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Approximate rough answer to how many pulls Options
 
Loveall
#1 Posted : 12/19/2018 3:28:00 PM

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There may be a simple way to asses the extraction of a substance after two pulls. Sorry of this has been posted before already (I cannot find it).

Let's assume that two pulls are done under your preferred condition (same conditions for both pulls) and you measure the yummy product you get from each, let's call M1 the mass obtained from the first pull, and M2 the mass obtained from the second pull.

Interestingly with only these two numbers some cool estimates can be done. For example, the total amount of total material (Mt) in the place your are pulling from (typically water) is,

Mt = M1^2/(M1-M2)


You would need infinite pulls to get it all. What if you want to get 90% of it, how many pulls do you need? The answer is in the plot below, where the fraction of product that moves into each solvent pull is:

p = 1- M2/M1


Just figure out what the p is, and look up how many pulls are needed to land in the green zone where the total fraction of product pulled (labeled Rn) is above 90%. For example, if you are pulling some bufotenine from dry Ca(OH)2 baked yopo seeds and M1=350mg and M2=210mg, then for p = 1 - 210/350 = 0.4, and five pulls are needed to get to the green area. You also know that there were Mt = 350^2/(350-210) = 875mg of extractable bufotenine in your yopo paste to begin with, and that after you complete 5 pulls you can expect to have a have a total of ~ 800mg Cool

In general you want to make M2 <<< M1, this means you have an efficient pulling technique. Try adding salt to the water, adjusting the pH, or even temp. and see how low you can get M2/M1. You may want to invest time to get to p = 1 - M2/M1 > 0.3 instead of just repeating the same pull over and over. Have fun!

Any credit goes to Elrik for posting this kind of data here.

Disclaimer: these formulas have not been thoroughly vetted and checked. I could have made an error. They will be updated if needed. Also, these are only gross approximations. After 3 or more pulls a fit to the data can be made which is more accurate. Also, they assume that the solvent is not saturated in the first pull and are only valid during the part of the extraction where solvent is not saturated with product.
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Elrik
#2 Posted : 12/19/2018 6:02:11 PM

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Great way to get a sneak preview while spinning the third pull Laughing
I finally accepted that I should limit myself to 400 gram cactus chip extractions until I find a 4 litre beaker, so my last one went very smoothly with consistent recovery of NPS on each pull [I was alternating between two nearly equal aliquots of xylene for the pulls so I could be pulling with one while titrating the previous] so lets try and fit the data.
1) 954mg 2) 556mg
3) 360mg 4) 249mg
5) 188mg
Total: 2.307g

From the first two pulls, using your equations, we could predict a theoretical total of 2.287g which is pretty close to my achieved yield after 5 pulls and from your p = 1- M2/M1 graph I had just entered the ≥90% yield zone which is pretty consistent with the best-fit curve I extrapolated from all 5 pulls.
So in a good situation this works pretty good.

I think your right and now that I have a stable procedure I should start tracking these extraction curves to see actual quantitative data on the effects of varying salt concentration, lye concentration, pull time, etc. Otherwise I'm basing my procedures on crude approximations.
 
Loveall
#3 Posted : 12/19/2018 7:42:17 PM

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Thanks for sharing the data Elrik, theory and practice seem roughly aligned to first order Thumbs up
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Excerpt from a McKenna talk transcript / audio.
 
 
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