Abstract = Spoilers

Although it feels like you're giving away the whole story -- because you are -- the Abstract should be all spoilers. — Although it feels like you’re giving away the whole story — because you are — the Abstract should be all spoilers.

I was trying to explain that the abstract of a lab report was a really brief summary of everything in the report.

One of my students said, “Oh, so it’s like spoilers.”

And they were right.

If you didn’t know, spoilers are little bits of information that give away the plot. While you don’t want any spoilers if you have not seen the movie, in a lab report you what to give away the plot because since it’s a short, concise summary, it’s what most people will read.

So now my middle-schoolers are submitting lab reports with a section called “Spoilers” on the cover page.

Bibme: A Nice Tool for Composing Bibliography References

While we were talking about what I expect from lab reports, one of my students pointed out the Bibme website. It helps put together bibliography references in a variety of styles: MLA, APA, Chicago, and Turabian.

Bending a Soccer Ball

Students from the University of Leicester have published a beautiful short research paper (pdf) on the physics of curving a soccer ball through the air.

It has been found that the amount a football bends depends linearly on the speed of the ball and the amount of spin.

— Sandhu et al., 2011: How to score a goal (pdf) in the University of Leicester’s Journal of Physics Special Topics

They derive the relationship from Bernoulli’s equation using some pretty straightforward algebra. The force (F) perpendicular to the ball’s motion that causes it to curl is:

$F = 2 \pi R^3 \rho \omega v$

and the distance the ball curls can be calculated from:

$D = \frac{\pi R^3 \rho \omega}{ v m } x^2$

where:

F = force perpendicular to the direction the ball is kicked
D = perpendicular distance the ball moves to the direction it is kicked (the amount of curl)
R = radius of the ball
ρ = density of the air
ω = angular velocity of the ball
v = velocity of the ball (in the direction it is kicked)
m = mass of the ball
x = distance traveled in the direction the ball is kicked

The paper itself is an excellent example of what a short, student research paper should look like. And there are number of neat followup projects that advanced, high-school, physics/calculus students could take on, such as: considering the vertical dimension — how much time it take for the ball to rise and fall over the wall; creating a model (VPython) of the motion of the ball; and adding in the slowing of the ball due to air friction.

↬ScienceDaily

Draining a Bottle Part 2: Linearizing Equations when you have to

Yesterday we used calculus to find the equation for the height of water in a large plastic water bottle as the water drained out of a small hole in the bottom.

Perhaps the most crucial point in the procedure was fitting a curve to the measured reduction of the water’s outflow rate over time. Yesterday, in our initial attempt, we used a straight line for the curve, which produced a very good fit.

q-v-t — Figure 1. The change in the outflow rate over time can be well approximated by a straight line.

The R² value is a measure of how good a fit the data is to the trendline. The straight line gives an R² value of 0.9854, which is very close to a perfect fit of 1.0 (the lowest R² can go is 0.0).

The resulting equation, written in terms of the outflow rate (dV/dt) and time (t), was:

$\frac{dV}{dt} = -0.0035 t + 3.9113$

However, if you look carefully at the graph in Figure 1, the last few data points suggest that the outflow does not just linearly decrease to zero, but approaches zero asymptotically. As a result, a different type of curve might be a better trendline.

Types of Equations

So my calculus students and I, with a little help from the pre-Calculus class, tried to figure out what types of curves might work. There are quite a few, but we settled for looking at three: a logarithmic function, a reciprocal function, and a square root function. These are shown in Figure 2.

type-curves-300 — Figure 2. Example curves that might better describe the relationship between outflow and time.

I steered them toward the square root function because then we’d end up with something akin to Torricelli’s Law (which can be derived from the physics). A basic square root function for outflow would look something like this:

$\frac{dV}{dt} = a \sqrt{t} + b$

the a coefficient stretches the equation out, while the b coefficient moves the curve up and down.

Fitting the Curve

Having decided on a square-root type function, the next problem was trying to find the actual equation. Previously, we used Excel to find the best fit straight line. However, while Excel can fit log, exponential and power curves, there’s no option for fitting a square-root function to a graph.

