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#1 |
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Registered User
Join Date: Oct 2007
Location: Finland
Posts: 6
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As a new TrainingPeaks WKO+ user I came to wonder how CTL is calculated? I had assumed it just to be a plain average TSS/day for period of CTL constant but obviously it is quite far from it.
I have total of about 3500 TSS for last 42 days and CTL constant 42 and I was aiming to build CTL to ~85 but to my surprise it is only a bit over 60. I had a 7 weeks break before starting to use WKO because of vacation and business so for the starting value I put only 10 but in my mind it should not affect if constant is 42 days and were looking 6 weeks back. Where does my thinking go wrong?
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"It takes all the running you can do, to keep in the same place." |
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#2 | |
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Registered User
Join Date: Oct 2004
Posts: 918
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Quote:
EWMA or ... http://www.cyclingpeakssoftware.com...agerscience.asp
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rmur |
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#3 | |
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Registered User
Join Date: Feb 2007
Location: Germany
Posts: 18
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Quote:
newCTL=oldCTL*(1-somethingsmall)+TSS/something The "somethings" are determined by the CTL constant. |
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#4 | |
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Registered User
Join Date: Oct 2006
Posts: 1,311
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Quote:
(TSS-CTL)*something small with something small ~0.024 for the default 42 day time constant That's kinda handy for rough CTL predictions based on daily load. IOW, if your CTL is currently at 100 and you take a total rest day your CTL will drop roughly 2.4 points. If your CTL is 100 and you do a 150 TSS ride your CTL will increase by roughly 1.2 (half the delta, half the change) points given the default 42 day constant. Sure makes it easy to see why CTL drops faster than it climbs, it's hard to do regular workouts at twice your CTL but easy to skip a day and get the big drop. -Dave |
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#5 |
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Registered User
Join Date: Oct 2007
Location: Finland
Posts: 6
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Thanks a lot for everyone! So in fact constant of 42 doesn't only mean that exercises for past 6 weeks affect the CTL but the starting value has a huge effect.
I made a little testing and calculating and you'll get pretty close to "somethingsmall" with 1/(no_of_days - 10). Probably this is all explained in detail somewhere but unfortunately work right now is taking way too much valuable training time so forgive the stupid questions for a beginner.
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"It takes all the running you can do, to keep in the same place." |
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#6 | |
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Registered User
Join Date: Oct 2004
Posts: 918
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Quote:
if you wanna play in Excel, the simplest (and quite accurate) way is: CTL_n = CTL_n-1 + (TSS_n - CTL_n-1)/(tc+0.5) CTL_n = current day's CTL (after training) TSS_n = current day's training load CTL_n-1 = yesterday's CTL tc = CTL time constant in days I'd also encourage you to read thru that info in the link I posted. And re-read it. And re-read it ...
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rmur |
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#7 | |
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Registered User
Join Date: Dec 2007
Posts: 43
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i had a similar quesiton a few months ago:
http://www.cyclingforums.com/t454327.html -Mike Quote:
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#8 | |
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Registered User
Join Date: Mar 2007
Location: Wisconsin
Posts: 109
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Quote:
todaysTSS*(1-EXP(-1/ctl_const))+yesterdaysCTL*EXP(-1/ctl_const) |
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#9 | |
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Registered User
Join Date: Oct 2004
Posts: 918
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Quote:
think of all the computing power going to waste there. I was reared on 4-bit processors ...
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rmur |
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#10 | |
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Registered User
Join Date: Oct 2007
Location: Finland
Posts: 6
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Quote:
Even I was somewhat familiar with the basic concepts from fitness-fatique/dual-factor theories used in gym training I had never before bothered to look under the hood. Very interesting also to see the practical application, how my fitness developes simply by raising CTL from practically zero (after a 7 weeks break).
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"It takes all the running you can do, to keep in the same place." |
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#11 | |
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Registered User
Join Date: Mar 2006
Location: Argentina
Posts: 71
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Quote:
I think this version, also quite simple, is a better match for PMC: CTL(n) = CTL(n-1)(1-1/Tc) + TSS(n)/Tc Differences are small anyway. |
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#12 | |
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Registered User
Join Date: Oct 2006
Posts: 1,311
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Quote:
The exponential form is easy to code into a spreadsheet and allows you to look out days, weeks or months if you do that sort of planning. -Dave |
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#13 | |
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Registered User
Join Date: Mar 2006
Location: Argentina
Posts: 71
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Quote:
The exponential form is more in-line with the original Banister equations but, that wouldn't be "CTL" (¿TM?), IHMO. |
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