G307 Biomathematics - UCLrmjbale/KAU/G307_lectures.pdf · 2010-10-29 · 8. Simple cell...
Transcript of G307 Biomathematics - UCLrmjbale/KAU/G307_lectures.pdf · 2010-10-29 · 8. Simple cell...
University College London
G307 Biomathematics
lecture materials
Stephen Baigent and Alexey Zaikin
2010
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Contents
1 General information and reading list. 1
1.1 Topics to be covered in lectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Reading list . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Using scaling arguments 5
2.1 First steps: Building a simple mathematical model . . . . . . . . . . . . . . . . . . . . 5
2.1.1 Example. Falling flea. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 Scaling arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2.1 Some basic physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2.3 Example: How fast can we walk before breaking into a run? . . . . . . . . . . . 9
2.2.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2.5 Models that involve Metabolic Rate . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2.6 Class Exercise (10mins): How long can a diving mammal stay under water on
one breath? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.7 Example: Why do large birds find it harder to fly? . . . . . . . . . . . . . . . . . 12
2.2.8 How to obtain lift law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2.9 Kleibers Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.2.10 Example. How does heart-rate scale with mass? . . . . . . . . . . . . . . . . . 15
2.2.11 Example: Thickness of fur . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.2.12 Class excercise: How long does it take to starve to death? . . . . . . . . . . . 16
2.2.13 Example: Swimming speed of a filter-feeder . . . . . . . . . . . . . . . . . . . . 16
2.3 Example: Ludwig von Bertalanffys Growth Model (1957) . . . . . . . . . . . . . . . . . 17
2.3.1 Case Study: Incubating Eggs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3 Oxygen transport and Insect respiration 21
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4 Bone 49
4.1 Revision of mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1.1 Linear motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.1.3 Conservation of impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.1.4 Conservation of kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.1.5 Boxing with and without gloves . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.1.6 Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.1.7 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
5 Chemotaxis 55
6 Brain 63
7 Blood 65
8 Bird flight 67
8.1 Basics of bird flight. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
8.2 Basics of lift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
8.3 Energy is required to counter: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
8.3.1 Parasitic drag Dp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.3.2 Induced drag Di . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
8.4 Steady level flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
8.5 Stable gliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
8.5.1 Linear stability analysis of a glide . . . . . . . . . . . . . . . . . . . . . . . . . . 76
8.5.2 Stable glide speeds and angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
8.5.3 Some crude estimates for stable glide speeds. . . . . . . . . . . . . . . . . . . 79
8.6 Soaring flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
8.7 Bounding flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
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Chapter 1
General information and reading list.
1.1 Topics to be covered in lectures
1. Using scaling arguments
2. Oxygen transport and Insect respiration
3. Strength of bones
4. Chemotaxis
5. Brain/Memory
6. Blood flood
7. bird flight
8. Simple cell electrophysiology
1.2 Reading list
Warning. This course is not (much) about learning methods and theorems and applying them tostandard problems. As such, there is no single book that you can read to cover the course. Thereare books that you might find helpful, or enjoy reading to supplement the lectures. A list of books thatgo with the course is the following:
1. Scaling Laws.
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CHAPTER 1. GENERAL INFORMATION AND READING LIST.
(a) Andrew A Biewener. Animal Locomotion. Oxford Animal Biology Series. CUP, 2003.[Good general reading, but particularly pages 10-14. Chapter 7 has an interesting sectionon jump perfprmance].
(b) Knut Schmidt-Nielsen. Scaling: why is animal size so important? CUP, 1984. [Does notbuild models, but is good background information on scaling in biology].
(c) D’arcy Wentworth Thomson. On growth and form. CUP. First published 1961. [Again, nomodel building, but excellent background and a real classic].
(d) Ludwig von Bertalanffy. General Systems Theory. 1969. george Braziller Inc. New York.[He discusses his growth model in pages 171-184].
(e) http://online.itp.ucsb.edu/online/pattern i03/west/ [For general interest,and also von Bertalanffy’s model].
2. Diffusion/Insect Respiration.
(a) Ove Sten-Knudson. Biological Membranes: Theory of transport, potentials and electricimpulses. Cambridge University Press, 2002.
(b) G.R. Grimmet and D.R. Stirzaker. Probability and random processes. Clarendon Press,Oxford. 1992.
(c) http://www.livescience.com/animals/061011 giant insects.html. [Read-able article on oxygen and insect size limitations].
(d) http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/T/Tracheal Breathing.
html
3. Bone.
(a) I.P. Herman, ”Physics of the human body”, Springer, ISBN-10: 3540296034, (2007).
4. Bird flight.
(a) Rayner.J.M.V. (2001). Mathematical modelling of the avian flight power curve, math. Meth.App.Sci., 24:1485–1514.
(b) Loghthill, M.J. (1974). Aerodynamics aspects of animal flight. Bulletin of the Institute ofmathematics and its applications, 10:369 393.
(c) from 1(a) - see above. Chapter 5. Sections 4.1.4.3 may also be useful background readingon fluids.
5. Electrophysiology.
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1.2. READING LIST
(a) see 2(a) above.
(b) J. Keener and J. Sneyd. Mathematical Physiology. Interdisciplinary Applied Mathematics8. Springer-Verlag, New York 1998. [Parts of Chapters 2,3 might be usefull].
6. Chemotaxis.
(a) J.D. Murray, Mathematical biology. I. Chapter 11, Springer 2001.
