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# First LHR Long Run. (Read 638 times)

Yes. Split your run in half, by time or distance. If you split your run by time: For each half divide avg HR into distance (distance/avg HR). Then subtract the first half quotient from the second half quotient. Then divide the difference by the first half quotient and you will get a percentage. If you split it by distance you can use time/avg HR. The difference is your answer will likely be a negative number, unless your aerobic fitness is that extremely high. A couple flaws with this. 1. Heat and high humidity can cause the percentage to be high, due to cardiac creep. 2. Try to run on a flat course or run 2 loops around the same course. If one half has more uphill then the other this can screw up results. 3. The warmup can cause the results to be higher. In WKO+, on a 2 hr+ run, if I counted the first 15min I got over 8% decoupling. By not includeing the first 15min, it was only a little over 6%. I am trying to find out if the 5% allowance is to account for warmup, some creep or both.
So... Today's run: 10:19 128 10:02 134 10:19 136 10:18 138 10:20 138 10:14 138 9:57 140 9:50 141 9:48 143 9:48 145 9:53 147 ((3253 seconds/143 avg 2nd half HR)-(3385 seconds/135 avg 1st half HR)) / (3385 seconds/135 avg 1st half HR) = n ((22.74825) - (25.07407)) / (25.07407) = -.09276 or 9.276% decoupling. Is that right? Since this was an 11 miler, I split mile six in half (307 seconds into ist half of run, and the same into 2nd half). What puzzles me is this....my HR was 5.926% higher for 2nd half, and my time was 3.9% faster. How could that rusult in a 9% decoupling? I must be calculating it incorrectly.
How about... (2nd half time-1st half time)/1st half time=x (2nd half Avg HR - 1st half Avg HR)/1st half Avg HR=y x+y=n n=decoupling In my above example (today's run): (3253 seconds-3385 seconds)/3385 seconds= < .039="" /> (143-135)/135 = .05926 <.039> + .05926 = .02026 or 2.026% decoupling </.039>

Future running partner.

In the article in paragraph 10, they actually use (((Power2/HR2) - (Power1/HR1)) / (Power1/HR1)) * 100 I should have been clearer before, but I have just started experimenting with this myself. This is really a measure of cardiac drift based on the relationship of speed/HR. It's assuming that there is always going to be some deteriorate in performance over time, within a single run of substantial duration, whether speed is decreasing or HR is increasing. The goal is for this deterioration to be less than 5%. I've been playing around in excel with this formula. Power is a metric used in cycling that increases as performance increases. Pace may not work in this formula because that metric decreases as performance increases. This changes the equation. In order to keep the equation consistent, I would use distance or avg. speed over HR. After breaking out my old Math major analysis skills( I'm such a geek! ), I would use this: ((( d1/hr1 ) - (d2/hr2)) / ( d1/hr1 )) * 100
I'll try that. First half= 5.42 miles, 135 Avg HR Second half= 5.61 miles, 143 avg HR (((5.42/135)-(5.61/143))/5.42/135))*100 = 2.29141% I forgot that I could look at my plotted route on Sport Tracks and determine my exact distance, halfway into the run (time).
I tried the numbers from last weeks long run as well. I thought THIS run was actually better, but according to the formula.... Split time Avg. HR 9:41 125 9:17 134 9:26 137 9:15 139 9:26 141 9:23 142 9:22 143 9:22 144 9:19 145 9:28 145 First half: 4.99 miles, 135 Avg HR Second half: 5.01, 143 Avg HR (((4.99/135)-(5.01/143))/(4.99/135))*100=5.032% The dew point reading was significantly lower that day. It was so low I didn't even bother to record it..I think low 50s. Temp was 69F.
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