(29-08-2026, 11:16 PM)oshfdk Wrote: You are not allowed to view links. Register or Login to view.However, I totally agree with the report that in the context of the results for the other folios it's overwhelmingly likely that this bifolio was produced around ~1420 and certainly no earlier than ~1390.
Again,
why? Even if the Author acquired that bifolio in the same buying/begging/stealing/rummaging transaction as the other folios, it does not follow that they were all manufactured at about the same time. Vellum is a non-perishable commodity, like timber, leather, copper, or vitriol, so a merchant's stock could well have new vellum mixed with decades-old vellum. Especially considering that the Author was happy to take off-cuts and grossly defective sheets. Are you perchance assuming that the Author made the vellum himself?
Quote:there is one more issue... how are [the probabilities computed exactly?
Refer to the following sketch:
[
attachment=17486]
Based on what I know of statistics (not specifically C14 dating), what they seem to have done was
- took several measurements of the C14/C12 ratio from each sample.
- computed the average and standard deviation of those values.
- assumed that the measurement errors had a Gaussian distribution with those parameters (red curve).
- cut that distribution into thin slices.
- mapped those slices through the calibration curve (blue line).
- assembled the mapped slices into a distribution for the times (black curve).
- computed ranges R,S that contained 68.4% or 95.2% of the total probability (of the gray area).
For a point on the vertical axis like A, that meets the blue curve at only one spot, step 5 is easy. They only had to pick the height of the mapped slice A' so that the its area (black) was the same as the area of the original slice (black). That would make the probability of the time being in the small range A' the same as the probability of the true C14/C12 ratio being in the small range A. The height of the black area at A' is the height at A times the slope of the blue curve at that point.
For a point like B, that meets the blue curve at three spots, they had to divide the small black area at B into three parts; that is, assume that the time could be in any of the three small intervals B1', B2', B3'. If I am not mistaken, the proper (Bayesian) way to split is to make each part inversely proportional to the slope at each of the three points. Anyway, the total small black area at B1', B2' and B3' should be equal to the small black area at B.
The result of of steps 4-6 is the probability distribution for the date of the object (gray area). The pink area being 1, the total gray area will be 1 too. If the blue curve were a straight line, the gray distribution would be a Gaussian too. But in general it is deformed, as in this case.
If the gray curve curve had a single hump, it could be summarized by finding an interval that includes 68.2% or 95.4% of the probability, and is either the smallest possible, or leaves out equal amounts of area on ether side, or is centered on the mean of the distribution. I don't know which of these is the "standard" criterion.
(Those specific percentages are conventionally used because, in a Gaussian distribution, they are the areas contained within one and two standard deviations away from the mean. For a non-Gaussian distribution, they are somewhat arbitrary.)
But in this case that single-interval summary would be rather uninformative, even misleading, because the distribution has two well-separated humps with non-trivial areas. Thus a better summary is to split the gray distribution into two curves and find the 68.2% or 95.4% intervals R,S for each hump. The areas of the humps for the f68 sample seems to be about 0.445 and 0.555. Then we can say that the date is within the R interval with 0.445 x 0.954 probability and within the S interval with 0.555 x 0.954 probability.
All the best, --stolfi