Comments on: More on Speleothem Dating

By: Geoff Sherrington

Geoff Sherrington — Tue, 23 Dec 2008 01:01:33 +0000

Re: Chris H (#20), Thank you Chris H. Both approaches are possible. My inquiry was to catch up with modern approaches. There is a particular case, seen for example with the several peaks commonly arising from isotope dilution in U work where one of the abundances is much, much less than others and so sets the error, as you noted in another way.

By: Chris H

Chris H — Mon, 22 Dec 2008 03:27:58 +0000

Re: Geoff Sherrington (#19), Both approaches are used. In the multiple-collector world, the canned control software tends to calculate ratios. These ratios are averaged over time and they are invariably treated as independent. However, strictly speaking they aren’t as usually there is a shared peak measurement. One should attempt to estimate the covariance of the ratios, but I doubt if this is normally done. From the mass spec software I’ve seen, any covariance is assumed to be zero. Of course, one can get around this by putting the best determined isotope in the denominator, which will minimize the correlation. In contrast, with noble gas machines (e.g. argon dating), which usually peak hop, one has to extrapolate peaks to gas inlet time (the isotopes get pumped by the machine itself during measurement). So each isotope tends to be treated as a separate measurement and they are essentially independent. However, one can then calculate the covariance expected in the ratio error estimates and this is normally done. At least I do it.

By: Geoff Sherrington

Geoff Sherrington — Mon, 22 Dec 2008 00:14:14 +0000

Quick question. If one uses an isotopic ratio (like from a mass spectrometer, measured at the same instant) of say U234 to U238, there will be instrumental errors with each isotope and so errors with the ratios. Is it permitted maths to treat the ratio as a single figure and to do customary statistics on a set of ratios, or does the error of each isotope have to be carried through separately to arrive at total error?

By: Chris H

Chris H — Thu, 18 Dec 2008 15:27:10 +0000

Re: Alan Wilkinson (#15), Yes, that’s what I’m talking about. The top equation above is a bit like a highly constrained line fit to a single point. Because of the constraints, you can’t just put any line through the single data point. Now let x=238U/234U and y=230Th/234U. Both the slope and intercept depend on the age t. The way I would approach this (I’m not a U-series person but an Ar-Ar dater), I would set up a minimization where we have to find not only t but also x(true) and y(true). x(true) and y(true) are exactly constrained to fit the top equation. Now, assuming that sigma(x) and sigma(y) are uncorrelated (we don’t have correlation info, and if they are correlated, it gets more complicated – like fitting a straight line through points with x and y errors and the errors are correlated), then you want to find x(true), y(true) and t such that ((x(meas)-x(true))/sigma(x))^2+((y(meas)-y(true))/sigma(y))^2 is minimized. The assumption of independence is equivalent to assuming that the error ellipse is aligned with the axes. Correlation tilts the error ellipse. This is graphically like finding t such that the error ellipse around the measured x and y isotope ratios is as small as possible when expanded to just touch the line. Now, one can probably do this in R, but most people in the biz would likely just get Ken Ludwig’s Isoplot public domain software (see http://www.bgc.org/isoplot_etc/software.html) over at the Berkeley Geochronology Center (BGC). I don’t use it myself as I tend to write my own software, but I’m pretty sure he has the whole thing worked out.

By: steven mosher

steven mosher — Thu, 18 Dec 2008 04:48:06 +0000

RE 16. Sharing source is like showering with other men. Some of us have no problem whatsoever, other’s are rightly ashamed.

By: Kevin

Kevin — Wed, 17 Dec 2008 17:28:17 +0000

I see it all the time in industrial engineering.

1) Results don’t correlate, send your test setup.

2) I can do that here. Send me your program.

3) Something funny seems to be happening, send your code.

4) I can’t see the function I want to look at, send me uncompiled code.

5) Please send me uncompiled code soon.

I don’t attribute it to deceit. Instead: hiding untidy code, or hiding issues unrelated to the point of contention, or the “author” never studied the code either but inherited or harvested it.

Even happens between people working for the same company at the same location. Sharing software can be a sensetive subject.

By: Alan Wilkinson

Alan Wilkinson — Wed, 17 Dec 2008 09:33:56 +0000

Re: Steve McIntyre (#11), Chris H is talking data point weighting, not errors or confidence levels of the derived parameter estimates? He is right that these must be applied if significantly unequal.

By: Chris H

Chris H — Tue, 16 Dec 2008 22:53:09 +0000

#11: In the words of the the Mythbusters when looking at a blown up car “well there’s your problem” (possibly). When you use the nonlinear optimization routine, it should be a weighted optimization where you minimize the chi-squared value. That will be calculated as ((the model isotope ratio minus the measured ratio)/(1 sigma measurement error))^2. Sum over all fitted isotope ratios. Please excuse the meta formula, but I think you’ll get the drift.

By: D. F. Linton

D. F. Linton — Tue, 16 Dec 2008 19:34:33 +0000

#9. I think you now have the adjustment equation correct. I changed mine to match but I haven’t had any luck replicating the BN6 values. I fell back to a simple check of the measured and corrected 234U to 238U ratio columns. These are related by equation (2). If you solve (2) for lambda_234 and plug in those values and the corresponding ages you get implied lambda_234 values ranging from 2.8274e-6 to 2.8423e-6. The mean of these implied lambda values is 2.8353e-6. I think that one problem in replicating Table 1 comes from the limited precision of the values given. If you use the given value for lambda_234 and compute the implied ages, you get, for lines 3 to the end:

Frank Age Computed Age Difference
105600 105871 -271
107500 107517 -17
105500 105444 56
102200 102249 -49
105400 105118 282
105600 105591 9
104600 104590 10
106800 106776 24
93600 93569 31
118300 118153 147
111200 111109 91
98600 98842 -242
114500 114402 98
97700 97620 80
105300 105354 -54
101900 101973 -73
101900 102005 -105
108400 108574 -174
116400 116302 98
114300 114380 -80

Given the exponentials, t might be quite sensitive to precision uncertainties.

By: Mike Davis

Mike Davis — Tue, 16 Dec 2008 18:41:32 +0000

Replicating methods: using a hat put 20 of each number from 0 thru 9 in hat. Then begin pulling numbers out one by one. What can I say? It has worked for me in the past.