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When we build equations and models, rather than representing the state of a system explicitly, we often represent state as a combination of variables, and compare these variables rather than states directly. The variables we include in our models have more meaning than just their numerical values. There is extra information attached that reminds us of what the variables mean and how their value should be interpreted. You'll find the most common examples of this when our model variables represent measurements. If we measure a football field, we don't say it is 100 long; we say a football field is 100 yards long. Similarly, the speed of light is \(3 \times 10^8\) meters per second. In physical chemistry, rather than describing the motion of every individual particle making up the gas, we might represent the state of a gas with numbers representing its temperature in degrees Celsius and pressure in Pascals.
In each of these cases, we state the number along with the its units (yards, meters per second, degrees Celsius, Pascals). The units of a measurement communicate both the measuringconvention employed and the underlying dimension of the thing being measured (length, speed, temperature). A dimension is an interesting numerical way or direction in which an object could differ from similar objects. This concept of "dimension" is closely related to the idea of a basis vector in linear algebra that can be use to represent a point on a line in space. We commonly distinguish between two different kinds of dimensions  fundamental dimensions and derived dimensions. There are many kinds of measurements with what we commonly consider fundamental dimensions. For example,
There are many other measurements that express quantities whose dimensions are derived in some way from these fundamental dimensions. For example,
Over history, our different cultures have accumulated many different units of measure for length (chain, foot, kilometer, angstrom,...), area (square footage, acres, hectares, ...), weight (talent, pound, Newton, ton,...), and money (drachma, dollar, euro, yen, ...). Units constrain how we use variables, similar to how a typesystem constrains variable use in some computer programming languages like C. It does not make sense to add together variables with different units  the number that comes out won't have any natural meaning. If the variables have the same dimensions, then we convert the units of the stray variables into the units we want, and then we can proceed with the addition. But if the variables have different dimensions, then they can not be converted to a common measurement, and so, the result of the sum won't have units that we can interpret.
The very simple constraints that systems of units place on our calculations have shockingly broad implications for the construction of model equations. In some cases, we can derive a nearly complete model based just on unit conversions and reasoning about the dimensions of the variables involved. This process is generally referred to as "dimensional analysis". Over the remainder of this chapter, we'll work through some preliminary examples of dimensional analysis, then lay out the Buckingham \(\Pi\)Theorem which encapsulates the core ideas we'll have found, and then conclude with a classic example application of dimensional analysis to the Trinity nuclear test in 1945.
Suppose we need to hang a cable between two telephone poles and we want to make sure the cable won't come down too close to garage roof. The more slack is in the cable, the farther down it will hang, but we cannot tell howfar down because the cable hangs with a curve. But we can use dimensional analysis to find a relation between the lowest point of a cable hanging between two anchor points of equal height. Call the depth \(d\) the height from the top of the anchor points to the lowest point on the cable. Well, \(d\) might depend on the length of the cable \(s\), the distance \(w\) between the anchors, the density of the cable \(\rho\), and the acceleration of gravity \(g\) through a general function relationship \[ f( d, w, s, \rho, g) = 0.\] The dimensions of a variables will be denoted square brackets [], and the fundamental units will be represented by letters (L)ength, (M)ass, and (T)ime. So,
Now, let's try to analyze this equation. It is a very general formula – it isn't even solved for one of the variables – but the variables do have units, and that helps. Maybe we can convert the units of these variables to something convenient. We can use conversion factors \(\lambda _ M\) for units of mass, \(\lambda _ L\) for units of length, and \(\lambda _ T\) for units of time. Then, whatever conversion factors we pick, \[ f( d \lambda _ L, w \lambda _ L, s \lambda _ L, \rho \lambda _ M \lambda _ L^{1}, g \lambda _ L \lambda _ T^{2}) = 0 \] Close inspection shows that time and mass each only appear once in this equation. With some thought, taking \(\lambda _ L = 1/s\) , \(\lambda _ M = {1}/{\rho s}\) , \(\lambda _ T = \sqrt{g/s}\), will lead to \[f\left( \frac{d}{s}, \frac{w}{s}, 1, 1, 1 \right) = 0.\] So, the cables depth \(d\) will only depend on the cable's length \(s\) and the distance between the anchors \(w\). The density \(\rho\) of the cable and the strength of gravity \(g\) have no effect. And this makes some sense  heavy chains and pieces of thread all seem to hang the same way in common experience. The other feature here is that the aspect ratios \(d/s\) and \(w/s\) are dimensionless numbers  for a given physical problem, these same aspect ratios will give the same number, no matter what units we choose for our measurements!
