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. Let's also assume that enough solvent was used to avoid saturation so water/solvent are in partition equilibrium.

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



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:



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 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 😎

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.

Great way to get a sneak preview while spinning the third pull
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.

Thanks for sharing the data Elrik, theory and practice seem roughly aligned to first order

Loveall, do you still find this to be reasonably accurate?

Did you use this for the pulls you did in your minimum polymer thread?

Can we use this (2 cielo pulls:Mt=M1^2/(M1-M2)) on 1g cactus samples to estimate % yield?

No need to reply:

Assuming reasonableness of IAL’s data, here, https://www.dmt-nexus.me...mp;m=1121536#post1121536 looks like 4 pulls for dmt is pretty good, based on your post above: Mt = (0.283)^2/(0.283 - 0.173) = 0.727g; 4 pulls gave IAL 0.0.624/0.727 => 86%, with 5th pull yielding only 0.013g, less than 2%.

I know, his method is not exactly the same as your prelim method, with diff solvent, but seems reasonable approach for naive decision on starting number of dmt pulls.

I will test out the minuscule amounts from pulls on 1g cielo run, see if this could be useful assay estimation technique.

Thanks for posting this.

thank you for this loveall. and thank you for bumping this cheelin, i have not stumbled across this yet.
i will dig out my numbers from some of my earlier extractions and see how they match up. just got to find them

I really like this as a potential estimator of % M Yield using Mt%= [M1^2/(M1-M2)]/P, where P=grams of cactus powder, M1= grams of M yielded from 1st set of consecutive pulls, M2 = grams of M yielded from 2nd set of consecutive pulls, and number of pulls is the same for each pull set.

For example, using data in the 2nd post above, assuming all pulls used similar volume and technique:
A. Using pull 1 vs pull 2: Mt% = [0.953^2/(0.954-0.556)]400 = 0.006 => 0.6%
B. Using pulls 1+2 vs pulls 3+4: Mt% = [{0.954+0.556}^2/({0.954+0.556}-{0.360+0.249})]/400 = .006 => 0.6%

This property would be particularly useful - if it holds up - when doing assays of very small amounts of sample material, where volumes from individual pulls are too tiny to properly salt/crystallize, but combined consecutive pulls could be properly salted/crystallized.

I am testing this now on 1g and 5g cactus powder samples of known CIELO tek % M Yield. Hope it works.

Good news/ bad news.

The good: two sets of combined consecutive pulls (e.g. treating pulls 1-3 and 4-6 as pulls M1 & M2 in the equation Mt=M1^2/(M1-M2)) seems to give relative results consistent with earlier calculations using post #2 data.

The bad: getting sufficient volumes of M1 and M2 to salt/crystallize with such small powder amounts is making it difficult to get the expected Mt levels. When salted as a single 5g powder sample of combined pulls 1-6, I get 1.1% yield; but when splitting the combined pulls into M1=pulls1-3, M2=pull 4-6, I am getting 0.6%.

Will work on this as tightening free time allows.

In my first max ion Tek extraction (100g MHRB), I got,

First pull: 950mg
Pulls 2-5: 500mg

So a 1.45% yield. I've seen higher yields reported (2%, even close to 3%). Wondering if I left something behind from not doing enough pulls.

I didn't do the latter pulls separately to redundantly calculate p because that's more work and I was worried of scraping losses on the smaller pulls.

If I understand the discussion in the thread that developed the simple plot here, using,

Mn = p*Mt*(1-p)^(n-1)

I can calculate p (see attached image where p*Mt has been canceled out, and x is used to represent p). So, therefore want to report p=0.65 for the max ion Tek in my case, well in the 90% region according to the plot. Using the formula from the development thread,

Rn = Tn/Mt = 1-(1-p)^n = 1-(1-0.65)^5 > 99%

So I have an indication here that I exhausted the bark (or at least what was available for the solvent to pull form the water). Any loses before pulling (e.g. leaving DMT locked in the bark tissue) are not accounted for with this check.

Thanks cyb for the TEK and to Loveall for ways to check how many pulls are needed from the water.

P.S: If anyone wants try this, they can edit the 0.95/0.5 part of the solution in Wolfram alpha.

Checking the 1st pull vs pulls 2-5 in Elrik's data above, p=0.37. Consistent with the p ~1-M2/M1 ~ 0.41 formula, both giving p ~ 0.4.

Barely going above 90% after 5 pulls. A much harder extraction for mescaline!

For the mescaline CIELO process I'm getting p = 0.48, and n = 5 pulls does indeed get > 90% (1-(1-p)^n = 96%) 🙂