Saturday, August 29, 2015

The Rate of Oxidation of Alcohol is a Constant

  In a 1936 paper by Neymark and Widmark¹ it was shown that the rate of oxidation of alcohol in the body is a constant. To avoid the delays encountered by diffusion, the alcohol was introduced by intravenous injection into the bodies of test animals and it was observed that the decrease in the blood followed a straight line descent. This is what one would expect if an enzyme was involved in the breakdown of the alcohol molecule. Since the alcohol has to combine with the catalyst, a limited amount of catalyst limits the rate of reaction. As the amount of alcohol in the blood increases above a critical limit the reaction rate tends to level off. This value is very small so the rate of decrease appears constant for typical concentrations of alcohol. It can be shown that the reaction rate is proportional to the product of the concentration of catalyst and the value of ρ in the formula below.

A group of enzymes, alcohol dehydrogenase, is required to metabolize alcohol.
¹ Neymark, M. and Widmark, E. M. P. (1936), Zur Frage der Kinetik des Aethylalkoholumsatzes im Organismus. Skandinavisches Archiv Für Physiologie, 73: 283–290. doi: 10.1111/j.1748-1716.1936.tb01471.x

Some Concerns About the Diffusion Model for the Widmark Fit

  I was concerned about the large deviation of the peak value of Widmark's data from the diffusion model curve for the best fit of the alcohol concentrations so I did a plot that included the three standard deviation error bounds the data and it just passes this statistical test.

The data used for the plot were the mean values shown in Table III of Widmark's 1914 paper. As is, one cannot reject the model based on just this experiment. One could try a more complicated diffusion model with a 5 compartment model including an eliminated compartment. The assumption is that the alcohol in the digestive tract flows into the blood before it flows into the remaining body tissues and the liver where it is eliminated.

The more complicated scheme may affect the peak value slightly since there will be a slightly greater concentration of alcohol in the blood than in the liver. Trying to find an equivalent circuit with fewer components will result in loss of information about the quantities of alcohol in each compartment and may confuse the situation somewhat.

To test more complicated models like the one above to see if they will product a slightly higher peak one could put together a number of electronic circuits and measure the voltage drop across capacitor CB. An alternative would be to simulate the response of a model using an electronic analog computer.

Friday, August 28, 2015

The Simple Model's Fit To Widmark's 1914 Alcohol Content Data And More

  In a 1914 paper by Widmark some more data is given for the concentration of alcohol in blood and urine versus time. The two are approximately the same as was shown in some earlier experiments. The data is from Table III. I tried fitting the simplified formula and wasn't impressed by the results.

So I tried a simple analogy that compares diffusion with electrical conduction.

For diffusion the concentration, c, of a substance acts like a potential which causes a quantity of mass m to move from one point to another. The rate of this flow is analogous to an electrical current. And there is an equivalent of Ohm's law I = ΔV/R which is dm/dt = Γ Δc. So we can draw an electrical circuit which will mimic the behavior of the diffusion.

The capacitors act like the compartments of the 3-compartment model representing the digestive tract, D, and the blood, B. One could add a extremely large capacitor to receive the eliminated alcohol but it would act like a short circuit since it would have a minute change in voltage with the addition of a small amount of charge q. The solution for the voltage across CB is the sum of two exponentials.

The new model gave a better fit. The way the circuit works is the charges on the capacitors first tend to equalize with the voltage across CD remaining slightly higher than CB and then a small current from each drains through RB. We would expect similar changes as the alcohol diffuses throughout the body.

Supplemental (Aug 29): The citation given for the article is,

Widmark, E. M. P. (1916), Über die Konzentration des genossenen Alkohols in Blut und Harn unter verschiedenen Umständen. Skandinavisches Archiv Für Physiologie, 33: 85–96. doi: 10.1111/j.1748-1716.1916.tb00107.x

Monday, August 17, 2015

Determining Recovery Times After Drinking Alcohol

  One can use the formula for BAC as a function of time to estimate the recovery time after drinking a quantity of alcohol. As an example suppose an 86 kg man drinks 2 liters of 5.5% beer. Using the data below...

...we can compute a BAC curve as a function of a parameter x = λt for a more general result and compare the results with the limiting BAC value of 0.08 gm/100ml. The equation that we need to solve is derived as follows. γ is the BAC converted to mass using the individual's effective fluid volume, Vfl, divided by the mass of alcohol consumed, Am.

Mathcad can easily solve the transendental equation for x given γ as follows. The advantage of using x is the curve is the same no matter what the individual's λ value is.

We can use the same method to compute a table to for the peak BAC value for given body mass and quantity of drink. For the 5.5% beer the peak BAC table using pounds and fluid ounces is,

Solving the corresponding BAC curves for right intersection point gives the recovery parameter x values for the BACs above the BAC limit. One may have to use Newton's method for approximating the zeros of a set of equations for each pair of mass and drink quantity.