Solving for \(d\), and using a new function \(\phi()\) to express the unknown dependence, \[ d = s \phi\left(\frac{w}{s} \right).\]
If \(s = w\), the cable's length would equal the distance between the poles, and the cable would have to be perfectly straight, so we should have \(d=0\), since the cable must be at least long enough to stretch between the two anchors. On the other hand, when \(w=0\), we expect \(d = s/2\). These imply \(\phi(0) = 1/2\), \(\phi(1) = 0\), and \(\phi()\) is a monotone decreasing function in between.
Note: We could have normalized by \(w\) instead of \(s\), in which case we would arrive at the alternative form of the dimensionless relationship \(d = w \psi\left(\frac{s}{w} \right)\). However, this form is indeterminate when \(w=0\), since we would get in that case \(0 \times \psi(\infty)\). Our version above avoids this issue since for all physically meaningful versions of the problem, the cable length \(s\) will always be positive.
Suppose we hang a lead sinker from the end of a long thread to make a pendulum, pull it to one side, and let it go so that it swings back and forth with a steady period. Can we predict the period (\(\tau\)) of the swing? There are a number of variables that we may want to measure to try to predict the period  the length of the pendulum (\(r\)), the mass of the sinker (\(m\)), the acceleration from gravity (\(g\)), and the initial angle (\(\theta _ 0\)). But the relationship between these variables and the period is unknown, so the best we can do atfirst is to write a general functional relationship \[\tau = f(r,g,m,\theta _ 0),\] where \(f()\) is an unknown function. To determine \(f()\), it looks like we need to explore 4 dimensions. However, we can greatly simplify this need for exploration based on consideration of the units of our variables.
If we change the units of time by \(\lambda _ T\), the units of length by \(\lambda _ L\), and the units of mass by \(\lambda _ M\), our function becomes \[\tau \lambda _ T = f( r \lambda _ L , g \lambda _ L \lambda _ T^{2}, m \lambda _ M, \theta _ 0 )\] Let \(\lambda _ L = 1/r\), \(\lambda _ T = \sqrt{g/r}\), and \(\lambda _ M =1/m\). Then \[\tau \sqrt{\frac{g}{r}} = f(1,1,1,\theta _ 0)\] If we rewrite the function of the right as \(H(\theta _ 0)\) and solve for the period, we find \[\tau = \sqrt{\frac{r}{g}} H(\theta _ 0)\] Note that the period is independent of the mass. But if we observed a swinging pendulum, we can say something about the gravitational accelerations in the place we observe it (like on the moon compared to earth) as long as we know the universal function \(H(\theta _ 0)\).
If we compare the prediction of this formula to some data, we find good agreement with at least one set of observations.
A soap bubble is wondrous object, so light it can float on the air, but strong enough to contract and trap a palmsize pocket of air in a nearly perfect sphere. We call this contraction by the bubble "surface tension". Air pressure from outside the bubble also pushes on it to contract. However, the air inside the bubble resists this contraction. As the bubble gets smaller, the pressure in the bubble must increase until it reaches a point where the difference in pressure from the inside and outside of the bubble balances the contraction forces created by the surface tension.
Without knowing any more, all we can say that the radius, pressure, and surface tension must satisfy an equation \[\chi(r, p, s)=0\] where
Instead of using the standard fundamental dimensions of mass and length, let us use force and length as our fundamental dimensions. And instead of using units of length and Newtons, let's convert our units to a new, more convenient system, where 1 meter becomes \(\lambda_L\) units of length, and 1 Newton becomes \(\lambda_F\) units of force. Then our model for a soap bubble becomes
\[\chi(r \lambda_L, p \lambda_F \lambda_L^{2}, s \lambda_F \lambda_L^{1})=0.\]
Taking \(\lambda_L = 1/r\) and \(\lambda _ F = 1/sr\), we find
\[\chi\left(1, \frac{p r}{s}, 1\right)=0.\]
Our function model which initially seemed to depend on 3 different arguments can now be reinterpreted as an unknown function of a single variable, \(X(pr/s) := \chi(1,pr/s, 1)\). So by substitution, \(X(pr/s) = 0\) and \(pr/s = X^{1}(0)\). If we define the proportionality constant \(C:=X^{1}(0)\), \[\frac{pr}{s} = C.\] Having a formula like this tells us several things already, even without knowing the constant \(C\). For a constant surface tension, then for the product \(pr\) to stay constant, the pressure difference between the inside and outside must decrease if the bubble is enlarged. Since the pop of a bubble depends on the pressure difference \(p\), small bubbles with greater pressure differences will pop more loudly than large bubbles. (Think champaign bubbles frothing vs. giant soap bubbles!) If we consider a latex balloon instead of a bubble, then the situation is different. The surface tension is not a constant, but rather increases as we inflate the balloon, creating a bigger pressure difference until things pop.
