Variability index info for April Classics?

mises said:

Anyone have numbers on VI for any pros in Flanders, LBL etc? I've seen HR info before which was extremely variable but am curious how the VI looks for the hillier races.

The Predictor-Lotto guys seem to be best about sharing their data, so I'd suggest checking back here periodically:

http://www.trainingpeaks.com/sites/predictor%2Dlotto/racetrainingdata.asp

acoggan said:

The Predictor-Lotto guys seem to be best about sharing their data, so I'd suggest checking back here periodically:

trainingpeaks.comracetrainingdata.aspOpen ↗

From the Bjarne Riis 1997 Amstel Gold Race data published by SRM (and available via CSV from Robert Chung), has a VI of 1.32, I calculate. This and other files are available:
http://mywebpage.netscape.com/rechung/wattage/

I get NP = 356, AP = 269

Dan

djconnel said:

Riis 1997 Amstel Gold Race data [...] has a VI of 1.32, I calculate. I get NP = 356, AP = 269

That came down in the last hour as AP rose (think about that). For most of the race, NP/AP was around 1.4.

RChung said:

That came down in the last hour as AP rose (think about that).

If I define a "generalized NPn" as follows:

NPn = ( integral { P*^n dt } / duration ) ^ (1/n)

where P* == smoothed power (30 second averaging window)m abd NP == NP4.

An improved variability index would be NP4 / NP1 (as opposed to NP4 / AP), as this prevents the possibility of NP < AP. A spike of power at the beginning or end of a workout, for example the acceleration phase of pursuit, can boost AP relative to NP, but not what is defined here as NP1. More on VI in a bit...

Consider NP2; the variance of P* is :

var P* = integral { (P* - NP1)^2 dt } / duration
         	    = integral { P*^2 dt } / duration -
         		  2 * NP1 * integral { P* dt } / duration +
         		  NP1^2
         
         	     = NP2^2 - NP1^2

the standard deviation (the integral form, not the sampled form) is then:

sigma P* = sqrt(var P*) = sqrt(NP2^2 - NP1^2)
      		 = NP1 sqrt( (NP2 / NP1)^2 - 1 )

I can define a generalized variability index:

VIn == NPn / NP1

Then

sigma P* = NP1 sqrt ( VI2^2 - 1)

or

VI2 = sqrt((sigma P* / NP1)^2 + 1)

So, for VI2, at least, if NP1 increases (which means AP1 increases, usually), and sigma P* stays the same,
then variability index drops.

A similar result (with more complicated math) may apply to VI == VI4.

Dan

djconnel said:

I can define a generalized variability index:

VIn == NPn / NP1

You know, I confess that the subtleties of the variability index still escape me (though I should probably also confess some antipathy to the coefficient of variation which, in spirit, had similar roots -- an antipathy made the worse because my own area of research was renewal theory and the CV attains perhaps its greatest role there). This "TSS problem" reminds me a bit of the dispute between Karl Pearson and R.A. Fisher over standard deviation (which, Pearson pointed out, scales with the mean) and variance (Fisher didn't give a rat's ass that it doesn't).

Bottom line, I think TSS is what it is and rather than "fix" it, I'd think about how to understand its characteristics and then to extend it. There's always room for an additional measure that captures what you're trying to capture, sort of like the (now) peaceful co-existence of variance and standard deviation.