Intramuscular determinants of the ability to recover work capacity above critical power

Abstract

Purpose

The primary purpose of this investigation was to compare the recovery of the W′ to the recovery of intramuscular substrates and metabolites using ³¹P- and ¹H-magnetic resonance spectroscopy.

Methods

Ten healthy recreationally trained subjects were tested to determine critical power (CP) and W′ for single-leg-extensor exercise. They subsequently exercised in the bore of a 1.5-T MRI scanner at a supra-CP work rate. Following exhaustion, the subjects rested in place for 1, 2, 5 or 7 min, and then repeated the effort. The temporal course of W′ recovery was estimated, which was then compared to the recovery of creatine phosphate [PCr], pH, carnosine content, and to the output of a novel derivation of the W′ _BAL model.

Results

W′ recovery closely correlated with the predictions of the novel model (r = 0.97, p = 0.03). [PCr] recovered faster $\left( {t \frac{ 1}{ 2}\;{ = }\; 3 8 {\text{s}}} \right)$ than W′ $\left( {t\frac{ 1}{ 2}\;{ = }\; 2 3 2 {\text{s}}} \right)$ The W′ available for the second exercise bout was directly correlated with the difference between [PCr] at the beginning of the work bout and [PCr] at exhaustion (r = 0.99, p = 0.005). Nonlinear regression revealed an inverse curvilinear relationship between carnosine concentration and the W′ t _1/2 (r ² = 0.55).

Conclusion

The kinetics of W′ recovery in single-leg-extensor exercise is comparable to that observed in whole-body exercise, suggesting a conserved mechanism. The extent to which the recovery of the W′ can be directly attributed to the recovery of [PCr] is unclear. The relationship of the W′ to muscle carnosine content suggests novel future avenues of investigation.

Appendices

Appendix 1

It is possible to derive the equation presented by Skiba et al. (2012) from first principles.

Here we conceptualize W′ in the framework of chemical kinetics. During periods of exertion above critical power (CP), W′ is depleted at a rate directly proportional to the difference between the power output and CP.

$$\frac{{dW^{\prime}}}{dt} = - (P - CP)$$

This first-order linear differential equation can be solved for a segment of time from u to t in which P exceeds CP, such that the amount of W′ remaining, W′(t), is calculated as follows:

$$W^{\prime}(t) = W^{\prime}(u) - (P - CP)(t - u)$$

During bouts of recovery in which P is less than CP, the rate of change of W′ depends on the amount of W′ remaining (i.e. recovery slows as W′ approaches the initial W′, W ₀ ′) and the power output relative to CP.

$$\frac{{dW^{\prime}}}{dt} = \left( 1 - \frac{W^{\prime}}{W^{\prime}_0} \right) (\text{CP} - P) $$

The first-order differential equation is solved using standard methods as follows.

$$\begin{gathered} D_{\text{CP}} = CP - P \hfill \\ \int \frac {dW^{\prime}} {\left(1 - \frac {W^{\prime}} {W^{\prime}_0} \right)} = \int D_{\text{CP}} dt \hfill \\ \end{gathered}$$

The integral is solved using the substitution rule. Note also that P is considered constant with respect to time, such that D _CP is constant.

$$\ln \left( {1 - \frac{{W^{\prime}(t)}}{{W^{\prime}_{o} }}} \right) = \frac{{D_{CP} }}{{ - W^{\prime}_{o} }}t + const$$

Here we state that for any time = u that follows the expenditure of W′, W′(t) = W′(u), which by definition is less than W ′ ₀. We substitute these values into the equation, and solve algebraically for W′(t) to obtain the final solution:

$$W^{\prime}(t) = W^{\prime}_{0} - (W^{\prime}_{0} - W^{\prime}\left( u \right))e^{{{{ - D_{\text{CP}} } \mathord{\left/ {\vphantom {{ - D_{\text{CP}} } {W^{\prime}_{0} }}} \right. \kern-0pt} {W^{\prime}_{0} }}\left( {t - u} \right)}}$$

We can also analyse the special case of a single segment of time in which the athlete exercises above CP, such that the initial value for W′(t) = W ′ ₀. The recovery after such a bout can be modelled using the following equation:

$$W^{\prime}\left( t \right) = W^{\prime}_{0} - W^{\prime}_{\exp } e^{{ - D_{\text{cp}} {t \mathord{\left/ {\vphantom {t {W0}}} \right. \kern-0pt} W^{\prime}_{0}}}}$$

where W′ _exp is the W′ expended during the prior segment in which P ≥ CP.

To calculate the time course of W′ for an entire power file, we compute W′ depletion for each segment of the power time course in which P ≥ CP and W′ recovery when P ≤ CP.

Appendix 2

Model of W′ recharge kinetics

We tested the notion that a number of linearly recovering entities (e.g. different synergistic muscles or individual muscle fibres or groups of fibres) sum to form the apparent nonlinear macroscopic recovery of a larger system (a muscle or muscle group).

$$\begin{gathered} \frac{{W^{\prime}\left( t \right)}}{{W^{\prime}_{0} }} = \left( {1 - \frac{{W^{\prime}_{\exp } }}{{W^{\prime}_{0} }}} \right) + \frac{{\left( {CP - P} \right)t}}{{W^{\prime}_{0} }},W^{\prime} \ge 0 \hfill \\ \frac{{W^{\prime}_{i} \left( t \right)}}{{W^{\prime}_{0i} }} = \left( {1 - \frac{{W^{\prime}_{\exp i} }}{{W^{\prime}_{0i} }}} \right) + \frac{{f_{i} \left( {CP - P} \right)t}}{{W^{\prime}_{0i} }},W^{\prime}_{i} \ge 0,0 \le f_{i} \le 1,\sum\limits_{i = 1}^{n} {k_{i} = 1} \hfill \\ \end{gathered}$$

For the purposes of the simulation, we assumed that the macroscopic W′ recharge rate was solely a function of the difference between CP and power output, and that 0 ≤ = W′ ≤ = W′ ₀, where W′ ₀ represents the fully charged W′ at rest. The macroscopic W′ was the arithmetic sum of multiple “microscopic” W′, one for each component of the system (Fig. 5a). The microscopic W′ recharge rate for a given component was a function of the amount of the total energy, (CP–P)*t, that it drew. This was defined as the fractional recharge, f _i. Different components of the muscle (fibres) or synergistic muscle group (individual muscles) in question may have different W′ ₀ and f _i values, with the properties of the distributions of these values determining the macroscopic W′ properties (Fig. 5b).