(b) J.D. Murray, Mathematical biology. II. Chapter 5.
7. Brain.
(a) J.D. Murray, Mathematical biology. I. Chapter 7, Springer 2001.
8. Blood.
(a) S.I. Rubinow, Introduction to Mathematical Biology, A Wiley-Interscience publication, NewYork. Chapter 4.
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CHAPTER 1. GENERAL INFORMATION AND READING LIST.
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Chapter 2
Using scaling arguments
2.1 First steps: Building a simple mathematical model
Warning: In this part of the course we make very simplistic assumptions about the biology. (Never-theless, our efforts are rewarded with answers that make broad sense.)
2.1.1 Example. Falling flea.
Why would a flea survive a fall from 30-storey building, whereas a human would probably not?
Is it:
1. because the human is much heavier?
2. because the flea has a stronger (exo)skeleton and hence can survive the impact?
3. because the fleas legs can absorb the impact (good shock-absorbers)?
4. some other (sensible) reason?
Galileo (or later Newton) tells us two cannon balls of different sizes reach the ground at the sametime - (by experience) this is not what we expect from fleas and humans, so whats missing?
Answer is friction - the drag on bodies due to air friction acts to decelerate a falling body. Overlong distances, bodies reach terminal velocity, which occurs where the frictional drag force balancesweight. So we need to understand how frictional drag depends on the size and shape of a body.
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CHAPTER 2. USING SCALING ARGUMENTS
Figure 2.1:
What is Drag?- It is a force due to friction (air/water/soil)We know:Force = change momentum / time,so we can find the drag force from
1. the bodys area in contact with the air
2. how much air is moved from standstill to (terminal)speed v in a given time t (which gives the momentumchange in time t)
force × time = changemomentum (of still air)force × t = air mass × speed= (density) × volume × v= ρ × (S × h) × v= ρ ×(S × v × t) × v= ρ S v2 thence drag force ∝ S v2
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2.2. SCALING ARGUMENTS
Assuming terminal velocity is reached by the flea and human:
weight = drag force = Mg ∝ Sv2 ⇒ v ∝
√MS
(2.1)
So how doesmass/area = M/S
differ for the flea and human?
Approximately:
Flea = 3mm long
Human = 2000mm long
Make the simplest assumption that there is a linear scale L such that
M ∝ L3, S ∝ L2 (2.2)
Then for each bodyM/S ∝ L (2.3)
Thus the terminal velocity varies with the bodys linear scale L as
v ∝√
L (2.4)
We say that the velocity scales as the square root of the body linear scale
For a flea and human we have (very approximately!)
Lflea/Lhuman ≈ 3mm/2000mm = 0.0015 (2.5)
and the terminal velocity of a human is approx 100mph, so
vflea = vhuman×√
Lflea/Lhuman ≈ 4mph (2.6)
This, combined with the strong exoskeleton of the flea, gives it a much better chance of survival!
2.2 Scaling arguments
The previous example is an example of a scaling argument -by making very simple assumptions wewere able to model how terminal velocity scales with linear scale L.
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CHAPTER 2. USING SCALING ARGUMENTS
The scaling argument summarizes:
weight = Mg ∝ L3 (2.7)
drag ∝ projected area × v2∝ L2v2 (2.8)
weight = drag ⇒ L3∝ L2v2 ⇒ v ∝
√L (2.9)
Notice that we dropped all boring constants to reach the essential point: v scales as square rootof L.
Inherent in our assumptions were that for the linear scale L that distinguishes bodies in the simi-larity class:
• length ∝ L1
• area ∝ L2 (so projected area ∝ L2)
• volume ∝ L3 (so mass ∝ L3)
In applying our model, we were also assuming that the model is being applied to two bodies ofthe same shape (not exactly the case for the flea and human, but this is a first approximation model!).
In the following models, we will assume that we are comparing between families of animals ofsimilar shape, i.e. are isometric, parameterised by linear scale L.
We will be interested in how the size of the animal effects its functions. If L is a length scale,then area scales as L2 and volume scales as L3. Since many of life’s processes depend on trans-port of substances across a surface area (e.g. lung surface), and that transport supplies a volume(e.g. blood), it is intriguing to ask how the fact that volume increases faster than area effects (limits)function.
2.2.1 Some basic physics
(M=mass, L=length, T=time)
1. force = mass × accn : force = MLT−2
2. work done = force × distance = ML2T−2, kinetic energy = M(LT−1)2
3. power = work done/time = ML2T−3
4. flux = amount/area/timee.g. mass flux = mass/area/time = ML−2T−1
heat flux = heat energy/area/time = MT−3
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2.2. SCALING ARGUMENTS
5. heat is energy transferred down a temperature gradient
2.2.2 Example 2
How high can an animal jump? Or more precisely: How does the height that an animal from the samesimilarity class vary with linear scale L?
Assumptions:
1. work done by leg muscles = gain in potential energy
2. muscle force ∝ cross-sectional area of muscle ∝ L2
(this is not obvious, but there are models that justifythis experimentally demonstrated fact).
3. height jumped = height gained h by centre of mass(good approx)
Potential energy gained = Mgh ∝ L3×h
Work done by muscles =
muscle force × vertical distance C.O.M. displaced
∝ (L2)× (L) = L3.
Hence, equating PE gained to Wk. Done by muscle
hL3∝ L3
Thus h ∝ L0. That is, the simple model predicts that, for animals in the same isometric class, theheight they can jump is independent of their size.
