Considering LEGO Robotics

lego-bots-02 — LEGO robots at the St. Louis Science Center.

There was a neat little conference today, organized by LEGO’s Education division. I’ve been trying to figure out a way to include robotics in my math and science classes, but since I haven’t had the time to delve into it, I was wondering if the LEGO Robotics sets would be an easy way to get started. It turns out that they have a lot of lesson plans and curricula available that are geared for kids all the way from elementary to high school, so I’m seriously considering giving it a try.

Pedagogically, there are a lot of good reasons to integrate robotics into our classes, particularly as the cornerstone of a project-based-learning curriculum.

The act of building robots increases engagement in learning. Just like assembling Ikea furniture makes people like it better, when students build something the accomplishment means more to them.
Working on projects builds grit, because no good project can succeed without some obstacles that need to be overcome. Success comes through perseverance. Good projects build character.
The process of building robots provides a sequence of potential “figure it out” moments because of the all steps that go into it, especially when students get ambitious about their projects. And students learn a whole lot more when they discover things on their own.
Projects don’t instill the same stress to perform as do tests. Students learn that learning is a process where you use your strengths and supplement your weaknesses to achieve a goal. They learn that their worth is more than the value of an exam.
Projects promote creativity, not kill it like a lot of traditional education.

In terms of the curriculum, Physics and Math applications are the most obvious: think about combining electronics and simple machines, and moving robots around the room for geometry. A number of the presenters, Matthew Collier and Don Mugan for example, advocate for using it across the curriculum. Mugan calls it transdisciplinary education, where the engineering project is central to all the subjects (in English class students do research and write reports about their projects).

I’ve always favored this type of learning (Somewhat in the Air is a great example), but one has to watch out to make sure that you’re covering all the required topics for a particular subject. Going into one thing in depth usually means you have to sacrifice, for the moment at least, some width. The more you can get free of the strictures of traditional schooling the better, because then you don’t have to make sure you hit all the topics on the physics curriculum in the seemingly short year that you officially teach physics.

The key rules about implementation that I gleaned from presentations and conversations with teachers who use the LEGO robotics are that:

Journaling is essential. Students are going to learn a lot more if they have to plan out what they want to do, and how to do it, in a journal instead of just using trial-and-error playing with the robots.
Promote peer-teaching. I advocate peer teaching every chance I get; teaching is the best way to learn something yourself.
2 kids per kit. I heard this over and over again. There are ways of making larger groups work, but none are ideal.

A Plan of Action

So I’m going to try to start with the MINDSTORM educational kit, but this requires getting the standard programming software separately. One alternative would be to go with the retail kit, which is the same price and has the software (although I don’t know if anything else is missing).

I think, however, I’ll try to get the more advanced LabVIEW software that seems to be used usually for the high school projects that use the more sophisticated TETRIX parts but the same microcontroller brick as the MINDSTORM sets. LabVIEW might be a little trickier to learn, but it’s based on the program used by engineers on the job. Middle and high students should be able to handle it. But we’ll see.

Since LabVIEW is more powerful, it should ease the transition when I do upgrade to the TETRIX robots.

The one potential problem that came up, that actually affects both software packages, is that they work great for linear learners, but students with a more random access memory will likely have a harder time.

At any rate, not I have to find a MINDSTORM set to play with. Since I’m cheap I’ll start by asking around the school. Rumor has it that there was once a robotics club, so maybe someone has a set sitting around that I can burrow. We’ll see.

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.

Morning Fog

lake-am-3815c2 — Early morning fog over the lake.

There’s a place on the road to school where you crest a little rise and the St. Albans golf course opens up before you. “Zen-like,” I’ve heard it described. On one lovely fall morning last week the view was absolutely ridiculous. I had to stop.

Resisting the coming winter, warmer air from down south just pushed over the hills overnight, trapping the cooler air in the valley, creating a thermal inversion that trapped a layer of fog just below the tops of the hills. Small tendrils of mist were rising off lake in the bottom of the valley, feeding the fog layer as the cooler valley air condensed the water vapor evaporating off the still warm lake.

Combine the fog, mists, early morning sunlight just beginning to reach into the valley, and the brilliant fall colors contrasting against the still-green lawns, and the result was absolutely amazing.

lake-am-b-300 — Not being able to leave well enough alone, I've also put together this "3d" gif. It does show the fog on the lake nicely though.

Laumeier Sculpture Park

Laumeier-3773 — At Laumeier Sculpture Park.

“It’s one of the places I’m most proud to bring people when they visit St. Louis,” commented (more or less) one of the other faculty on our field trip to the Laumeier Sculpture Park. My hope was that this trip, combined with our visit to the Leonardo Da Vinci Exhibition, would be a nice way to demonstrate that the distance between art and science isn’t so large after all.

Laumeier-3737-s — I believe this sculpture is called, "Balance".

I required all of my students (physics and middle school science) to identify their favorite piece and sketch it. We’ll be covering forces, balance and mechanics in the coming quarter, making this part of the spark-the-imagination part of the lesson.

