This has been discussed in at least a couple other threads, but it doesn't quite seem to have been resolved. If SWIM is not mistaken, cactus is generally cored, skinned, and despined before it is cut and dried, so to determine the dry yield per size of cactus (it generally sold by size, rather than weight), one would have to subtract the weight of that excess material. The following (found here: http://www.anoniem.org/?http://www.dmt-nexus.me/forum/default.aspx?g=posts&t=4001&p=2) seems to be fairly accurate in terms of its sources, but leaves out that one piece of information.
About here, we'd have subtract the weight of the core and other excess material.
According to the findings from the recently developed extraction techniques, dried trichocereus mescaline yields generally vary upwards of 1%. So if one were to assume 66.3g of dried material from an average 12in. cutting, that cutting could easily contain 1g or more, far exceeding an acceptable dosage of mescaline.
Anyway, all that aside, what's needed is a general formula that subtracts the average weight of the excess material. Because of the rising popularity of the trichocereus, it is probably advisable that those who are able to, grow their own. Those who engage in this would probably benefit from knowing what kind of yield to expect from their expenses and the work they put into it, and knowing how well they can sustain.
69ron said:Kannamate said:found an estimate on erowid not sure if it's accurate should be close I would guess
1.5 - 2.0 inch diameter section = 1.0 ounces per inch of length
2.0 - 2.5 inch diameter section = 1.5 ounces per inch of length.
2.5 - 3.0 inch diameter section = 2.0 ounces per inch of length
3.0 - 3.5 inch diameter section = 2.5 ounces per inch of length
3.5 - 4.0 inch diameter section = 3.0 ounces per inch of length
4.0 - 4.5 inch diameter section = 3.5 ounces per inch of length
Ok, that's at least something to start with.
Kannamate said:Ok that makes sense is there anyway you can know about how much it would weigh based on measurements like 12" tall 3.5 wide?
About here, we'd have subtract the weight of the core and other excess material.
69ron said:Trichocereus pachanoi is 93.5% water by weight according to Poisson (1960). 100% - 93.5% = 6.5% dry matter.
So, with that above data, a 12" tall 3.5" wide piece of San Pedro would weigh about 36 oz (1020 grams). You multiply 1020 grams by 0.065 and you get a dry weight of 66.3 grams.
According to the findings from the recently developed extraction techniques, dried trichocereus mescaline yields generally vary upwards of 1%. So if one were to assume 66.3g of dried material from an average 12in. cutting, that cutting could easily contain 1g or more, far exceeding an acceptable dosage of mescaline.
Anyway, all that aside, what's needed is a general formula that subtracts the average weight of the excess material. Because of the rising popularity of the trichocereus, it is probably advisable that those who are able to, grow their own. Those who engage in this would probably benefit from knowing what kind of yield to expect from their expenses and the work they put into it, and knowing how well they can sustain.