Dynamics of dinosaurs
MATH0011
Questions:
Numbers and Patterns in Nature and Life
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Lecture 1
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Dynamics of Dinosaurs
How heavy are
dinosaurs?
Would sand support a
big dinosaur just as
well as a small one?
How about clay?
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Basic Facts about Dinosaurs
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Basic Facts about Dinosaurs (cont.)
Dinosaurs were a particular subclass of
reptiles, lived in the Mesozoic geological era
(230 million – 65 million years ago). (Human
appeared only about 1 million years ago.)
Although they are reptile-like, dinosaurs are
not ancestors of modern reptiles (such as
crocodiles and lizards). In fact, dinosaurs are
ancestors of birds.
Not all dinosaur species lived at the same time.
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Basic Facts about Dinosaurs (cont.)
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Basic Facts about Dinosaurs (cont.)
The largest known dinosaur is
Argentinosaurus (35 - 45 m long). Another
well-known big one is Diplodocus (25 m long).
The smallest dinosaur fossils
found so far are 23mm and
25mm in length, found in
Guiyang of China in 2007.
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The most famous is
the Tyrannosaurus
(T-rex, > 13 m long).
It is the largest known
land predators.
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Diplodocus
Argentinosaurus
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Basic Facts about Dinosaurs (cont.)
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Basic Facts about Dinosaurs (cont.)
Some dinosaurs walked on two legs (bipedal),
some walked on four legs (qradrupedal).
stegosaurus
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Iguanodon
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Dinosaurs are studied
mainly through their
fossils, which are mainly
bone (including teeth)
fossils. Over the millions
of years, organic matters
in those bones were
replaced by minerals,
forming complex
compounds (fossilization).
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How Heavy Are Dinosaurs?
Basic Facts about Dinosaurs (cont.)
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Therefore, the remains of dinosaurs that we
see in museums which resemble bones are
actually some kind of stones.
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Difficulty: No live specimens
available, nor even dead
ones with their whole body
(including blood and body
fluids, etc.) intact.
Remark: One may use
a weighing bridge to
weigh large mammals
such as elephants.
Weighing an elephant
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Weighing Dinosaurs Using Scale Models
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Weighing Dinosaurs Using Scale Models (cont.)
Method 1: Using Archimedes’ Principle with
scale models.
Apparatus: anatomically accurate dinosaur
model, beaker, beam balance, water.
Example: Suppose a model of Triceratops, of
scale 1:40, is used. Suspend it from one arm
of a beam balance over a beaker. (If the model
is not as dense as water then hang a metal
weight from its tail.)
Setup of apparatus
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Weighing Dinosaurs Using Scale Models (cont.)
Weighing Dinosaurs Using Scale Models (cont.)
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By Archimedes’ principle, upthrust is equal to the
weight of water displaced. Therefore:
Make sure that the model is not touching the
bottom or the side of the beaker.
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Pour water into the beaker so that the tail weight (if
any) is submerged in water but the model is above
the water. Enough weights are put on the pan to
balance the system. Record the weight as Wa (kg).
Wa – Wb = upthrust on model = mass of volume of
water occupied by the model.
Hence volume of model = (Wa – Wb )/(density of water).
Suppose Wa – Wb = 0.095 kg. Since density of water
is 1,000 kg / m3, volume of the model is 0.000 095 m3.
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Add water until the model is completely submerged.
The upthrust of water on the model will put the
system unbalanced. Remove enough weight on the
pan until the system is balanced again. Record the
weight as Wb (kg).
Since the (linear) scale of the model is 1:40, the
dinosaur had 40x40x40=64,000 times the volume of
the model.
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Weighing Dinosaurs Using Scale Models (cont.)
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Weighing Dinosaurs Using Scale Models (cont.)
Hence volume of the Triceratops was 0.000 095 x
64,000 = 6.08 m3.
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Theorem 1 Suppose a model of scale 1: α is
used. If an upthrust of c kg (= Wa – Wb) is
observed, and a body density of d kg/m3 is
assumed, then this method will give an estimate of
the dinosaur’s body mass (in kg) as
To get the mass of the dinosaur from its volume,
we must estimate its density. Since most animal
have about the same density as water (because
they either just float in water, with very little body
parts above water surface, or they just sink), we
may assume dinosaurs had a density equal to that
of water, which is 1,000 kg/m3.
