Lab 1 - Measurement

This one is easy. Put identifying info on your records. You do not need to write a paragraph. A descriptive name that points to supporting resources, along with a date, is often all that is necessary.

• Phy 221 - Lab 1 Measurement - Sept 08 2021 - P Malsom

If you move to a different part of the experiment, please make a new section in your notes. Again you do not need to write a paragraph, but please be descriptive.

Always make a sketch of the experiment (a picture is often worth a thousand words).

• Take your time making a sketch. You will want something clear to reference later.
• Make the sketch large enough that you can label dimensions and dynamics, but leave room to record measurements. I find around a quarter of the page is a nice size for a detailed drawing.
• Be deliberate in your handwriting style. Each of your variables should be distinguishable and unique (make sure \(u\) and \(\mu\) are distinguishable in your handwriting).

This is a core part of the good measurement habits. Recording to paper is faster and more reliable than typing into a computer. /If you only use a computer, you will eventually lose data. This can be CATASTROPHIC if some of your data is interdependent. Please hear my warnings and record your data to a piece of paper. You can always re-enter data from paper if the computer loses it.

• Always record the measurements on the same page as the sketch. Data means nothing without a sketch to reference.
• Label your sketch with variables rather than actual numbers. (use \(\ell\) instead of \(0.1\))
• Record your error. Usually this is done with a “plus-minus”, but could also be a list of measurements used to find a standard deviation.
• Record your unit of measurement (please use SI units).

\[\ell=0.1 \pm 0.02~m\]

Before performing the actual analysis (preferably on a computer), you should evaluate some or all of the critical calculations with a calculator. Try to examine if this is what you expect the value to be. If it is not, review your measurements to find any mistakes!

This course will encourage you to use Python as a calculator and bugs are bound to show up in your code. This hand calculation serves as a double check to see if there is a mistake in your programming. It is very hard to find errors without a hand calculated estimate to use for comparison.

In the first experiment, you will make some basic measurements and estimate the volume of the piece of wood. This estimation will yield a quick result with marginal accuracy and gives no estimation of uncertainty.

Before we begin, I want to refresh your understanding of the normal distribution. The normal distribution is commonly written as the following function:

\[f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} ~ \exp \left(- \frac{(x-\mu)^2}{2\sigma^2} \right)\]

When you examine this function, please remember that l\(\sigma\) (sigma) and \(\mu\) (mu) are fixed values, so the mathematical function looks like \(f(x)=e^{-x^2}\), which is a “bell curve”. The term in the front (\(1/\sqrt{2\pi\sigma^2}\)) is a normalization factor, which is a constant because \(\sigma\) is a known value.

• \(\mu\) - Mean - most probable measurement
• \(\sigma\) - Standard Deviation - spread of the measurements

Approximately 68% of all measurements will fall into the “one standard deviation” range. A measurement is said to have a value of the mean (\(\mu\)) and an error of one standard deviation (\(\sigma\)). You should quote your value as:

\[E[x] = \mu \pm \sigma\]

import matplotlib as mpl
import matplotlib.pyplot as plt
import numpy as np

mu = 0.0
sigma= 1.0

x=np.linspace(-3*sigma,3*sigma,num=500)
f=1/np.sqrt(2*np.pi*sigma**2) * np.exp(- (x-mu)**2/(2*sigma**2))

xfill=np.linspace(-sigma+mu,sigma+mu,num=500)
zfill=np.zeros(500)
ffill=1/np.sqrt(2*np.pi*sigma**2) * np.exp(- (xfill-mu)**2/(2*sigma**2))

fig, ax = plt.subplots()  # a figure with a single Axes

plt.title("The Normal Distribution")
ax.text(-0.2, 0.2, "~68%", fontsize=15)
ax.text(-1.1, -0.05, r'$-\sigma$', fontsize=15)
ax.text(0.9, -0.05, r'$+\sigma$', fontsize=15)
ax.text(-0.1, -0.05, r'$\mu$', fontsize=15)

