Order of magnitude, worked slowly

The full tutorial version of lecture 1. Every derivation done rather than sketched, every number traced, and the failures left in.

CCBS · EST 5053 · Fall 2026 · lecture 1 · exposition · the lecturer's model of the artifact each team produces

This is the third of lecture 1's three pages, and the longest. The core is the spine of the ninety-five minutes. The lineage is where the method came from. This page is the one to read afterwards, with a pen, when you actually want to be able to do it.

1How to read this

Slowly, with paper, and out of order if you like.

Parts 2 to 4 are the three moves of the method, one part each, and they are independent enough to read in any order. Part 5 carries the method into cells, which is what the rest of the course is about. Part 6 is practice.

The material is 2025's, rebuilt rather than transcribed. That lecture left two records, a handwritten note and a scribe note taken in the room, and both are on file.¹ They disagree in nine places. Where this page departs from either, it says so where it departs.

Everything numeric here is produced by analysis/oom_numbers.py in the lecture's research folder, which runs with no arguments and prints all ninety-two numbers with each input marked as either a value read from a source or a standard constant.¹⁴ If a number in this page is not in that output, it is a bug and worth telling me about.

2Your first estimate, badly

Before any technique: guess, then find out what your guess was worth.

Here is a question. How many seconds have you been alive?

Do not read on. Answer it in your head, now, and write the number down.

Most people do one of two things. Either they freeze, because they cannot remember how many seconds are in a year, or they produce something within a factor of three, quickly, by chaining round numbers: about 20 years, about 400 days a year, about 25 hours a day, about 4000 seconds an hour. That chain gives $2\times 10^{1} \times 4\times 10^{2} \times 2.5\times 10^{1} \times 4\times 10^{3} = 8\times 10^{8}$, and the true answer for a twenty-year-old is $6.3\times 10^{8}$.

Every rounding in that chain was wrong. 400 is not 365, 25 is not 24, 4000 is not 3600. Each error is 5 to 10 per cent, and there are three of them, and the answer is still right to 25 per cent, which is far inside the factor of ten we were aiming at.

Now compare the two failures. Freezing produced no answer. Rounding produced an answer good to 25 per cent in four seconds. That gap is the entire skill, and everything below is a systematic version of it.

3What "order of magnitude" buys

A factor of ten is not a weak claim. It is usually the claim that settles the question.

Students often find the deliberate imprecision uncomfortable, so it is worth being concrete about what a factor of ten decides.

The pattern is that questions about whether a mechanism is needed at all have answers separated by many orders of magnitude, and questions about which of two similar things is better do not. Order-of-magnitude reasoning answers the first kind completely and the second kind not at all. Knowing which kind you have is the first decision.

4Units as a constraint

A physical law cannot know what units you chose, and that ignorance is worth exploiting.

Start from something obvious. If a relation among physical quantities is true, it stays true when you switch from metres to feet. The law does not know about your ruler.

Now make that useful. Suppose a relation involves a length $L$, a time $T$ and a speed $v$. Change your unit of length so that every length is multiplied by $\lambda$. Then $L \to \lambda L$ and $v \to \lambda v$, while $T$ is untouched. Any relation surviving this must be built from combinations in which the $\lambda$'s cancel, and $vT/L$ is one such combination.

So the real content of a physical law is a relation among dimensionless quantities. Everything else is an artefact of your choice of units. This is the observation that Buckingham turned into a theorem.²

5The Π theorem, worked slowly

The counting, done in full, so you can see where each number in it comes from.

Write your quantities as $Q_{1}, ..., Q_{n}$. Each has dimensions expressible in $k$ fundamental units, which in mechanics are mass M, length L and time T, so $k = 3$. Write the dimensions of $Q_{i}$ as

A product $\Pi = Q_{1}^{p_{1}} \cdot \cdot \cdot Q_{n}^{p_{n}}$ is dimensionless exactly when the exponents of M, L and T each vanish:

That is a homogeneous linear system: three equations in the $n$ unknowns $p_{i}$. Its solution space is a vector space, and if the three equations are independent its dimension is $n - 3$. In general, with $k$ independent units, the dimension is $n - k$.

