Strength of People of Different Sizes
MATH0011
Questions :
Numbers and Patterns in Nature and Life
„
Lecture 8
Allometry: Effects of Scaling on
Living Things
„
„
http://147.8.101.93/MATH0011/
If an average man of
bodyweight 50Kg can lift
30Kg, how heavy
should a man of 100Kg
be expected to lift?
Should it be 60Kg?
Will the weight lifted be
proportional to the
bodyweight of the lifter?
0
1
Name
Bodywt(Kg)
Twoliftstotal(Kg)
48
WANGMingjuan
47.82
213.0
53
LIPing
52.84
224.0
58
GUWei
57.40
241.0
63
THONGSUKPawina
62.54
256.0
69
KASAEVAZarema
68.94
275.0
75
LIUChunhong
71.85
285.5
75+
JANGMiͲran
115.12
300.0
Snatch
The two Olympic lifts
2005 Women Weightlifting Record (2 lifts total)
Total wt lifted / body wt
Class(Kg)
Total wt lifted (Kg)
Source: www.iwf.net
2005 Weightlifting World Championships (Women)
Clean & Jerk
Body wt (Kg)
2
Body wt (Kg)
3
„
„
„
Since the ratio weight liftedбbody weight does not
keep constant (it actually decreases) as the body
weight of the athlete increases, the data does not
support the assumption that
F v M (i.e., F = kM for some constant k)
where F = maximum force generated,
M = body mass.
Note that, in theory, body weight and body mass are
two different concepts. But since
(body weight) = (body mass) x g
where g is the Earth’s gravity constant, we may treat
them as equal for convenience.
In fact, it is well known that
muscle strength is
proportional to the crosssection area of the muscle.
Structure of a skeletal muscle
„
„
Let L denote the (linear) size of a
person. Then
Volume of the person: V v L3.
Mass of the person: M v V.
Therefore M v L3.
Cross-section area: A v L2
Muscle force: F v A, therefore F v L2
Hence
L
F v L2 v M2/3
4
Because F v M2/3, we have F = k M2/3 (k is constant)
Therefore log F = log k + (2/3) log M,
i.e., slope of log-log curve is 2/3 § 0.67.
Total wt lifted F (Kg)
„
5
„
Possible reason for the outlier observation: superheavy weight class athletes usually have higher
body fat percentage.
blue line has slope 1, which suggests F v M,
does not fit data well.
red line has slope 0.67, fits data better.
WANGMingjuan (48Kgclass)
Body mass M (Kg)
2005 World Weightlifting Championships (Women)
6
JANGMiͲran(75+Kgclass)
7
Walking Speed
The situation for the Men’s records is similar.
Total wt lifted F (Kg)
„
„
blue line has slope 1, which suggests F v M,
does not fit data well.
red line has slope 0.67, fits data better.
„
Suppose the average walking speed of an adult
person of average height is around 5 km/hr.
Question: What would be the natural walking
speed of a child whose height is half of that of an
average adult?
Body mass M (Kg)
2005 World Weightlifting Championships (Men)
8
9
„
„
„
To simplify analysis,
assume the adult and
the child are geometric
similar (i.e., the child is
just a scaled-down
version of the adult).
Let:
L = body height
H = hip height
S = stride length
„
„
10
Because:
S v H, H v L
it follows that S v L, i.e.,
stride length is proportional
to body height.
This would suggest that
walking speed is
proportional to body height,
which implies that the child
whose height is half of the
adult would walk at half of
the speed of the adult.
However this is not true!
11
„
„
„
In fact walking speed depends on both stride length and
the number of steps made within a given period of time.
„
We can model the leg as a pendulum.
For a pendulum of length L, if the
amplitude of swing is not too big (e.g.,
less than 40 degree) then the frequency
w of swinging is about
1 g
2ʌ L
where g is gravity constant of Earth.
Therefore number of steps (N) that a
person of height L can walk within a fixed
duration of time while walking will obey
the relation N v LЁ1/2.
„
„
Because
(number of steps walked in fixed time): N v LЁ1/2,
(stride length): S v L,
(walking speed): v v SN
therefore v v L Lí1/2 = L1/2,
Hence a child whose height is half of an adult will walk
at (0.5)1/2 § 0.71 times the speed of the adult.