So far, we've studied a few problems, each with their own dimensionless groups. The procedures we've invoked can be summarized as follows.
Buckingham \(\Pi\) theorem: If there are \(M\) dimensional parameters involving \(N\) kinds of units, then system has \(MN\) dimensionless groups. For any dimensioned equation \[f(q _ 1,q _ 2, ... ,q _ M) = 0\] where the \(q _ i\)'s are the measurable variables with \(N\) fundamental dimensions, then the equation can be stated as \[F(\Pi_1,\Pi_2, ... ,\Pi _ {MN})=0\] where each \(\Pi _ i\) is a dimensionless group constructed from the \(q\)'s and has the form \[\Pi_i = q_1^{a_{i1}} q_2^{a_{i2}} \ldots q_M^{a _ {i,M}}\] where the exponents \(a _ i\) are rational numbers (they can always be taken to be integers: just raise it to a power to clear denominators).
Consider a ship moving across the surface of the ocean. One of the main sources of drag for the ship is it's wake, which consists of surface waves moving under the influence of gravity, as well as the ship.
Parameters:
If we conveniently arrange our table of variables and units,
r 
g 
v 
\(\mu\) 
\(\rho\) 

length, \(L\) 
1 
1 
1 
1 
3 
time, \(T\) 
0 
2 
1 
1 
0 
mass, \(M\) 
0 
0 
0 
1 
1 
Applying the Buckingham \(\Pi\) theorem, there are 5 variables and 3 fundamental dimensions
So, from the \(\Pi\) theorem, we deduce that the row nullspace must be spanned by two linearly independent basis vectors, each corresponding to a dimensionless group of the model. In reduced roweschelon form (RREF), the dimension matrix \[\left[\begin{matrix} 1 & 1 & 1 &1 & 3 \\ 0 & 2 & 1 &1 & 0 \\ 0 & 0 & 0 & 1 & 1 \end{matrix}\right]\] becomes \[\left[\begin{matrix}1 & 0 & \frac{1}{2} & 0 &  \frac{3}{2}\\0 & 1 & \frac{1}{2} & 0 &  \frac{1}{2}\\0 & 0 & 0 & 1 & 1\end{matrix}\right]\] \[\left[\begin{matrix}2 & 0 & 1 & 0 & 3\\0 & 2 & 1 & 0 & 1\\0 & 0 & 0 & 1 & 1\end{matrix}\right]\]
\[C_1 \begin{bmatrix} 1 \\ 1 \\ 2 \\ 0 \\ 0 \end{bmatrix} + C_2 \begin{bmatrix} 3 \\ 1 \\ 0 \\ 2 \\ 2 \end{bmatrix} \]
from sympy import *
from sympy.abc import *
vs=(r,g,v,mu,rho)
A = Matrix([[1,1,1,1,3],[0,2,1,1,0],[0,0,0,1,1]])
ns = A.nullspace()
F = lambda k : Mul( * [i**j for i,j in zip(vs, ns[k])])**2
print latex((F(0),F(1)))
There are two free variables (columns 3 and 5), which we can use to construct basis vectors of the row nullspace. This leads to an equation \[F\left( \frac{v^{2}}{g r}, \frac{g r^{3}\rho^{2}}{\mu^{2}} \right)=0\] This is a perfectly good dimensionless formula, but intuitively, we find it easier to work with dimensionless groups that have small integer exponents and sparse overlap of variables. If we multiply the second by the first and take a square root, we get the more standard (and slightly simpler) version \[F\left ( \frac{v^{2}}{g r}, \quad \frac{r \rho v}{\mu} \right )=0\] The first dimensionless term is called the Froude number (\(Fr:=\dfrac{v^{2}}{g r}\)) while the second dimensionless term is called the Reynolds number (\(Re:=\dfrac{r \rho v}{\mu}\)).
