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1 Energy Costs of Physical Activity. 2 Are the activities appropriate, in term of exercise intensity, to achieve the target heart rate? Is the combination.

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Presentation on theme: "1 Energy Costs of Physical Activity. 2 Are the activities appropriate, in term of exercise intensity, to achieve the target heart rate? Is the combination."— Presentation transcript:

1 1 Energy Costs of Physical Activity

2 2 Are the activities appropriate, in term of exercise intensity, to achieve the target heart rate? Is the combination of intensity and duration appropriate for achieving an energy expenditure goal to balance or exceed caloric intake? When Health Fitness Instructors (HFI) recommend specific physical activities to participants, they are usually concerned the following 2 questions. To answer these questions the HFI should become familiar with the energy costs of various activities

3 3 1. Ways to measure energy expenditure 2. Ways to express energy expenditure 3. Formula for estimating the energy cost of activities 4. Energy requirements of walking, running, cycle ergometer, and stepping Oxygen cost of walking walking on a horizontal surface walking up a grade walking at different speeds Oxygen cost of jogging and running jogging and running on a horizontal surface jogging and running up a grade jogging and running at different speeds Oxygen cost of cycle ergometry leg ergometry arm ergometry Oxygen cost of bench stepping

4 4 Ways to measure energy expenditure Direct calorimetry A chamber insulated water flowing through walls Heat given off subject warm Water temperature change Volume of water flowing through the wall per min Example Vol = 20 L/min, Tem increase = 0.5 C 20L/min x 1 kcal/C x 0.5 C = 10 kcal/min All the energy released from CHO, Fat ???

5 5 Indirect calorimetry Estimate energy production by measuring oxygen consumption Measurement Carbohydrate Fat Protein Caloric density (kcal/g) 4.0 9.0 4.0 Caloric equivalent of 1L of O 2 5.0 4.7 4.5 (kcal /L) Respiratory quotient (RQ) 1.0 0.7 0.8 (cell level) Respiratory exchange ratio (R) (gas exchange)

6 6 Indirect calorimetry Closed-circuit spirometry inhale 100% O2 exhaled CO2 is absorbed Is it good for measuring exercise O2 consumption? Open-circuit spirometry inhale room air exhaled air is collected then what?how to measure the following: rest O2 consumption exercise O2 consumption maximal O2 consumption

7 7 Ways to measure energy expenditure VO2 (L / min) Example: submaximal run on a treadmill, 80-kg man, ventilation = 60 L/min, inspired O2 = 20.93%, expired O2 = 16.93% VO2 (L / min) = 60 L/min x (20.93% - 16.93%) = 60 L/min x 4% = 2.4 L/ min The energy required for an activity is calculated on the basis of a subject’s steady-state oxygen uptake (VO2) measured during an activity.

8 8 VO2 (ml/kg/min) Example: 80-kg man with a VO2 = 2.4 L/min VO2 (ml/kg/min) = 2.4 L/min x 1000 ml/L ÷ 80 kg = 2400 ml/min ÷ 80 kg = 30 ml/kg/min METs (metabolic equivalents) Example: 30 ml/kg/min 1 MET = 3.5 ml/kg/min METs = 30ml/kg/min ÷ 3.5 ml/kg/min = 8.6 METs

9 9 Conversion to appropriate unites & knowledge of common equivalents Weight 1 kg =2.2 lb 1 kg = 1 kp = 9.8 N Speed 1 mi/h = 26.8 m/min Distance 1 mi = 1.6 km Work 1 L O2 = 5 kilocalories (kcal) 3500 kcal = approximately 1 lb of fat gain or loss 1 kg.m = 1.8 ml O2 Power 1 MET = 3.5 ml/kg/min = 1 kcal/kg/hr = 1.6 km/hr = 1.0 mi/hr 1 watt = 6.0 kg/m/min

