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Measuring Energy Expenditure. Direct Calorimetry Indirect Calorimetry. Caloric Equivalents Carbohydrate – 5 Kcals/LO 2 Fat – 4.7 Kcals/LO 2 Protein – 4.5 Kcals/LO 2. Expressing Energy Expenditure VO 2 LO 2 /min and mLO 2 / kg bwt /min LO 2 /min is absolute expression
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Direct Calorimetry • Indirect Calorimetry
Caloric Equivalents • Carbohydrate – 5 Kcals/LO2 • Fat – 4.7 Kcals/LO2 • Protein – 4.5 Kcals/LO2
Expressing Energy Expenditure • VO2 LO2/min and mLO2/kgbwt/min • LO2/min is absolute expression • mLO2/kgbwt/min is relevant expression • Expressed relevant to body weight • Kcals/min – 5 Kcals/LO2 • METs – equivalent to 3.5 mLO2/kgbwt/min • Kcal/kg/hr – 1 Kcal/kg/hr = 1 MET
Metabolic Equations • Used to estimate oxygen cost vO2for certain activities • Typical standard deviation of 7-9% • Can be used to estimate maximal oxygen consumption from the last stage of a graded exercise test • Certain pre-reqs apply if these equations are applied to a GXT – rate of progression must be slow, stage changes must be suitable to the subject. Need subject to reach steady state at each stage.
Total oxygen cost = net cost of activity + 3.5 mkm • Net cost is considered voluntary cost • In some equations broken down into horizontal and vertical costs vO2 = (0.1 mkm/m/min . m/min) + (1.8 mkm/m/min.m/min. %grade) + 3.5 mkm Horizontal component Vertical Component
Walking Equation vO2 = (0.1 mkm/m/min . m/min) + (1.8 mkm/m/min.m/min. %grade) + 3.5 mkm • 0.1 mkm/m/min is the expression of oxygen cost for every meter per minute of horizontal movement • 1.8 mkm/m/min is the expression of oxygen cost for every meter per minute of vertical movement • Vertical movememtn is the product of horizontal speed (m/min and the grade expressed as a decimal fraction) • Equation losses accuracy at higher walking speeds. As walking speed exceeds 4 mph the mechanical efficeincy of walking for most indivudals decreases.
Running Equation vO2 = (0.2 mkm/m/min.m/min) + (0.9 mkm/m/min.m/min. %grade) + 3.5 mkm • Equation is appropriate to use any time the subject is truly running • Oxygen cost of horizontal movement is doubled • Vertical oxygen cost is decreased by half due to flight phase of running
Leg Ergometry Equation vO2 = (10.8 ml/Watt/min . Watt ÷ kgbwt) + 7 mkm • 10.8 mLO2 per Watt is the oxygen cost associated with leg ergometry • METs are doubled to account for one resting MET and one MET for unloaded cycling or the oxygen cost for turning the pedals with no load on the flywheel
Arm Ergometry vO2 = (18 ml/Watt/min . Watt ÷ kgbwt) + 3.5 mkm • 18 mLO2 per Watt is the oxygen cost associated with arm ergometry
Stepping vO2 = (0.2 mkm/step/min . steps/min) + (1.33 . 1.8 mkm/m/min.m/Step . Steps/min) + 3.5 mkm • 0.2 mkm is the horizontal oxygen cost for every step • Oxygen cost for every m/min of vertical work is 1.8 mkm • 1.33 represents total work • 1 is positive work or work against gravity • 0.33 is an estimate of negative work of work with gravity; the downward phase of the step • Negative work is somewhere between 20 and 40% of positive work, ACSM uses 33%
To effectively use these equations you must know the following: 1 mph = 26.8 m/min 1Watt = 6 Kp-m/min kcals/min = mkm * kgbwt / 200 mLO2 /kcal
Tom is walking at 3mph on a 2% grade. What is his oxygen consumption and energy cost in Kcals/min? Tom weighs 70 kg. • 3 mile/hr x 26.8 m/min/mile/hr = 80.4m/min vO2 = (0.1 mkm/m/min . 80.4m/min) + (1.8 mkm/m/min.80.4m/min. 0.02) + 3.5 mkm vO2 = 8.04 mkm + 2.89 mkm + 3.5 mkm = 14.43 mkm
14.43 is the oxygen cost to estimate the caloric expenditure per minute use this equation: kcals/min = mkm * kgbwt / 200 mLO2 /kcal kcals/min = 14.43 mkm * 70 kg / 200 mLO2 /kcal Kcals/min = 1010.1 mLO2/min / 200 mLO2 /kcal Kcals/min = 5.05
Tom wants to expend 350 Kcals every time he walks. How long will he have to walk at this pace and grade to achieve this caloric expenditure? 375 Kcals ÷ 5.05 Kcals/min = 74.25 min