2. Predicting Energy Expenditure ACSM Metabolic Equations

3. Why useful?  Exercise prescription  ACSM certification exam

4. Basic Algebra Review  Solve for Y when you are given X E.g. Y = 800 + (3.2x) Solve for Y when x = 47

5. Y = 950.4  Y = 800 + (3.2x)  Y = 800 + (3.2 x 47)  Y = 800 + 150.4  Y = 950.4

6. More than one X  Solve for Y when x 1 and x 2 are given  Y = (3.7x 1 ) + (47x 2 ) + 300  When x 1 = 4.5 and x 2 = 10.5

7. Y = 810.15  Y = (3.7x 1 ) + (47x 2 ) + 300  Y = (3.7 x 4.5) + (47 x 10.5) + 300  Y = 16.65 + 493.5 + 300  Y = 810.15

8. Solve for X when given Y  E.g. Y = 3.5 + (0.1x 1 ) + (1.8 x 0.075x 1 )  When Y = 25  25 = 3.5 + (0.1x 1 ) + (0.135x 1 )

9. Y = 91.5  25 = 3.5 + (0.1x 1 ) + (0.135x 1 )  25 = 3.5 + 0.235x 1  21.5 = 0.235x 1  x 1 = 21.5  0.235  x 1 = 91.5

10. Basic Energy Expenditure Principles  Mass – def. – the weight of an object at rest  Force – def. – the weight of an object in motion

11. Work  Work – def. – the application of force through a distance  Work = force x distance  E.g. A 75 kg man walks 10 meters. He has done 750 kgm of work.

12. Power  Power – def. – work divided by time  Power = worf x d  t t

13. 2 Types of Power  Mechanical power Weight lifting  Metabolic power E.g. aerobic power or oxygen uptake (VO 2 ) VO 2 units – ml. kg. min.

14. If Power is f x d, then… t  If a person pedals a Monark cycle ergometer with 1.5 kg of resistance on the flywheel and is pedaling at 50 rpm, then that person has a power output of 450 kg. m. min. Note: Monark cycle ergometers have a flywheel travel distance of 6 meters per revolution.

15. P = f x d t  1.5 kg x 6 m. rev. x 50 rpm = 450 kg. m. min.  1 min.

16. Energy  Energy – def. – capability to produce force, perform work, or generate power. Units: Energy expenditure – kcal Cycle workrates – kg. m. min. Aerobic power – ml. kg. min.

17. Aerobic Power (VO 2 )  Absolute and relative energy expenditure  Absolute VO 2 – total amount of O 2 used (L. min. or ml. min.)  1 L of O 2 burns 5 kcal

18. Relative VO 2  Relative VO 2 = total O 2  body weight  E.g. – Man weighs 110 kg and has VO 2 of 3.0 L. min.  Boy weighs 50 kg and has VO 2 of 3.0 L. min.  Who has higher relative aerobic power?

19. Man – 110 kg = 27.3 ml. kg. min 3000 ml  Boy – 50 kg = 54.5 ml. kg. min.  3000 ml  Therefore, the boy has higher relative aerobic power.

20. METS  MET – def. – 3.5 ml. kg. min. of aerobic power  How many METS does a woman with 45 ml. kg. min. have?

21. Answer – 12.9 Mets  45/3.5 = 12.857 Mets

22. Example of Energy Expenditure Calculation  John walks at 2.5 mph up a 2% grade on a treadmill. He weighs 75 kg. How many kcal is he expending? Which equation? What VO 2 units will answer give? Which VO 2 units do you need to calculate kcal?

23. Step 1 – Determine VO 2 using the walking equation  Convert speed from mph to m. min. 26.8 x 2.5 = 67 m. min.  VO 2 = (speed x 0.1) + (speed x grade x 1.8) + 3.5  = (67 x 0.1) + (67 x 0.02 x 1.8) + 3.5  = 6.7 + 2.4 + 3.5  VO 2 = 12.6 ml. kg. min.

24. Step 2 – Convert units to L. min.  Ml. min. = ml. kg. min. x body weight (kg)  = 12.6 x 75  = 945 ml. min.  Convert ml. min. to L. min.  = 945/1000  = 0.945 L. min.

25. Step 3 – Convert L. min. into kcal. min.  Kcal. min. = L. min. x 5  = 0.945 x 5  = 4.7 kcal. min.

26. How many minutes would it take for John to lose a pound of fat?  1 pound of fat = 3500 kcal.  3500/4.7 kcal. min.  = 744.7 min.