For a ship in water, \(\rho \approx 10^3\) kg/\(m^3\), \(\mu \approx 10^{3}\) kg/s/m, \(r \approx 10\) m, and \(v \approx 1\) m/s, so the Reynolds number is about \(10^7\), very large. If the Reynolds number is large, we can asymptotically expand around its reciprical being small, and hope to find
\[v \propto \sqrt{gr}\]
Thus, the speed is proportional to the square root of waterline length, and longer boats will be faster that shorter boats, all else equal. Also, ships will be faster on planets will lower gravity, and slower on planets with higher gravity.
On the Froude number: In 1857, Froude was consulted on the performance of the Great Eastern, a huge iron ocean liner that was, in spite of great efforts, was so slow she wasn't worth the cost of fuel to run. He began towing models, first in a creek, then in a constructed tank. He noticed that geometrically similar small and large hulls produced different wave patterns  but when larger hulls were towed at great speeds (making \(Re\) large!), he could find a particular speed at which wave patterns were almost identical. That is, when \(v^2/g\ell\) were the same for the small and large ships! When using scale models to predict what a prototype will do, typically the Froude number is kept constant so the surface wave pattern will remain the same.
(from Bridgeman, 1922) A drop of water falls from a faucet. It oscillates from ovoid to sphere to ovoid as it falls, as the surface tension holding it together fights with the momentum of the water inside it. (The oscilations of big water balloons are similar, though less regular). Can we find scalinglaw equations for this frequency?
Very important note: \(f\) is an dependent variable, much like tension in the hanging cable equation above. But the linear algebra doesn't care... if the matrix approach gives us only one dimensionless variable that includes \(f\) we can derive from that a scaling relationship for \(f\). So for now also include
f 
V 
s 
\(\rho\) 

length, \(L\) 
0 
3 
0 
3 
time, \(T\) 
1 
0 
2 
0 
mass, \(M\) 
0 
0 
1 
1 
1 dimensionless group, by the \(\Pi\) theorem.
from sympy import *
from sympy.abc import *
vs=(f,V,s,rho)
A = Matrix([[0,3,0,3],[1,0,2,0],[0,0,1,1]])
ns = A.nullspace()
F = lambda k : Mul( * [i ** j for i,j in zip(vs, k)])
print latex([ F(i) for i in ns])
This leads to the single group symmetry equation \[\frac{V \rho f^{2}}{s} = C\] Thus, as drops get bigger in volume, their frequency of oscillation decreases.
Alternative approach 1
Just as valid to write \(f \propto V^a\rho^b s^c\) \(\Rightarrow\) \([f]=[V]^a[\rho]^b[s]^c\) \(\Rightarrow\) \(T^{1}=L^{3a}M^bL^{3b}M^cT^{2c}\) which gives \(a=1/2\), \(b=1/2\), \(c=1/2\). Thus \(f=C \sqrt{s/V\rho}\) for some constant \(C\).
Alternative approach 2
Same as 1 but with explicit linear algebra, pulling the dependent variable \(f\) out of the dimensional matrix. Take \(f\propto V^a\rho^b s^c\).
\[\left[\begin{matrix}3&0&3\\0&2&0\\0&1&1\end{matrix}\right] \left[\begin{matrix}a_L\\a_T\\a_M\end{matrix}\right] = \left[\begin{matrix}0\\1\\0\end{matrix}\right]\]
This time, since we've removed \(f\), the dimension of the nullspace is zero, since the matrix is square and invertible! From the \(\Pi\) theorem, we deduce we must have 0 linearly independent dimensionless parameters that combine \(V\), \(s\), \(\rho\); equivalently, the matrix equation has no homogeneous solutions.
Solving this matrix equation we find the particular solution \([a_L,a_T,a_M]^T=[1/2,1/2,1/2]^T\). Therefore,\[f=C \sqrt{\frac{s}{V\rho}}.\]
Trinity site, New Mexico, 1945
These photographs were declassified and published in 1947 in Life Magazine and other news outlets as part of a public relations campaign on nuclear weapons and nuclear power. It was thought that the pictures were benign and would not reveal any more about America's topsecret nuclear weapons program than the public already knew. That was wrong. In 1950, G.I. Taylor, an English fluid dynamicist who had worked independently of the Manhattan project, published a pair of papers (one,two) explaining a 1941 report and using the pictures to calculate the bomb's energy yield.