If λ = 0.33/hr, the value for nondrinkers, the corresponding number of hours that it will take for an individual's BAC to drop below BAC limit are as follows. The zeros in the tables are for BACs below the BAC limit so there are no "recovery times."

For comparison note that three 12 fl.oz. bottles of beer is 36 fl.oz. and a gallon is 128 fluid ounces.

Supplemental (Aug 18): The tables above were computed for men with a Widmark ρW of 0.7 liter/kg. Women can use the same tables if they use a body mass reduced by 6/7 = 0.86. A gender neutral table would replace the body mass with the effective fluid volume Vfl.

Supplemental (Aug 20): The fit to Schweisheimer's data was fairly good considering the simple model assumed but all the subjects consumed the same amount of alcohol and therefore we can't really say that Widmark's formula for the peak BAC was validated. The amount of data was also limited and as a result we didn't have accurate knowledge of true shape of the curve to be fit. Consequently, we shouldn't be surprised if our estimates of the recovery times are off somewhat.

Saturday, August 15, 2015

Drink Tables

  An easy way to condense the information about how much one can safely drink and stay under a given limit is to compute a table which I have done for a peak alcohol content of 0.10 gm/100ml using Widmark's rho factors of 0.7 liter/kg for men and 0.6 liter/kg for women. The columns are computed for a number of alcohol percentages of the drinks. The rows are for the given body masses in kilograms.

For a lower peak BAC value like 0.08 gm/100ml one can multiply the volumes given by 0.08/0.10 = 0.8 to get the reduced volume that one can safely drink.

Friday, August 14, 2015

Estimating the Drink Limit for a Chosen Peak BAC

  If one wants to limit one's peak blood alcohol content one can use Widmark's formula to estimate the volume of a drink of a certain percentage that will just reach this limit. The two examples below show sample calculations for a 80 kg man and a 50 kg woman and a BAC limit of 0.10 gm/100ml.

Wednesday, August 12, 2015

Expected Correlation Between Effective Fluid Volume and Body Mass

  We can use the Widmark factor to estimate the expected correlation of the effective fluid volume with the body mass.* This is probably best done with actual data since the peak blood alcohol content and consequently Widmark's factor may depend on the t = 0 values of D and B. A calculation for the abstainers will illustrate this.

Using Widmark's formula we find that ρWMb = 57.05 liters. Vfl = 1/α so using the previous result for α we can compute a Vfl and get the result found in the last blog. We can combine the ρW/e into a single constant β = 0.258 liter/kg for men. The result for women will be β = 0.221 liter/kg.

*edit (Aug 12): meant the proportionality factor instead of the correlation coefficient.

edit (Aug 13): Made some minor changes to the calculation. One can compare the correlation of Widmark's "volume" and the effective fluid volume with mass and see if there is less dispersion for the latter. Eating is likely to affect the effective fluid volume. The saying is one shouldn't drink on an empty stomach. This is something that should be taken into account while doing a study. It is not included in this simple model of the blood alcohol content.

Monday, August 10, 2015

Estimating the Blood Fluid Volume

  If one knows the mass of alcohol in the blood and also the blood alcohol content at some point in time one can make an estimate of the blood fluid volume for an individual. One can then use this constant to convert any mass of alcohol to its equivalent concentration in the blood. From the curve fits we know the concentrations Dc0 and Bc0 at time t = 0. The formula for B(t) tells us that at some time in the past, say -Δt, we will have B(-Δt) = 0. The quantity of 10.35% alcohol consumed was 1 liter from which we can determine the mass Am and the effective blood fluid volume Vfluid. Note that the subscripts m and c indicate mass and concentration respectively.

Schweisheimer's data indicates a size variation among his test subjects as can be seen from the following calculations. Note the units that were used for the measurements.

Since the sum Dc0+Bc0+Ec0 = Ac we can estimate the amount of alcohol eliminated, Ec0, at t = 0.

Thursday, August 6, 2015

Conclusions Drawn From Schweisheimer's Blood Alcohol Measurements

  The 3-compartment model predictions for blood alcohol content gives a fairly good explanation for Schweisheimer's 1913 observations. The dependence of the peak time on the transport constant λ explains the decreased peak time with increasing alcohol usage. The magnitude of the peak blood alcohol just depends on the initial values of D0 and B0. When someone chugs a drink B0 = 0 and we get the largest value for the peak alcohol, Bmax = D0/e. The exponential base, e, in the divisor would have to be included in Widmark's rho factor, ρW, which corrects the BAC value for the ratio of the mass of alcohol to the body mass. The conclusion is that the divisor in Widmark's formula is just proportional to the body water content and the expected blood alcohol vs time is given by the formula,

Notice that only the mass of alcohol in the digestive tract, D0, is multiplied by t in the expression in parenthesis. So sipping a drink over time increases B0 and reduces one's peak BAC which is somewhat intuitive.