Is this a good model? How high can a flea jump?
2.2.3 Example: How fast can we walk before breaking into a run?
Consider this (very) simplified picture of the human gait (figure 2.2). The human walks with straightlegs, so the the COM moves in a series of circular arcs. The front foot leaves the ground if the
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CHAPTER 2. USING SCALING ARGUMENTS
component of weight is not strong enough to provide the centripetal acceleration which increases asv2. This gives us a limit on the walking speed.
10/7/10 17
€
a =v 2
Rcentripetalacceleration
€
v
€
L
€
mg
friction for grip
€
To keep foot on groundduring arc :
mg cosθ > m × v 2
R, so vmax = gR
€
θ
Figure 2.2:
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2.2. SCALING ARGUMENTS
2.2.4 Example
Minimum nerve speed required to make it possible for a animal to balance (e.g. flamingo )
toppling of animal scales like free fall :
∝ t =
√2hg
∝√
L
nerve speed required = distance from brain to muscle / timetaken
∝ L/t
Thusnerve speed scales as L√
L=√
L
2.2.5 Models that involve Metabolic Rate
Metabolism is the set of chemical reactions that occur in living organisms in order to maintain life.Animals use ATP to fuel their metabolic demands, e.g. in growth, locomotion, maintenance, immuno-logical defence, etc. The cells power plants” are organelles called mitochondria which generate mostof the cell’s supply of ATP.
food + oxygen ⇒ ATP⇒ ADP + Phosphate + ENERGY ⇒ muscle force.
The metabolic rate (rate of energy metabolism) of an organism (using aerobic respiration) can beassumed to be equated with the rate of oxygen consumption.
A simple (isometric) scaling argument for variation of metabolic rate (assuming a resting state andafter a period of fasting) with size is as follows:
Metabolic rate B = rate of oxygen consumption ∝ area of lungs supplying oxygen to mitochondria∝ L2
Body Mass M ∝ L3
Thus B ∝ L2 = (L3)2/3 ∝ M2/3 (Rubners Law).
Another argument, at least for warm-blooded animals, put forward by Rubner, is that a warm-blooded animal maintains a constant body temperature, and so their metabolism runs at a rate suchthat this temperature is maintained. Since the body loses heat energy at a rate proportional to theirbody surface area, which scales as L2, the metabolic rate ought to scale as L2.
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CHAPTER 2. USING SCALING ARGUMENTS
Whatever the argument we (might) accept for the L2 law, we take the law as fact for now. (However,we will see later that it can be improved upon with the experimentally determined B ∝ M3/4.)Thus until stated otherwise, we assume that
B ∝ L2∝ M2/3.
2.2.6 Class Exercise (10mins): How long can a diving mammal stay under water onone breath?
A diving mammal (e.g. whale) stores oxygen in it blood before diving. When that oxygen is exhaustedit must surface for more air.
amount of stored oxygen ∝ lung volume ∝ blood volume ∝ L3
Metabolic Rate - rate at which mammal uses stored oxygen is ∝ L2
Thus duration of a dive scales as L3/L2 = L.
Thus the larger you are, the longer you can dive.
NB: We have ignored any specialisation that makes it more efficient for the animal to dive. Thusfor example, whales slow their heart beat and blood flow to their muscles is reduced; these factorsenable whales to dive for longer. When we build our simple models, we keep them simple by ignoringsuch specialisations. We are interested in broad statements about how things typically vary withscale.
2.2.7 Example: Why do large birds find it harder to fly?
Facts/assumptions:
1. Drag ∝ L2v2 (see flea/human model)
2. (Not obvious!) Maximum lift during gliding and wingflapping ∝ Awv2 where Aw = wing area ∝ L2
3. Metabolic rate ∝ L2 = rate at which energy is available.
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2.2. SCALING ARGUMENTS
2.2.8 How to obtain lift law
Bernoulli’s Theorem (Sketch)
For steady flow of inviscid incompressible fluid
ρv2
2+ p+ρgz = const
along streamline. Here p = pressure, z = fluid depth, v = fluid speed, ρ = density.Proof.
F = MA :
ρdvdt
=−ρgk−∇p
(Here d/dt means rate of change following a fluid particle, the so-called material derivative.)
ρ
2d |v|2
dt=−ρgk · v−∇p · v =− d
dt(ρgz+ p)⇒ integrate
(use :ddt
p(x(t)) = ∇p · ddt
x(t) = ∇p · v etc)
Figure 2.3:
Back to wing lift now. Air particles moving around the wing profile start and end at same time, sotop particle must move faster. By Bernoulli, this generates a lift ∝ v2, where v is the wind speed, thatis the speed of the air relative to the wing.
Max lift must just overcome gravity, so minimum flying speed v is given by
Awv2∝ Mg ∝ L3,
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CHAPTER 2. USING SCALING ARGUMENTS
so that since Aw ∝ L2, v ∝ L1/2.
The required power (for flapping wings to get lift) is Power (work done /time = force x distance/time)
Flying power = drag× v ∝ L2v3∝ L2+3/2 = L7/2
Metabolic power ∝ L2, so that required power exceeds supplied power for larger L (L7/2 > L2 for Llarge enough), and hence there is an upper limit on bird size.
2.2.9 Kleibers Law
But experimentally B ∝ M2/3 is not observed!
Figure 2.4:
Instead one finds experimentally (by measuring oxygen consumption of animals in a resting state
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2.2. SCALING ARGUMENTS
and after they have fasted for sufficient period) that
metabolic rate B ∝ m3/4.