plt.yticks([])
plt.xticks([])
plt.ylim([-0.07,0.42])


ax.plot(x,f,color='black')
ax.plot([-1,-1],[-0.01,0.01],color='black',alpha=0.2)
ax.plot([-3*sigma,3*sigma],[0,0],color='black',alpha=0.2)
ax.plot([0,0],[-0.01,0.01],color='black',alpha=0.2)
ax.plot([1,1],[-0.01,0.01],color='black',alpha=0.2)
ax.fill_between(xfill,zfill,ffill, facecolor='black', alpha=0.05)

#plt.show()
plt.savefig("img/01-normalcurve.png",bbox_inches='tight')
plt.close()

For this experiment, you have two quantities to measure: the radius and the length. Lets start by measuring the radius, and then we can use a shortcut for the measurement of the length.

The radius of the branch is obviously non-uniform. Every time you measure, you will get a slightly different answer. This is what it means to have an Error in a measurement. But the question remains: how should you report the correct amount of error? This question is not easily answered, and a full treatment would be seen in a statistics or mathematics course on probability distributions. If the data is normally distributed (we will assume this for the entire course), the correct value for the error is the standard deviation.

The standard deviation is the \(\sigma\) (sigma) in the above mathematical equation. 68% of all of the measurements will fall in this range (sometimes called “one sigma”). The standard deviation is easily calculated using python or other programs such as excel or even some calculators. Below is some code which you can use to calculate the mean and standard deviation of a set of radius measurements.

Please make many measurements of the radius of your non-uniform object. Aim for 20-40 measurements from different locations. In general, the more measurements you make, the better your results will be, but there are diminishing returns (as \(1/\sqrt{N}\)). The list of measurements can just be written out in a list or table in your lab notes.

Now lets use a computer to calculate the standard deviation of your set of measurements. Please enter your list of radii measurements into the following code to find the standard deviation:

import numpy as np

diameter=[0.0283,0.0296,0.0249,0.0293,0.0318,0.0274]

radii=np.array(diameter)/2

print(radii)
print("Mean  = ", np.mean(radii))
print("StdDev= ", np.std(radii))

#+BEGIN_EXAMPLE
[0.01415 0.0148  0.01245 0.01465 0.0159  0.0137 ]
Mean  =  0.014275000000000001
StdDev=  0.0010593826818797198
#+END_EXAMPLE

Thus, the value for the radius would be quoted as \(r=0.286 \pm 0.021 ~m\).

A visualization of this process may be useful. Please insert your data into this code if you want to see a histogram (binning) of your data:

import matplotlib as mpl
import matplotlib.pyplot as plt
import numpy as np

radii=[0.0283,0.0296,0.0249,0.0293,0.0318,0.0274]

mu = np.mean(radii)
sigma = np.std(radii)
num_bins=10

normalization = np.max(np.histogram(radii,num_bins)[0])

x=np.linspace(mu-3*sigma,mu+3*sigma,num=500)
f= normalization * np.exp(- (x-mu)**2/(2*sigma**2))

plt.xlabel("Radius Measurements (m)")
plt.ylabel("Number of times measured")
plt.plot(x,f,color='black')
plt.hist(radii,num_bins)
#plt.show()
plt.savefig("img/01-radiusmeas.png",bbox_inches='tight')
plt.close()

I want to be direct about this. This analysis is somewhat crude. There is an entire field devoted to this topic (statistics). I am just giving you the simplified (and most used) formula. You can read more on this topic as well as complete some example problems on the Lab Resources - Error Propogation course webpage.

There are three main assumptions which are built in to the simplified error propagation which is often used by science and engineering analyses:

1. The randomness is normally distributed (symmetric Gaussian distribution).
2. Variables are independent of each other (error in length does not impact the error in radius).
3. The function is approximately linear over the range of the error (necessary for Taylor expansion).