Trying to recreate Taylor's thought process might go something like this. (It did not, if you look at the above papers.)
Taylor knew his thermodynamics very well. In particular, he knew that the ratio expansion of a gas as it absorbs energy (under slowly changing conditions!) depends on the specific heat of the gas at constant pressure \(C _ p\) and the specific heat of the gas at constant volume \(C _ v\). The specific heat is the change in the internal energy per change in temperature per amount (measure in mass). In the case of constant volume, all of the energy from an increase in temperature must be absorbed through increases in pressure due to acceleration of gas molecules. In the case of constant pressure, an increase in temperature corresponds to an increase in volume, so molecular acceleration is offset by a decrease in density.
So, to calculate the energy of the Trinity bomb test from the pictures in Life, Taylor needed a formula that involved the energy \(E\), the time \(t\), the radius \(R\), the initial atmospheric density \(\rho _ 0\), and the specific heats \(C _ p\) and \(C _ v\), \[f( E, t, R, \rho _ 0, C _ p, C _ v) = 0,\] for some unknown function \(f\). The fireball is moving so fast, the atmospheric density has no chance to change in the short term. The specific heat has units of joules per kilogram per degree Kelvin.
\(E\) 
\(t\) 
\(R\) 
\(\rho _ 0\) 
\(C _ p\) 
\(C _ v\) 

distance 
2 
0 
1 
3 
2 
2 
time 
2 
1 
0 
0 
2 
2 
mass 
1 
0 
0 
1 
0 
0 
temp 
0 
0 
0 
0 
1 
1 
We can save ourselves some work in the rowreduction by rearranging the rows and columns of the matrix ...
t 
R 
E 
\(\rho _ 0\) 
\(C _ p\) 
\(C _ v\) 

time 
1 
0 
2 
0 
2 
2 
distance 
0 
1 
2 
3 
2 
2 
mass 
0 
0 
1 
1 
0 
0 
temp 
0 
0 
0 
0 
1 
1 
Now, by rowreduction,
t 
R 
E 
\(\rho _ 0\) 
\(C _ p\) 
\(C _ v\) 

time 
1 
0 
0 
2 
0 
0 
distance 
0 
1 
0 
5 
0 
0 
mass 
0 
0 
1 
1 
0 
0 
temp 
0 
0 
0 
0 
1 
1 
Taylor was very suspicious of the role of the functional dependence on the adiabatic constant \(\gamma\)  a nuclear bomb explosion does not seem like a situation where things change slowly. And he was unsure which value to use for \(\gamma\). If the the atmosphere was acting like it's usual diatomic self, \(\gamma = 1.4\), but if the explosion was so hot that it split nitrogen and oxygen molecules up into individual atoms, \(\gamma = 1.67\). But he didn't give up and was able to use some small explosive experiments to estimate \(S(\gamma)\approx 1\), and obtained some formulas.
Now, in 1950, he had the pictures from Life magazine with which to work.
[ Data : hide , shown as table , shown as CSV shown as QR ]
# time (milliseconds), radius (meters)
#
# Original data used by GI Taylor to estimate
# the energy of the Trinity test's atomic bomb.
# Data represent the radius of the atmospheric
# shockwave of the bomb explosion at sequential
# time points, and were extracted from timestamped
# and scaled pictures released with the announcement
# of the bomb.
#
0.1,11.1
0.24,19.9
0.38,25.4
0.52,28.2
0.66,31.9
0.80,34.2
0.94,36.3
1.08,38.9
1.22,41.0
1.36,42.8
1.50,44.4
1.65,46.0
1.79,46.9
1.93,48.7
3.26,59.0
3.53,61.1
3.80,62.9
4.07,64.3
4.34,65.6
4.61,67.3
15.0,106.5
25.0,130
34.0,145
53.0,175
62.0,185
From this data, we can determine the yintercept. Although he was somewhat surprised how straight the line really was, G. I. Taylor was then able to solve for energe E, and estimate the explosion size at 16.8 kilotons, with a range from 9.5 to 34 kilotons. With other methods, the strength was estimated to be the same as 20 kilotons of TNT.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 

Notes: (1) plot on the left is from Taylor's 1950 paper. (2) The yaxes are different since the plot on the left measures radius in centimeters, while the plot on the right measures it in meters.