A zoologist R. McNeill Alexander (1976) took
measurements of many species of animals and found a
very general formula v § (0.25)g í0.5S1.67H í1.17 which
can estimate walking speed of most animals (human,
dog, elephant, camel, ostrich, … ). Here S is stride
length and H is hip height.
12
13
Imagine a Giant Ant
The Claim That “Ants Are Super Strong”
„
„
„
Some people believe that ants are
stronger than human, because they are
capable of carrying an object 5 times as
heavy as their own weight, while an
average human can carry only about
her/his own weight.
„
„
Thus, if giant ants that are as large as
human exist, then we human beings
are doomed. (Hey, this sounds like a
good plot for a science fiction movie!)
The fact that “muscle strength is proportional to the
cross-section area of the muscle” holds not just for
mammals, but also for most living things.
Suppose that a giant ant, whose body length is equal to
the height (say, 1.75m) of an average man, exists.
Suppose this giant ant is geometric similar to (i.e., it is
just a scaled up version of) a normal ant, which is 0.005m
(5mm) long.
a typical ant
„
Question: how valid is the above reasoning?
14
15
Some SciFi Movies – More Fiction Than Science
„
„
„
„
Let ө denote the ratio of corresponding lengths of the
giant ant and the normal ant (ө is also called the scale
factor). In this case, ө = 1.75/0.005 = 350.
body mass of giant ant = ө3 x body mass of normal ant
For the normal ant, let M, F be its body mass and the
weight it can support, respectively; for the giant ant, let M’,
F’ denote the corresponding quantities.
Since the giant ant and the normal ants are made of the
same material, we have M’ = ө 3 M . Thus
F’бM’ = (ө2 F)б(ө3 M) = (1бө ) (FбM)
= (1б350 ) 5 = 1б70 § 1.43%
„
Thus the giant ant, which can carry at most 1.5% of its
body weight, will be too weak to stand on its legs.
„
„
This 1958 movie is about
a woman of 50 ft tall. Will
the scenario in the movie
be possible?
Notice that this tall
woman is just a scaled up
version of a normal
woman. Assuming a
height of 5.5 ft for an
average adult female, the
scale factor in this case is
ө = 50б5.5 § 9.
16
Jumping Height of Animals
Now let M be the body weight (body mass) of the
normal person, and F be her muscular strength. The
stress on her, due to carrying her body weight with
her muscular strength, is the ratio MбF. Let M’ and
F’ denote the corresponding quantities of the 50 ft
tall women. Then
(ө3
Antelopes can jump 2m.
Fleas can jump 4cm.
But
2m
4cm
height of antelope
height of flea
>>
„
M)б(ө2
M’бF’ =
F)
= (ө) (MбF) = (9M)бF
„
17
This means that the stress on the tall woman is as if
a normal person feels that her body weight is
suddenly increased to 9 times as before. Therefore
the tall woman could never stand up in the first place.
18
2m
Does it mean fleas are
stronger than antelopes?
4cm
19
„
Work done (i.e., energy
produced) by a contracted
muscle is:
W=Fd
„
„
F
„
„
Suppose an animal has linear size L and body mass M.
Then its muscular force F and the distance d through
which the muscle shortens are related to L as:
F v L2, d v L
Thus the energy produced by the animal when it performs
a certain muscular action is
W = F d v (L2)L, i.e., W v L3
„
d
here F = force exerted by muscle
d = distance through which the muscle shortens
Suppose, when the animal jumps, all
the energy produced is turned into
potential energy, raising the animal’s
body mass through a height h:
W = M g h (= potential energy)
(here g is gravitational constant)
h
20
„
„
„
„
„
„
From W = M g h, we obtain h = W / Mg, which means
h v W / M (because g is constant).
On the other hand, since W vL3 and M v L3, we have
h v L3 / L3 = 1, which means h is a constant, independent
of L.
This implies that when an animal is scaled to a different
(either smaller or bigger) size, the height it can jump will
not be changed!
