(represented by dots). (b) The volume curve represented by Equation (6), for

*ti* = 1.18 s

is maximum and *V* is zero. This figure is adopted from our work in Ref [1].

Now, let us obtain its order-of-magnitude by adopting representative values of *Ra* and *Ca* in normal and disease states. Let us take the above computed values of *Ra* = 2.275 (cmH2O)sL-1 and *Ca* =0.2132 L(cmH2O)-1 and *BR* = 12 m-1 or 0.2 s-1, computed by matching Equation (6) to the data of Figure 3.

Then, in a supposed normal situation, the value of *LVI*1 is of the order of 33.88. In the case of obstructive lung disease (with increased *Ra*), let us take *Ra* = 5 (cmH2O)sL-1*, Ca* = 0.12 L(cmH2O)-1 and *BR* = 0.3 s-1; then we get *LVI*<sup>1</sup> = 118.6*.* For the case of emphysema (with enhanced *Ca*), let us take *Ra* = 2.0 (cmH2O)sL-1*, Ca* = 0.5 L(cmH2O)-1 and *BR* = 0.2 s-1**;**  then we obtain *LVI*<sup>1</sup> = 144. In the case of lung fibrosis (with decreased *Ca*), we take *Ra* = 2.0 (cmH2O)sL-1*, Ca* = 0.08 L(cmH2O)-1 and *BR* = 0.2 s-1; then we obtain *LVI*<sup>1</sup> = 3.7.

We can, hence summarize that *LVI*1 would be in the range of **2-5** in the case of fibrotic lung disease, **5-50** in normal persons, **50-150** in the case of obstructive lung disease and **150-200** for the case of emphysema,. This would of course be needed to be verified by analyzing a big patient population.

Now, all of this analysis requires pleural pressure data, for which the patient has to be intubated. If now we evaluate the patient in an outpatient clinic, in which we can only monitor lung volume and not the pleural pressure, then let us develop a non-invasive method for determining lung compliance (*C*), resistance-to-airflow (*R*) and ventilatory index.

## **4. For non-invasive assessment of lung status and determination of lung compliance and resistance-to-airflow**

Our primary need is to be able to determine lung pressure *PN*(*t*) non-invasively. If we observe the *PN*(*t*) curve in Figure 3(a), we can note that, during the period of time from *t* = *ti* = 1.18 s to *t* = *tv* = 2.02 s, we can represent it as:

$$P = P\_N = P\_k \sin \alpha \rho\_p (t - t\_i) + P\_0 \tag{11.-a}$$

$$\mathbf{r} \equiv P\_k \sin \alpha\_p (t - 1.18) + 2.5 \tag{11-b}$$

where (i) *Pk* =*Pm* – *P*0, and *Pk* = 0.5 cmH2O in Figure 3(a), (ii) *t* = *ti*, the inflection point on the lung volume curve (= 1.18 s in Figure 3(b)), and (iii) *P*0 = 2.5 cmH2O.

We can determine the value of ω*<sup>p</sup>*, by invoking the condition that the pressure *P* becomes maximum (= *Pm*) at *t* = *tm* = (*tv* + *ti*)/2. Hence, at *t* = *tm*,

$$P\_m = P\_k \sin \alpha \rho\_p (t\_m - t\_i) + P\_0$$

But since *Pk* = *Pm* – *P*0, we get

$$P\_m - P\_0 = (P\_m - P\_0) \sin \alpha\_p (t\_m - t\_i)$$

or

$$\sin a\_p(t\_m - t\_i) = 1$$

$$\text{whereein} \qquad \qquad \mathbf{t}\_m = (\mathbf{t}\_v + \mathbf{t}\_i) / \,\,\mathbf{2} = (2.02 + 1.18) / \,\,\mathbf{2} = 1.6\,\,\mathbf{2}$$

Therefore, we have

$$
\alpha\_p(t\_m - t\_i) = \alpha\_p(1.6 - 1.18) = \pi \,/\, 2 = 1.57\tag{12}
$$

In Equation (12), *tm* = (*tv* + *ti*)/2 and both *tv* (= 2.02 s) and *ti*(= 1.18 s) can be known from the lung volume curve. Hence, ω*p* can be determined. For Figure 3 data, we get ω*<sup>p</sup>* = 3.73 rad/sec.

