Power output of the lower limb during variable inertial loading: a comparison between methods using single and repeated contractions

Abstract

The power–inertial load relationship of the lower limb muscles was studied during a single leg thrust using the Modified Nottingham Power Rig (mNPR) and during cycling exercise in nine young male subjects. The relationship between peak power and inertial load showed a parabolic-like relationship for mNPR exertions, with a peak [937 (SD 246) W] at 0.158 kg m², this being significantly (P <0.05) different from the power generated at both the lowest [723 (162) W] and highest [756 (206) W] inertial loads. In contrast, for cycling exercise power output did not differ significantly between inertial loads, except at the lowest inertia where power output was significantly (P<0.05) less compared with all other inertial loads. Maximum peak power output during cycling was 1,620 (336) W, which was significantly (P <0.05) greater than that recorded on the mNPR. However, a close association was observed between the mean power generated by each method (r=0.84, P<0.05). The results suggest that during a single contraction a range of inertial loads is required to allow peak power to be expressed. Above a certain critical value, this is unnecessary during cycling movements where the load can be repeatedly accelerated.

Appendix

Polynomial fitting

In order to determine power output, a second-order polynomial (mNPR) and third-order polynomial (Cycle) were fitted to the power velocity data. These specific polynomials were chosen based on the best fit to the data (r ² value). The differentiated equation was then used in order to calculate the maximal velocity and subsequent power.

Linear torque–velocity relationship

From the torque–velocity relationship shown in Fig. 5 the equation of the line can be represented by: T=−a×v+b. Where T is instantaneous torque, v is instantaneous velocity and both a and b are constants. Power can then be calculated as the product of torque and velocity which, if the above equation is substituted for torque, yields: P=av ² +bv. Thus maximal power will be achieved when the gradient of the power curve is zero, i.e. when 2×a×v+b=0. From this formula it can be seen that the velocity at peak power (v _opt) is $\raise0.7ex\hbox{${ - b}$} \!\mathord{\left/ {\vphantom {{ - b} {2 \times a}}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${2 \times a}$}.$ Knowing that torque is zero at v _max, v _max can be written as 0=a×v _max+b. v _max is therefore –b/a, half this value equates to $\raise0.7ex\hbox{${ - b}$} \!\mathord{\left/ {\vphantom {{ - b} {2 \times a}}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${2 \times a}$}$ or v _opt.