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.

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 D_c0 and B_c0 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 A_m and the effective blood fluid volume V_fluid. 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 D_c0+B_c0+E_c0 = A_c we can estimate the amount of alcohol eliminated, E_c0, at t = 0.

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, B_max = D₀/e. The exponential base, e, in the divisor would have to be included in Widmark's rho factor, ρ_W, 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, ρ_W, 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.

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 = t_f. 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 t_f? 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 t_f the consumption rate drops to zero and the amounts in the digestive tract and blood are D_f and B_f. The solution consists of two segments that are equal at t = t_f. 

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.

Removing Rust From Tools

  I had a pair of vernier calipers that had some annoying rust spots it and decided to look into ways of removing them Sunday evening. One can clean silver by using a tarnish remover but that is essentially a method of polishing the silver. Wikipedia indicated that electrolysis can be used to remove rust so I decided to try this method. There are a number of YouTube videos that show the process in action.

  When two electrodes are placed in water the voltage difference between them produces chemical changes in the water. Dissolving a salt, such as washing soda, in the water results in some of it being broken up into its positive and negative ions increasing the conductivity of the water. The positive electrode is at a higher electrical potential and is known as the anode while the negative electrode is at a lower potential and is known as the cathode. Negative ions are attracted to the positive anode and are therefore known as anions. Similarly, positive ions are attracted to negative cathode and are known as cations. A water solution has H ⁺ and OH ⁻ ions in equilibrium with the water present so the positive hydrogen ions are attracted to the negative electrode and the negative hydroxide ions are attracted to the positive electrode where they are respectively released as hydrogen and oxygen bubbles.

  The negative electrode is attached to the tool from which the rust is to be removed. The hydrogen ions attracted to this cathode combine with the oxygen in the rust to produce iron which stays on the tool and hydroxide ions which go into solution. Chemically this is a reduction reaction. At the positive electrode oxygen is combined with a sacrificial piece of steel, producing rust in the process, via an oxidation reaction.

  I used supplies from my kitchen to clean the vernier calipers. I placed some water in a casserole dish and the calipers and a used joist bracket as by piece of sacrificial metal. Instead of washing soda I added baking soda to the water which worked fairly well. Technically the resulting hydroxides of iron in the waste water should probably be neutralized before disposal. A variable Radio Shack dc power supply was connected to the electrodes and the current was set to half and amp.

  Two and an half hours later I removed the calipers and rinsed them off. The surface was somewhat lackluster so I used some steel wool to polish them after disassembly. Then I oiled them with cooking spray to prevent rust before reassembly. I felt the cooking spray would be less volatile than WD40 and its nonstick properties would allow the parts to slide easier. The cooking spray has a greasy/waxy feel to it which is probably due to the lecithin in it.

  The method used to refurbish the calipers was probably not the optimal method. It would probably take some testing to arrive at the best way of doing this. One could compare different oils to see which prevents rust from reforming the best. The Wikipedia article indicates that lecithin is an antioxidant. An electrochemical method known as cathodic protection is used to prevent corrosion. Just placing tools on an aluminium, zinc or magnesium foil might also provide some protection against rust formation.