One of my students showed me this game, [Fantastic Contraption](http://fantasticcontraption.com/). The basic idea is to use a couple of different "machine" parts to build something that will move an object into a target area. Not a bad game. But what do I do when I look at a game? I think - hey! I wonder what kind of physics this "world" uses. This is very similar to [my analysis of the game Line Rider](http://scienceblogs.com/dotphysics/2008/09/the-physics-of-linerider/) except completely different.

Fantastic Contraption gives the unique opportunity to build whatever you want. This is great for creating "experiments" in this world.

The first step is to "measure" some stuff. The game includes three types of "balls" and two types of connectors. The balls are:

- Clockwise rotating
- Counterclockwise rotating
- Non-driven

Connectors:

- wood lines - these can not pass through each other
- water lines - these can pass through each other, but not the ground

First question: Do the different balls have the same mass? This can be tested by creating a little "balance"

![Screenshot 05](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Now, I can test this by adding two of the same balls on each side (well, one on each side). It is still balanced. Now for two different types of balls:

![Screenshot 06](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Note: the blue ball does not spin and the yellow is a clockwise spinner. They look balanced. What about a blue and a conterclockwise spinner? Still balanced. So, it appears all the balls have the same mass.

What is the linear mass density for the two types of sticks? To measure this, I created a device with a ball at one end and the pivot NOT in the center, but it still balances:

![Screenshot 10](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Here you can see three forces acting on the device: the gravitational force on the ball, the gravitational force on the stick, and the pivot point pushing up. Since the stick is clearly not a point object, I have to draw it's gravitational force at the center of the stick. (I am not going to derive that right now, you will just have to trust me).

Newton's laws says that the forces must add up to the zero vector if the object is staying at rest. This means (in the y-direction, where y is up):

![Screenshot 11](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Here m_{s} is the mass of the stick and m_{b} is the mass of the ball. This would make the the gravitational pull on the ball -m_{b}g (notice it is the y-component, so I can have it negative). From all of this, I could solve for the force the pivot pushes on the balance, but what good is that? What I am really looking for is the mass of the stick. To do this, I need to consider torque. Here is the real definition of torque:

![Screenshot 12](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

This definition is a little more complex than I want to go into (but I had to say it). The torque is technically a vector resulting from the cross product of a force and a vector from the point of rotation to the point the force is applied. The scalar version of torque can be written as:

![Screenshot 13](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Here, r is the distance from the point that you want to calculate the torque about (I chose the pivot point) and the point where the force is applied. ? is the angle between the force and the distance to point about which to calculate the torque. In this case, the angle is 90 and sin(90) = 1. Another important consideration is the sign of the torque. I will arbitrarily call counterclockwise torques positive and clockwise torques negative.

So, how do I use torque? Well, I need to know the distance from the pivot point to the center of the ball and from the pivot point to the center of the stick. I can use [my favorite free video anlaysis program, tracker,](http://www.cabrillo.edu/~dbrown/tracker/) to do this (even though it is just an image)

I will use the diameter of one of the balls as my unit (from the center of an attachment point circle to another one). Doing this, I get the distance to the ball and the center of the stick as:

![Screenshot 15](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

- Here I am using "U" as my distance unit - described above.

- To find the distance from the pivot to the center of the stick required some trickeration. I measured the length of the stick. I then used half that distance and measured from the one end of the stick to find the center. Knowing that point, I could then measure to the pivot point. Using these measurements in the torque equation:

![Screenshot 16](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Note that the torque due to the pivot does not contribute at all. This is because I calculated the torques about the pivot point. The distance from the pivot point to the pivot point is zero (thus zero torque).

So, I have the mass of the stick in terms of the mass of the ball. I can also get the linear mass density of the stick:

![Screenshot 17](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Cool - I should stop here. No!!!! I am on a roll. I will now calculate the linear mass density for the "water" stick. I can't do quite the same thing because the water would fall through the pivot. Instead, I will do the following. First, I will make a stick with two ball (one on each end) balance. Then I will replace one of the balls with "hanging" water so that it is still balanced. At this point, the mass of the water stick will be the same as the ball (I could have done this with the wood stick if I had thought of it then).

![Screenshot 18](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

You may not be able to tell, but this is two overlapping full water sticks and one shorter one. I will have to combine the length of all of these. This gives a total length of water = 8.5 U. So, the linear mass density for water is:

![Screenshot 19](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

Interesting. The linear density is half that of the sticks. Must be dense sticks. I tried putting a wood stick versus a water stick that was twice as long - they balanced.

**Acceleration of falling objects**

Do things accelerate? Is there air resistance? I created an engine that just kind of "flung" a ball up. I used [copernicus](http://www.danicsoft.com/projects/copernicus/) to capture the video from the screen. Then [tracker video](http://www.cabrillo.edu/~dbrown/tracker/) to get position time data. Here is what I found:

![Screenshot 20](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

This shows that it does indeed acceleration. Using [the ideas from a previous post on graphing](http://scienceblogs.com/dotphysics/2008/09/basics-making-graphs-with-ki…), the acceleration of the object is twice the coefficient in front of the squared term, this means that:

![Screenshot 21](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

If this is on Earth, then this acceleration should be 9.8 m/s^{2}. With this assumption, I can find the conversion from U to m:

![Screenshot 22](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screensho…)

**What is left?**

Questions to answer:

- Is there air resistance? From the above data, maybe not. To test this, I need to launch a ball with very high speed. If the horizontal velocity changes, then there is likely air resistance
- Make a pendulum, does it oscillate at the expected rate (assuming the dimensions from here)? I already started to set this up, but there is CLEARLY some type of frictional force slowing it down.
- Friction - what is the coefficient of friction? Does this game follow the model for friction where the frictional force is some coefficient times the normal force?
- What kinds of torque are these rotating balls capable of
- What is the moment of inertia of these balls? Are the cylinders or spheres?

I will probably answer some of these questions - but if someone ones to answer them first, I will gladly link to your results OR post them here.

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Wow, nice work. This is why I like this blog, as opposed to many others. There's a genuine interest in the science.

There is a list of physics parameters for this game here: http://fc.therisenrealm.com/wiki/List_of_physics_parameters

BadarZ,

I am not sure if I want to look at that link. I want to look, but I don't want to spoil the fun.

Hi, Can you give a clue about how to justify that on the balance with one rotating and one non-rotating ball there will be equilibrium? Thanks.

Zeynel,

From my understanding, the rotating ball does not rotate if it is not connected to the center of the ball. If it does rotate, clearly angular momentum is not conserved (which is something I did not think about before - but clearly the rotating balls do not have angular momentum due to their rotations).

"This shows that it does indeed acceleration."

Maybe you should spend time brushing up on your English instead of wasting your obvious Physics talent with blogs like this...

@Gfunk - Looks like you should be the one writing a grammar blog. Nice work, Sherlock, finding the single grammatical error in the entire document.

If you like this, look up the game "Armadillo Run"

@Gfunk you are so cool. so so so cool. your obvious coolness is so cool.

Very very cool. I played Fantastic Contraption for ages, it's cool to see how they apply actual physics to the game and how you can still calculate these things. I made a post with some cool solutions, and looking through them again it's cool to see some of these effects:

http://xatal.com/miscellaneous/50-more-unique-fantastic-contraption-sol…

I like the investigation =] It's really nice to see other people interested in the derivation. Check out the tutorials I'm starting currently on particle-based physics. We try to delve a little more into the math and science than most. I'll keep subscribed to you!

http://blog.brandonpelfrey.com/?p=58