The proof is linear algebra and the content is a rank computation, which is why the count occasionally surprises you: if two of your quantities have proportional dimension vectors, the matrix drops rank and you get one more group than you expected.

6Three worked Π problems

Small, complete, and increasing in interest. Do these before the blast wave.

(a) The pendulum

Quantities: period $T$, length $L$, gravity $g$, bob mass $m$. So $n = 4$. Dimensions used: M, L, T, so $k = 3$. Therefore $n - k = 1$, one group.

Find it. Mass appears in exactly one quantity, $m$, and nothing else can cancel it, so its exponent must be zero. The period cannot depend on the mass, and we knew that before doing any mechanics. What remains is $T$, $L$, $g$ with dimensions T, L, LT⁻², and the only combination is

The true constant is $2\pi \approx 6.3$, which the theorem cannot supply. Note that it is not of order one, and this is the honest counterexample to the rule of thumb in §5 of the lineage page. Constants involving a full revolution tend to carry a $2\pi$, so when your estimate is out by about six, suspect a missing angle.

(b) Diffusion, which is the one you will use most

Quantities: a time $t$, a distance $x$, a diffusion coefficient $D$ with dimensions L²T⁻¹. So $n = 3$, and only L and T appear, so $k = 2$. One group:

Stop and notice what just happened. Section 14 derives that same result by writing down the diffusion equation, discretising a flux balance, and taking a scaling limit, which is most of a page of work. Here it took two lines and no physics beyond knowing the units of $D$. The most useful relation in this entire course follows from counting.

(c) Drag on a sphere, where the count comes out differently

Quantities: drag force $F$, sphere radius $R$, speed $v$, fluid density $\rho$, viscosity $\eta$. So $n = 5$, $k = 3$, and $n - k = 2$. Two groups, so the theorem does not settle it. The standard choice is

and all the theorem yields is $\Pi_{1} = f(Re)$ for some unknown function. That is genuinely less than an answer, and it is also exactly the right amount of structure: it says every sphere in every fluid at every speed collapses onto one curve, which is why a wind-tunnel model predicts a full-size aircraft. Reducing five variables to one curve is what "physically similar systems" means, and it is the title of Buckingham's paper.

7The blast wave, start to finish

The famous one, with every step shown and the answer checked against the real data.

A large energy $E$ is released at a point in air of density $\rho_{0}$. A shock expands. How does its radius $R$ grow with time $t$?

Step 1: choose the list

This is the step that is physics rather than arithmetic, and the step that can be wrong.

• $E$: it is the only thing driving the motion.
• $t$, $R$: the question.
• $\rho_{0}$: the air must be pushed, and inertia is what resists.

And now the omissions, which need justifying. Ambient pressure is left out because the shock is violent, so $E \gg p_{0}R^{3}$. This is why the answer fails at late times, when the blast weakens to a sound wave. Viscosity is left out because the Reynolds number is enormous. The bomb's own mass is left out once the swept-up air outweighs it, which is Taylor's condition $R > (\rho_{s}/\rho_{a})^{1/3}R_{\mathrm{bomb}}$.

Step 2: count

$n = 4$ (E, t, R, ρ₀), $k = 3$ (M, L, T), so one group.

Step 3: find it

Seek $\Pi = E^{a}t^{b}R^{c}\rho_{0}^{d}$. Dimensions: $[E] = ML^{2}T^{-2}$, $[t] = T$, $[R] = L$, $[\rho_{0}] = ML^{-3}$. Collect:

Set $a = -1$. Then $d = 1$ from M, $b = -2$ from T, and $c = 3d - 2a = 3 + 2 = 5$ from L. So

The whole derivation is solving three linear equations. And it delivers a hard prediction: on a log-log plot of $R$ against $t$, the slope is exactly $2/5$, no matter what the bomb was made of.

Step 4: test it against the actual explosion

Taylor read twenty-five radii and times off declassified photographs of the Trinity test and tabulated them.⁴ Fitting his numbers gives a slope of 0.406 against the predicted 0.400.

