Predicting Energy Expenditure ACSM Metabolic Equations

Why useful?  Exercise prescription  ACSM certification exam

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

Y =  Y = (3.2x)  Y = (3.2 x 47)  Y =  Y = 950.4

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

Y =  Y = (3.7x 1 ) + (47x 2 )  Y = (3.7 x 4.5) + (47 x 10.5)  Y =  Y =

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

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

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

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.

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

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.

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.

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

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.

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

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?

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.

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

Answer – 12.9 Mets  45/3.5 = Mets

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?

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

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  = L. min.

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

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.  = min.

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

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

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

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

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

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

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

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 = m. min.  VO 2 = (179.6 x 0.2) + (.10 x x 0.9)  =  = 55.6 ml. kg. min.

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 x 0.9)  = 19.4  VO2 = = 58.8 ml. kg. min.

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

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

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)  =  = 1553 ml. min. (absolute VO 2 )  1553/78 = 19.9 ml. kg. min. (relative VO 2 )

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?

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

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]

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

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]  =  = 31.4 ml. kg. min.

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.

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]  =  = 21.4 ml. kg. min.