To get around this we linearized the square-root function. The equation, after all, looks a lot like the equation of a straight line already, the only difference is the square root of t, so let’s substitute in:

$x = \sqrt{t}$

to get:

$\frac{dV}{dt} = a x + b$

Now we can get Excel to fit a straight line to our data, but we have to plot the square-root of time versus temperature instead of the just time versus temperature. So we take the square root of all of our time measurements:

Time	Square root of time	Outflow rate
t (s)	t^1/2 = x (s^1/2)	dV/dt (ml/s)
0.0	0	3.91
45.5	6.75	3.52
97.8	9.89	2.94
140.9	11.87	3.52
197	14.04	3.21
257	16.05	3.01
315.1	17.75	2.81
380.1	19.50	2.53
452.9	21.28	2.23
529.6	23.01	1.92
620.7	24.91	1.69
742.7	27.25	1.45

We can now plot the outflow rate versus the square root of time (Figure 3).

q-v-sqrt(t) — Figure 3. Linear trend relating the outflow rate to the square root of time. The regression coefficient (R²) of 0.9948 is better than the simply linear trend of outlfow rate versus time (which was 0.9854).

The equation Excel gives (Figure 3), is:

$\frac{dV}{dt} = -0.1395 x + 5.21$

and we can substitute back in for x=t^1/2 to get:

$\frac{dV}{dt} = -0.1395 \sqrt{t} + 5.21$

Getting back to the Equation for Height

Now we can do the same procedure we did before to find the equation for height.

First we substitute in V=πr²h:

$\frac{d(\pi r^2 h}{dt} = -0.1395 \sqrt{t} + 5.21$

Factor out the πr² and move it to the other side of the equation to solve for the rate of change of height:

$\frac{dh}{dt} = \frac{-0.1395}{\pi r^2} \sqrt{t} + \frac{5.21}{\pi r^2}$

Then integrate to find h(t) (remember $\sqrt{t} = t^{1/2}$ ) :

$\int \frac{dh}{dt} dt = \int \left( \frac{-0.1395}{\pi r^2} t^{1/2} + \frac{5.21}{\pi r^2} \right) dt$

gives:

$h = \frac{-0.1395}{(3/2) \pi r^2} t^{3/2} + \frac{5.21}{\pi r^2} t + c$

which might look a bit ugly, but that’s only because I haven’t simplified the fractions. Since the radius (r) is 7.5 cm:

$h = -0.000526 t^{3/2} + 0.029 t + c$

Finally we substitute in the initial value (t=0, h=11) to solve for the coefficient:

$c = 11$

giving the equation:

$h = -0.000526 t^{3/2} + 0.029 t + 11$

Plotting the equations shows that it matches the measured data fairly well, although not quite as well as when we used the previous linear function for outflow.

h-v-t_sqrt — Figure 4. Integrating a square root function for the outflow rate gives a modeled function for the changing height over time that slightly undermatches the measured heights.

Discussion

I’m not sure why the square root function for outflow does not give as good a match of the measurements of height as does the linear function, especially since the former better matches the data (it has a better R² value).

It could be because of the error in the measurements; the gradations on the water bottle were drawn by hand with a sharpie so the error in the height measurements there alone was probably on the order of 2-3 mm. The measurement of the outflow volume in the beaker was also probably off by about 5%.

I suspect, however, that the relatively short time for the experiment (about 15 minutes) may have a large role in determining which model fit better. If we’d run the experiment for longer, so students could measure the long tail as the water height in the bottle got close to the outlet level and the outflow rate really slowed down, then we’d have found a much better match using the square-root function. The linear match of the outflow data produces a quadratic equation when you integrate it. Quadratic equations will drop to a minimum and then rise again, unlike the square-root function which will just continue to sink.

Conclusions

The linearization of the square-root function worked very nicely. It was a great mathematical example even if it did not produce the better result, it was still close enough to be worth it.

The Draining of a Plastic Bottle: Integrating a Physics Experiment into Calculus

Abstract

I punched a small hole (about 1mm radius) in a one gallon plastic bottle and had my students measure the rate at which water drained. Even though the apparatus and measurement technique was fairly rough, we were able to, with a little calculus, determine the equation for the height of the water in the bottle as a function of time.

calculus_jug_1110 — Figure 1. A student uses a stopwatch to measure the outflow rate of water from the plastic bottle.