10 10 Formula for estimating the energy cost of activities Total O2 cost = sum of energy components = net O2 cost of activity + 3.5 ml/kg/min Total cost of grade walking = net O2 cost of the horizontal walk + net O2 cost of the vertical + resting metabolic rate (1 MET = 3.5 ml/kg/min) Subject must follow instructions carefully - do not hold on to the treadmill railing; maintain the pedal cadence Work instruments (treadmill; ergometer) must be calibrated

11 11 Energy requirements of walking, running, cycle ergometer, and stepping O2 cost of walking on a horizontal surface * walking speeds of 50 and 100m/min or 1.9 to 3.7 mi/hr (mi/h x 26.8 = m/min or m/min ÷ 26.8 = mi/hr) * ACSM (1995) the formula can be used for speed faster than 3.7 mi/hr as long as the person is truly walking, not jogging or running. VO2 = 0.1 ml/kg/min x (horizontal velocity, m/min) + 3.5 ml/kg/min m/min Net O2 cost of walking 1 m/min on a horizontal surface is 0.100 to 0.106 ml/kg/min. A value of 0.1 ml/kg/min is used in ACSM equation.

12 12 What are the estimated steady-state VO2 and METs for a walking speed of 90 m/min (3.4 mi/hr x 26.8 = 90 m/min) ? VO2 = 90 m/min x 0.1 ml/kg/min + 3.5 ml/kg/min m/min = 9.0 ml/kg/min + 3.5 ml/kg/min = 12.5 ml/kg/min METs = 12.5 ml/kg/min ÷3.5 ml/kg/min = 3.6 MTEs Can the formula be used to predict the level of activity required to elicit a specific energy expenditure? YES. See the next example

13 13 An unfit participant is told to exercise at 11.5 ml/kg/min to achieve the proper exercise intensity. What walking speed would you recommend? 11.5 ml/kg/min = ? m/min x 0.1 ml/kg/min + 3.5 ml/kg/min m/min 11.5 ml/kg/min - 3.5 ml/kg/min = ? m/min x 0.1 ml/kg/min m/min 8 ml/kg/min = ? m/min x 0.1 ml/kg/min m/min ? m/min = 8 ml/kg/min ÷ 0.1 ml/kg/min = 80 m/min m/min 80 m/min ÷ 26.8 = 3.0 mi/hr

14 14 O2 cost of walking up a grade sum of O2 cost of horizontal walking O2 cost of vertical component resting metabolic rate *O2 cost of moving (walking or steeping) 1 m/min vertically = 1.8 ml/kg/min *Vertical velocity = grade (?%) x speed m/min VO2 = 0.1 ml/kg/min x (horizontal velocity, m/min) m/min + 1.8 ml/kg/min x (vertical velocity) + 3.5 ml/kg/min m/min

15 15 What is the total O2 cost of walking 90 m/min up a 12% grade? Horizontal component = 90 m/min x 0.1 ml/kg/min = 9.0 ml/kg/min m/min Vertical component = 12% x 90 m/min x 1.8 ml/kg/min m/min = 19.4 ml/kg/min VO2 (ml/kg/min) =9.0 (horizontal) + 19.4 (vertical) + 3.5 (rest) = 31.9 ml/kg/min METs = 31.9 ml/kg/min ÷ 3.5 ml/kg/min = 9.1 METs

16 16 To estimate the setting needed to elicit a specific O uptake Set the treadmill grade to achieve an energy requirement of 6 METs (21.0 ml/kg/min) when walking at 60 m/min Net O2 cost of the activity = 21 - 3.5 = 17.5 ml/kg/min Horizontal component = 60 m/min x 0.1 ml/kg/min = 6.0 ml/kg/min m/min Vertical component = 17.5 - 6.0 = 11.5 ml/kg/min 11.5 ml/kg/min = grade x 60 m/min x 1.8 ml/kg/min m/min = grade x 108 ml/kg/min Grade = 11.5 ÷ 108 = 0.106 x 100% = 10.6%

