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# Measurement of Work, Power, and Energy Expenditure

Berry College - Dept. of Kinesiology
by

## David Elmer

on 6 September 2016

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#### Transcript of Measurement of Work, Power, and Energy Expenditure

79.04% nitrogen
20.93% oxygen
0.03% carbon dioxide
English vs. Metric systems
plus... metric system is standard for most scientists and scientific journals
so...
Let's learn the metric system!
or at least be comfortable using it.
meter
(distance)
1 x 10 = 1
0
decimeter
decimeter
decimeter
decimeter
decimeter
decimeter
decimeter
decimeter
decimeter
decimeter
1 x 10 = 0.1
-1
centimeter
centimeter
centimeter
centimeter
centimeter
centimeter
centimeter
centimeter
centimeter
centimeter
1 x 10 = 0.01
-2
millimeter
millimeter
millimeter
millimeter
millimeter
millimeter
millimeter
millimeter
millimeter
millimeter
1 x 10 = 0.001
-3
(dm)
(cm)
(mm)
(m)
micrometer
1 x 10 = 0.000001
-6
(um)
nanometer
(nm)
1 x 10 = 0.000000001
-9
picometer
(Angstrom = )
1 x 10
-10
1 x 10 = 0.000000000001
-12
(pm)
+
System International (SI) units
Metric System
=
easier comparisons of data
Defining Work and Power
Work = force x distance
Work (J) = force (N) x distance (m)
Work expressed in joules or newton-meters
example (from book):
Lift 10 kg up 2 m
1 kg = 9.81 N
10 kg = 98.1 N
so...
wrong units!
kg is mass, we need force
Work = 98.1 N x 2 m
= 196.2 newton-meters or joules
and sometimes you'll see it this way:
Lift 10 kg up 2 m
10 kg x 2 m = 20 kgm
"kilogram meters"
1 kgm = 9.81 joules
20 kgm = 196.2 joules
so...
Before we move on to power
1 kcal = 4186 J
1 pound of fat = 3500 kcal
so to burn 1 pound of fat by this calculation, you need to do
3500 kcal x 4186 J = 14,651,000 J of work
on top of your normal daily activity
Power =
Time
Work
or
Force x Distance
Time
or
Force x Velocity
For power:
Force is important ( force, power)
Velocity is important ( velocity, power)
Time is important ( time, power)
example:
2 people separately lift a 50 kg weight 1m, 10 times
both people do 4905 J of work
person 1 takes 60 sec to do it
4905 J
60 sec
= 81.75 watts
person 2 only takes 30 sec to do it
4905 J
30 sec
= 163.5 watts
the two people have equal strength, but person 2 is more powerful
there's a difference!
kilometer
1 x 10 = 1000
3
Measuring Work and Power
ergometry
- the measurement of work output
ergometer
- the device used to measure
a specific type of work
Bench step
Cycle ergometer
a bench you step on.
known height (distance)
known body weight (mass - )
set step up cadence, or count # of step ups
time - either set time or open ended
convert to force
example:
70 kg person
0.3 m step
30 steps/min
10 min
work =
686.7 N
(70 kg x 9.81 N/kg)
force
x
(0.3 m/step x 30 steps/min x 10 min)
90 m
distance
remember, watts = joules/sec
=
61,803 J
power =
61,803 J
600 sec
=
103 watts
work
time
(60 sec/min x 10 min)
stationary bike.
Friction braked or magnetically braked
flywheel travels a known distance (6 meters) per pedal revolution
resistance is set by adding weight (kg) to the rack
set or count number of revolutions per minute
set or measure time of activity
example:
1.5 kg resistance
6 m per revolution
60 revolutions per minute
10 min
work =
(1.5 kg x 9.81 N/kg)
52,920 J
x
14.7 N
force
3600 m
=
(6m/rev x 60 rpm x 10 min)
distance
power =
52,920 J
600 sec
work
time
=
88.2 watts
measuring mechanical work/power on a treadmill is complicated
unless there is incline, introducing a vertical component
- "amount of vertical rise per 100 units of belt travel"
10% = 10 m rise every 100 m
calculated by multiplying the sine of the treadmill angle by 100
SOH CAH TOA anyone?
example:
60 kg person
200 m/min
10 min
(200 m/min x 0.075 x 10 min)
work =
588.6 N
x
150 m
(60 kg x 9.81 N/kg)
force
distance
=
88,290 J
power =
88,290 J
600 sec
work
time
=
147 watts
Measuring Energy Expenditure
very useful for understanding or prescribing weight loss plans or dietary needs for athletes
Direct calorimetry
Indirect calorimetry
foodstuffs + O
2
Heat + CO + H O
2
2
Produced by:
Cellular respiration (bioenergetics)
Cell work
Foodstuffs + O
2
heat + ATP
cellular work
(which produces)
S.I. unit for heat is the , but the common term is the
joule
calorie
amount of heat required to raise the temperature of one gram of water by one degree Celsius.