Step 5: get the number

Rearranging (7), $E = K(\gamma ) \rho_{0} R^{5}/t^{2}$. Taylor computed $K = 0.856$ for $\gamma = 1.4$ in the first of the two papers, by integrating the similarity solution numerically.³ With with $\rho_{0} = 1.25\times 10^{-3}\,\mathrm{g}/\mathrm{cm}^{3}$ and $R^{5}t^{-2} = 6.67\times 10^{23}$ in CGS:

Taylor published 16,800 tons. The lineage page (§9) explains why comparing that to the radiochemical 21 kilotons requires care, and it is worth reading before you quote either number.⁵

8When counting is not enough

Two failure modes, both common, both avoidable once named.

The list is wrong. Include an irrelevant quantity and you get an extra group and lose the sharp answer. Omit a relevant one and you get a confident falsehood with no warning. There is no formal defence. What helps is the habit of writing the omissions down as in step 1, because an omission you have written down is an assumption you can test, and an omission you have not is a mistake waiting.

Two quantities share dimensions. If your list has both a length $L$ and a length $d$, their ratio $d/L$ is already dimensionless and the theorem tells you nothing about it. That is a real limitation and also a real signal: the ratio is a shape parameter, and the theorem is telling you that shape matters and units cannot decide it.

9Hangzhou, all the way down

The core page gives this in half a page. Here is the version with the objections answered.

Evacuate $10^{7}$ people. By air, one runway at a departure every two minutes and a hundred people a plane gives $7\times 10^{4}$ people a day, so about 140 days. By road, a hundred outbound lanes each carrying $Nv/L$ = $(4)(20)/(40)$ = 2 people a second gives $1.7\times 10^{7}$ a day, so about half a day.

Now the objections, which is where the actual lesson is.

The substitution $L = v\tau$ turns $Nv/L$ into $N/\tau$ and kills the first two objections outright. What survives is that the answer is proportional to the number of lanes and to the occupancy, and those are the two things to go and check. The estimate has told you what to measure.

And then the conclusion that sounds absurd and is not: since $\tau \approx 2\,\mathrm{s}$ for people on foot as well as for drivers, walking evacuates the city at a quarter of the rate of driving, not a twentieth. In an emergency where fuel or vehicles are short, telling people to walk out along the highways is a defensible plan. That is a real policy conclusion from four lines of arithmetic.

10Bombs: terminal velocity

From the 2025 problem set.¹³ A nice case where the bottleneck is a balance of two forces.

An object falls. Gravity pulls with $\rho_{s}gL^{3}$. At high Reynolds number, air resistance pushes back with roughly $\rho_{a}v^{2}L^{2}$, because the object sweeps out a tube of air of cross-section $L^{2}$ and accelerates it to about $v$. Set them equal:

For a person, $\rho_{s}/\rho_{a} \approx 800$ and the transverse size in a stable belly-down fall is $L \approx 0.3\,\mathrm{m}$:

The measured figure for that posture is about 200 km/h. Within 15 per cent, from a force balance with no drag coefficient.

Two more numbers fall out for free. The time to reach terminal velocity is $v/g \approx 5\,\mathrm{s}$ and the distance is $v^{2}/2g \approx 120\,\mathrm{m}$. So a bomb released from 5 km is at terminal velocity long before impact, and if it is small it never approaches free fall at all. Its impact speed would be $v = \sqrt{2gh} = 313\,\mathrm{m}/\mathrm{s}$ only if it stayed in free fall, which is Mach 0.92, so the sound barrier is genuinely in play. Requiring the terminal velocity of steel to exceed that needs $L > v^{2}\rho_{a}/(\rho_{s}g) \approx 1.6\,\mathrm{m}$. Real free-fall bombs are about that size, and the estimate has just explained why.

11Why bombs got smaller

A two-line scaling argument with a strategic conclusion.