17 17 O2 cost of walking at different speeds Beyond walking speed of 50 - 100 m/min (1.9 - 3.7 mi/hr), O2 requirement for walking increases in a curvilinear manner.’ Many people chose to walk at a fast speed rather than jog. Table 7.2 p133

18 18 Energy cost of walking expressed in kcal/min Energy cost of walking increases with the speed of the walk, the rate of increase is higher at the higher speed. Example: 170-lb participate speed 2 to 3 mi/hr 4 to 5 mi/hr O2 cost ___ to ____ Kcal/min ___ to ____ Kcal/min T7.3 p133

19 19 O2 cost of jogging and running for speed of 130 - 350 m/min (4.85 - 13 mi/hr) O2 cost of jogging and running on a horizontal surface VO2 = 0.2 ml/kg/min x (horizontal velocity, m/min) + 3.5 ml/kg/min m/min Net O2 cost of jogging or running 1 m/min on a horizontal surface is about twice that of walking, 0.2 ml/kg/min. What is the O2 requirement for running a 10K race on a track in 60 min? 10K = 10,000 m 10,000m ÷ 60 min = 167 m/min VO2 = 167 m/min x 0.2 ml/kg/min + 3.5 ml/kg/min m/min = 36.9 ml/kg/min METS = 36.9 ml/kg/min ÷ 3,5 ml/kg/min = 10.5 METS

20 20 A 20-year-old female distance runner with a VO2max of 50 ml/kg/min wants to run intervals at 90% of VO2max. At what speed should she run on a track given that one mile equals 1610m? 90% x 50 ml/kg/min = 45 ml/kg/min VO2 = 0.2 ml/kg/min x (horizontal velocity, m/min) + 3.5 ml/kg/min m/min 45 ml/kg/min = 0.2 ml/kg/min x (horizontal velocity) + 3.5 ml/kg/min m/min 45 ml/kg/min - 3.5 ml/kg/min = 0.2 ml/kg/min x (horizontal velocity) m/min 41.5 ml/kg/min = 0.2 ml/kg/min x (horizontal velocity) m/min speed = 41.5 ml/kg/min ÷ 0.2 ml/kg/min = 207 m/min m/min 1610 m ÷ 207 m/min = 7.78 min, (60”x 0.78 = 47”), 7:47” mi pace

21 21 O2 cost of jogging and running up a grade VO2 = 0.2 ml/kg/min x (horizontal velocity, m/min) m/min + 0.9 ml/kg/min x (vertical velocity) + 3.5 ml/kg/min m/min Net O2 cost of jogging or running 1 m/min on a horizontal surface is about twice that of walking, 0.2 ml/kg/min. O2 cost of running up a grade is about one half that of walking up a grade, O2 cost of running 1 m/min vertically is about 0.9 ml/kg/min What is the O2 cost of running 150 m/min up a 10% grade? Horizontal components = 0.2 ml/kg/min x 150 m/min = 30 ml/kg/min m/min Vertical component = 0.9 ml/kg/min x (10% x 150 m/min) m/min = 13.5 ml/kg/min VO2 = 30.0 (horizontal) + 13.5 (vertical) + 3.5 (rest) = 47 ml/kg/min =13.4 METs

22 22 The O2 cost of running 350m/min on a flat surface is about 73.5 ml/kg/min. What grade should be set on a treadmill for a speed of 300 m/min to achieve the same VO2? Horizontal components = 0.2 ml/kg/min x 300 m/min = 60 ml/kg/min m/min Vertical component = 73.5 (total) - 60 (horizontal) - 3.5 (rest) = 10.0 ml/kg/min Also, Vertical component = 0.9 ml/kg/min x (vertical velocity) m/min Hence, 10.0 ml/kg/min = 0.9 ml/kg/min x (vertical velocity) m/min = 0.9 ml/kg/min x ( grade x speed ) m/min = 0.9 ml/kg/min x ( grade x 300m/min ) m/min = 270 ml/kg/min x grade Grade = 10 ml/kg/min ÷ 270 ml/kg/min =.037 = 3.7 % grade

