# Dimensional analysis

( previous, home, next )

(3 days of lecture)

Topics

• Basics of dimensional analysis
• Soap bubbles
• Pendulum period
• Nuclear explosions

## Introduction

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 measuring-convention 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,

• length
• angle
• time
• mass
• charge
• temperature
• money
• people

There are many other measurements that express quantities whose dimensions are derived in some way from these fundamental dimensions. For example,

• area has dimensions of square length
• electrical current has dimensions of charge per time
• force has dimensions of mass by length per square time

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 type-system 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 lecture, 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.

## Preliminary examples

### Scope of the hanging cable

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 how-far 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,

• $$[d] = L^1 M^0 T^0$$,
• $$[w] = L^1 M^0 T^0$$,
• $$[s] = L^1 M^0 T^0$$,
• $$[\rho] = L^{-1} M^1 T^0$$,
• $$[g] = L^{1} M^0 T^{-2}$$,

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 ratios $$d/s$$ and $$w/s$$ are dimensionless numbers – for a given physical problem, these 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.

### Pendulum period

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 at-first 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.

### Pressure in a soap bubble

A soap bubble is wondrous object, so light it can float on the air, but strong enough to contract and trap a palm-size 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

• $$r$$ is the radius of the bubble, $$[r] = L^1 T^0 M^0$$.
• $$p$$ is the pressure difference inside to outside, in Newtons per square meter, $$[p] = L^{-1} T^{-2} M^1$$.
• and $$s$$ is the surface tension, in Newtons per meter, $$[s] = L^0 T^{-2} M^1$$.

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 re-interpreted 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.