(1) Accuracy of the model, and
Thus, in our case, the mass of the Triceratops is
estimated as 6.08 m3 x (1,000kg/m3) = 6,080 kg.
(2) Accuracy of the estimate on the body density of
dinosaurs.
α3 cd ∕ 1000
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Remark: Accuracy of this method relies on:
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Weighing Dinosaurs with Leg Bone Measurement
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Weighing Dinosaurs with Leg Bone Measurement (cont.)
Method 2: This method requires leg bone
measurements. Fortunately, major leg bones are
often well preserved in fossils.
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Dinosaurs stood and
moved with legs
underneath their bodies,
not splayed out to each
side like modern
reptiles such as
Footprints of (a) a lizard, (b) a bird,
and (c) a mammal.
crocodiles and lizards.
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Anderson et al. (1985) measured the circumferences
of the humerus and femur (upper bones in the fore leg,
and hind leg, respectively) of several dinosaur fossils
and some modern mammals.
They added the two
circumferences
together, and plotted
these against the body
mass of modern
mammals.
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Weighing Dinosaurs with Leg Bone Measurement (cont.)
Weighing Dinosaurs with Leg Bone Measurement (cont.)
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They observed that an equation of
body mass in kg = a (total circumferences in mm) b
where a = 0.000084, b = 2.73, would fit the data of
modern quadrupedal mammals very well.
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They then used the above equation to estimate the
body mass of quadrupedal dinosaurs.
For bipeds, they used the formula
body mass in kg = a (circumference of femur in mm) b
where a = 0.00016, b = 2.73.
Log-log plot of data (quadrupedal mammals) obtained by Anderson et al.
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Estimated body masses of Dinosaurs
Weighing Dinosaurs with Leg Bone Measurement (cont.)
(in tonnes, 1 tonne = 1000 kg)
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1st method
2nd method
(Alexander, 1985)
(Anderson et al., 1985)
Apatosaurus
–
37.5
Brachiosurus
46.6
31.6
Diplodocus
18.5
5.8
Tyrannosaurus
7.4
4.5
Anatosaurus
–
4.0
Triceratops
6.1
–
Iguanodon
5.4
–
Stegosaurus
3.1
–
Dinosaur species
This formula
underestimates
the masses of
kangaroos and
overestimates the
masses of
ostriches.
They used this formula to estimate body masses
of bipedal dinosaurs.
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Masses of Some Modern Mammals
(in tonnes)
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Weighing Dinosaurs Using Scale Models (cont.)
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Here are some data on modern mammals for comparison.
Males
Females
Blue whale
91
110
African elephant
5.5
2.8
Hippopotamus
2.5
2.1
Black rhinoceros
1.2
1.1
African buffalo
0.75
–
Lion
0.18
0.15
Human
0.07
0.05
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Note that the second method consistently
gives smaller estimates than the first method,
and in the case of Diplodocus the discrepancy
is large.
Which of these two methods give more
accurate estimates is debatable.
Although there are uncertainties about the
masses of dinosaurs, from these estimates it
is quite clear that the largest dinosaurs were
exceedingly heavy.
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Trace Fossils of Dinosaurs
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Trace Fossils of Dinosaurs
Apart from body fossils, there are also trace fossils
of dinosaurs, which include footprints, trackways,
or even teeth marks, nests, and droppings.
Footprints of dinosaurs show that dinosaurs had
walked with their feet directly under their body, like
modern birds and mammals, and not splayed out to
the sides like crocodiles.
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Tracks of dinosaur tails are not usually found. This
suggests that dinosaur tails may not have dragged
along the ground. Hence the whole body weight of the
dinosaur was supported by only its legs.
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As large dinosaurs are heavy, one may wonder
whether they got bogged down, or would sunk in
soggy wetland.
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Trace Fossils of Dinosaurs
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Stress Acting on Ground
As we noted in the previous lecture, the body
weight and the body mass are two different
concepts.
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The body weight is a force W, exerted towards the
center of the earth, due to earth’s gravity g acting
on the body mass M. The equation is
W (newton) = M (kg) x g (m/s2)
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The value of g is customarily taken as 9.8 m/s2.