Introduction

Questions about water draining from a tank are a pretty common in calculus textbooks, but there is a significant difference between seeing the problem written down, and having to figure it out from a physical example. The latter is much more challenging because it does not presuppose any relationships for the change in the height of water with time; students must determine the relationship from the data they collect.

The experimental approach mimics the challenges faced by scientists such as Henry Darcy who first determined the formulas for groundwater flow (Darcy, 1733) almost 300 years ago, not long after the development of modern calculus by Newton and Leibniz (O’Conner and Robertson, 1996).

Procedure

I punched a small hole, about 1mm in radius, in the base of a plastic, one-gallon bottle. I chose this particular type of bottle because the bulk of it was cylindrical in shape.

Students were instructed to figure out how the rate at which water flowed out (outflow rate) changed with time, and how the height of the water (h) in the bottle changed with time. These relationships would allow me to predict the outflow rate at any time, as well as how much water was left in the container at any time.

Data collection:

To measure height versus time, we marked the side of the bottle (within the cylindrical region) in one centimeter increments and recorded the time it took for the water level to drop from one mark to the next. There were a total of 11 marks.
To measure the outflow rate, we intercepted the outflow using a 25 ml beaker (not shown in Figure 2), and measured the time it took to fill to the 25 ml mark.

Results

Time elapsed since last measurement	Height of water in container	Time to fill 25ml beaker
Δt (s)	h (cm)	t_f (s)
0.0	11	6.4
45.5	10	7.1
52.3	9	8.5
43.1	8	7.1
56.1	7	7.8
60.6	6	8.3
57.6	5	8.9
65.0	4	9.9
72.8	3	11.2
76.7	2	13.0
91.1	1	14.8
122.0	0	17.3

Table 1: Outflow rate, water height change with time.

To analyze the data, we calculated the total time (the cumulative sum of the elapsed time since the previous measurement) and the outflow rate. The outflow rate is the change in volume with time:

$\text{outflow rate} = \frac{volume}{time} = \frac{dV}{dt} = \frac{25 ml}{t_f}$

So our data table becomes:

Time	Height of water in container	Outflow rate
t (s)	h (cm)	dV/dt (ml/s)
0.0	11	3.91
45.5	10	3.52
97.8	9	2.94
140.9	8	3.52
197	7	3.21
257	6	3.01
315.1	5	2.81
380.1	4	2.53
452.9	3	2.23
529.6	2	1.92
620.7	1	1.69
742.7	0	1.45

Table 2. Height of water and outflow rate of the bottle.

The graph of the height of the water with time shows a curve, although it is difficult to determine precisely what type of curve. My students started by trying to fit a quadratic equation to it, which should work as we’ll see in a minute.

h-v-t — Figure 3. The decrease in height with time is not a straight line (is non-linear).

The plot of the outflow rate versus time, however, shows a pretty good linear trend. (Note that we do not use the first three datapoints (in Table 2), which we believe are erroneous because we were still sorting out the measuring method.)

Although I’ll note here that the data should ideally be modeled using a square root equation (Torricelli’s Law), that is beyond the present scope of the problem (we’ll try that tomorrow as a follow exercise).

Plotting the data in Excel we could add a linear trendline.

q-v-t — Figure 4. The outflow rate decreases linearly with time.

For the linear trendline, Excel gives the equation:

$y = -0.0035 x + 3.9113$

The y-axis is outflow rate (the change in volume with time), and the x axis is time (t), so the linear equation becomes:

$\frac{dV}{dt} = -0.0035 t + 3.9113$

Notice that this is a differential equation.

To determine the rate of at which the height of water in the container is changing, we need to recognize that the container is cylindrical in shape, and the volume (V) of a cylinder depends on its radius (r) and height (h):

$V = \pi r^2 h$

which can be substituted into differential in the rate equation:

$\frac{d(\pi r^2 h)}{dt} = -0.0035 t + 3.9113$

since π and the radius (r) are constants (since the jug’s shape is a cylinder), they can be pulled out of the differential:

$\pi r^2 \frac{dh}{dt} = -0.0035 t + 3.9113$

Dividing through by πr² solves for the rate of change of height:

$\frac{dh}{dt} = \frac{-0.0035 t + 3.9113}{\pi r^2}$

Isolating the coefficients gives:

$\frac{dh}{dt} = \frac{-0.0035}{\pi r^2} t + \frac{3.9113}{\pi r^2}$

This equation should give the rate at which the height changes with time, however, if you look at it carefully you’ll realize that for the time range we’re using (less than 800 seconds) the value of dh/dt will always be positive. We correct this by recognizing that the outflow rate is a loss of water, so a positive outflow should result in a negative change in height, therefore we rewrite the equation as:

$-\frac{dh}{dt} = \frac{-0.0035}{\pi r^2} t + \frac{3.9113}{\pi r^2}$

which gives:

$\frac{dh}{dt} = \frac{0.0035}{\pi r^2} t - \frac{3.9113}{\pi r^2}$

Now comes the calculus

Now, we can use this rate equation to find the equation for height versus time by integrating with respect to time:

$\int \frac{dh}{dt} dt = \int \left( \frac{0.0035}{\pi r^2} t - \frac{3.9113}{\pi r^2} \right) dt$

to get:

$h = \frac{0.0035}{2 \pi r^2} t^2 - \frac{3.9113}{\pi r^2} t + c$

And all we have to do to the find the constant of integration is substitute in a known point, an initial value. As is often the case, the best point to use is the starting point where t=0 makes the rest of the calculations easier. In our case, when t=0, h=11:

$11 = \frac{0.0035}{2 \pi r^2} (0)^2 - \frac{3.9113}{\pi r^2} (0) + c$

so:

$c = 11$

And our final equation becomes:

$h = \frac{0.0035}{2 \pi r^2} t^2 - \frac{3.9113}{\pi r^2} t + 11$

which is a quadratic equation as my students guessed before we did the calculus.

Does it work?

You will notice that in the math above, we never use the height data in determining equation for height versus time; all the calculations are based on the trendline for the outflow rate versus time.

As a result, we can compare the results of our equation to the actual measurements to see if our calculations are even close. Remarkably, they are.

h-v-t-model-v-measured — Figure 5. The results from our equation (modeled) match the measured heights so well, the data points are difficult to distinguish on the graph.

Discussion

Despite all the potential for error, particularly, our relatively crude measurement techniques, and the imperfect cylindrical shape of our plastic bottle, the experiment went remarkably well.

Students found it quite challenging, and required some assistance even though this is a problem they have seen before in their textbook.

The problems in the textbook use Torricelli’s Law, which should much better describe the draining of a tank than the linear equation we find for dV/dt.

Torricelli’s Law:

$\frac{dV}{dt} = a \sqrt{2gh}$

where a is the area of the outlet hole, and g is the acceleration due to gravity.

In our actual experiment it is difficult to tell that a square root function would work better. Excel does not have an option for matching a square root function, so the calculations would become more involved (although it could be set up using Excel’s iterative solver or Goal Seek function).

Conclusion

Our experiment to use calculus to determine the rate of change of the height of water in a leaking plastic water bottle was a successful exercise even though the roughness of the data collection did not permit identification of the square root law for leakage.

How to Write a Research Paper

An excellent series by the American Chemical Society starts with the basics of, “How to Write a Paper to Communicate Your Research,” but also addresses the question of why publish your research. It ought to help my students understand why I’m so insistent on lab reports.

ᔥ ACS Publishing ↬ Pale Blue Blog

Projectile Motion

Abstract

A series of still photographs of a projectile (soccer ball) in motion were used to determine the equation for the height of the ball (h(t) = 4.9 t² + 14.2 t + 1.25), the initial velocity of the ball (14.2 m), the maximum height of the ball (11.6 m), and the time between each photograph (0.41 s). The problem was solved numerically using MS Excel’s Solver function. There are much easier ways of doing this, which we did not do.

Introduction

projectile-diagram — Figure 1. Calculated elevation of the soccer ball after launch.

projectile-3832c2-800 — Figure 2. Animation of the soccer ball projectile.

One of physics lab assignments I gave my students was to see if students could use a camera to capture a sequence of images of a projectile, plot the elevation of the projectile from the photographs, determine the constants in the parabolic equation for the height of the projectile, and, in so doing, determine the velocity at which the projectile was launched.

I offered my old, digital Pentax SLR that can take up to seven pictures in quick sequence and be set to fully manual. A digital video camera with a detailed timestamp would have been ideal, but we did not have one available at the time.