Take equation (7) and eliminate time in favour of the expansion speed, which stays near the speed of sound while the destructive phase lasts. From $R \propto (Et^{2}/\rho )^{1/5}$ and $v \approx R/t$:

Sublinear. Doubling the yield does not double the area, it multiplies it by $2^{2/3} = 1.6$. So split the same total energy into $N$ pieces and the total area becomes

Dividing a warhead into a hundred smaller ones covers $100^{1/3} \approx 4.6$ times the area for the same energy. That is the entire physics case for multiple independently targetable warheads, and it is one exponent.

12Jump height, derived properly

The classic result, with the muscle model made explicit so you can see which assumption carries the weight.

Energy at the top of the jump is $E_{g} = mgh$. Energy available is whatever the muscle did. Model the muscle as doing work $w$ per unit volume per contraction, occupying a fraction $f_{m}$ of a body of density $\rho$ and length $L$:

with no $L$ on the right. Put in numbers: $f_{m} \approx 0.15$, $w \approx 10^{5}\,\mathrm{J}/\mathrm{m}^{3}$, $\rho = 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}$:

Haldane's figure for a man, from 1926, is "about five" feet, which is 1.5 m.⁸

Read (15) as a ceiling rather than a prediction. It charges the animal every joule its whole muscle mass can do, and not all of that muscle points upward. A standing human jump lifts the centre of mass about half a metre, a high jumper with a run-up gets most of the rest, and both sit under 1.5 m. So the claim the balance actually supports is that jumpers of every size should reach within a factor of a few of the same height, which is a claim you can go and test.

13The power wall, and the spring

The energy is available at every size. The time in which to deliver it is not, and that is what forces a mechanism.

If $h$ is the same for all jumpers then so is the takeoff speed, $v = \sqrt{2gh} \approx 3\,\mathrm{m}/\mathrm{s}$. The animal accelerates over roughly its own length, so

which is linear in size while $h$ is constant. Now convert to the quantity muscle is actually limited by, which is power per unit volume:

The flea's row is over by three decades, and the size of the excess is not the point: any excess at all settles it. It cannot be a muscle-powered jump. The resolution is a spring: contract slowly against an elastic element, hold with a latch, release fast.

This pattern, load slowly and release through a latch, is now known to be general across small fast biological systems. It is also the reason the estimate was worth making: it did not tell us the flea uses a spring, it told us that the flea must use something other than muscle, which is a question sharp enough to go and answer.

Someone went and answered it. Sutton and Burrows filmed ten hedgehog fleas at 5000 frames per second, and their Table 2 is worth setting beside the estimate line by line.¹⁵

Read the last column downward. The geometry agrees, the two kinematic quantities each differ by about a factor of two, and the power differs by twenty-four. That is not four independent discrepancies. It is one, entering three times.

The one is $v$, and this section is what predicted it. Every row of Table 3 assumed the flea reaches the flat line, so $v = \sqrt{2gh} \approx 3\,\mathrm{m}/\mathrm{s}$. But the flat line is what a jumper reaches when it can spend its stored energy, and the right-hand panel of Figure 2 says the flea cannot. It leaves at $1.3\,\mathrm{m}/\mathrm{s}$ at $39^\circ$, which lifts its centre of mass about $3.4\,\mathrm{cm}$: a factor of twenty under the line, exactly where the time argument says it should be.

The relations themselves come through intact. $\tau = L/v$ is exact: put the measured $1.8\,\mathrm{mm}$ and $1.3\,\mathrm{m}/\mathrm{s}$ in and you get $1.4\,\mathrm{ms}$, the measured value to two figures. Equation (17) then carries the same factor twice, because $P/V = wv/L$ is linear in $v$ and the $\tau$ in the denominator carries another. A scaling argument that gets its relations right and one input wrong produces exactly this signature, which is why you propagate the uncertainty and not only the value.

14Heat: deriving t = x²/κ

The long way, so that you know what the short way in §6(b) was standing on.

Two physical inputs. First, heat flows down a temperature gradient in proportion to it, which is Fourier's law with conductivity $K$:

Second, storing heat raises temperature in proportion to the heat capacity per volume $C_{v}$.