### Logistic Growth

Dimensional analysis does not just apply in the abstract, but can also be applied to specific systems. For example, suppose the growth of an E. coli population’s size $$N(t)$$ is described by the logistic equation $\frac{dN}{dt} = r N \left( 1 - \frac{N}{K} \right).$ Then the population’s size at a specific time $$t$$ depends on the initial population size $$N _ 0$$, the time $$t$$, the growth rate $$r$$, and the carrying capacity $$K$$. In abstract function form, $0 = \Gamma(N, N _ 0, t,r,K).$ Using $$\lambda _ N$$ for the units of population size and $$\lambda _ t$$ for the units of time, $0 = \Gamma(N \lambda _ N, N _ 0 \lambda _ N, t \lambda _ t , r \lambda _ t ^{-1} ,K \lambda _ N).$ And so, if we choose to measure population size in fractions of the carrying capacity ($$\lambda _ N = 1/K$$) and to measure time in units of growth rate (_ t = r$), then $0 = \Gamma\left(\frac{N}{K}, \frac{N _ 0}{K}, r t, 1, 1 \right).$ Thus, there are 3 dimensionless groups in logistic growth: the current population size relative to carrying capacity, the initial population size relative to the carrying capacity, and growth-rate scaled time. This suggests we consider our differential equation under the change of variables $$t = \tau / r$$ and $$N = K u$$. Under these substitutions, the logistic equation becomes $\frac{K du}{(1/r) d \tau} = r (K u) \left( 1 - \frac{(K u)}{K} \right).$ After simplifying, we are left with simplified equation $\frac{du}{d \tau} = u \left( 1 - u \right).$ Thus, to understand the logistic equation, we don’t have to solve it for all different values of the parameters $$r$$ and $$K$$. Instead, it will suffice to solve the logistic equation when $$r=1$$ and $$K=1$$ and to use that solution. By setting the initial condition $$u(0) = N(0)/K$$ and integrating, the size of the population at time $$t$$ will be $$N(t) = K u(r t)$$. ## Dimensionless symmetry groups So far, we’ve studied a few problems, each with their own dimensionless groups. The procedures we’ve invoked can be summarized as follows. 1. Fundamental units are chosen 2. The units of all variables are expressed in terms of a product of powers of the fundamental units 3. Hypothesis: If the world is a continuum, its laws should not depend on the units with which we describe it. 4. By picking convenient units, we can exploit unit symmetry to arrive at general symmetry forms for our laws. Buckingham $$\Pi$$ theorem: If there are $$M$$ dimensional parameters involving $$N$$ kinds of units, then system has $$M-N$$ 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 $F(\Pi_1,\Pi_2, ... ,\Pi _ {M-N})=0$ 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). ### Linear algebra approach to construction of dimensionless groups #### Ship-speed example 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: • $$L$$ = ship length • $$v$$ = velocity has dimensions of length per time • $$g$$ = gravity’s acceleration has dimensions of length per time per time • $$\rho$$ = density of water has units of mass per length cubed • $$\mu$$ = viscosity has dimensions of mass per length per time. 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 • The dimensions of the row space = 3 • The dimensions of the row nullspace = 2 • The dimensions of the col space = 3 • The dimensions of the col nullspace = 0 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 row-eschelon 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 $\left[\begin{matrix}1 & 0 & \frac{1}{2} & 0 & - \frac{3}{2}\\0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2}\\0 & 0 & 0 & 1 & 1\end{matrix}\right]$ [Show code] 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 $F\left( \frac{v^{2}}{g r}, \frac{g r^{3}\rho^{2}}{\mu^{2}} \right)=0$ 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 water-line 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 behavior of the Great Eastern, a huge iron ocean liner that was, in spite of great efforts, just too big for her power: she 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. For more details see On Size and Life" (1983) by T.A. McMahon and J.T. Bonner. ### Frequency of oscillations in a water drop (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 scaling-law equations for this frequency? • $$f$$ is frequency of oscillations • $$\rho$$ is the mass density • $$V$$ is volume of the drop • $$s$$ is the surface tension 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 • $$f$$ is frequency of oscillations.  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. [Show code] 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 $$f\propto V^a\rho^b s^c$$. • $$\rho$$ is the mass density, $$[\rho]=M/L^3$$ • $$V$$ is volume of the drop, $$[V]=L^3$$ • $$s$$ is the surface tension, $$[s]=M/T^2$$ • … combined to recover $$[f]=1/T$$ $\left[\begin{matrix}3&0&-3\\0&-2&0\\0&1&1\end{matrix}\right] \left[\begin{matrix}a_L\\a_T\\a_M\end{matrix}\right] = \left[\begin{matrix}0\\-1\\0\end{matrix}\right]$ 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,$f=C \sqrt{\frac{s}{V\rho}}.$ ## GI Taylor’s analysis of shock-wave speeds Trinity site, New Mexico, 1945 This photographs were 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 top-secret nuclear weapons program than the public already knew. That was wrong. In 1950, G.I. Taylor, an English fluid dynamicist who had not worked on the American project, published papers (one,two) using the pictures to predict energy yield. Trying to recreate Taylor’s thought process might go something like this. 1. These are great pictures! Can we learn anything from them? 2. Plot the data (regular and log-log) 3. Create a hypothesis – scaling law relationship between radius and time! 4. Derive scaling law by dimensional analysis 5. Check that the curve slope matches our prediction!! 6. Use a little physics to estimate the total energy based on the intercept of the fitted curve. Radius $$R$$ depends on time $$t$$, energy $$E$$, density of the atmosphere $$\rho_0$$, and the specific heat of the atmosphere $$\gamma$$, so $$R = f( E, t, \rho_0, \gamma )$$ 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 $$\gamma$$ has units of joules per kilogram per degree Kelvin.  $$E$$ $$t$$ $$R$$ $$\gamma$$ $$\rho _ 0$$ distance 2 0 1 2 -3 time -2 1 0 -2 0 mass 1 0 0 0 1 temp 0 0 0 -1 0 We can save ourselves some work in the row-reduction by re-arranging the rows and columns of the matrix …  t R E $$\rho _ 0$$ $$\gamma$$ time 1 0 -2 0 -2 distance 0 1 2 -3 2 mass 0 0 1 1 0 temp 0 0 0 0 -1 Now, by row-reduction,  t R E $$\rho _ 0$$ $$\gamma$$ time 1 0 0 2 0 distance 0 1 0 -5 0 mass 0 0 1 1 0 temp 0 0 0 0 1 So, there is a 1-dimensional column nullspace with the spanning basis $$\{ [-2,5,-1,1,0]^T \}$$. After dimensional analysis, $F(t^{-2} R^{5} E^{-1} \rho_0^{1}, \gamma) = 0$ $R = E^{1/5} t^{2/5} \rho_0^{-1/5} S(\gamma)$ $\log R = \frac{1}{5}\left( \log E + 2 \log t - \log \rho_0 \right) + \log S(\gamma)$ $\frac{5}{2} \log R = \log t + \frac{1}{2} \left(\log E - \log \rho_0 \right) + \frac{5}{2} \log S(\gamma)$ [ 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 time-stamped # 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, we can determine the y-intercept. Knowing something about specific heats and air density, G. I. Taylor was able to solve for energe E, and estimate the explosion size at 16.8 kilotons. The actual strength was estimated with other methods as equivalent to 20 kilotons of TNT. [Show code] from numpy import * from pylab import loglog, show, xlabel, ylabel, savefig, text, subplot, plot # data copied from webpage, saved as 'trinity.csv' bomb_data = loadtxt('trinity.csv', delimiter=',') bomb_data[:,0] = bomb_data[:,0]/1000 # converting milliseconds to seconds bomb_data[:,1] = bomb_data[:,1]*100 # converting meters to centimeters # conversions to match Taylor's paper. # First, we log-transform our data set log_data = log10(bomb_data) # use log base 10 log_data[:,1]=log_data[:,1] # Now, we construct A and b for A x = b. b = log_data[:, 1:2]*5./2. A = hstack([ones((len(b), 1)), log_data[:, 0:1]]) # solve A x = b for x using gaussian elimination with partial pivoting x = linalg.solve( A.T.dot(A), A.T.dot(b) ) k, m = float(x[0]), float(x[1]) # float calls convert 0-d arrays to scalars print (k, m) x = linspace(-4.1,-1,100) figure(figsize=(10,8)) plot(log_data[:,0],log_data[:,1]*5./2.,'ro',markersize=8) plot(x,m*x+k,'b',linewidth=2) legend(['Data','Best fit,$(5/2)\log_{10}R=%.3f\log_{10}t+%.3f$'%(m,k)],loc='upper left', fontsize=20) xlabel('$\log_{10}t$(s)',fontsize=24) ylabel('$(5/2)\log_{10}R\$ (m)',fontsize=24)
xlim([-4.1, -1])
#ylim([7.5, 11])
xticks(fontsize=20)
yticks(fontsize=20)
savefig('Taylor.png')
show()

