Manual Resolution Calculation

[lee majors]

Heya all

I stumbled onto this site while trying to find the answer to my following question and hope somebody here could help me

I am revising basic hplc principles in readiness for a job interview tomorrow and for the life of me cannot remember how to manually calculate the resolution factor.....I know the equation is

R = RTb - RTa /0.5 (PWa +PWb)

but i seem to remember from calculating these years ago that it is not just a case of putting in the RT in min and the PW in cm. I THINK i have to convert everything to seconds by measuring the baseline and seeing what 1cm equates to in terms of time e.g 1cm = 60secs of chromatogram, and measuring the PW in cm and multiplying this by the value in time of one cm and then solving.......am i right?????? Also the same question goes for calculating plates

N = 5.54 (RT/PW)sq

would i have to do the same type of calculation for this or can i just plug the numbers in as they are.......AND do i solve the brackets first then times by 5.54 then square it or square it first then times by 5.54

Sorry for the lack of introduction and stuff in this my first post but i have to get revising!!!

thanks guys

[Consumer Products Guy]

Wow, $6M man, I also haven't done resolution calcualtions manually, like "ever". Tell them you'd look it up in USP <621> chromatography/system suitability, or use the instruments chromatography software to calculate that automatically, like we do with agilent Chemstations. Knowing WHERE to look up stuff is really MORE important that just reciting such.

Sorry about Farrah.

[danko]

Hi Lee

You are going to use a ruler – right? So, use millimeters to measure both retention times and peak widths. That’s all.
As for the plate count equation; the power has the first priority. So you square first and the brackets are thus solved. Then multiply by 5.54.

Good luck tomorrow.

[tom jupille]

Be careful!!

The peak width in your resolution calculation is *baseline* width. The peak width in the formula you are using for plates is *width at half-height*.

The correct formula using baseline width is:

N = 16 (tR/Wb)^2 (note that the "fudge factor" is 16 instead of 5.54

If you have a perfectly Gaussian peak, the baseline width is approximately 1.7 x the width at half-height.


And, just to clear up an mathematical ambiguity, the resolution formula is

Rs = (tR2 - tR1)/[0.5*(Wb2+Wb1)] where 1 and 2 refer to the respective peaks.

"Knock 'em dead" in the interview!

[PSand]

Has anyone had any experience or knowledge of calculating resolution (R) using the baseline peak width equation but employing peak widths defined by the automated integration events/times (ie. Drop lines or start of the peak at baseline) instead of extrapolation of tangents to the baseline?

Are there any resources, articles, or supporting abstracts which discuss more unconventional resolution calculations the people may know of?

[tom jupille]

That is a large, economy-sized can of worms.

First of all, the usual definition of resolution is built on three assumptions:
1. (very important) the peaks are Gaussian
2. (less important) the peaks are already reasonably well separated.
3. (kind of important) the peaks are comparable in size (at least within the same order of magnitude).

The distribution of a Gaussian can be characterized by it's variance (sigma-squared) and thus by sigma. Baseline width (where the tangents intersect the baseline) is 4 * sigma. Width at half-height is 2.35 * sigma, etc. The traditional measure of Rs = 1.5 as "99% baseline resolution" comes from the the fact that it represents a 6-sigma separation between peak centers, and thus has just under 1% overlap between two equal-sized Gaussian peaks. The simplest answer to your question is: if the peaks are reasonably Gaussian, then sigma can be obtained from a peak width measurement made at any fraction of the peak height; all you need is a table of the normal distribution (or a copy of Excel). If the two peaks are moderately well separated, you can characterize the first peak by the front half of it's width (i.e., the distance from the leading edge of the peak to it's "centerline") and the second peak by the back half of its width. As the resolution decreases, of course, this approach becomes less accurate.

Once you get away from Gaussian peaks, things become hazy. If the first peak tails, then the overlap between the two peaks will be larger than it would be if both were Gaussian. How much larger depends on the disparity in the peak sizes and on the details of the tailing peak shape. The latter is the real problem because you generally don't have any a priori knowledge about those details.

Overlaying all of that is the question of exactly how the data system determines peak start/end and how it allocates the split between partially resolved peaks. If you want to get into the details, get a copy of Dyson's "Chromatographic Integration Techniques" (http://www.amazon.com/Chromatographic-I ... 0854045104 ).

Unless you're designing a data system, I wouldn't worry too much about the details. Resolution is an artificial construct that we use to evaluate the quality of a separation (there are others, by the way, such as valley/height ratios). Pick a measurement that gives you consistent results, set a target that allows you to meet the goals of your method, and get on with it. In that respect, the traditional definition of resolution (despite its flaws) has the advantage of being familiar to auditors and reviewers.

[PSand]

Thanks for the swift reply. Certainly a'customized approach' is open to higher level of regulator scrutiny since it deviates from typical industry practice. However, wouoldn't the R values generated using this approach still carry some meaning/correlation to the more common/accepted approach?

Also, could you drill down a bit more with regards to the impact of using a drop line (integration event) approach if the "actual" resolution were less than 1? Specifically, in a case where the smaller of the two peaks in the pair is ~1:3 to 1:5 ratio to the larger, and the critical pair elute at the first two of a 6 peak system.

[mbicking]

Most data systems will offer choices about how to calculate resolution and theoretical plates. For example, the ChemStation provides calculations for plate count and resolution using tangents, half height, five sigma, and statistical methods. As tailing increases, the differences in the values also increase, as you would expect, since each is making the measurement a different way.

Which one you choose is probably less important than being consistent in which numbers you use, and how.

[tom jupille]

However, wouoldn't the R values generated using this approach still carry some meaning/correlation to the more common/accepted approach?

PSand and I have been corresponding "off-line" on this, but for general benefit:

No, actually the correlations are all over the map. Imagine a set of 6 peaks, each resolved from its neighbor with Rs = 0.75 (that's a 3-sigma separation between peak centers). Assuming they are all Gaussian and evenly spaced, the vertical drop from the valleys will assign to peaks 2, 3, 4, and 5 a width of 3 x sigma (instead of the correct 4 x sigma), and the resolutions will calculate out to 1.0 (instead of the correct 0.75).

Now, imagine another set of 6 peaks, but each resolved from its neighbor with Rs = 1.25 (a 5-sigma separation). Using the perpendicular drops will make the widths of peaks 2 - 5 look like 1.25 x sigma, and the resolution will again calculate to 1.0 (instead of the correct 0.25).

Those are the simple cases . With varying peak heights, varying spacing, and non-Gaussian shapes, any correlation between the calculated and "actual" resolution values becomes little better than a guess.