Now the easy way of getting the velocity data would be to estimate the heights (h) of the ball from the image using some sort of known reference (in this case the whiteboard), and determine the time between each photograph (Δt) by photographing a stopwatch using the same shutterspeed settings. After all, the average velocity of the ball between two images would be:

$! \bar{v} = \frac{\Delta h}{\Delta t}$

The reference whiteboard is four feet tall (1.22 m) in real life, but 51 pixels tall in the image. Using this ratio (i.e. 1.22 m = 51 px) we can convert the heights of the ball from pixels to meters:

elevations — Table 1. Table showing the conversion of the height of the ball in pixels to elevation (in meters).

Unfortunately, I think my students forgot to do the pictures of the stopwatch to get Δt, the time between each photograph. Since the lab reports are due on Monday, and it’s the weekend now I’m curious to see what they come up with.

However, I was wondering if they could use just the elevation data to back out the Δt. So I gave it a try myself. Even the easiest way of solving this problem is not trivial, in fact, I ended up resorting to Excel’s iterative solver to find the answers. While this procedure probably goes a little beyond what I expect from the typical high school physics student, more advanced students who are taking calculus might benefit.

Procedure

We took the reference whiteboard (1.21 m tall), a soccer ball, and the camera outside. The whiteboard was leant vertically against the post of the soccer goal. The ball was thrown vertically by a student standing next to the whiteboard (see Figure 1) while pictures were taken. The camera’s shutterspeed was 1/250th of a second. The distance from the camera to the person throwing the ball (and to the whiteboard) were not measured.

The procedure was repeated several times, but only one trail was used in this analysis.

The images were loaded onto a computer, and the program GIMP was used to determine the distance, in pixels, from the ground to the projectile. The size of the reference whiteboard, in pixels, was used to calculate the height of the soccer ball in meters.

The elevations measured off the photographs were then used to calculate the release velocity, time between snapshots, and maximum height of the ball.

The Equation for Elevation

I started with the fact that once the ball is released, the only force acting on it is the force of gravity. Since the mass of the ball does not change we only have to consider the acceleration due to gravity (-9.8 m/s²). I also neglect air resistance to make things easier.

Finding the Velocity Equation

Start with the fact that, acceleration is the rate of change of velocity with time. You can write it in the differential form:

$! a = \frac{dv}{dt} = -9.8$

so we integrate with respect to time to get the equation for velocity as a function of time:

$! v(t) = \int -9.8 dt$

$! v(t) = -9.8 t + c$

where c is an unknown constant. What we do know though, is that at the beginning, when the ball is just launched, time is zero (t = 0) so c_v becomes the initial velocity (v₀) at which the ball is thrown:

at t = 0, v(0) = v₀:
$! v_0 = -9.8 (0) + c$

$! v_0 = c$

So our velocity equation becomes:

$! v(t) = -9.8 t + v_0$

Finding the height equation

Now since we know that velocity is the rate of change of distance (in this case height) with time:

$! v(t) = \frac{dh}{dt} = -9.8 t + v_0$

so we integrate again to find the height equation:

$! \frac{dh}{dt} = -9.8 t + v_0$
$! h(t) = \int (-9.8 t + v_0) dt$
$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + c$

Similar to what we did with the velocity equation, to find the new constant c we consider what happens at the start time, when the ball is launched, and t = 0 and h(0) = h₀;

$! h_0 = \frac{-9.8 (0)^2}{2} + v_0 (0) + c$

so:
$! h_0 = c$

The constant is equal to the initial height of the ball — the height of the ball when it’s thrown. So we end up with the final equation:

$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + h_0$

Results

Solving all the unknowns

At this point, although we have an equation for the height of the ball, we don’t know the initial velocity (v₀), nor do we know the initial height of the ball when it’s released (h₀). And we still don’t know the time when the ball is at each position.

With that many unknowns we’d need the same number of independent equations to be able to solve for them all. It may be possible, but instead of analytically solving the equations I opted to take a numerical approach, and use Excel’s Solver function.