Figure 8.12 illustrates the remarkable range of scales over which the 3/4 law holds.Aside: An new argument for 3/4 power law was recently published by West (1997) under the
following assumptions:
• mammalian energy distribution networks (circulatory system, lungs) are fractal-like in structure;
• systems have evolved to maximise their metabolic capacity by maintaining networks that occupya fixed percentage of the volume of the body.
So from now on we will acknowledge experimental data and assume the allometric scaling (asopposed to isometric scaling) law for metabolic rate with mass:
B = B0m3/4. (Klieber’s Law).
2.2.10 Example. How does heart-rate scale with mass?
Assume that the heart beats fast enough to supply enough oxygen for the organism’s metabolism.Facts:
1. Metabolic rate ∝ m3/4.
2. Blood volume ∝ L3 ∝ m.
The rate of oxygen is transport around body is ∝ r×L3 ∝ r×m where r = heart rate (assume pumpvolume ∝ body volume). Thus r×m ∝ m3/4 giving r ∝ m−1/4. Smaller bodies have faster heart rates:
e.g. masked shrew (0.003kg) has r = 600, whereas elephant (4000kg) has r = 30. A human(80kg) has r = 80.
2.2.11 Example: Thickness of fur
Consider a class of similar animals in a cold environment. How does their fur thickness scale withmass?
Recall: Heat is energy transferred down a temperature gradient ∆T/∆x. heat flux = heat en-ergy/area/time = k(∆T/∆x), k = thermal conductivity of material (independent of scale).
To maintain body temperature (in surrounding of constant temperature, so that ∆T is constant) wethus need metabolic rate ∝ heat flux ∝ surface area × (temp difference / fur thickness h)
m3/4∝ L2/h ∝ m2/3/h,
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CHAPTER 2. USING SCALING ARGUMENTS
so that fur thickness h ∝ m2/3−3/4 = m−1/12. Hence larger animals tend to have thinner fur.
2.2.12 Class excercise: How long does it take to starve to death?
Using Kliebers law, power ∝ m3/4
And energy reserves ∝ mass m, energy used up to starvation ∝ power × time to starve = m3/4× t.Thus t ∝ m×m−3/4 = m1/4.
2.2.13 Example: Swimming speed of a filter-feeder
rate of gain of stored energy ∝ food energy input rate - metabolism of stores
food input rate = F0U ,metabolism = basal rate + power to overcome (speed-dependent) drag= P0 + drag ×U
= P0 +(P1U2)×U = P0 +P1U3.
Thus need to look at the functionG(U) = F0U −P0−P1U3.G′(U) = F0−3P1U2 = 0 where U =
√F0/3P1 .
Not viable if Gmax < 0, i.e. (after some algebra) P0 > F0
√F03P1
(1− 1
3P1
).
Figure 2.5:
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2.3. EXAMPLE: LUDWIG VON BERTALANFFYS GROWTH MODEL (1957)
2.3 Example: Ludwig von Bertalanffys Growth Model (1957)
von Bertalanfy was one of the founders of ”General Systems Theory” (http://en.wikipedia.org/wiki/Ludwig von Bertalanffy) Here is a (very) simple model he developed to study growthof an organism.
He assumed that all an organism’s available energy is channeled into :
1. Growth of the organism - building new cells, all taking the same energy to generate
2. Maintenance of the existing cells - keeping existing cells alive by supplying resources and re-moving waste products.
We have:
incoming power (metabolic rate [B]) = number of cells in body Nc(t)× metabolic rate of one cell [= Bc]+ energy required to create new cell [= Ec]× rate of increase in number of cells Nc(t)
⇒ B = NcBc︸︷︷︸maintenance
+EcdNc
dt︸ ︷︷ ︸growth
Now:body mass m = Ncmc, where mc =mass of 1 cell (assumed identical for all cells). Take B = B0m2/3
(isometric scaling, i.e. ∝ L2 [see exercise sheet 1 for the m3/4 case]). Thus
B0m2/3 =Bcmmc
+Ec
mc
dmdt
rearrange ⇒ dmdt
= αm2/3−βm,
where α = mcB0/Ec, β = Bc/Ec.Solve:
dmdt
= αm2/3−βm, m(0) = m0 (small, since organism starts small!).
Write asdmdt
= m2/3(α−βm1/3)
and substitute u = m1/3. Then 3du/dt = m−2/3dm/dt which gives
dudt
=13(α−βu), u(0) = m1/3
0 .
This has general solution u(t) = α
β+Aexp−βt/3. To find A, use initial data:
m1/30 =
α
β+A.
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CHAPTER 2. USING SCALING ARGUMENTS
Hence we obtain:u(t) =
α
β(1− exp−βt/3)+m1/3
0 exp−βt/3,
and finally in terms of m:
m(t) =(
α
β(1− exp−βt/3)+m1/3
0 exp−βt/3)3
.
(Sketch similar in Q2, sheet 1).
2.3.1 Case Study: Incubating Eggs
• An egg is a self-contained unit. It has all the nutrients it needs for the embryo to develop -except for oxygen.
• But must also be rid of waste products, such as carbon dioxide and water. The shell is mainlycalcium carbonate with pores that allow influx and outflux of nutrients and waste products(gases and water).
• The shell must be strong enough to withstand roosting, but weak enough to allow chick to breakout when hatching.
