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次最大速度和最大速度下皮划艇运动的能量学

Energetics of kayaking at submaximal and maximal speeds.

作者信息

Zamparo P, Capelli C, Guerrini G

机构信息

Dipartimento di Scienze e Tecnologie Biomediche, Sezione di Fisiologia, Università di Udine, via Gervasutta 48, I-33100 Udine, Italy.

出版信息

Eur J Appl Physiol Occup Physiol. 1999 Nov-Dec;80(6):542-8. doi: 10.1007/s004210050632.

DOI:10.1007/s004210050632
PMID:10541920
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).

摘要

在八名中高级运动员(三名男性和五名女性;体重45 - 76千克;身高1.50 - 1.88米;年龄15 - 32岁)中,评估了皮划艇运动每单位距离的能量消耗(C(k),千焦×米⁻¹),测试速度包括亚最大速度和最大速度。在亚最大速度下,C(k)通过将稳态耗氧量(VO₂,升×秒⁻¹)除以速度(v,米×秒⁻¹)来测量,假设能量当量为20.9千焦×升⁻¹ O₂。在最大速度下,C(k)由总代谢能量消耗(E,千焦)与距离(d,米)的比值计算得出。E被假定为三项之和,这是威尔基(1980年)最初提出的:E = AnS + αVO₂max×t - αVO₂max×τ(1 - e⁻(t×τ⁻¹)),其中α是O₂的能量当量(20.9千焦×升⁻¹ O₂),τ是在肌肉水平运动开始时达到VO₂max的时间常数,AnS是无氧能量利用产生的能量量,t是运动时间,VO₂max是净最大VO₂。个体VO₂max通过在最大强度的1000米或2000米跑步的最后一分钟测量的VO₂获得。从最短距离(250米)到最长距离(2000米),平均代谢功率输出(E,千瓦)分别相当于个体最大有氧功率(VO₂max)的141%和102%。氧化过程提供平均(标准差)功率随行驶距离增加而增加[从250米时的0.64(0.14)千瓦增加到2000米时的1.02(0.31)千瓦],而无氧来源提供的功率则呈现相反趋势。在2.88至4.45米×秒⁻¹的整个速度范围内,净C(k)是速度的连续幂函数:C(k) = 0.02×v².²⁶(r = 0.937,n = 32)。

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