It may seem like nature prefers one system fundamental dimensions (length, time, mass, ...), but there are actually many ways to choose fundamental units. For example, instead of using time, length, and mass as fundamental dimensions, we could use time, length, and force, in which case mass would have derived dimensions of force by time squared per length. In the traditional American system of units, for example, we regularly use pounds (a unit of force) instead of slugs or kilograms (units of mass). And in the theory of quantum mechanics, it is common to use momentum rather than mass as a fundamental unit.
Dimensional analysis is largely viewed as much art as science. While there are relevant mathematical theorems like Buckingham's Pi theorem, there is still an element of arbitrariness in how we do dimensional analysis, and this arbitrariness plays into how we think about a problem  particularly when we start to consider the relative magnitudes of quantities for approximations.
In 1672, as part of a scientific expedition to Cayenne in French Guyana, Jean Richer observed that his Paris clocks were losing 148 seconds each day. Giovani Cassini guess that this lose was a result of weaker gravity. If so, how much stronger was the gravity in Paris?
The Deborah number is a dimensionless number used in rheology to describe the "solidness" of a "fluid". It is the ratio of the "relaxation time" of a fluid to the observation time of interest. If the Deborah number is very large, then a fluid moves very slowly relative to the observation time and acts like a solid. Deborah numbers near one indicate a normal fluid, and really small Debora numbers indicate a "hydrostatic" situation where the fluid moves so quickly, we can assume it is always near equilibrium. Estimate your Debora numbers the following scenarios, and explain your reasoning.
A bicyclist traversing a flat circular race track leans into the turn to keep her bicycle balanced.
Find 4 dimensional variables that together should determine the angle at which she should lean.
Determine the units of each variable, in terms of the fundamental units of length, time, mass, charge, and temperature.
According to the Buckingham \(\Pi\) theorem, how many dimensionless groups can be constructed from your 4 dimensional variables.
Find a general formula for a functional relationship between your variables and the angle of lean, expressed in terms of dimensionless groups.
Which of your initial 4 variables does not actually effect the angle of her lean.
In Principa Mathematica, Isaac Newton was the first to attempt to calculate the speed of sound in air. Let's replicate his result.
Determine three commonly known properties of an ideal gas might effect the speed of sound (\(c\)) in that gas.
Now, use dimensional analysis to come up with a formula for the speed of sound \(c\) as a function of these three values.
(McMahon and Bonner, 1983, p. 7678) When you cook a roast, the cooking time (\(T\)) is defined as the time needed for the center of the roast to reach a predefined temperature. The cooking time actually depends on the thermal conductivity (\(k\)) of the roast, its density (\(\rho\)), the radius of the roast (\(R\)), and the specificheat capacity at constant pressure (\(s_p\)).
Find a dimensionless group relating these five variables.
If we halve the size of the roast, how should the cooking time be changed, assuming everything else stays the same?
The drag force \(D\) on a ship depends on the water density \(\rho\), the water viscosity \(\mu\), the ship length \(L\), and the ship velocity \(v\).
What's the number of dimensionless terms needed to express the functional relationship among these 5 variables?
Find two different formulas formulas relating these 5 variables to each other. One formula should make use of the Reynolds number \(L \rho v / \mu\), while the other formula should make use of the drag coefficient \(D/(\rho L^2 v^2)\).
Nuclear fission occurs as neutrons bounce around atoms of plutonium, shattering some nuclei and releasing more neutrons in a chain reaction. Whether or not a chain reaction will keep going or die out depends on how much bouncing occurs before the neutron escapes from the plutonium or are absorbed. In a simple theory of the chain reaction, the minimum mass of a sphere of uranium needed to sustain a nuclear chain reaction (called the "critical mass") depends on the following three variables.
Use dimensional analysis to find a formula for the critical mass \(m\) as a function of these three variables.
(From Continuum Modeling in the Physical Sciences by E. van Groesen and Jaap Molenaar) Consider a train travelling through a light rain shower, where raindrops accumulate on the window of the train and trail down in diagonal streaks. Use dimensional analysis to find a formula for the speed of the train as a function of the raindrop size and the angle of descent of the drop.
(modified from Schmidt, 1977) Meteorite and asteroid impacts create craters. While these are eroded over time on earth, they are easily seen still on the moon and mars. There are two different hypotheses for what controls the impact crater size, which we derive below.