Consider a spherical Egg!
Figure 2.6:
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2.3. EXAMPLE: LUDWIG VON BERTALANFFYS GROWTH MODEL (1957)
Questions:
1. How does the egg incubation time scale with the mass of the egg?
2. How does the shell thickness vary with egg size?
Assumptions/Notes
1. Loss of water is the limiting effect - must not be too rapid else the embryo dehydrates. Oxy-gen and CO2 diffuse across embryo and egg shell faster than water (so can be consideredinstantaneous on the time scale of water movement).
2. Rate of water production ∝ metabolic rate
3. Would expect shell thickness to increase with shell size, since shell has to contain and protectyolk.
4. Total water loss ∝ size of egg
5. Water is lost via pores length = shell thickness d and with total area Apores over the shell.
Total water lost in Tinc = daily water loss ×Tinc
Tinc ∝Megg
daily water loss
and daily water loss ∝ metabolic rate ∝ M3/4egg . Hence
Tinc ∝Megg
M3/4egg
= M1/4egg ,
which agrees quite well with the experimentally observed M0.217.But how might the shell thickness d change with egg size?
water flux ∝ pore area × concentration gradientpore length
(see later lectures on diffusion).
∝density of pores in shell ×area
shell thickness×∆C
Since the area of the shell is proportional to R2, and ∆C is constant,
⇒ M3/4egg ∝
density pores×R2
d∝
density pores×M2/3egg
d
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CHAPTER 2. USING SCALING ARGUMENTS
density poresd
∝ M3/4−2/3egg = M1/12
egg .
Hence if pore density is a constant, independent of egg size,
d ∝ M−1/12egg
which would mean that eggs get thinner with increasing egg size!So the density of pores must be size dependent. In fact, experimentally it is observed pore density
∝ M4/3 and the shell thickness scales as d ∝ M1/2egg .
PS: It is also experimentally established that for birds Megg ∝ M3/4.
20
Chapter 3
Oxygen transport and Insect respiration
21
CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
22
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
24
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
26
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
30
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
32
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
34
35
36
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
38
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
44
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
46
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CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION
48
Chapter 4
Bone
4.1 Revision of mechanics
4.1.1 Linear motion
Suppose a location is described by r(t) or x(t). dxdt is a velocity. A motion of a rigid bidy is then
described by
md2x(t)
dt2 = ∑F, (4.1)
where ∑F is a sum of force’s projection on the axis x.
4.1.2 Example
A mass m hitting a spring
0 x
V0
Figure 4.1: A mass hitting a spring.
is described by the following equation
md2x(t)
dt2 =−kx(t)−δdx(t)
dt(4.2)
with initial conditionsdx(0)
dt= v0, x(0) = 0. (4.3)
49
CHAPTER 4. BONE
4.1.3 Conservation of impulse
For the point an impulse p = mv, where m is the mass and v is the velocity.
The II Newton lawd pdt
= ∑ F (4.4)
orm
dvdt
= ∑ F (4.5)
If ∑ F = 0 then d pdt = 0 and p = 0.
4.1.4 Conservation of kinetic energy
Let A be a mechanical work, i.e. the amount of energy transferred or changed by a force actingthrough a distance. Then
A = ∑ Fdr = ∑ F vdt = mvdv (4.6)
Change of kinetic energy W
dW = A = mvdv = vd p (4.7)
without force d p = 0 and dW = vmdv = 0, so
W = const = mZ
vdv =m2
Zdv2 =
mv2
2. (4.8)
4.1.5 Boxing with and without gloves
Boxing with gloves is inelastic collision.
f
Vfist,ibefore:
after:
V
Figure 4.2: With gloves due to inelastic collision, after the punch the fist moves together with thehead.
50
4.1. REVISION OF MECHANICS
Using the law of the impulse conservation we can write:
marmv f ist,i
2=
marmv f ist, f
2+mheadv f , (4.9)
using v f ist, f = vhead, f = v f and vhead,i = 0.
We getv f =
v f ist,i
1+ 2mheadmarm
≈ 0.236v f ist,i, (4.10)
using approximate values marm = 0.05mb,mhead = 0.081mb, where mb is the mass of the body. Here ’theindexes are ’f” - final, ”i”-initial.
Boxing without gloves is elastic collision:
head,f
Vfist,ibefore:
after:
VVfist,f
Figure 4.3: Without gloves due to elastic collision, after the punch the fist moves in the opposite asthe head.
We need one more conservation law as we get one more variable - the speed of the fist after thepunch. We use conservation of kinetic energy and get:
marmv f ist,i
2=
marmv f ist, f
2+mheadvhead, f (4.11)
marm
2
(v f ist,i
2
)2=
marm
2
(v f ist, f
2
)2+
mhead
2v2
head, f (4.12)
This givesvhead, f =
v f ist, f
1+ mheadmarm
= 0.382v f ist,i, (4.13)
if the same values are used. Notice that this value is 1.6 larger then the one with gloves. Boxingwithout gloves lead to broken knuckles or even can lead to death! Notice also that a good boxer willhave large speed of the fist, and the mass of the head should be much less than the mass of the arm.
51
CHAPTER 4. BONE
4.1.6 Rotation
The torque τ about some axis z is defined as the
τ = r× F , (4.14)
where the distance vector r defines the point where the force F is applied. it follows that τz = rF sin(Θ).if Θ = 0 then τ = 0.