  People participating in drinking parties should show some social responsibility and not allow individuals to poison themselves. Fraternity pledging rituals often involve drinking and from time to time we hear about a pledge dying of alcohol poisoning. The same could be said about people passing out during Spring Break.

  One needs to be careful about definitions. The e in the divisor could be moved to modify the numerator and the exponent would become 1-λt. The formula for BAC above needs to be checked out more thoroughly and it might be useful to have an app for one's cell phone which would compute the percentage blood alcohol content vs time or predict the peak BAC so one could avoid getting fined for a DUI violation. One should check to see what formula an app uses for best results.

Supplemental (Aug 6): Widmark's formula appears to have been modified in the English Wikipedia. A simple relation defining the Widmark rho factor would be the proportionality factor connecting the amount of alcohol consumed with the product of the peak BAC and body mass. If we define alpha as the conversion factor between B and BAC we can show that the formula for BAC vs time is,

Supplemental (Aug 7): In the previous supplement it was assumed that the chugging procedure was used and so the amount of alcohol consumed equals to D0. To be consistent we would need to set B0 = 0. With steady drinking the peak BAC can fluctuate since some of the alcohol enters the blood while drinking and so B0 does not zero. In addition D0 does not equal the amount of alcohol consumed at the end of the drinking period and so doesn't cancel out in the conversion factor. The net result is the Widmark factor, ρW, is not well defined for the steady drinking case and consequently one cannot make accurate predictions of the BAC according to the 3-compartment model.

Simpler Equations for the Blood Alcohol Content

  When the absorption rate, λ, and the elimination rate, μ, for the 2-compartment model for the blood alcohol content are equal the solution becomes degenerate but simplified alternative formulas can be found.

For this B(t) there is only one search parameter, λ. B0 and D0 can be determined by the method of least squares if λ is known. This was done for the abstainers and the resulting curve is practically indistinguishable from the previous result.

Tuesday, August 4, 2015

Comparison of the Blood Alcohol Model with Observations

  I found some blood alcohol content data in measurements made by Schweisheimer in 1913 to test the 3-cell Blood Alcohol Model on. The data was rather sparse so I separately combined the results for Abstainers, Moderates and three of the Heavy Drinkers to obtain a nominal least squares fit for each group. One can use the formula for the chugged drink curve to do the fit since it approximately the same shape as the steady drinker curve. One does not know the total amount of alcohol in the blood but one can assume that it is proportional to the blood alcohol content and use X and Y as instead of G and H. One does not know when the drinkers started drinking so a B0 constant has to be included in Y. If λ and μ are known one can solve for X and Y using ordinary least squares. One can then search a grid of λμ-values to find the minimum variance for each set of data. Here are the results for the fits for the Abstainers, Moderates and Heavy Drinkers.

The values for the absorption factor, λ, and the elimination factor, μ, are approximately equal in each case and increases with the amount of drinking. The sum of D0 and B0 are approximately independent of the amount of drinking. Each test subject drank one liter of wine whose alcohol content was 10.35 percent.

  The abstainers showed the highest peak blood alcohol content. They were also the slowest to recover. The length of time for which the blood alcohol content was 0.10 decreased with the amount of drinking.

  Widmark's formula relates the blood alcohol content to the amount of alcohol consumed and the body's water content.

Modeling Blood Alcohol Content Over Time

  Suppose you were drinking and you wondered how your Blood Alcohol Content changed over time. We can use a 3-cell model that tracks the amount of alcohol in the digestive tract, in the blood and the amount that has been eliminated over time. The model and the equations expressing the changes over time is shown below.

We can solve the rate of change equations for different situations such as chugging a drink at t = 0 or drinking at a steady rate over a time interval from an initial time t = 0 to a final time t = tf. The general solution for blood alcohol for drinking rate R0, the amount in the digestive tract, D0, and the amount in the blood, B0, at t = 0 is a function involving simple exponential functions,

The constants λ and μ depend on the individuals tolerance for alcohol. 

  What is the solution if one drinks over time tf? Initially there is no alcohol in the digestive tract and blood so D0 = 0 and B0 = 0 so the constant coefficients, F, G and H, in the solution only depend on R0. At time tf the consumption rate drops to zero and the amounts in the digestive tract and blood are Df and Bf. The solution consists of two segments that are equal at t = tf

If one chugs a drink at t = 0 then R0 and B0 equal zero and D0 equals the amount of alcohol drunk. So the solution is,

One can plot the two solutions and compare the results. The two solutions are similar in shape but slightly shifted in time. 

The values for λ and μ chosen are similar to those of someone who rarely drinks.