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Published byDominick Johns Modified about 1 year ago

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Direct Calorimetry Indirect Calorimetry

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Caloric Equivalents ▪ Carbohydrate – 5 Kcals/LO 2 ▪ Fat – 4.7 Kcals/LO 2 ▪ Protein – 4.5 Kcals/LO 2

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Expressing Energy Expenditure VO 2 LO 2 /min and mLO 2 /kg bwt /min ▪ LO 2 /min is absolute expression ▪ mLO 2 /kg bwt /min is relevant expression ▪ Expressed relevant to body weight Kcals/min – 5 Kcals/LO 2 METs – equivalent to 3.5 mLO 2 /kg bwt /min Kcal/kg/hr – 1 Kcal/kg/hr = 1 MET

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Metabolic Equations Used to estimate oxygen cost vO 2 for 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.

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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 vO 2 = (0.1 mkm/m/min. m/min) + (1.8 mkm/m/min. m/min. %grade) + 3.5 mkm Horizontal componentVertical Component

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vO 2 = (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.

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vO 2 = (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

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vO 2 = (10.8 ml/Watt/min. Watt ÷ kg bwt ) + 7 mkm 10.8 mLO 2 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

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vO 2 = (18 ml/Watt/min. Watt ÷ kg bwt ) + 3.5 mkm 18 mLO 2 per Watt is the oxygen cost associated with arm ergometry

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vO 2 = (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%

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To effectively use these equations you must know the following: 1 mph = 26.8 m/min 1Watt = 6 Kp-m/min kcals/min = mkm * kg bwt / 200 mLO 2 /kcal

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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 vO 2 = (0.1 mkm/m/min. 80.4m/min) + (1.8 mkm/m/min. 80.4m/min. 0.02) + 3.5 mkm vO 2 = 8.04 mkm + 2.89 mkm + 3.5 mkm = 14.43 mkm

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14.43 is the oxygen cost to estimate the caloric expenditure per minute use this equation: kcals/min = mkm * kg bwt / 200 mLO 2 /kcal kcals/min = 14.43 mkm * 70 kg / 200 mLO 2 /kcal Kcals/min = 1010.1 mLO 2 /min / 200 mLO 2 /kcal Kcals/min = 5.05

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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

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