A torque leads to change in the angle φ which defines the position of the body, and in the angularfrequency Ω = dΘ
dt as
τ = IdΘ
dt, (4.15)
where I is a moment of inertia. The moment of inertia around some axis is
I = ∑i
miR2i =
Zρ(r)R2dV, (4.16)
where R is the distance from axis, r distance vector from the axis, ρ is a density, V - volume.
4.1.7 Example
A force F is applied to the surface of a cylinder with height h, radius R and density ρ. Find equationof motion which describes the rotation. The angle Θ = 90, and the cylinder rotates without friction.
Solution: First we calculate the moment of inertia:
I =Z
r2ρdV (4.17)
Consider integration with respect of r from 0 to R (see fig.4.4):
I =Z R
0r2
ρ(π(r+dr)2−πr2)h = ρπhZ R
0r2(r2 +2rdr+(dr)2−r2)= ρπh
Z R
0r22rdr = 2ρπh
Z R
0r3dr = 2ρπh
R4
4=
mR4
4,
(4.18)where m is the cylinder mass. Hence we have as the equation of motion:
r dr
Figure 4.4: Integration to find a moment of inertia.
52
4.1. REVISION OF MECHANICS
FR =mR2
4d2Θ
dt2 ,d2Θ
dt2 =4F
mR3 , (4.19)
where Θ is the angle of the rotation position.Below please find pictures for the handout:
53
CHAPTER 4. BONE
Figure 4.5: For more details please read separately available material
54
Chapter 5
Chemotaxis
For additional reading please refer to separately distributed material. Below please find figures forthe handout.
55
CHAPTER 5. CHEMOTAXIS
56
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CHAPTER 5. CHEMOTAXIS
58
59
CHAPTER 5. CHEMOTAXIS
60
61
CHAPTER 5. CHEMOTAXIS
62
Chapter 6
Brain
For additional reading please refer to separately distributed material.
63
CHAPTER 6. BRAIN
64
Chapter 7
Blood
For additional reading please refer to separately distributed material. Below please find figures forthe handout.
65
CHAPTER 7. BLOOD
66
Chapter 8
Bird flight
8.1 Basics of bird flight.
Figure 8.1:
In planes: wings produce lift. Propeller or jet engine produces thrust. In birds: wings must produceboth thrust and lift.
Birds have evolved so that they
• have light weight skeleton - have smaller porous bones, hollow bones with strengthening struts,skull ≈ 1% body weight
• efficient respiratory system to provide for high metabolic rate required for flight. Anterior airsacs lungs, posterior air sacs.
• eat beries and other high energy foods
67
CHAPTER 8. BIRD FLIGHT
8.2 Basics of lift
The lift force is always perpendicular to the forward motion and the drag is along the line of forwardmotion.
Figure 8.2: Vu > Vl ⇒ pressure difference by Bernoulli ⇒ lift force.
Figure 8.3: Angle of attack.
Why evolve an airfoil shape? Why not just use the angle of attack for lift?Asymmetry of wing
1. produces lift at zero angle of attack
2. produces more lift than any symmetric wing at any angle of attack
3. produces less (pressure) drag.
68
8.2. BASICS OF LIFT
Stalling: large angle of attack leads to large drag.
Figure 8.4: Turbulent flow ⇒ less lift, more drag.
The Kutta-Joukowski theorem is a fundamental theorem of aerodynamics. It states that the lift perunit of the wing length is a product of density of the fluid, velocity at some distance, and the circulationaround the wing.
Figure 8.5: Calculation of the velocity circulation.
L =−ρu×Γ (8.1)
where Γ is a velocity circulation, i.e.Γ =
Iγ
V cosθdθ
69
CHAPTER 8. BIRD FLIGHT
Figure 8.6: Actual flow around wing is the mathematical sum of flow yielding no lift and a flow of purecirculation
As the wing starts from rest in stationary air, a vortex with circulation −Γ appears behind the wingand is matched by a circulation Γ around the wing itself (conservation of vorticity). The circulationcomes off the tips of the wings and leaves as vortices that trail behind the wings. The circulationaround the wings, in the form a bound vortex, gives rise to lift given by equation (8.1). However, theenergy used to create the trailing vortices manifests as “induced” drag on the wing. If there is no liftthen there is no induced drag.
8.3 Energy is required to counter:
• weight
• parasitic drag = frictional drag (body surface drag) + pressure drag (low pressure ”suck” behindwing)
• induced drag ∝ L2, where L2 =lift generated. This is from the energy in the trailing vortices ordisturbances in a large region of air in the wake of the wing
70
8.3. ENERGY IS REQUIRED TO COUNTER:
Figure 8.7: Vortex wake leads to a trailling vortex behind the wing. The vortices dissipate eventually,but may persist for some time.
8.3.1 Parasitic drag Dp.
We assume that the pressure drag is small in comparison to the frictional drag . Dp = rate of transferof momentum from the surface of the bird to the thin mass of air (boundary layer) which is draggedforward with speed u.
Figure 8.8: Boundary layer. Zero velocity on surface ⇒ air speed in a small distance.
frictional drag force =ddt
(momentum of air displaced by wing) ∝ u2
Define dimensionless drag coefficient
Cd =D
12 ρSu2
,
where u = air speed of wing.
Dp =12
CdρSu2,
where ρ is air density and S wing area.