For estimation purposes, one may simply take g =
10 m/s2.
We define stress s as force per unit area.
Thus, when a bipedal dinosaur whose
weight is W N (N stands for newton) is
standing symmetrically, each foot will
support ½ W. The compression stress s
acting over a foot or a cross-section of leg
whose area is ½ A cm2 is
s = (½ W) ∕ (½ A) = W/A (N/cm2)
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Stress Acting on Ground
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Stress Acting on Ground
Theorem 2 A big dinosaur, which was α times
the linear size of a similar shaped small one with
the same body density, would exert α times the
stress on the ground.
Proof (cont.) Since body density kept unchanged,
body weight is proportional to body volume. Hence
W’= α3 W
Proof. Let W, A, s denote the body weight, the total
cross-section area of the feet touching the ground,
and the amount of stress exerted on the ground,
respectively, of the small dinosaur; let W’, A’, s’
be those of the big dinosaur. Now
On the other hand, A’ = α2A. Therefore
s‘ = W’∕A’ = (α3 W)∕(α2A) = αW∕A = α s .
□
(volume of big dinosaur) = α 3 (volume of small
dinosaur)
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Stress Acting on Ground
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Stress Acting on Ground
The strength of ground is measured by its yield stress,
which is the particular value of stress which causes it to
collapse. In other words, the ground will support an
animal (or a building, etc.) standing on it as long as s < y,
where s is the stress the animal exerted on the ground,
and y is the yield stress of the ground.
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y = a Ab, where b = 0,
i.e. y = a, where the constant a depends on the water
content.
Let A be the area over which weight is being applied. It is
found that the yield stress follows an equation
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Y = a Ab,
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For clay, its strength depends on cohesion between
particles, and its yield stress follows the formula
For dry sand, its supporting strength depends on
friction between particles, and its yield stress follows
formula
y = a A1/2
where a and b are constants whose values depend on
the type of the ground.
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Stress Acting on Ground
Stress Acting on Ground
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In general, the stress exerted by an animal of
weight W standing on an area A is W∕A, so the
ground is safe if its yield stress
Proof. Using b = ½, the maximum weight of dinosaur
which can be supported on an area A of sand is
y > W∕A, i.e., if Ay > W.
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Theorem 3 Sand would support a big dinosaur as well
as a small one.
Wmax = aA3/2
On the other hand, the ground is ready to yield if
(†)
If this dinosaur is scaled up by a length factor of α, then
its body weight is scaled up α3 times, i.e., LHS of
equation (†) is scaled up α3 times. On the other hand,
its feet area A will be scaled up α2 times; thus A3/2 will
be scaled up α3 times, which means RHS of equation
(†) is scaled up α3 times, and the equality in (†)
remains unchanged. This means that the larger
dinosaur is still safe.
□
W = Ay = A (a Ab) = aAb+1 = Wmax,
where Wmax is the maximum weight of animal
which can be supported on an area A of ground.
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Stress Acting on Ground
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Stress Acting on Ground
The table below shows stress exerted by different animals on
the ground.
Theorem 4 Clay would not support a big dinosaur as
well as a small one.
Mass
(kg)
Total foot
area A
(m2)
Ground
stress W/A
(N/m2)
Apatosaurus
35,000
1.2
290,000
Tyrannosaurus
7,000
0.6
120,000
Iguanodon
5,000
0.4
125,000
African elephant
4,500
0.6
75,000
Human
70
0.035
20,000
Proof. Using b = 0, the maximum weight of dinosaur
which can be supported on an area A of sand is
Wmax = aA
(‡)
If this dinosaur is scaled up by a length factor of α, then
its body weight is scaled up α3 times, i.e., LHS of
equation (‡) is scaled up α3 times. On the other hand,
its feet area A will be scaled up α2 times; thus RHS of
equation (‡) is scaled up α2 times, and the equality in
(‡) will become “>”, which means that the larger
dinosaur is not safe.
□
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Reference
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Mathematics Masterclasses, M. Sewell (ed.), Oxford
University Press, 1997.
Dynamics of Dinosaurs & Other Extinct Giants, R.M.
Alexander, Columbia University Press, 1989.
WWW Resources
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http://www.journal-for-young-scientists.net/content/view/24/48/
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