Predicting Energy Expenditure ACSM Metabolic Equations.

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 = 800 + (3.2x)

Solve for Y when x = 47

Y = 950.4

Y = 800 + (3.2x) Y = 800 + (3.2 x 47) Y = 800 + 150.4 Y = 950.4

More than one X

Solve for Y when x1 and x2 are given

Y = (3.7x1) + (47x2) + 300

When x1 = 4.5 and x2 = 10.5

Y = 810.15

Y = (3.7x1) + (47x2) + 300 Y = (3.7 x 4.5) + (47 x 10.5) + 300 Y = 16.65 + 493.5 + 300 Y = 810.15

Solve for X when given Y

E.g. Y = 3.5 + (0.1x1) + (1.8 x 0.075x1) When Y = 25 25 = 3.5 + (0.1x1) + (0.135x1)

Y = 91.5

25 = 3.5 + (0.1x1) + (0.135x1)

25 = 3.5 + 0.235x1

21.5 = 0.235x1

x1 = 21.5

0.235 x1 = 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 = w or f x d t t

2 Types of Power

Mechanical power Weight lifting

Metabolic power E.g. aerobic power or oxygen uptake (VO2)

VO2 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 – kcalCycle workrates – kg.m.min.Aerobic power – ml.kg.min.

Aerobic Power (VO2)

Absolute and relative energy expenditure

Absolute VO2 – total amount of O2 used (L.min. or ml.min.)

1 L of O2 burns 5 kcal

Relative VO2

Relative VO2 = total O2

body weight E.g. – Man weighs 110 kg and has VO2 of

3.0 L.min. Boy weighs 50 kg and has VO2 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 = 12.857 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 VO2 units will answer give?

Which VO2 units do you need to calculate kcal?

Step 1 – Determine VO2 using the walking equation

Convert speed from mph to m.min. 26.8 x 2.5 = 67 m.min.

VO2 = (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 VO2 = 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 = 0.945 L.min.

Step 3 – Convert L.min. into kcal.min.

Kcal.min. = L.min. x 5 = 0.945 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. = 744.7 min.

ACSM Walking Equation

VO2 = 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 VO2 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. 26.8 x 3 = 80.4

VO2 = (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.

What is the VO2 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 O2?

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] + 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

ACSM Running Equation

VO2 = horizontal component + vertical component + resting component

VO2 = [speed (m.min.) x 0.2] + [grade x speed (m.min.) x 0.9] + 3.5

Problem: What is Frank’s VO2 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.

VO2 = (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.

Problem: What is the VO2 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.

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

VO2 = resistive component + resting component

VO2 = (power output x 2) + (3.5 x body weight)

Problem: What is the VO2 for a cyclist who pedals at a power output of 640 kgm.min.

and who weighs 78 kg?

VO2 = (640 x 2) + (3.5 x 78) = 1280 + 273 = 1553 ml.min. (absolute VO2)

1553/78 = 19.9 ml.kg.min. (relative VO2)

ACSM Arm Cycling Equation

VO2 = (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 VO2 if he weighs 68 kg?

Answer: 16.7 ml.kg.min.

VO2 = (300 x 3) + (3.5 x 68) = 900 + 238 = 1138 ml.min. 1138/68 = 16.7 ml.kg.min.

ACSM Stepping Equation

VO2 = horizontal component + vertical component

VO2 = [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 VO2?

VO2 = (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.

Problem: Marianne steps up and down a 6 inch bench at a step rate of 30 steps per

minute. What is her VO2?

Don’t forget to convert step height to meters.

Answer: 21.4 ml.kg.min.

VO2 = (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.