Take a slab from $x$ to $x + \delta x$, of unit cross-section. Heat in minus heat out over time $\delta t$ equals the energy stored:

Divide by $\delta x \delta t$, and let both go to zero. The left side becomes $-\partial J/\partial x$, and with (18):

Two material constants have become one, and $\kappa$ has dimensions L²T⁻¹. Now the scaling step: over one characteristic time $t$ and one characteristic length $x$, replace $\partial /\partial t$ by $1/t$ and $\partial^{2}/\partial x^{2}$ by $1/x^{2}$. The temperature cancels:

Which is (4), obtained in §6(b) by counting units. The long derivation was not wasted: it produced the equation itself, the identification $\kappa = K/C_{v}$, and therefore a way to compute $\kappa$ from tabulated properties instead of measuring it. That is exactly what catches the error in §15.

15The potato, and how to check it

Move 3 fixes the one constant in $t \approx x^{2}/\kappa$ from something you have cooked. The same line then reaches the moon, and the routine that keeps it honest is the most transferable thing on this page.

Equation (21) contains one material constant and nothing else, so the whole method now rests on getting $\kappa$. Move 3 says to fix it from personal experience rather than from a table. Bake a potato. Two numbers, both of which you already have: a whole potato is about 10 cm across and the heat has to reach the middle, so the distance is a few centimetres, call it 3 cm; an oven takes more than an hour and less than an afternoon, so call it two. Run (21) backwards:

Now check it against an independent route to the same number, because move 3 is the only one of the three that can betray you quietly. Section 14 gave $\kappa = K/(\rho c_{p})$, and a potato is mostly water, whose three constants are all tabulated:

Fifteen per cent apart. The agreement is better than the estimate deserves, and what earns your trust is not the agreement but the fact that it does not depend on the guesses being good. Push both inputs to the edge of anything a person would claim, 3 to 5 cm and one to three hours, and $\kappa$ runs from $8\times 10^{-8}$ to $7\times 10^{-7}$: a box a factor of eight wide with water's value inside it. An anchor is not a claim to know the number. It is a claim to bound it.

The relation reads forwards as well as backwards. With water's $\kappa$ and 3 cm to the centre, (21) gives 6300 s, about 105 minutes, which is a baked potato. And the check is not idle, because the same experience supports a second anchor that is also right: chopped potato in boiling water travels about a centimetre and is done in ten minutes, giving $(0.01)^{2}/600 = 1.7\times 10^{-7}$. Two different kitchen memories, two independent anchors, both within twenty per cent of the tabulated value.

The moon

Now spend the constant somewhere nobody has been, which is what an anchored scaling law is for. Rock conducts about seven times better than water:

The moon's radius is 1737 km, and nothing else about the moon is needed. Equation (21) gives the time for its centre to feel its surface:

Twenty times the age of the solar system, so the question was the wrong question: conduction has not cooled the moon and is nowhere near finishing. The answerable question is how far it has got, and (21) inverted answers it:

The outer 22 per cent of the radius, with the interior yet to hear about the surface at all. That is a checkable statement about a body nobody has touched, it took two lines, and the constant in it came from dinner.

16Life at low Reynolds number

From here on the object is a micron across, so the question is which forces run that scale. One dimensionless number answers it, and the answer deletes inertia and most of your physical intuition with it.

The first thing to establish about an unfamiliar scale is which forces dominate it. Two act on anything moving through a fluid, they scale differently with size, and so their ratio is a property of the scale rather than of the object. Estimate both with the move from section 14: replace every derivative by its variation over one characteristic scale. Take a body of size $L$ at speed $v$ in a fluid of density $\rho$ and viscosity $\eta$.

Inertia is mass times acceleration. The mass is $\rho L^{3}$, and the body changes its own velocity by about $v$ in the time $L/v$ it takes to move its own length, so the acceleration is $v/(L/v) = v^{2}/L$:

Viscosity is a stress acting on a surface. The stress is $\eta$ times a velocity gradient, the gradient is about $v/L$, and the surface is about $L^{2}$:

Divide, and everything about the fluid and the body collapses into a single dimensionless group:

Section 5's counting says that group is the only one there is: four quantities, three dimensions, one dimensionless product. So dimensional analysis guarantees that some function of $\rho vL/\eta$ is the whole answer, and the force estimate tells you what the number physically is. Do both. The theorem gives uniqueness and the estimate gives meaning, and neither is a substitute for the other.