I started by setting up the equations to calculate the height of the ball at six different times to correspond with our six height measurements. It was necessary therefore to create a set of variables:

Time when we started taking pictures (t₁): Since we don’t know how long after we threw the ball we started taking pictures, I made this a variable called t₁.
The time between each picture (dt): I made the assumption that the time between each picture would be constant. The shutter speed was constant (1/250th of a second) so there is no obvious reason why the time should be different.
Initial velocity (v₀): The initial upward speed at which the ball was thrown. Obviously, the faster the initial speed the higher the ball goes, so this is a fairly important parameter.
Initial height (h₀): We also don’t precisely know how high the ball was when it was released, so this also needs to be a variable.

By defining the time between each picture as dt, we can write the time that each picture was taken in terms of the time of the initial picture (t₁) and dt. After all the second picture would have been taken dt seconds after the first for a total time of:

$! t(P2) = t_0 + dt$

similarly for all the pictures:

photo-times — Table 2. Table of expressions giving the time when each of the six photos were taken.

Now I set up an Excel spreadsheet and gave all the unknown variable values and initial value of 1:

excel-solver — Table 3. Table in Excel for determining the height of a projectile. All of the unknown variables' values are highlighted in green and have been given an initial value of 1.

Now I just had to run Solver and tell it that I wanted the Total Residual, which gives the difference between the h(1) equation’s values for height and the actual, measured values, to be as close to zero as possible. A perfect fit of the equation to the data would have a total residual of one, but that’s not possible when you’re dealing with real data.

solving — Table 4. Parameters set in Solver to determine the values of the unknown constants.

Even so, I had to goose Solver a bit for it to produce reasonable numbers. I put in a few constraints:

dt >= 0: We could not have a negative time between pictures.
h₀ <= 1.25: 1.25 meters seemed reasonable for the height at which the ball was released.
t₁ <= 1: It also seemed reasonable that the time when the first picture was taken was less than one second after the ball was thrown.

I ran the Solver a few times, and had to reset dt to 0.5 at one point when it had become zero, but the final result looked remarkably good: the total difference between the modeled line and the actual data was only 0.113 meters.

solver-after — Table 5. Solver's solution for the unknown constants in the height equation.

So we found that:

Initial velocity: v₀ = 14.2 m/s
Height at release: h₀ = 1.25 m
Time between pictures: dt = 0.41 s
Time when the first picture was taken: t₁ = 0.44 s

Which makes the height equation:

$! h(t) = \frac{-9.8 t^2}{2} + 14.2 t + 1.25$

Using these constants in the height equation, we could see how good fit the height equation was to the data:

projectile-graph — Figure 3. Graph comparing the modeled heights (from the h(t) equation) to the actual data.

Maximum Height of the Ball

Finally, the maximum height of the ball can be read off the graph, but it can also be determined using the equation for the height of the ball:

$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + h_0$

We know that the maximum height is reached when the ball stops moving upward and starts to descend. At that point, the vertical velocity of the ball is zero. Since the velocity of the ball is the rate of change of height ( $v = \frac{dh}{dt}$ ) we can differentiate the height equation to get an equation for velocity.

$! h(t) = \frac{-9.8 t^2}{2} + v_0 t + h_0$

$! v = \frac{dh}{dt} = -9.8 t + v_0$

since we’ve determined that the initial velocity of the ball is 14.2 m/s we get:

$! v = -9.8 t + 14.2$

when the velocity is zero (v = 0):

$! 0 = -9.8 t + 14.2$

which can be solved for t to find that the time the ball reaches it’s maximum height is:

$! t = 1.45 seconds$

Putting this into the height equation:

$! h(1.45) = \frac{-9.8 (1.45)^2}{2} + 14.2 (1.45) + 1.25$

gives:

$! h_{max} = 11.58 meters$

Discussion

I’m quite happy with the way this project turned out. The fit between the modeled heights (h(t)) and the actual heights was very good.

My primary concern going into the project was that the distortion from the camera lens would make this technique impossible, but that appears not to be a significant problem.

Most of this calculation, including the somewhat tricky numerical solution using Solver could have been avoided if I’d calibrated the camera, simply by pointing it at a stopwatch (using the same shutterspeed as in the experiment) and measuring the time between snapshots. It will therefore be interesting to see if the actual time between shots (dt) is close to the dt of 0.41 seconds calculated by the model.

Finally, as noted above, a video camera with a timestamp would possibly be a more useful technology for this experiment.