Notes: (1) plot on the left is from Taylor’s 1950 paper. (2) The y-axes are different since the plot on the left measures radius in centimeters, while the plot on the right measures it in meters.

Model validated! With that, knowing something about specific heat (which we’ve neglected so affects $$C$$), and taking air density $$\rho _ 0=1.25\times 10^{-3}$$ grams per cm$$^3$$, G. I. Taylor was able to solve for energe E, and estimate the explosion size at $$7.14\times10^{20}$$ ergs, equivalent to 16.8 kilotons of TNT. The actual strength was estimated with other methods as equivalent to 20 kilotons of TNT.

## Concluding remarks

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.

# Exercises

1. 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?

2. 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.

1. A balloon popping.
2. Pouring maple syrup onto blueberry pancakes.
3. The errosion of the Appalachian mountain chain.
3. A bicyclist traversing a turn on a road leans into the turn to keep her bicycle stable.

1. Find 3 dimensional variables that together should determine the angle at which she should lean.

2. Determine the units of each variable, in terms of the fundamental units of length, time, mass, charge, and temperature.

3. According to the Buckingham $$\Pi$$ theorem, how many dimensionless groups can be constructed from your 3 dimensional variables.

4. Find a general formula for a functional relationship between your variables and the angle of lean, expressed in terms of dimensionless groups.

4. In Principa Mathematica, Isaac Newton was the first to attempt to calculate the speed of sound in air. Let’s replicate his result.

1. Determine three commonly known properties of an ideal gas might effect the speed of sound ($$c$$) in that gas.

2. Now, use dimensional analysis to come up with a formula for the speed of sound $$c$$ as a function of these three values.

5. McMahon and Bonner, 1983, p. 76-78) When you cook a roast, the cooking time ($$T$$) is defined as the time needed for the center of the roast to reach a pre-defined 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 specific-heat capacity at constant pressure ($$s_p$$).

1. Find a dimensionless group relating these five variables.

2. If we halve the size of the roast, how should the cooking time be changed, assuming everything else stays the same?

6. 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 $$v$$.

1. What’s the number of dimensionless terms needed to express the functional relationship among these 5 variables?

2. 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 $$D/(\rho L^2 v^2)$$.

7. 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.

• The density of uranium $$\rho$$ (mass per meters cubed)
• The diffusion rate of neutrons in uranium $$D$$ (area per time)
• The rate of nuclear collisions $$c$$ (per time).

Use dimensional analysis to find a formula for the critical mass $$m$$ as a function of these three variables.

8. (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.

9. (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.

1. 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.

2. 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.

10. (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 an 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.

1. What parameters besides the distance $$r$$ does the angle $$\theta$$ depend on?

2. 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.)

11. 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 $$\theta$$.

1. Use dimensional analysis to derive an equation for the attack angle $$\theta$$ as a function of a dimensionless product of the other variables.

2. Solve your equation for the lift force $$F$$.

3. 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 whirling-arm set-up
#
# Extracted by Tim Reluga, 2014-04 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

1. 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 $h = \sqrt{ (x+1) (y-1) }.$ Use the theory of dimensional analysis to critique this formula.

2. Find a formula for the volume of an n-dimensional sphere in terms of the sphere’s radius $$r$$.

3. 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 space-time 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.

1. Rewrite our functional relationship between radius and volume to incorporate this conversion factor.
2. Differentiate your new functional relationship with respect to $$\lambda$$ to get a partial differential equation.
3. Evaluate this partial differential equation at $$\lambda = 1$$. This PDE is the

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