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.
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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 C_B. An alternative would be to simulate the response of a model using an electronic analog computer.

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 C_B 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 C_D remaining slightly higher than C_B and then a small current from each drains through R_B. 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

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, V_fl, divided by the mass of alcohol consumed, A_m.

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 ρ_W 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 V_fl.

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.

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.

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.

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 ρ_WM_b = 57.05 liters. V_fl = 1/α so using the previous result for α we can compute a V_fl and get the result found in the last blog. We can combine the ρ_W/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.