Note the exponents. $Re$ carries one power of $L$ and one of $v$, and a small swimmer is small in both, so the ratio does not decline gently as you shrink. It collapses.

For a swimming bacterium, $\rho = 10^{3}$ kg/m³, $v = 30\,\mu\mathrm{m}/\mathrm{s}$, $L = 1\,\mu\mathrm{m}$, $\eta = 10^{-3}$ Pa·s:

Purcell's own statement is "about $10^{-4}$ or $10^{-5}$. For these animals inertia is totally irrelevant. We know that F = ma, but they could scarcely care less."⁹

Making the number physical

A number is not an intuition. Here is the conversion. Stop pushing the bacterium and ask how far it coasts. Its momentum is $mv$ and Stokes drag is $6\pi \eta av$, so the stopping time is $m/(6\pi \eta a)$ and the distance is that times $v$:

Purcell gets 0.1 Å, and the factor of three between us is what an order-of-magnitude calculation is for. Either way: a bacterium that stops swimming stops within a fraction of the width of an atom. There is no coasting, no momentum, no gliding. If it is moving, something is pushing it right now.

The structural consequence is the scallop theorem. With inertia gone, the flow depends only on the sequence of shapes, not on the rate, so any stroke that retraces itself nets zero displacement however violently performed. Escaping it needs a stroke that traverses a loop in shape space, which is why bacteria rotate helical flagella.

17Diffusion, and why it decides

Equation (4) again, now with the numbers that run the rest of the course.

The diffusion coefficient for a sphere of radius $a$ follows from balancing thermal energy against Stokes drag, which is the Stokes-Einstein relation:

The $1/a$ is the honest version of the lecture's "D scales as one over size". For a 30 kDa protein, $a \approx 2$ nm:

Now put it in (4) and read off the whole course.

That table is one of the most useful things in the course. It says a bacterium gets mixing free and a neuron cannot have it at any price, and the entire apparatus of kinesin and dynein is the consequence of an exponent of two.

18The null model, and its limit

In biology the estimate is usually not the answer, and that is the point. Except when it is.

The standard use is as a null hypothesis. Compute what plain physics does with no regulation, compare, and interpret the gap. Table 5 has both of the common outcomes in it: the axon row is "physics is far too slow, so a mechanism must exist", and the E. coli row is "physics is fast enough, so stop modelling it".

The third outcome is the one that complicates the framing, and it deserves its own statement. Berg and Purcell asked how precisely a bacterium could possibly measure a chemical concentration, given that molecules arrive by diffusion and arrival is a counting process with its own noise.¹⁰ They derived a bound that no receptor design can beat, then compared it with what bacteria achieve in chemotaxis, and found them close to it.

There is no gap. Nothing is left to attribute to design, because physics already forbids doing better. That is a stronger result than any gap would have been, and it changes the question from "why is the cell built this way" to "the cell is built the only way that works".

19The checklist

What to actually do, in order, when you meet a quantitative question.

Two book-length versions of this routine exist, and either is worth owning. Goldreich, Mahajan and Phinney's Caltech notes organise it by method rather than by subject, so a single chapter runs from astrophysics to biology.⁶ Mahajan's Street-Fighting Mathematics makes the approximations themselves the subject and names the moment each one becomes safe.⁷ The lineage says where both came from.

20Ten questions

Three minutes each, a factor of ten, and the routine above every time.

1. How many ATP molecules are in one E. coli at any instant?
2. How long does it take to transcribe an average gene?
3. What is the total mass of RNA in a human cell, compared with protein?
4. How many ribosomes does a bacterium need to double in twenty minutes?
5. How far apart are two molecules of a protein present at 1 µM?
6. How many times a second does a given enzyme meet its substrate at 1 mM?
7. What fraction of a cell's energy goes into making protein?
8. How thick is the cell membrane, from the length of a lipid?
9. How many cells are there in your body, and how do you know?
10. How much would the answer to (9) change if you were wrong about cell size by a factor of two?