23 23 O2 cost of jogging and running at different speeds In contrast to walking, the energy cost of jogging and running increases in a linear and predictable manner with increasing speed T7.4-7.5 p136

24 24 O2 cost of cycle ergometry * body weight is supported by the cycle * work rate is determined primarily by the pedal rate and the resistance on the wheel * O2 requirement in liters per minute is approximately the same for people of different sizes for the same work rate. Thus, when a light person is doing the same work rate as a heavy person, the relative VO2 (ml/kg/min), or MET level, is higher for the lighter person. * The O2 cost of doing 1kpm of work is approximately 1.8 ml. During cycle exercise, energy must be expended to overcome the friction (unmeasured work)in the drive train of the cycle. This additional work requires about 10% of the O2 required to do the measured work, so 0.2 ml/kpm is added to the 1.8 ml/kpm to get the net cost per kilopond meter of work on a cycle ergometer, 2 ml/kpm.

25 25 VO2 (ml/min) = work rate (kpm/min) x 2ml/kpm + (3.5 ml/kg/min x body weight in kg) Since 6 kpm/min = 1 W (watt), VO2 (ml/min) = work rate (W) x 6 x 2ml/kpm + (3.5 ml/kg/min x body weight in kg)

26 26 What is the O2 cost of doing 600 kpm/min (100W) on a cycle ergometer for 50-kg and 100-kg subjects? VO2 = 600 kpm/min x 2 ml/kg/min + (3.5 ml/kg/min x kg wt) For the 50-kg subject VO2 = 1200 ml/min + (3.5 ml/kg/min x 50 kg) = 1200 ml/min +175 ml/min = 1375 ml/min 1375 ml/min ÷ 50 kg = 27.5ml/kg/min = 7.9 METs For the 100-kg subject VO2 = 1200 ml/min + (3.5 ml/kg/min x 100 kg) = 1200 ml/min + 350 ml/min = 1550 ml/min 1550 ml/min ÷ 100 kg = 15.5 ml/kg/min = 4.4 METs

27 27 A 70-kg participant must work at 6 METs to match the intensity of his walking program. What force (load) should be set on a Monark cycle ergometer at a pedal rate of 50 rev/m? 6 METs =6 x 3.5 ml/kg/min x 70 kg = 1470 ml/min VO2 (ml/min) = work rate (kpm/min) x 2ml/kpm + (3.5 ml/kg/min x body weight in kg) 1470 ml/min = kpm/min x 2 ml/kpm + (3.5 ml/kg/min x 70 kg) Net cost of cycling = 1470 - (3.5 ml/kg/min x 70 kg) = 1470 - 245 = 1225 ml/min 1225 ml/min = ? kpm/min x 2 ml/kpm Work rate (kpm/min) = 1225 ml/min ÷ 2 ml/kpm = 612 kpm/min 612 kpm/min = (50 rev/m x 6m/rev) x force = 300 m/min x force Force = 612 kpm/min ÷ 300 m/min = 2.04 or 2.0 kp

28 28 O2 cost of bench stepping Total O2 cost is the sum O2 cost of stepping up 1.8 ml/kg/min O2 cost of stepping down 0.33 x 1.8 ml/kg/min O2 cost of moving back and forth on a surface at the specified cadence step rate x 035 ml/kg/min 1.33 x 1.8 ml/kg/min VO2 = step height (m) x lifts/min x 1.33 x 1.8 ml/kg/min m/min + step rate x 0.35 ml/kg/min

29 29 What is the O2 requirement for stepping at a rate of 20 steps/min on a 20-cm bench? VO2 = step height (m) x lifts/min x 1.33 x 1.8 ml/kg/min m/min + step rate x 0.35 ml/kg/min VO2 = 0.2 m/lift x 20 lifts/min x 1.33 x 1.8 ml/kg/min m/min + 20 x 0.35 ml/kg/min = 9.6 ml/kg/min + 7.0 ml/kg/min = 16.6ml/kg/min or 4.7 METs


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