27. ACSM Walking Equation  VO 2 = horizontal component + vertical component + resting component  = [speed (m. min.) x 0.1] + [grade x speed x 1.8] + 3.5

28. Problem: What is Sue’s VO 2 if she walks at 3 mph up a 7.5% grade on the treadmill?

29. Answer: 22.4 ml. kg. min.  Convert speed into m. min. 26.8 x 3 = 80.4  VO 2 = (80.4 x 0.1) + (0.075 x 80.4 x 1.8) + 3.5  = 8.04 + 10.85 + 3.5  = 22.4 ml. kg. min.

30. What is the VO 2 expressed in Mets?  1 Met = 3.5 ml. kg. min.  22.4 ml. kg. min. = 22.4/3.5  = 6.4 Mets

31. Problem: At what speed would Jim need to walk at 7.5% grade on the treadmill to use 20 ml. kg. min. of O 2 ?  20 ml. kg. min. = (0.1x) + [(0.075x) x 1.8] + 3.5

32. Answer: 2.6 mph  20 = (0.1x) + [(0.075x) x 1.8] + 3.5  16.5 = 0.1x + 0.135x  16.5 = 0.235x  X = 16.5= 70.2 m. min. 0.235 70.2/26.8 = 2.6 mph

33. ACSM Running Equation  VO 2 = horizontal component + vertical component + resting component  VO 2 = [speed (m. min.) x 0.2] + [grade x speed (m. min.) x 0.9] + 3.5

34. Problem: What is Frank’s VO 2 if he runs at 6.7 mph up a 10% grade on the treadmill?  Convert speed into m. min.  26.8 x 6.7 = 179.6 m. min.  VO 2 = (179.6 x 0.2) + (.10 x 179.6 x 0.9) + 3.5  = 35.9 + 16.2 + 3.5  = 55.6 ml. kg. min.

35. Problem: What is the VO 2 for the previous example if the grade were increased to 12%?  Only change is the vertical component; thus you can use the horizontal and resting component values from previous example.  Horizontal component = 35.9  Resting component = 3.5  Vertical component = (.12 x 179.6 x 0.9)  = 19.4  VO2 = 35.9 + 19.4 + 3.5 = 58.8 ml. kg. min.

36. Leg Cycling  Power Output = resistance x rpm x m. rev.  E.g. 3kg x 60rpm x 6m/rev. = 1080 kgm. min.  1 Watt = 6 kgm. min.  1080/6 = 180 watts

37. ACSM Leg Cycling Equation  VO 2 = resistive component + resting component  VO 2 = (power output x 2) + (3.5 x body weight)

38. Problem: What is the VO 2 for a cyclist who pedals at a power output of 640 kgm. min. and who weighs 78 kg?  VO 2 = (640 x 2) + (3.5 x 78)  = 1280 + 273  = 1553 ml. min. (absolute VO 2 )  1553/78 = 19.9 ml. kg. min. (relative VO 2 )

39. ACSM Arm Cycling Equation  VO 2 = (power output x 3) + (3.5 x body weight)  E.g. Pete arm cycles at a power output of 300 kgm. min. What is his VO 2 if he weighs 68 kg?

40. Answer: 16.7 ml. kg. min.  VO 2 = (300 x 3) + (3.5 x 68)  = 900 + 238  = 1138 ml. min.  1138/68 = 16.7 ml. kg. min.

41. ACSM Stepping Equation  VO 2 = horizontal component + vertical component  VO 2 = [step rate (steps. min.) x 0.35] + [step height(m) x step rate x 1.33 x 1.8]

42. Convert step height from inches to meters  Inches x 2.54/100  E.g 12 inch step converts to:  12 x 2.54/100 = 0.3 m

43. Problem: Anne steps up and down at a rate of 24 steps per minute on a 16 inch bench. What is her VO 2 ?  VO 2 = (24 x 0.35) + [(16 x 2.54/100) x 24 x 1.33 x 1.8]  = 8.4 + 22.9  = 31.4 ml. kg. min.

44. Problem: Marianne steps up and down a 6 inch bench at a step rate of 30 steps per minute. What is her VO 2 ?  Don’t forget to convert step height to meters.

45. Answer: 21.4 ml. kg. min.  VO 2 = (30 x 0.35) + [(6 x 2.54/100) x 30 x 1.33 x 1.8]  = 10.5 + 10.9  = 21.4 ml. kg. min.