Find a functional relationship for crater radius \(r\) depending on the weight of the asteroid \(W\) and the density \(\rho\) of the substrate being moved by the explosion.
If the crater radius also depends on the strength of gravity \(g\) (which is different on each planet and moon), as evidence suggests it does for large explosions, find a new relationship for the crater radius.
(From The Art of Approximation in Science and Engineering by Sanjoy Mahajan) When Einstein proposed the theory of relativity, one of his tests of it relied heavily on dimensional analysis. Suppose a small object (comet, photon, derilict spaceship) enters the solar system at high speed and passes within a minimum distance \(r\) from the sun. The sun's gravity bends its path and the object lives the solar system travelling in a different direction than it entered. Let \(\theta\) be the angle between the old path and the new path. This angle depends on the minimum distance \(r\); the smaller the distance, the sharper the turn.
What parameters besides the distance \(r\) does the angle \(\theta\) depend on?
Using dimensional analysis, derive a formula for the angle \(\theta\) in terms of \(r\) and the other parameters.
Under Newton's theory of gravity, the proportionality constant in this formula should be 2 while in Einstein's theory, the constant should be 4. Einstein's value was eventually confirmed using radio astronomy measurements.)
In Principia Mathematica, Isaac Newton discusses the resistance a fluid poses to an object moving through it. These ideas were subsequently applied to the calculation of the lift force created by an inclined plate moving through the air. The lift force \(F\) depends on the density of the fluid \(\rho\), the surface area of the plate \(S\), the velocity of the plate through the fluid \(v\) and the angle of attack of the plate \(\theta\).
Use dimensional analysis to derive an equation for the attack angle \(\theta\) as a function of a dimensionless product of the other variables.
Solve your equation for the lift force \(F\).
Newton believed that when the angle of attack was small, the lift force scaled like the square of the angle. In 1804, George Cayley tested Newton's idea. How does Newton's prediction compare with Cayley's data?
[ Data : hide , shown as table , shown as CSV shown as QR ]
# velocity (ft/sec), angle of attack (deg), lift fource (ounces)
#
# Early data on lift coefficient of a square plate
# moved using a whirlingarm setup
#
# Extracted by Tim Reluga, 201404 from
# Aerodynamics in 1804, the pioneering work of Sir George Caley
# by A. H. Yates, Flight, page 612.
#
# two data sets, the first at 15 feet per second velocity
# and the second at 21.8 feet per second velocity
#
# column 1: velocity of plate
# column 2: angle of attack, in degrees
# column 3: force of lift, in ounces
#
15.0,3.0310,0.1324
15.0,6.0190,0.1620
15.0,8.9630,0.2918
15.0,11.906,0.4542
15.0,14.894,0.6315
15.0,17.882,0.6787
21.8,3.0750,0.1139
21.8,6.0190,0.1752
21.8,9.0070,0.2408
21.8,11.950,0.3179
21.8,14.894,0.4759
21.8,17.838,0.5996
21.8,19.903,0.6222
Suppose a classmate told you that the formula for the time \(t\) a projectile is in the air (ignoring air resistance) is given by the formula \(t = \dfrac{v + \sqrt{v^2 + 2 y}}{g}\) where \(v\) is the initial vertical velocity, \(y\) is the initial height above the ground, and \(g\) is the acceleration of gravity. Based on principles of dimensional analysis, how can you tell this equation must be wrong? Can you guess your classmate's mistake?
A crank has suggested that the height \(h\) of a child can be predicted from the height \(x\) of the mother and the height \(y\) of the father using the formula \[h = \sqrt{ (x+1) (y1) }.\] Use the theory of dimensional analysis to critique this formula.
Find a formula for the volume of an ndimensional sphere in terms of the sphere's radius \(r\).
Buckingham's Pi theorem is related to legendary mathematician Emmy Noether's famous theorem that all conservation laws in physics are actually statements the symmetries of spacetime described using partial differential equations. While recovering Noether's theorem is to ambitious for us here, we can show how the Pi theorem leads to PDE's. Suppose we wish to find a formula for the volume of a 4 dimensional sphere. We expect there to be a formula \(f(r,V) = 0\), where \(r\) is the sphere's radius and \(V\) is the sphere's volume. Let \(\lambda\) be a conversion factor for units of distance.