Another half-year has come and gone and with the Holliday’s just around the corner you know something great is almost here. The Winter Exhibition. What, you thought I was talking about Christmas (or whatever you celebrate at this time of year)? HA. I laugh at your misjudgment.
If you read my post about last years exhibition, you should know how this goes. We choose an inquiry question based around a topic for your grade. This year, Star Wars! The grade 8’s and the grade 9’s worked together on this one (already an improvement from last year), and we had a light side and a dark side. The light was in the PLP room, and the dark was in the library.
![](https://www.blog44.ca/owenh/files/2017/12/1C1C15F2-5640-43B9-9EB4-F7781D1D9124-1vvx0rm-300x181.png)
We were chosen for each side based on our questions that we came up with. There were questions about the morality of using a Jedi mind trick on people, or real life lightsabers, or wether or not Star Wars weaponry would overrun our current weaponry, stuff like that. I chose an incredibly unique question (at least I hope). My question was “Could primitive technology destroy a modern armoured vehicle?”. I looked at the scene in Return of the Jedi when the ewoks destroyed an AT-ST by smashing it with logs. Would that be possible in real life?
To find out this answer, I need to calculate a few numbers. The first is the joules of energy, requiring the mass and velocity of the logs. The second is the joules per square centimetre, calculated by finding the square area of the surface of the logs. Finally, the joules per square centimetre the armour of the AT-ST can take before it breaks.
JOULES
-Mass
The very first thing I needed to do was find out the mass of the logs. To find that, I needed to know how big they were. Here’s where things get interesting. To find the size of the logs, I needed to know how big something in both of the frames I was looking at is. I chose the window of the AT-ST. But to find that, I needed to find the height of the man I could see in the frame. I took the average height of a Canadian male, assumed their height were the same, and went from there. The average height of a Canadian male is 176 cm. We can only see half of the man in the shot, so we half the height to 88 cm.
![](https://www.blog44.ca/owenh/files/2017/12/E2F764D5-4C92-435B-9811-672BC9525D17-1vb0k7s-284x300.jpeg)
Now, stick with me here, I need to make my own unit of measurement. I call it “IRL” standing for In Real Life. An IRL cm is one cm on my ruler, and IRL is used when measuring something on a picture.
Now then, back to business. The 88 cm of the man is equal to 4 IRL cm, meaning that 1 IRL cm is equal to 22 cm. The windows side is 2 IRL cm, or about 50 cm. From there, we can move to a different picture with different scaling.
On our new picture, the window side is still equal to 50 cm, but is equal to 2.5 IRL cm. The log in this frame is 5 IRL cm, making it 100 cm wide, or one meter. Knowing this, we can calculate the circumference of the circle and the length of the circle. First is the length. Similar to how we calculated the diameter of the log, we can find the length of the log by taking the 5 IRL cm equalling 100 cm, then find how many IRL cm wide the log is in our new frame.
![](https://www.blog44.ca/owenh/files/2017/12/EFA431A6-2433-4475-931F-6EFAEDD215C6-2ia121v-300x211.jpeg)
On this frame, the logs 100 cm width is now 4 IRL cm. If 4 IRL cm is equal to 100 cm, then the logs length of 260 cm is equal to 8.6 IRL cm.
OK, now that we know the length, we need the circumference. With the width (diameter) being 100 cm, we can find the circumference by using 2 times PI times Radius. With the radius being 50 cm, multiplying 50 by PI gives us 157 cm. Doubling that gives us the circumference of 314 cm^2
And finally, the log is a cylinder, so we need to calculate the volume. The log is 100 cm in diameter by 260 cm in length with a circumference of 314 cm^2. Using this we can calculate that the volume is 2.04^3, or 8.489 m^3.
To find the mass, we need to take the weight of Redwood (the weight in KG/M^2) and use that to calculate the weight of the log.
Velocity
Alright, now we know the mass to be 918 kg, now we need velocity. To calculate velocity, we need to do some simple frame by frame measurements. After doing just that, i found that the log travels 1.4 IRL cm per frame, or 8.4 IRL cm over 6 frames. In the 30 Frames Per Second of the movie, the log travels at 14.58 M/S. Finally! Now we can calculate the last bit, Joules.
Energy
To calculate the Joules of the log, we need to fill in this formula “E=1/2M(V^2)” with our numbers. Half of mass is 459 kg. Velocity squared is 212.576 M/S. 459 times 212.576 is equal to…
97,572.56 Joules of energy.
Holy jeez, thats five times the energy of a normal armour piercing round! But is it enough when applied across one square meter? Who knows?
Well, I do. We need to find the Joules per square centimetre. To find that, we divide the 97k Joules of energy by 100, then divide by 100 again. Doing that gives us a slight problem. The log isn’t a square, so we cant say the log is a square meter. A perfect circle is about 75% of a square, so we make the 9.7 J/cm^2 into a whopping 12 J/cm^2. To put that into perspective, a single joule is similar to moving an apple 1 meter in one second. Now move 12 apples 1 meter in one second. That’s so lame, im not even going to se if the armour can withstand it. A bullet from a hand gun has more J/cm^2, and most bullets aren’t even armour piercing.
https://youtu.be/XZBElpLN7GQ
What would it take?
A 9mm round from a handgun hits with about 692 J/cm^2. The smaller area and much higher velocity boost the Joules up tremendously. So what would it take for the log to pierce the armour? We could do one or two things. We could decrease the surface area of the log by making it into more of a cone shape. Doing this would increase the J/cm^2. And/or we could increase the velocity by dropping it from a much greater height. Both of these would increase the J/cm^2 by quite a bit, and would prove more effective at destroying the AT-ST.
THE ROOM
Light side
I was on the light side in the PLP room, and the dark side was in the library. I dont know much about the dark sides decor, but the light side was pretty awesome. We split the room into three sections. The first was the outside of a rebel base on Hoth, and the other two section we inside the base. In these sections, we were with people that had a similar topic to our own. We had environment (outside on Hoth), Tech (Inside the base) and philosophy(Also inside). I was with Robin, Isabelle H, Isabelle L, Sam, and Kiera in the Tech section. In out section we had our snack on “The Admiral Snackbar”, and a game we called “pin the lightsaber on Yoda”.
We did have a schedule for people to be running the game, but my project was right next to the game, so I was in charge of running it when someone else couldn’t. We also had skits that certain people would act out. Two were scenes from the movies, and the rest were improv. The two scenes were Leia’s “Help me Obi-Wan Kenobi” transmission and when Jyn Erso first meets Chirrut in Rouge One.
During the preparation for this exhibition, I learned some new and useful things. Obviously I learned all the things I needed to to calculate the answer to my question. But I also learned a few other things. First, I learned to be more accepting of others ideas. When the Grade 8’s suggested the improv, I was worried at first. I felt that it would end up going wrong and they would do horribly at it. But after seeing the improv, I thought it was good. While others definitely don’t share my opinion, the grade 8’s did a lot better than I would have done.
I also learned that having a lot of team members can help a group go along way. Comparing last years exhibition to this years, we did a lot worse last year. The grade 9’s (last year) didn’t help us at all, so we had no idea what to be doing.
I think that with that knowledge and with each of the grades helping, the June exhibition is going to be incredible!