Hence, in our case of Figure 3 data, we can represent lung pressure *PN* (= *P*) between *ti* and *tv* as:

$$P = P\_k \sin \alpha \rho\_p (t - t\_i) + P\_0 \tag{13-a}$$

$$= 0.5\sin 3.7\Im(t - 1.18) + 2.5\tag{13-b}$$

and

$$\dot{P} = 1.87 \cos 3.73(t - 1.18) \tag{13-c}$$

So, in general, the parameters of the lung pressure curve are *Pk* and *P0*, since *ti* and ω*<sup>p</sup>* can be determined in terms of *tv* and *ti*, as per Equation (12) and Figure 3. Likewise, we can also represent lung volume between *ti* and *tv* as:

$$V = V\_T \sin \alpha\_v (t - t\_i) + V\_0 \tag{14-a}$$

where*VT* is the tidal volume and *V*0 is the lung volume at *t* = *ti*. Based on Figure 3(b) data, we can rewrite Equation (14-a) as follows:

$$V = 0.25 \sin a\_v (t - 1.18) + 0.31$$

At *t* = *tv* = 2.02 s, *V* = *VT* = 0.55 L. Hence, (2.02 1.18) 1.57 ω*<sup>v</sup>* − = (or π/2), so that 1.87 ω*<sup>v</sup>* = rad. So then Equation (14-a) can be written as:

$$V = 0.25 \sin 1.87(t - 1.18) + 0.3\tag{14.b}$$

$$
\dot{V} = 0.47 \cos 1.87 (t - 1.18) \tag{14-c}
$$

$$
\ddot{V} = -0.88 \sin 1.87(t - 1.18) \tag{14-d}
$$

Now based on Equation (13), we can represent the governing lung ventilation model Equation (1-b) as

$$R\dot{V} + \frac{V}{C} = P\_K \sin 3.7\% (t - 1.18) + P\_0 \tag{15.-a}$$

or as,

838 Biomedical Science, Engineering and Technology

Now, all of this analysis requires pleural pressure data, for which the patient has to be intubated. If now we evaluate the patient in an outpatient clinic, in which we can only monitor lung volume and not the pleural pressure, then let us develop a non-invasive method for

determining lung compliance (*C*), resistance-to-airflow (*R*) and ventilatory index.

**compliance and resistance-to-airflow** 

We can determine the value of

But since *Pk* = *Pm* – *P*0, we get

Therefore, we have

rad/sec.

and

*ti* and *tv* as:

lung volume curve. Hence,

or

*t* = *ti* = 1.18 s to *t* = *tv* = 2.02 s, we can represent it as:

maximum (= *Pm*) at *t* = *tm* = (*tv* + *ti*)/2. Hence, at *t* = *tm*,

lung volume curve (= 1.18 s in Figure 3(b)), and (iii) *P*0 = 2.5 cmH2O.

ω

wherein *t tt m vi* =+ = + = ( ) / 2 (2.02 1.18) / 2 1.6

ωω

ω

**4. For non-invasive assessment of lung status and determination of lung** 

Our primary need is to be able to determine lung pressure *PN*(*t*) non-invasively. If we observe the *PN*(*t*) curve in Figure 3(a), we can note that, during the period of time from

> <sup>0</sup> *PP P tt P* = = −+ *Nk p i* sin ( ) ω

sin ( 1.18) 2.5 ≅ −+ *P t k p* ω

where (i) *Pk* =*Pm* – *P*0, and *Pk* = 0.5 cmH2O in Figure 3(a), (ii) *t* = *ti*, the inflection point on the

