Chapter 1, Problem 6SQ

Summary Introduction

To calculate: The estimated O₂ cost of horizontal treadmill running for a 70Kg subject at 150, 200, and 235 m.min^-1.

Introduction: Energy expenditure is basically useful in exercise associated weight loss programs. This is measured in terms of consumption of oxygen in ml. kg^-1.min^-1. The energy expenditure is denoted as the volume of oxygen consumed (VO₂).

Explanation of Solution

Given data;

Treadmill speed = 150 m.min^-1

Subject’s weight = 70 Kg

The volume of oxygen consumed (VO₂):

$\begin{array}{c}{\text{VO}}_{\text{2}}\text{=}\text{Weight}\text{×}\text{0}\text{.2}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{/}\text{(m}{\text{.min}}^{\text{-1}}\text{)}\text{×}\text{speed}\text{(m}{\text{.min}}^{\text{-1}}\text{)}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}\text{70}\text{×}\text{0}\text{.2}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{/(m}{\text{.min}}^{\text{-1}}\text{)}\text{×}\text{150(m}{\text{.min}}^{\text{-1}}\text{)}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}\text{2100ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}2103.5ml.k{g}^{-1}.mi{n}^{-1}\end{array}$

Given data;

Treadmill speed = 200 m.min^-1

Subject’s weight = 70 Kg

The volume of oxygen consumed (VO₂):

$\begin{array}{c}{\text{VO}}_{\text{2}}\text{=}\text{Weight}\text{×}\text{0}\text{.2}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{/}\text{(m}{\text{.min}}^{\text{-1}}\text{)}\text{×}\text{speed}\text{(m}{\text{.min}}^{\text{-1}}\text{)}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}\text{70}\text{×}\text{0}\text{.2}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{/(m}{\text{.min}}^{\text{-1}}\text{)}\text{×}\text{200 (m}{\text{.min}}^{\text{-1}}\text{)}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}\text{2800 ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}2803.5ml.k{g}^{-1}.mi{n}^{-1}\end{array}$

Given data;

Treadmill speed = 235 m.min^-1

Subject’s weight = 70 Kg

The volume of oxygen consumed (VO₂):

$\begin{array}{c}{\text{VO}}_{\text{2}}\text{=}\text{Weight}\text{×}\text{0}\text{.2}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{/}\text{(m}{\text{.min}}^{\text{-1}}\text{)}\text{×}\text{speed}\text{(m}{\text{.min}}^{\text{-1}}\text{)}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}\text{70}\text{×}\text{0}\text{.2}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{/(m}{\text{.min}}^{\text{-1}}\text{)}\text{×}\text{235 (m}{\text{.min}}^{\text{-1}}\text{)}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}\text{3290 ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\text{+}\text{3}\text{.5}\text{ml}{\text{.kg}}^{\text{-1}}\text{.}{\text{min}}^{\text{-1}}\\ \text{=}3293.5ml.k{g}^{-1}.mi{n}^{-1}\end{array}$

Conclusion

The estimated O₂ cost of horizontal treadmill running for a 70Kg subject at 150, 200, and 235m.min^-1 is 2103.5ml.kg^-1.min^-1, 2803.5ml.kg^-1.min^-1, and 3293.5ml.kg^-1.min^-1 respectively.