71
CHAPTER 8. BIRD FLIGHT
8.3.2 Induced drag Di
Induced drag = rate of transfer of momentum to the trailing vortices
= Kinetic Energy per unit length in the trailing vortex system.
mass M in vortex region×air speed UT
=MU2
TU=
MU2
d
Also we have
Momentum transferred per unit length to vortices = Lift force L× (1/U)
HenceLU
=momentum of vorticeslength of vortex system
mass per unit length of vortex system ∝ ρ× wing area ∝ ρb2,
where b is wing semi-span. So the induced drag
Di = K. E. per unit length
=1/2(momentum)2/masslength d of vortex system
= 1/2(momentum/length d)2/mass/length d
∝ (L/u)2/(ρ×b2) =L2
ρb2u2 .
Define
Di =KL2
12 ρb2u2
,
where K is some factor for wing shape, etc. Hence total drag
D(u) =12
CdρSu2 +KL2
12 ρb2u2
8.4 Steady level flight
We have L = Mg
drag =KL2
12 ρb2u2
+12
ρSu2Cd (8.2)
72
8.5. STABLE GLIDING
Figure 8.9:
In level flight L = Mg so in equation (8.2)
D =KM2g2
12 ρb2u2
+12
ρSu2Cd
is the drag under these conditions. The bird must produce a thrust T = D to counteract the drag.The rate at which the bird does work to counter the drag, the power=P = thrust × velocity = force×distance
time , i.e.
Power = P = U ×D(U) =KM2g2
12 ρb2u
+12
ρSCdu3.
8.5 Stable gliding
This is where the bird glides in a downward straight line without flapping its wings (at least betweenwing movements that allow it to change direction). Birds with suitable wing characteristics, e.g.buzzards, albatrosses, can travel large distances using this mode of flying. Here the power to counterthe drag and provide lift comes from the gravitational pull on the bird, and we are interested in a glidedownwards at a constant angle θ to the horizontal.
D =12
ρu2SCd +KL2
12 ρu2b2
Write asD = αu2 +β/u2
where α = 12 ρSCd and β = 2KL2
ρb2 . To find Dmin, differentiate w.r.t. u and set the gradient of D to zero:
2αu− 2β
u3 = 0 ⇒ u4 = β/α, umd = (β/α)1/4
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CHAPTER 8. BIRD FLIGHT
Figure 8.10: Forces by gliding. D as a function of u.
Dmin = α
√β
α+β
√α
β= 2√
αβ
So needmgsinθ ≥ 2
√αβ
But at critical angle θmin,
12
mgsinθmin =(
1/2ρu2SCdKL2
1/2ρu2b2
)1/2
=(
SCdKL2
b2
)1/2
12
mgsinθmin = (SCdK)1/2 Lb
=(SCdK)1/2
bmgcosθmin
tanθmin =2b(SCdK)1/2
θmin = tan−1(
2b(SCdK)1/2
)θmin is the minimum glide angle possible.
What is the air speed umd (md=minimum drag) at this minimum angle?
74
8.5. STABLE GLIDING
umd =(
β
α
)1/4
=
(KL2
12 ρb2
/12
ρSCD f
)1/4
=(
4KL2
ρ2b2SCD f
)1/4
=(
4Km2g2 cos2 θmin
ρ2b2SCD f
)1/4
Figure 8.11: But this is not a stable glide.
Figure 8.12:
For steeper angles, θ > θmin, a stable glide is possible. mgsinθ = D(u) has more than one root u∗
(see figure 8.10) but only one of them is stable, namely u > umd . To see why consider how the birdresponds to a small perturbation in its speed when at the higher speed uhigh where D f dominates. Ifthe speed slightly increases, the drag increases and reduces the speed. If the speed decreases, thenthe drag is less than the weight component mgsinθ and so there is an acceleration and the speedincreases back to uhigh. On the other hand, if the speed is ulow where Di dominates, then a slightincrease in speed will decrease the drag, and lead to an acceleration, increased speed and so on, sothat the speed would build up to uhigh where it is stable. If at ulow the speed slightly decreased then itwould continue to decrease until the lift becomes too small and the bird would stall.
To confirm these ideas we may do the linear stability analysis:
75
CHAPTER 8. BIRD FLIGHT
Figure 8.13:
8.5.1 Linear stability analysis of a glide
Consider a small perturbation ξ away from the glide speed u∗: u = u∗+ξ.
mu = mgsinθ−D(u∗+ξ)
m(ξ) = mgsinθ− [D(u∗)+D′(u∗)ξ+ ...]− [(mgsinθ−D(u∗)) = 0(glide)]−D′(u∗)ξ+ ...
mξ =−D′(u∗)ξ, to first order in ξ.
Henceξ =−D′(u∗)
mξ
giving
ξ(t) = ξ(0)exp(−D′(u∗)
mt
)for growth of the perturbation.
Hence the glide is stable only if D′(u∗) > 0, i.e. the bird needs the larger velocity uhigh of the twofor a stable glide at a glide angle θ.
Thus for a stable glide with air speed u, we need D′(u) > 0, which is equivalent to D f > Di, i.e. atspeed u,
u2SCD f >4KL2
ρ2u2b2
b2 >4KL2
ρ2SCD f u4
b >
(KSCD f
)1/2 L12 ρu2S
76
8.5. STABLE GLIDING
Hence at lower speeds, need b larger. That is the bird spreads its wings.
Figure 8.14:
See fig. 8.15
Figure 8.15:
L = mgcosθ, D = mgsinθ. Need D f > Di so see fig 8.15 (middle and right).