<sup>0</sup> *PP ttP m k pm i* = −+ sin ( ) ω

0 0 *m m* ( )sin ( ) *PP PP tt* −= − −

sin ( ) 1 *pm i* ω

( ) (1.6 1.18) / 2 1.57 *pm i p*

In Equation (12), *tm* = (*tv* + *ti*)/2 and both *tv* (= 2.02 s) and *ti*(= 1.18 s) can be known from the

Hence, in our case of Figure 3 data, we can represent lung pressure *PN* (= *P*) between

<sup>0</sup> sin ( ) *PP tt P* = −+ *k pi* ω

*t t* − =

ω*pm i*

> π

*p* can be determined. For Figure 3 data, we get

*t t* −= − = = (12)

= −+ 0.5sin 3.73( 1.18) 2.5 *t* (13-b)

(11-a)

ω*<sup>p</sup>* = 3.73

(13-a)

(11-b)

*<sup>p</sup>*, by invoking the condition that the pressure *P* becomes

$$R\,\dot{V} + \frac{V}{C} = 0.5\sin 3.7\%(t - 1.18) + 2.5\tag{15-b}$$

Let us employ this equation to determine the values of *C* and *R* at some specific points in the ventilation cycle. At *t* = *ti* (the inflection point) = 1.18 s, *V* = 0 Ls-2, *V* = 0.48 Ls-1 and *V* = 0.3 L. Now, we can differentiate Equation (15) as:

$$
\vec{R'} + \frac{\dot{V}}{C} = (P\_K \alpha\_p) \cos 3.73(t - 1.18) \tag{16-a}
$$

or as,

$$
\dot{R}\ddot{V} + \frac{\dot{V}}{C} = (0.5)(3.73)\cos 3.73(t - 1.18)
$$

$$
= 1.86\cos 3.73(t - 1.18) \tag{16-b}
$$

Hence, from this Equation (16), we get:

$$\frac{\dot{V} \text{(= 0.48)}}{\text{C}} = 1.86 \text{ , or } \text{C} = 0.258 \text{ L (cmH} \text{\_2O)} \cdot 1.8$$

Then, upon substituting this value of *C* into Equation (15), we get:

$$R(0.48) + \frac{0.3}{0.258} = 2.5 \text{, or } R = 2.79 \text{ (cmH}\_2\text{O)} \text{sL}^{-1}.$$

Hence at *t* = *ti* =1.18 s, *C* = 0.258 L(cmH2O)-1 and *R* = 2.79 (cmH2O)sL-1. (17) Let us now evaluate *C* and *R* at *t* = *tk* =1.6 s, the time associated with the peak lung pressure. From Equations (15) and (16), we can put down:

$$R\dot{V} + \frac{V}{C} = 0.5\sin 3.73(1.6 - 1.18) + 2.5 = 3\tag{18-a}$$

$$R\,\dot{V} + \frac{\dot{V}}{C} = 1.86\cos 3.73(1.6 - 1.18) = 0\tag{18-b}$$

At *t* = *tk* =1.6 s, we get from Equations (13) and (14), *V* = 0.48 L,*V* = 0.33 Ls-1, *V* = −0.622 Ls-2, *P* = 3 cmH2O and *P* = 0 (cmH2O)s-1.

Substituting these values into Equations (18-a) and (18-b), we get:

$$0.33R + \frac{0.48}{C} = 3\tag{19-a}$$

$$-0.622R + \frac{0.33}{C} = 0\tag{19-b}$$

from which we obtain for *t* = *tk* =1.6 s,

$$\text{C} = 0.22 \,\text{(cmH}\_2\text{O)} \,\text{s} \,\text{L}^{-1}, \text{R} = 2.51 \,\text{(cmH}\_2\text{O)} \,\text{s} \,\text{L}^{-1} \tag{20}$$