Therefore, if an antelope can jump a height of 2m, then a
smaller (but geometric similar) version of it can also jump a
height of 2m.
As a result, the claim that fleas are stronger than antelopes
because they can jump a greater height (relative to their
body size) is not scientifically justified.
22
21
„
„
Note that in the above cases, we assume that the two
living beings under comparison are geometric similar,
i.e., one of them is scaled from the other by the same
scale factor, no matter in length-wise, width-wise, or
“depth”-wise. This is mostly the case when we are
considering individuals from a same species. But this
may not be true when comparing individuals of different
species.
In fact, due to the same reasons we deduced earlier,
species of large body size usually have much thicker
bones (and much thicker muscles) than species of
smaller sizes.
23
Both the siamang and the gorilla are primates. But
the siamang, whose height is only half of that of the
gorilla, has bones that are relatively more slender.
„
„
Therefore, the lion is not
a simple enlargement of
the cat, although both of
them are members of the
cat family.
Siamang
Gorilla
24
„
„
Hence, for species
of different sizes,
the skeleton mass
S is not
proportional to the
body mass M.
It is found (Kayser
and Heusner, 1964)
that
25
„
Proportional equations of the form
y v xb , i.e., y = k xb (k is some constant) ,
which are called power rules (or power laws),
appear in a lot of places in the study of animal
physiology.
„
S v M1.13
26
One interesting example is on metabolic rates,
which can be measured in terms of consumption
of oxygen per unit time (liter of O2 per hour), or
alternatively, amount of heat produced per unit
time (kcal per hour), of various organisms.
27
Metabolic Rate
„
It is observed that the metabolic rates per unit body mass
increases tremendously when the size of the animal
becomes very small.
„
28
Body surface (cm2)
„
„
For mammals, energy is required for generating
body heat to keep warm. Since body heat is
dissipated at a rate proportional to the surface
area of the animal, the German physiologist Max
Rubner (1883) proposed that metabolic rate r, at
least for warm blooded animals, should be
proportional to the surface area A of the animal:
rvA
By measuring the surface area of the entire body
in a series of vertebrates, it is observed (see the
graph on the next slide) that
A v M0.63
29
Since A v M0.63 , if Rubner’s theory that r v A is
correct, then one would have r v M0.63.
However, real data on metabolic rates
(Hemmingsen, 1960) shows that r v M0.75 is
more likely. Moreover, this is true not only for
warm-blooded vertebrates, but also for coldblooded vertebrates and micro-organisms as
well (see graph on next slide).
Body mass (g)
Surface area of the entire body in a series of vertebrates
30
31
Real data shows r v M0.75 (Hemmingsen, 1960).
Metabolic rate (kcal б h)
„
ld co
ve
ed
d
o
blo
rm
tes
wa
ra
b
te
er
v
ed
od
o
l
b
tes
ra
b
rte
„
„
m
m
nis
a
g
or
or
c
i
s
Body mass (g)
„
„
The law r v M0.75 , which was first proposed by
the American veterinary scientist Max Kleiber in
1932, is now taken as true universally, and is
called the Kleiber’s Law.
Thus the theory proposed by Rubner, that
metabolic rate is mainly depending on surface
area of the animal, could not explain the actual
data.
In fact, since the Kleiber’s law holds also for
cold-blooded vertebrates, the theory based on
heat dissipated through surface area seems
irrelevant.
32
However, people are still debating about the
reason behind the exponent 0.75.
33
Reference
„
„
One theory on this is that, since oxygen passes
through the lung, r should be proportional to the
surface area of the lung. It turns out that the
lung’s surface is actually not 2-dimensional, but
is a fractal object of fractal dimension d slightly
bigger than 2. Thus r v Ld v Md/3, and d/3 is
close to 0.75.
„
„
„
34
Mathematics in Nature, J.A. Adam, Princeton
University Press, 2003.
Newton Rules Biology, C.J. Pennycuick, Oxford
University Press, 1992.
On Size and Life, T.A. McMahon & J.T. Bonner,
Scientific American Books, 1983.
How Animal Works, K.Schmidt-Nielsen, Cambridge
University Press, 1972.
35