These assumptions are often very close to correct in real world experiments, and this error formula is frequently used in science and engineering. If you are running a very sensitive experiment, you may find these assumptions are false, and you may have to use more advanced statistics to estimate the error correctly (ex: the correlation matrix and non-symmetric probability distributions).

The procedure to calculate the error uses a new type of derivative known as the partial derivative (\(\partial\)), in which all variables except for the variable used for the derivative are regarded as constant. You will see this type of derivative used frequently in Calc III. Partial derivatives are nice because you can take the derivative of any function in the same way you would with derivatives in Calc 1 (normal differentiation rules). Again see the Lab Resources - Error Propogation course webpage for details and examples.

The formula to calculate the error in a function \(f(A,B,\ldots)\) with independent variables \(A\), \(B\), \(\ldots\) is given by:

\[E[ f(A,B,\ldots) ] = \sqrt{\left( \frac{\partial f}{\partial A} E[A] \right)^2+\left( \frac{\partial f}{\partial B} E[B] \right)^2+\cdots}\]

Lets apply this to the volume formula: if we set \(f(A,B) = V(r,\ell) = \pi r^2 \ell\), the error in the volume becomes: \[E[V] = \sqrt{ \left( \frac{\partial V}{\partial r} E[r] \right)^2 + \left( \frac{\partial V}{\partial \ell} E[\ell] \right)^2}\] Please do not let this formula scare you, there are two derivatives, but they are not very difficult to evaluate (remember \(V = \pi r^2 \ell\)) \[\frac{\partial V}{\partial r} = \frac{\partial}{\partial r} \pi r^2 \ell = \pi \ell \frac{\partial}{\partial r} r^2 = \pi \ell (2\cdot r) = 2 \pi r \ell\]

\[\frac{\partial V}{\partial \ell} = \frac{\partial}{\partial \ell} \pi r^2 \ell = \pi r^2 \frac{\partial}{\partial \ell} \ell = \pi r^2\] This gives the final error formula as: \[E[V] = \sqrt{ \left( (2\pi r \ell) E[r] \right)^2 + \left( (\pi r^2) E[\ell] \right)^2}\] This is the final result for the error in the volume based off of the spherical model. This formula looks complicated, but you know the right-hand-side and can simply calculate!

import numpy as np

radius = 0.0286
Eradius = 0.0021
length = 0.37
Elength = 0.01

V= np.pi * radius**2 * length
EV=np.sqrt( (2*np.pi*radius*length*Eradius)**2 + (np.pi*radius**2*Elength)**2 )

print("Volume = ", V, "+-", EV)

#+BEGIN_EXAMPLE
Volume =  0.0009507879369642138 +- 0.00014197115977463345
#+END_EXAMPLE

Please note that a computer can do all of this work for you! Please refer to Lab Resources - Error Propogation course webpage for an example.

Calculate the error estimate using your own measurements. You can perform the calculation by hand, or you can reuse the code above in Jupyter.

Now is a very good time to discuss systematic error. Systematic error is an error which is present but unaccounted and skews your results consistently in one direction. They commonly arise from oversimplified models. This type of error is often hidden from view and you must be careful to reassess your measurement techniques to minimize these errors.

As an example, lets imagine your object has another cylinder embedded into it (maybe where a branch was growing out). By making our model a uniform cylinder, we ignore any other radial bulges which directly impacts our calculation in one direction. It does not matter how many times you measure the branch, the result will be skewed towards more volume because of the protrusion. Systematic errors can lead to incorrect interpretation of data, because the error is often buried in assumptions which are easy to overlook. It is a good habit to write down any possibly dubious assumptions in your notes (paper or typed) for discussion or thought later.

Checkpoint: Please think about any systematic errors which could possibly effect the experiment. What assumptions are dubious? Try to identify which error is the largest and highlight your thoughts in your lab notes.