So how is the experiment doing? What questions did I formulate? What do the erlenmeyer’s look like today? I have not allowed air in for a fresh Oxygen supply.
The Images are taken with my smartphone and the lighting conditions are not balanced and all that. The photo on the left is of the 14th of May 2017 and the one to the right is taken on the 30th of April 2017.Observations:
I’m running 4 experiments in numbered erlenmeyer’s
- Tab-water and oak (way to much)
- 100 ml Alcohol / 400 ml distilled water / 3 grams of oak
- 400 ml Alcohol / 100 ml distilled water / 3 grams of oak
- 450 ml distilled water / 3 grams of oak
- The saturation of all the colour seems to have deepened for erlenmeyers 1, 2 and 3,
- Erlenmeyer 4 is doing something, but very very slowly. Apparently (?!) there is nothing for the oak to react with in the distilled water? Not drawing any conclusions yet.
- The colour in both 2 and 3 seems to get darker. With that I could mean either lightness (scale from “white” to “black”) or “saturation”. The images below illustrate the difference for NCS S 0580-Y60R. (Yes I did a colour management course for my work)
- The question is: Does the colour deepen because the “pigment” deepens in lightness or does the amount of the same colour “pigment” get more so the saturation increases. Or is it a combination of both.
- Is just one chemical component responsible for the colour, or is there a mix of components acting separately from each other creating a mix of colours.
- Is this non-enzymatic browning caused by Maillard reactions?
- Is the colouration in erlenmeyer 1, enzymatic browning because of the lack of Ethanol?
- Can I make a supposition that the colouration in the 20ABV is not enzymatic browning since this would mean all wines depend on this process for colour change? Does that happen when wine turns sour / brown?
Erlenmeyer number 1 contains tab water and lots of oak chips. The colour seems less “earthy” now then 2 weeks ago. There is evaporation and condensation of water on the inside of the erlenmeyer. I have not allowed air in for a fresh Oxygen supply.
The 20% ABV mixture is allowing water to evaporate and condensate on the inside of the Erlenmeyer. The colour has not changed very significantly compared to when the experiment began. There is a hint of a copper colour starting to develop. But honestly it is very clear that low alcohol maturation of wood clearly goes at a significantly slower pace.
Erlenmeyer with the 80% ABV is looking almost bourbon/sherry like in deep copper. Notice how I combine bourbon/sherry in the previous sentence. I want to test a theory with this experiment. My theory being:
- Some sherry producers use new American oak barrels for maturation.
- Bourbon producers use new American oak barrels for maturation.
- Lets say they both use a medium charr.
- The bourbon maturation (erlenmeyer 3) takes out the colour and congeners as seen in erlenmeyer number 3.
- The Sherry (erlenmeyer 2) takes out “only” the colour and congeners as seen in erlenmeyer number 2.
- Both barrels move to Scotland as first fill ex bourbon or first fill ex sherry.
- Logic would suggest the congeners NOT taken out by the sherry are still available for the Scotch to draw out when filled at 63,5 ABV or higher.Ex-Bourbon Scotch Congeners = New American Oak – Bourbon Maturation + bourbon crossoverEx-Sherry Scotch Congeners = New American Oak – Sherry Maturation + sherry crossover
- To illustrate my theory: As an example only. The charts below are not based on any research besides this experiment I am running.
- A new Cask holds 100% of the congeners in its oak
- Lets say 80% of congeners are drawn out by Bourbon Maturation
- Lets say 20% of congeners are drawn out by Sherry Maturation.
- Lets say, in the final scotch, 10% of the bourbon or 10% sherry congeners are transferred to the Scotch. The % really doesn’t matter for me to illustrate the influence on the cask congeners available.
Available to drawn out
100 % available
100% of the new cask congeners are available in “Virgin” American Oak matured Scotch. Hence the pie chart shows 100% availability of congeners.
Available to draw out
80 % still available
20 % taken out
10 % crossover
For a Sherry Cask 20% of the congeners are taken out, but 80% is still available. So the cask still holds 80% of the congeners and some new ones due to the sherry. The light blue part of the pie chart is no longer available. In a 100% piechart this looks rather odd by resulting is 80% > 73% … sorry!
First fill ex bourbon
Available to draw out
20 % still available
80 % taken out
10 % crossover
In an Ex bourbon barrel 80% of congeners (green) are taken out of the barrel. Only 20% of the original Cask congeners are still available for the scotch plus some bourbon crossover (if any)
So when Distillers say “Sherry” maturation they really mean:
“Sherry Matured Scotch”= “New Cask – Sherry Maturation + Sherry crossover” ….. or
“Sherry Matured Scotch” = 80% “bourbon” + 20 % left over cask + “sherry crossover”
Did that make any sense? If I am horribly wrong in my thinking please let me know!
Erlenmeyer 4 is showing very little activity. There is a very slight colouration. There is condensation of water, but other than that nothing is happening. Of it is happening VERY slowly. Does this mean pure water is the least reactant, since there is nothing to react with?
More on this experiment here: https://iladdie.wordpress.com/category/experiment/