Depletion of  the Anaerobc reserve

The decreasing of the anaerobic reserve when riding at power P during unit time t can easily be calculated  as 

Loss of anaerobic reserve = Power x Anaerobic Fraction x time

This plot shows the anaerobic power fraction vs. total power P for a  particular cyclist.

For all power between 0 and his RT of 346 W the anaerobic contribution or loss is zero, and recovery may occur.

Between 346 W and his SCP of 477 W his anaerobic fraction  increases linearly from zero to approximately 6 % .These orange points indicate the hard exercise or Slow Death zone. In this graph the yellow zone might appear as almost negligible  but it is the most important zone for eventually winning the Tour de France because all long climbs and time trials are mainly ridden in this power zone.

At power higher than 477 W (red data ) we enter the severe exercise, or Fast Death zone. Here the anaerobic power  is simply proportional to the total power because all efforts are higher that the maximal aerobic power, and thus

Loss of anaerobic reserve = (Power – Maximal Aerobic Power) x time

In cycling slang  “shooting a cartouche" or “burning a match” means depleting your anaerobic reserve to near zero in one continuous effort.

Recovering anaerobic reserve W'bal

After may-be 10 or 20 minutes a new cartouche may be shot.
In order to recover from this depletion we must reduce power to below the recovery threshold RT, and the rate of recovery is proportional to how deep under RT we are

The rate of recovery is proportional to (RT – Power)

This recovery rate is also proportional to the inverse of the state of depletion. The deeper the state of depletion,   the lower the reserve, the faster the restore rate.

Think of it as if the anaerobic reserve were a small vessel with liquid connected to a second huge vessel with a tap that can be open or closed. On opening the tap liquid flows from the big into the small vessel. The flow depends on how far you open the tap i.e. (RT-P) and on the difference of the liquid levels in  both vessels e.g.   W'bal

 Let λ be the rate of restoring  then λ = (RT - P)/W'bal and the half-life time for recovery is T1/2 = 0,693/λ

To clarify a little , suppose a cyclist is 50% depleted from a total anaerobic capacity W' of 18000 J and he reduces his power to 45 W below his recovery threshold RT then
Recovery rate λ = 45 /9000 = 0,005 s-1
The half-life time T1/2 = 0,693 /λ = 138,6 s
thus after approximately 140 sec his anaerobic reserve has recovered from 9000 Joule to 13500 Joule.

Correct ? Not quite because  when W'bal increases, T1/2 also increases and the recovery slows down when it progresses. Therefore the recovery has to be calculated from second to second with an ever changing half-life T1/2

This Exhaustion-Recovery process can be coded into computer applications such as
1. TYPHOON : As a standalone program, taking its power data input from a recorded power file.
2. SUPERCYCLE : As Android or IOS app that allows  its  recording and visualization  live during the ride

Typhoon and Supercycle are twins, they use exactly identical mathematical models and yield identical results.

Both programs or apps need to know the basic parameters from the extended CP-model i.e. RT, CP, W’, W” and SCP and in order to compute these it is necessary to test and introduce some 5 reference power-duration data. For practical reasons we obtain these record values from tests to exhaustion after 2 min, 5 min, 10 min, 20 min and 40 min