Finally, let us evaluate *C* and *R* at *t* = *tv* =2.02 s. From Equation (15), we get:

$$R\dot{V} + \frac{V}{C} = 0.5\sin 3.73(2.02 - 1.18) + 2.5\tag{21}$$

Now at *t* = *tv* =2.02 s, *V* = 0.55 L, *V* = 0 Ls-1. So then, from Equation (21), we obtain:

$$\frac{0.55}{\text{C}} = 2.5 \text{, or } \text{C} = 0.22 \text{ (cmH}\_2\text{O)} \text{s} \text{L}^{-1} \tag{22}$$

It can be noted that the values of *C* and *R* given by Equations (17), (20) and (22) are similar to their average values *Ca* = 0.218 L(cmH2O)-1 and *Ra* = 2.275 (cmH2O)sL-1. This lends a measure of confidence to our Equation (15), for which *V* and *V* are given by Equations (14-b) and (14-c).

Let us now proceed to how we can determine the values of lung pressure function parameters *Pk* and *P*0 along with *R* and *C* from the monitored values of lung volume. At *t* = *ti* =1.18 s, we have from Equation (15-a)

$$
\Delta \dot{V} + \frac{V}{C} = P\_\text{K} \sin 3.73(t - 1.18) + P\_0 \tag{23}
$$

so that by substituting *V* = 0.3 L, *V* = 0.48 Ls-1, we get:

840 Biomedical Science, Engineering and Technology

<sup>=</sup> <sup>=</sup> , or *C* = 0.258 L(cmH2O)-1.

*R* + = , or *R* = 2.79 (cmH2O)sL-1.

Let us now evaluate *C* and *R* at *t* = *tk* =1.6 s, the time associated with the peak lung pressure.

0.5sin 3.73(1.6 1.18) 2.5 3 *<sup>V</sup> RV*

1.86 cos3.73(1.6 1.18) 0 *<sup>V</sup> RV*

At *t* = *tk* =1.6 s, we get from Equations (13) and (14), *V* = 0.48 L,*V* = 0.33 Ls-1, *V* = −0.622 Ls-2,

0.48 0.33 3 *<sup>R</sup> C*

0.33 0.622 0 *<sup>R</sup> C*

0.5sin 3.73(2.02 1.18) 2.5 *<sup>V</sup> RV*

It can be noted that the values of *C* and *R* given by Equations (17), (20) and (22) are similar to their average values *Ca* = 0.218 L(cmH2O)-1 and *Ra* = 2.275 (cmH2O)sL-1. This lends a measure of confidence to our Equation (15), for which *V* and *V* are given by Equations

Let us now proceed to how we can determine the values of lung pressure function

parameters *Pk* and *P*0 along with *R* and *C* from the monitored values of lung volume.

Now at *t* = *tv* =2.02 s, *V* = 0.55 L, *V* = 0 Ls-1. So then, from Equation (21), we obtain:

() () 1 1

+ = − += (18-a)

+ = − = (18-b)

+ = (19-a)

− += (19-b)

2 2 *C R* 0.22 cmH O sL , 2.51 cmH O sL − − = = (20)

+ = − + (21)

*<sup>C</sup>* <sup>=</sup> , or *C* = 0.22 (cmH2O)sL-1 (22)

( 0.48) 1.86 *<sup>V</sup> C*

Then, upon substituting this value of *C* into Equation (15), we get:

*C*

*C*

Finally, let us evaluate *C* and *R* at *t* = *tv* =2.02 s. From Equation (15), we get:

*C*

0.55 2.5

Substituting these values into Equations (18-a) and (18-b), we get:

From Equations (15) and (16), we can put down:

*P* = 3 cmH2O and *P* = 0 (cmH2O)s-1.

from which we obtain for *t* = *tk* =1.6 s,

At *t* = *ti* =1.18 s, we have from Equation (15-a)

(14-b) and (14-c).

