Who doesn't love Angry Birds? Well, according to The Oatmeal, the green angry bird is the least likable. Here is how the green bird is described (click through to read the full comic - it is pretty funny)

Ok, let me assume you don't know about the green angry bird. Basically, when you throw it, it goes like any other bird. However, when you tap it the bird changes its motion in a way that it usually comes back somewhat. I like to call it the boomerang bird, but it isn't actually a boomerang.

How does this thing work? What is its motion like after it is tapped? Does it have a constant acceleration? Does the post-tap acceleration depend on when it was tapped? To explore these questions, I made some videos of the green bird in action. The trajectory of the bird can be determined using Tracker Video analysis. On to the physics.

Perhaps I should point out something important and useful. From my previous analysis, I found that the height of the big sling shot was 4.9 meters. I will need that.

Here is my first plot. This shows vertical position vs. time. It includes a fit to the part of the motion before the tap. Note that the green bird went off screen for a short bit.

This shows an acceleration of around 10 m/s2. Close enough for me (there did seem to be excessive zooming for this shot - that makes things a little bit more challenging). Now, what about the motion after the tap?

This looks like a fairly constant vertical velocity after the tap. In this case, the y-velocity seems to be -11 m/s. In the x-direction, the green bird seems to have a constant velocity before the tap (in this case, about 16.9 m/s). Maybe there is now a constant horizontal acceleration after the tap. This quadratic equation seems to sort of fit.

This would give a horizontal acceleration of -32.8 m/2. I think I need to look at some more trials. For this next "throw", the bird once again has a constant y-velocity after the tap (-8.65 m/s). In the x-direction, the bird has a velocity of 23.7 m/s before the tap. Here is a fit after the tap.

This looks like a constant acceleration of 54.4 m/s2. Ok, here is my guess. After the tap, the green bird has zero acceleration in the y-direction and the x-direction has an acceleration of - 2*v x that the bird had before the tap. This seems to work for these two runs, but I should look at some more runs.

Let me go ahead and admit that collected data on this dumb is not trivial. First, the zooming of the screen is a pain in the rear. You can "zoom out" before you shoot the bird (and I don't mean 'shoot the bird'). However, if you download examples from online you can't control the zoom. Also, it turns out that making a video of my iPod is not as easy as it sounds. It is a careful balance of getting the right distance and having the video camera on the right settings. Here is a collage of some of my shots. (oh, these were from Angry Birds Seasons - Halloween)

From all this data, I collected:

Initial x and y-velocity.

The time after the launch that I tapped the bird (not sure if I will need it).

The x-acceleration of the bird in the time around the "tap".

The x and y-velocity of the bird a long time after the tap (because it seems constant).

Let me go ahead and point out something obvious. The motion of the green bird can be broken into three parts. First, there seems to be a pretty normal projectile motion phase (before the tap). Then, when you tap it, there is some x-acceleration. For example, just look at this plot of a typical green bird in motion.





From both of those, it looks like a constant velocity some time after the "tap".

So, what about the tap part? Here is the graph that took so long for me to create. This is a plot of the initial (pre-tap) x-velocity vs. the during the tap x-acceleration. Oh, this is for 10 different green bird shots (on the same level).

As you can see, the data seems a little rough. What helps a bunch is to have a 'backwards shot' green bird. For those two data points, the initial x-velocity is negative. This makes the fitting function seem to work. The fit linear fit for this data gives a slope of 2.3 s-1 (yes, that is the correct units) with an intercept of 0.06 m/s2. Not too bad. So, if this model works, then during the tap, the x-acceleration is:^^

I am not too sure about the duration of this tap acceleration, but I have a way to find that out. Also, I am not too sure about the y-acceleration during this time. Is it still -9.8 m/s2? Or maybe it is 0 m/s2. If I know the time of duration for the tap acceleration, I can find the y-acceleration. Here is the plan. First, I know v x-1 v x-3 (where the "1" is for before the tap and 3 is after). From the definition of acceleration, I get:

If the time for the tap is constant, then I should be able to plot v x1 vs. v x3 - v x1 and it should be a straight line. Here is that plot.

The linear function that fits this data has a slope of -0.42 (let me say no units). According the relationship above, this would mean the slope is:

Note: I know the units look weird. That is because that "2" actually has units with it. Remember that the acceleration was -2 times the velocity. So the "2" must have units of 1/s. This gives the correct units for time of seconds. So, putting all this together I get a tap-time of 1.19 seconds. Well, I was hoping for some nice number like "1".

Now, back to the y-velocity and y-acceleration. Let me assume the change in motion in the y-direction takes the same amount of time as the x-direction. That is to say that the tap time is the same for both x and y. If this is the case, then I can use the tap time to find a y :

Clearly, I should plot v y1 vs. v y2 . If the acceleration is constant, then the slope of this plot should be 1 and the intercept will tell me something about the acceleration. Before I make this plot, I need to know the y-velocity right before the tap. I recored the initial y-velocity as well as the time the "tap" took place. From this, I can find the velocity right before (which I will call v 2 ) with:

For this case, t is the time from the launch to the tap. Now, making the plot (with v y2 instead of v y1 ):

This doesn't look very linear. I guess I will have to look at this with some better data. Ok, but that will be another post.

Testing the Model —————–

Going back to the horizontal acceleration model. Here is a test. If my model works, then what should happen if I shoot the bird straight up with no x-velocity? Unfortunately, you can't actually do this in Angry Birds since the sling shot gets in the way. You can sort of do it though. I am not going to tell you the answer to this question, you will have to try it out for yourself. Oh, also just so you know. I am going to have to also come up with a model of how to effectively use this angry green bird. Maybe then it won't be so hated.

Summary ——-

For the green bird, the motion consists of a projectile motion phase, a "tap" phase, and a post-tap phase.

For the tap phase, it seems the horizontal acceleration has a value of twice the initial x-velocity (but in the opposite direction).

This tap acceleration lasts for about 1 second.

After the tap, the bird moves with a constant velocity (constant in both the x- and y-direction)

I am still uncertain of the y-acceleration during the tap.

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