Also, detailed sketches are not easy. An accurate drawing requires a lot more careful observation than even taking a picture. By trying it themselves they’d get a much greater appreciation of Da Vinci.

The large pieces are quite impressive. The brightly painted metal tank combination at the top of this post just towers over everything. However, one of the neatest was carved out of one enormous piece of wood. It’s made to emulate the distinctive, ridged bark of the cottonwood tree. The artist accentuates the ridges and valleys quite elegantly, making a wonderfully warm and organic abstraction.

Laumeier-3791 — Looking up at the cottonwood tree sculpture.

There’s also an indoor museum (which was closed while we were there), and, apparently, pieces are added and taken away so the park is worth revisiting. The only problem is that you’re not allowed to climb on the sculptures. This is a quite understandable precaution to protect the pieces, but, as some of the students observed, the sculptures “invite” you into and onto them. It’s a stark contrast to the St. Louis City Muesum, which is designed specifically to be played on.

Science Starts with Careful Observation

jar-notes-0274 — Page 1 of my notes: Diagram of our mysterious mixture.

jar-0257 — Sealed jar with a number of unknown substances.

The middle school started science this week with a mysterious jar of unknown substances: a couple immiscible liquids; some plastics and metals of different densities.

As they try to separate and identify the mixture they’ll be learning about handling potentially hazardous materials, material physical and chemical properties (like density and pH), and a little chemistry.

But the first thing they need to learn is how to take notes. Science starts (and ends) with observation. Careful observation. And most middle schoolers need mentoring to make sure their notes are rigorous.

To this end, after they wrote up their observations of what was in the jar, I put together my own notes as a general reference. This is not the only way to take notes, but I’m going to have them amend their own notes to make sure they’re neater and have as much detail as possible.

jar-notes-0275 — Page 2 of notes on the mysterious jar.

jar-notes-0276 — Page 3 of notes on the mysterious jar.

What does, “Good for the Environment” mean?

Recycling-Rates — Recycling rates for drink containers in the United States.

A number of my middle-school students seemed to believe that recycling is the be-all and end-all of environmentalism.

h01_25367113 — In October, 2010, toxic red mud broke through a holding dam and flooded several towns and flowed into the Mercal River. Red mud is a waste product produced when extremely corosive sodium hydroxide is used to dissolve aluminum out of bauxite. In this picture, "A Hungarian soldier wearing chemical protection gear walks through a street flooded by toxic sludge in the town of Devecser, Hungary on Tuesday, Oct. 5, 2010. (AP Photo/Bela Szandelszky)" (image via The Boston Globe)

We were trying to determine what type of material would make for the best drink bottles.

I have a deep reluctance to reflexively consider anything, “good for the environment,” considering that the environmental impact of any particular product is a complex thing to assess. My students, on the other hand, seem to think that recycling is good and all the rest of it can go hang.

I’d want to add up all the environmental costs: the raw materials; the energy input; the sources of the energy input; and the emissions to the air and water, especially all the other external costs of pollutants that people tend not to want to pay for. To my students, these things have been invisible.

Perhaps it’s the success of the environmental movement that’s pushed things to the background. We’re not struggling through smog everyday – although we’ve had some bad days this summer – and even big issues, like the BP oil spill, are a bit remote and seem so far away.

So, I tried showing the Story of Stuff today. It’s definitely a piece with a “point-of-view”, but I was hoping it would be provocative.

h16_25373407 — At least 4 people and many animals were killed. Many of the 120 injuries from the red mud spill were from chemical burns. "Tunde Erdelyi rescues a cat from the toxic sludge in the village of Devecser, Hngary on October 5, 2010. (REUTERS/Bernadett Szabo)" from The Boston Globe.

And it was.

It certainly got a lot of the students agitated, ready to challenge its assertions about just how bad pollution problems really are today, which created a nice opening for me to point out the need for skepticism in the face of any information received. Of course, at that point they were probably a little skeptical about me too, but reasoned skepticism is at the heart of the scientific perspective I’d like them to learn as “apprentice” scientists.

I’d like them to read Orwell too, but that’s another battle.

One student was stimulated enough that, I hope, they’ll actually do a little research into the facts presented in the video and present their findings to the class.

I’ll also have to do a little follow-up on how to argue. In particular I’ll need to post a picture of Paul Graham’s Hierarchy of Disagreement and point out that it’s better to try to refute the actual argument rather than attack the messenger.

We’ll see how it goes tomorrow.

Global Warming Art

Annual_Average_Temperature_Map-art — Global temperatures (averaged from 1961-1990). Image created for Global Warming Art by Robert A. Rohde.

Talk about evoking conflicting emotions. The image is astoundingly beautiful – I particularly like the rich, intense colors – but the subject, global warming, always leaves me with sense of apprehension since it seems so unlikely that enough will be done to ameliorate it.

The source of the image, Global Warming Art has a number of excellent images, diagrams and figures. The National Oceanic and Atmospheric Administration also has lots of beautiful, weather-related diagrams. I particularly like the seasonal temperature change animation I made from their data.