0.3 (0.48) 2.5 0.258

Hence at *t* = *ti* =1.18 s, *C* = 0.258 L(cmH2O)-1 and *R* = 2.79 (cmH2O)sL-1. (17)

$$R(0.48) + \frac{0.3}{C} = P\_0 \tag{24}$$

Also, from Equation (16), we get by substituting the values of *V* = 0.48 Ls-1 and *V* = 0 Ls-2

$$\frac{0.48}{C} = 3.73P\_k \tag{25}$$

At *t* = *tm* =1.6 s, we have from Equations (15) and (16) as well as by substituting the values of *V* = 0.48 L,*V* = 0.33 Ls-1 and *V* = −0.622 Ls-2, we get:

$$0.33R + \frac{0.48}{C} = P\_k + P\_0 = P\_m \tag{26}$$

$$-0.622R + \frac{0.33}{C} = 0 \text{ , or } R = 0.53/C \tag{27}$$

Then at *t* = *tv* =2.02 s, we get from Equation (15), along with *V* = 0.55 L, *V* = 0 Ls-1,

$$\frac{0.55}{C} = P\_0 \tag{28}$$

We hence have Equations (24), (25), (26), (27) and (28) to solve and determine the best values of the four unknowns *R*, *C*, *Pk* and *P*0. For this purpose, we define the ranges of these four terms, as:

*R*: 2.1, 2.2, 2.3 (cmH2O)sL-1; *C*: 0.20, 0.21, 0.22 L(cmH2O)-1; *Pk* = 0.4, 0.5, 0.6 (cmH2O); *P*0 = 2.4, 2.5, 2.6 (cmH2O)

Then, in order to satisfy these equations, we obtain for the best values of *R*, *C*, *Pk* and *P*0, based on the solution in Appendix, as

$$R = 2.29 \text{(cmH}\_2\text{O)} \text{sL}^{-1}, \text{ C} = 0.22 \text{L} \text{(cmH}\_2\text{O)}^{-1}, \text{ P}\_k = 0.58 \text{(cmH}\_2\text{O)}, \text{ P}\_0 = 2.47 \text{(cmH}\_2\text{O)} \text{ (29)}$$

As can be noted, these values of *C* and *R* correspond to the average values of *C* and *R* given by Equation (8). This then lends credibility to our procedure for non-invasive determination of *C* and *R*, for lung disease detection. This procedure enables us to in fact determine lung pressure toward evaluation of *C* and *R*.

Now since this procedure enables us to determine maximum lung driving pressure *Pm* = *Pk* + *P*0, we can also formulate the non-dimensional lung ventilatory index as:

$$LVI\_2 = \frac{R}{C} \frac{\left(TV\right)^2}{\left(P\_m\right)^2} (BR) (60)^2 \tag{30}$$

wherein *BR* is in s-1. For our case, *R* = 2.275 (cmH2O)sL-1, *C* = 0.2132 L(cmH2O)-1, *Pm* = 3 cmH2O, and *TV* = 0.55 L. This gives *LVI*2 = 25.8. By using this *LVI*2 index, we can expect its value to be of the order of 30 for normal subjects, 300 for COPD patients, 5 for emphysema patients, and 100 in the case of lung fibrosis.

Here again, we need to determine *LVI* for normal lung states as well as for different lung disease states. We can then compare which of the formulas (10) or (30) enable better separation of lung disease states from the normal state.

*Comments related to values of the ranges of the parameters:* Before proceeding to the next section, let us address the basis of providing the above indicated ranges of parameters. The lung ventilation volume and driving pressure curves in Figure 3 are for a normal case. By carrying out this procedure for other normal subjects, we can define and confirm the above mentioned normal ranges for these parameters, for obtaining their best values.

Now then how do we distinguish subjects with disorders, such as obstructive lung disease (with increased value of *Ra*), emphysema (with enhanced value of *Ca*), lung fibrosis (with decreased value of *Ca*)? This can be made out from the shape and values of the lung ventilation volume curve. So then by repeating this procedure for subjects with these disorders, we will be able to characterize the shapes of the lung ventilation curves for normal subjects and for prescribing appropriate ranges of the parameters, for obtaining the best values of these parameters.
