Energetics of kayaking at submaximal and maximal speeds.

P Zamparo, C Capelli, G Guerrini
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引用次数: 64

Abstract

The energy cost of kayaking per unit distance (C(k), kJ x m(-1)) was assessed in eight middle- to high-class athletes (three males and five females; 45-76 kg body mass; 1.50-1.88 m height; 15-32 years of age) at submaximal and maximal speeds. At submaximal speeds, C(k) was measured by dividing the steady-state oxygen consumption (VO(2), l x s(-1)) by the speed (v, m x s(-1)), assuming an energy equivalent of 20.9 kJ x l O(-1)(2). At maximal speeds, C(k) was calculated from the ratio of the total metabolic energy expenditure (E, kJ) to the distance (d, m). E was assumed to be the sum of three terms, as originally proposed by Wilkie (1980): E = AnS + alphaVO(2max) x t-alphaVO(2max) x tau(1-e(-t x tau(-1))), were alpha is the energy equivalent of O(2) (20.9 kJ x l O(2)(-1)), tau is the time constant with which VO(2max) is attained at the onset of exercise at the muscular level, AnS is the amount of energy derived from anaerobic energy utilization, t is the performance time, and VO(2max) is the net maximal VO(2). Individual VO(2max) was obtained from the VO(2) measured during the last minute of the 1000-m or 2000-m maximal run. The average metabolic power output (E, kW) amounted to 141% and 102% of the individual maximal aerobic power (VO(2max)) from the shortest (250 m) to the longest (2000 m) distance, respectively. The average (SD) power provided by oxidative processes increased with the distance covered [from 0.64 (0.14) kW at 250 m to 1.02 (0.31) kW at 2000 m], whereas that provided by anaerobic sources showed the opposite trend. The net C(k) was a continuous power function of the speed over the entire range of velocities from 2.88 to 4.45 m x s(-1): C(k) = 0.02 x v(2.26) (r = 0.937, n = 32).

皮划艇在次最大和最大速度下的能量学。
对8名中高水平运动员(男3名,女5名;体重45-76公斤;1.50-1.88米高;15-32岁)以次最大和最大速度。在次最大速度下,通过将稳态耗氧量(VO(2), l x s(-1))除以速度(v, m x s(-1))来测量C(k),假设能量相当于20.9 kJ x l O(-1)(2)。在最大速度下,C(k)由总代谢能量消耗(E, kJ)与距离(d, m)之比计算,假设E为三项之和,最初由Wilkie(1980)提出:E = AnS + alphaVO(2max) x t-alphaVO(2max) x tau(1-e(-t x tau(-1))),其中alpha是O(2)(20.9 kJ x l O(2)(-1))的能量当量,tau是在肌肉水平运动开始时达到VO(2max)的时间常数,AnS是来自无氧能量利用的能量量,t是表现时间,VO(2max)是净最大VO(2)。个人VO(2max)由1000米或2000米最大跑最后一分钟测量的VO(2)获得。从最短距离(250米)到最长距离(2000米),平均代谢能输出(E, kW)分别达到个体最大有氧能(VO(2max))的141%和102%。氧化过程提供的平均(SD)功率随着距离的增加而增加[从250 m处的0.64 (0.14)kW增加到2000 m处的1.02 (0.31)kW],而厌氧源提供的趋势相反。净C(k)是在2.88至4.45 m x s(-1)的整个速度范围内速度的连续幂函数:C(k) = 0.02 x v(2.26) (r = 0.937, n = 32)。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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