this is a very small amount, so when you see "calorie" on a food label, etc. it actually means...
kilocalorie
1 kcal = 4186 J
determines metabolic rate by measuring heat production
respiration
amount of heat required to raise 1 kg of water 1 degree Celsius
(rpm)
force
distance
direct relationship between O consumed and heat production
2
measuring O consumption provides estimate of metabolic rate
2
type of food influences the estimation due to different amounts of energy released
Carbohydrates
(sugars)
5.05 kcal/liter O
2
Fats
4.7 kcal/liter O
2
(Proteins)
4.83 kcal/liter O
2
*not a primary source of metabolic energy
general estimation can be used
5 kcal/liter O
2
How do you measure O consumption?
2
open circuit spirometry
modern-day systems are automated
give an automatic read-out of oxygen consumption and carbon dioxide production
"gold standard" is the
Douglas bag
balloon that collects expired air
use O and CO analyzers and gasometer to determine O consumption and CO production
2
2
2
2
Haldane equation:
based on the concept of nitrogen balance
nitrogen is not produced or consumed by the body
oxygen consumption =
amount of oxygen inspired - amount of oxygen expired
Volume of air
Percentage of that air that is oxygen
each of these consists of
VO
= (V x F O )
2
I
I
2
-
(V x F O )
E
E
2
oxygen consumption
volume of air inspired
fraction of inspired air that's oxygen
volume of air expired
fraction of expired air that's oxygen
Air:
(measured)
(measured)
(calculated)
("known")
you can get both of these things from a Douglas bag
how?
- nitrogen balance
(V x F N ) = (V x F N )
I
I
2
E
E
2
must add up to 100% (or 1.0)
rearrange...
V
=
(V x F N )
F N
E
E
2
I
2
I
You're already measuring O and CO
2
2
Expressions of Energy Expenditure
Absolute VO
2
(liters/min)
Relative VO
2
(ml/kg/min)
kcal/min
METs
~ 5 kcal/liter O
2
Relative VO
2
3.5 ml/kg/min
Absolute VO (L/min) x 1000 ml/L
2
Body weight (kg)
What if I don't have a way to measure energy expenditure?
estimate it!
How much energy is required to do a certain amount of work?
Net efficiency
it depends on:
% net efficiency =
work output
energy expended
x 100
(above rest)
*units must be the same on top and bottom
example:
cycle ergometer
Resistance = 2 kg
50 rpm
resting VO = 0.25 L/min
exercising VO = 1.5 L/min
Distance = 6 m/rev
2
2
(2 kg x (50 rpm x 6 m/rev))
600 kgm/min
/
426.8 kgm/kcal
(1 kcal = 426.8 kgm)
1.41 kcal/min
(
)
6.25 kcal/min
5 kcal/L O x 1.25 L/min
(1.5 L/min - 0.25 L/min)
(
)
2
x 100
=
22.6%
only 22.6% of the energy burned was used to do mechanical work
Walking:
horizontal component =
(0.1 x speed) + 3.5
vertical component =
m/min
1.8 x vertical gain
Running:
horizontal component =
(0.2 x speed) + 3.5
VO =
2
VO =
2
VO =
2
vertical component =
VO =
0.9 x vertical gain
Cycling:
VO =
2
1.8 x work rate
body weight
+ 7
ml/kgm
kgm/min
kg
Factors influencing efficiency:
Work rate:
efficiency decreases as work rate increases
Movement speed:
at lower power outputs, lower speed = greater efficiency
at greater power outputs, greater speed = greater efficiency
This also depends on the individual...
e.g. Lance Armstrong vs. Jan Ullrich
~110 rpm
~85-90 rpm
Muscle fiber type:
"slow twitch" muscle fibers are more efficient than "fast twitch" muscle fibers
depends on genetics and training
Running Economy:
since you don't normally measure work during flat running, you don't normally measure running efficiency
measure oxygen cost of running at certain speeds
More economical = less O used at a given speed
2
men and women typically have similar running economies, except for at higher speeds
(men become more efficient)
and you know the % nitrogen in the air
Friction braked:
Charles' Law
Boyle's Law
proportional relationship between gas temperature and gas volume
at constant temperature, number of gas molecules in a given volume varies inversely with pressure
what if you want to compare VO2max tests that were collected at different temperatures and/or altitudes?
(which you always do)
Standardize the temp and pressure
T
1
T
2
=
V
1
V
2
P V = P V
1
1
2
2
0
o
C
,
760
mmHg
V
2
=
V
T
x
1
2
T
1
V
2
=
P V
1
1
P
2
DRY
but you have to account for water vapor as well, which is temperature dependent...
V
2
=
(P - PH O)
1
2
V
1
P
2
(273 )
K
V
STPD
=
V
ATPS
273
(273 + T)
[
]
V
STPD
=
V
ATPS
P
1
- PH O
2
760 mmHg
[
]
V
STPD
=
V
ATPS
[
273
(273 + T)
]
[
P
1
- PH O
2
760 mmHg
]
Full transcript