Estimate all ten before looking anything up. The place to check afterwards is BioNumbers, which Milo and colleagues built for exactly this,¹² and the argument for why a biologist should carry such numbers at all is Phillips and Milo's.¹¹

Question 10 is not a filler. Propagating an uncertainty through an estimate is the step that turns a number into a claim, and it is the one most often left out.

21Eight things worth extending

Every one of these is a real question with a reachable answer. None is a reading exercise. Each names where to start and what a finished answer looks like.

An extension mini-essay is short, so pick one and go all the way down rather than surveying three. The mark of a good one is a number at the end that was not in the lecture, with an honest statement of what it rests on.

1. Where the jumping peak actually sits. Ilton et al. argue that shrinking stops helping once the spring itself becomes the limit, but they do not give the threshold for any one lineage.¹⁶ Take resilin at $E \approx 1\,\mathrm{MPa}$ and chitinous cuticle at $1-40\,\mathrm{GPa}$, the flea's measured $0.6\,\mu\mathrm{J}$ over $1.4\,\mathrm{ms}$, and predict the body mass at which take-off speed peaks. Check it against froghoppers at $5-6\,\mathrm{m}/\mathrm{s}$ and $20-50\,\mathrm{mg}$. A finished answer is one curve, one predicted mass, and a sentence saying which input you trust least.
2. Galileo's bone, against real bones. The square-cube law says a supporting bone's cross-section must grow like $L^{3}$, so its diameter like $L^{1.5}$. Go and find what measured limb bones actually do across a few decades of mammal mass. If the exponent is not 1.5, then either the argument has a hidden assumption or the animals are compensating some other way, and both possibilities have a literature. This is the course's founding example, and checking whether its prediction holds is more useful than repeating it.
3. Two thirds against three quarters. A pure heat-loss argument gives metabolic rate $\propto M^{2/3}$. The number usually quoted is $3/4$, and the gap has been argued over for eighty years. Get one published dataset, fit the exponent yourself, and put an honest interval on it. The interesting question is not which value is right. It is whether the data can distinguish the two at all, which is a question about error bars and about the range of masses you need.
4. The constant the theorem withholds. Π gives the form and never the prefactor, and §7 gets away with it because Taylor's constant sits near 1. Find a published dimensional-analysis result whose dimensionless prefactor is nowhere near 1, and explain what physical fact is hiding inside it. Where a large constant comes from is usually a geometry or a small parameter that the dimension count could not see.
5. Is the 0.1 s wall a wall? The whole jump argument turns on "fast muscle needs roughly 0.1 s to shorten by half its length." Find where that number comes from and how much it varies. Insect asynchronous flight muscle oscillates at hundreds of hertz, which looks like a counterexample until you ask what is actually shortening. A finished answer says whether §13 is describing a wall or a slope.
6. A third correction. Two numbers on these pages were wrong and were caught by going to the source: the potato anchor, and Haldane's flea. There will be others. Find one, in these three pages or in the 2025 notes, say what the number should be, and say how you know. This is the most valuable essay on the list and the one most likely to be publishable inside the course.
7. Dimensional analysis with more dimensions. Huntley's refinement treats length along a direction as dimensionally distinct from length across it, which raises $k$ in the $n - k$ count and leaves fewer Π groups. Apply it to one problem from §5 or §12. Does it sharpen the answer, or does it quietly smuggle in physics you have not justified? Both outcomes make a good essay, and the second is the more instructive.
8. Estimate something nobody has measured. The point of the method is not to re-derive known numbers. Pick a quantity in your own research area with no published value, estimate it two independent ways, and compare. If the two agree to a factor of three you have learned something. If they disagree by a hundred, you have learned more, and the essay is about which input carries the discrepancy.

†References

Items marked PDF have their full text in the course's literature/ archive. The lineage page discusses most of them in context.