Conclusion

It is possible to analyze the projectile path of an object using a series of snapshots, to determine the initial velocity of the projectile, its release height, and the time between snapshots, if you can assume that the time between snapshots is identical. There are, however, much easier methods of solving this problem.

References

None, but this is where they’d be if I had any.

Appendix

The Excel spreadsheet where all the calculations were done is here.

How to Write Lab Reports

If I have seen further it is only by standing on the shoulders of giants
— Isaac Newton (1676) via Wikiquotes.

Science advances when scientists share their results. If someone tests an hypothesis and finds that it’s wrong, if they share their results, others won’t have to waste time by repeating the same experiments. If someone makes a breakthrough and publishes what they found, then scientists all around the world can use that information to develop new experiments and new applications of that newly discovered principle. Sharing is essential, so it’s important for students to learn how to share well.

Scientists usually communicate their results by giving presentations to other scientists at conferences and publishing articles in scientific journals. Often these presentations are full of the specialized language different types of scientists use with each other, so sometimes science journalists will translate that into regular English news articles that everyone can read and understand. The New York Times and the BBC have good science sections, but what they present comes first from scientists’ formal presentations and articles.

As a result, good presentations and good lab reports are a great way to start learning how to communicate like a scientist.

Lab Reports

A good way to figure out what should go into a lab report is to look at a published article. We have a bunch of copies of Science, which has research articles toward the middle and the back. Articles in Science tend to be brief and fairly dense because it’s one of the premiere journals, so the outlines are not as explicit as you’d find in other places; an Open Access Journal might provide better examples, especially if you’re looking them up online.

Based on our observations, we decided on the following parts for a good lab report:

Title: Be short, but unique to give a good idea of what your project is about. Since my classes seldom all do the same experiment, this is very useful. Answer the questions: What did you do? Why did you do it? and What did you find?
Authors: Who gets the credit for the work. Usually authors are listed by who did the most work first, but since everyone’s expected to work equally on their group projects you can choose some random or arbitrary order.
Abstract: A brief summary of the work, include: what is the problem you’re trying to solve; what you did to solve the problem; and what results you came up with. The abstract should contain all the spoilers.
Introduction: Go into some more detail about what the problem is you’re working on, and why it’s important. State your hypothesis and how you’re going to test it. Overview previous work your project is based on.
Procedure/Methods: Describe, in detail, what you did, what apparatus you used. Both words and diagrams are useful here.
Results: Tell us what you found. Graphs, charts and tables will be very useful here.

evaporative-heat-loss-exp-0038a — Figure 1. An example of a diagram. In this case labels have been placed on a photograph of the apparatus. Notice also the caption, which you are reading at this very moment, that goes with the figure.

Note on Figures: You should have figures, charts, diagrams and tables in your Procedure and Results sections, but you can have them anywhere they’re appropriate. Each figure needs to have a caption explaining the figure. A useful approach to figures and captions is to try to write them so that someone could understand the entire article by only looking at the figures and reading their captions. One of my students says that popular magazines, like People, are written that way (or at least that’s how they’re read).
Analysis and Discussion: To paraphrase a student, “Explain why you think you got those results.” Even if the results are unexpected, or especially if they’re unexpected, you need to explain them. This is also your chance to explain why all of your critics are wrong and you were right all along. If you do that though, it should be written in scientific, passive-aggressive language.
Conclusions: Summarize. In the abstract you’re telling them what you’re going to tell them. In the Introduction, Procedure, Results and Discussion sections you’re telling them. In the Conclusion, you’re telling them what you told them. Hopefully by that time they’ll have had enough chances to figure out what you were trying to tell them.

citing_websites — Figure 2. An example of a citation for a website.

References: Be sure to include a list of the references you used to do your work. This is how you give credit to the people who’s work you are building on. The Yale Library has an excellent page on citing sources. There are a different citation styles you can use but remember the purpose: to give credit where it’s due, and to allow others to be able to find those references easily. All citations should have the author, the date published (or when you accessed it if it is a website), the title, and a way to track down the work.

Note that scientific magazines, like Science and Nature, are very different from a popular magazine like People, for one thing, as was pointed out to me today, the pages don’t smell like perfume (instead they smell like science).

Updates

This paper, on how to bend a soccer ball, is a good example of a student research paper.