8.5.2 Stable glide speeds and angles
For a stable glide, we have the force balance
mgsinθ =KL2
12 ρb2u2
+12
SCdρu2
Hence
mgsinθ =2Km2g2 cos2 θ
ρb2u2 +12
SCdρu2
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CHAPTER 8. BIRD FLIGHT
Figure 8.16:
D(u) =α
u2 +βu2
α
β=
4Km2g2 cos2 θ
ρ2b2SCd
We solve for u:α
u2 +βu2 = γ(= mgsinθ)
βu4− γu2 +α = 0
u2 =γ
2β± 1
2β(γ2−4αβ)1/2
For stability we take the larger root:
u2 =mgsinθ
2 12 ρSCd
+mgsinθ
ρSCd
(1−4cot2 θ
KSCd
b2
)1/2
u2 =mgsinθ
ρSCd
1+(
1−4cot2 θKSCd
b2
)1/2
Notice that since θ > θmin,tanθ > tanθmin =
2b
√KSCd
the speed u is real.
78
8.6. SOARING FLIGHT
8.5.3 Some crude estimates for stable glide speeds.
As we have seen, for stable glide we need D f > Di. This gives D = D f + Di < 2D f ⇒ D f > 12 D and
D = D f +Di ≥ D f > 12 D. Since D = mgsinθ, we then have mgsinθ ≥ D f > 1
2 mgsinθ which gives
mgsinθ ≥ 12
ρu2SC f >12
mgsinθ
√2mgsinθ
ρSC f≥ u >
√mgsinθ
ρSC f(8.3)
which is a crude estimate for the stable glide speed at a glide angle θ > θmin.
8.6 Soaring flight
This is where the bird glides, but takes advantage of upward air currents to maintain or gain theirheight, such as from the windward side of cliffs, or from a thermal). All that is required is that verticalrate of descent is less than the upward component Uair of the rising air:
Uair sinφ ≥U sinθ.
For a seagull to remain stationary with respect to the cliff, then, it needs to adjust its stable glidespeed U and angle θ (see figure 8.17) to satisfy
U sinθ = Uair sinφ
U cosθ = Uair cosφ.
Hence the seagull chooses θ = φ and U = Uair if it is remain stationary. That is, the seagull alignsitself head on into the wind and adjusts its wings so that it remains stationary. The balance betweenthe air speed U , angle θ satisfies equation (8.6), provided that u exceeds the stalling speed of theseagull and the full-span (maximised b) minimum drag speed.
Some birds, such as vultures and buzzards, use rising air currents known as thermals for lift.They soar in circles using the energy in the thermal to provide lift. If the upward air speed is v (soφ = π/2 here) then they can fly in a circle at the same altitude and at speed u when v = usinθ. Fromequation (8.6) this gives that u satisfies√
2mg(v/u)ρSC f
≥ u >
√mg(u/v)
ρSC f
79
CHAPTER 8. BIRD FLIGHT
Figure 8.17: Soaring flight. The bird takes advantage of rising air currents to remain aloft for longperiods of time.
so that (mgvρSCd
) 13
< u ≤(
2mgvρSCd
) 13
.
8.7 Bounding flight
This is the mode of flying used by many small birds that do not have the wing area or span for glid-ing. The bounding flight consists of periodic change between essentially parabolic projectile motion,where the wings are folded, and flapping in order to regain height. Here we are concerned with theefficiency of flapping fight over pure glides.
What scope is there for energy saving for birds flying at speeds well in excess of the minimumdrag speed umd? At such speeds the induced drag (∝ 1/U2 = energy cost per unit distance to supportweight with lift) is relatively small, but the parasitic drag from the loss of momentum to the boundarylayer around the outstretched wings and body is large. Some small birds have evolved a mode offlying called “bounding flight” where they reduce the drag cost by fold their wings for part of their flightpattern. Thus bounding flight consists of periods where the bird flaps and glides (so the wings areoutstretched) to gain height alternating with periods when their wings are folded (so no lift) and theirflight path is parabolic (projectile motion). Since on average, for level flight, the lift has to balance theweight, if the wings are to be folded for some of the time, the lift has to be enhanced when the wings
80
8.7. BOUNDING FLIGHT
Figure 8.18: Bounding flight. The bird flies with alternating periods of flapping/gliding (wings out-stretched) and free fall (wings folded).
are outstretched. Thus (as suggested by Mr S. B. Furber [unpublished]) if the wings are outstretchedfor a fraction f of the time, then the enhanced lift must be
L =mgf
.
This increases the induced drag when the wings are outstretched, but they are not outstretched forall of the time. So is there any benefit from bounding flight?
For normal flight ( f = 1) let Di be the induced drag and D f = Db + Dw be the frictional drag splitinto body drag Db (which is always present) and Dw the drag from the outstretched wing. The averagedrag is
D = Db + f
(Dw +
K(mg/ f )2
12 ρb2u2
)= Db + f Dw +
Di
f.
How does this drag vary with f ?dDd f
= Dw−Di
f 2 ,
so that D is extremal at f =√
DiDw
which is a minimum. So bounded flight is less costly when the birdflies at speeds such that Di/Dw < 1. Indeed, at the minimum D = Db + 2
√DwDi and comparing with
the drag for normal flight f = 1 we obtain
Db +Dw +Di− (Db +2√
DwDi) = Dw +Di−2√
DwDi > 0,
81
CHAPTER 8. BIRD FLIGHT
by the arithmetic-geometric mean inequality.
82