#  Direct Calorimetry  Indirect Calorimetry  Caloric Equivalents ▪ Carbohydrate – 5 Kcals/LO 2 ▪ Fat – 4.7 Kcals/LO 2 ▪ Protein – 4.5 Kcals/LO 2.

## Presentation on theme: " Direct Calorimetry  Indirect Calorimetry  Caloric Equivalents ▪ Carbohydrate – 5 Kcals/LO 2 ▪ Fat – 4.7 Kcals/LO 2 ▪ Protein – 4.5 Kcals/LO 2."— Presentation transcript:

 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 ▪ 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

 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.

 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

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.

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

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

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

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%

 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

 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

 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

 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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