Dot Physics

Physics of Linerider III: Air Resistance

There is no air resistance in line rider. Sorry to spoil the suspense.

To test for the presence of an air resistance force, a track was created that let the rider fall.

![linerider air 1](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/linerider-air-1.jpg)

(note the markers on the side. These are used to keep track of how the origin is moving).

Below is the y position of the rider as a function of time:

![linerider falling](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/linerider-falling.jpg)

In this situation, the rider falls about 100 meters. A quadratic line is fit to the data and an acceleration is obtained that is very similar to the previous case (where air resistance was assumed to be negligible). If there had been air resistance, this graph would have become more linear as the rider fell. Perhaps 100 meters is not far enough to fall, but in real life this should be far enough to detect the presence of an air resistance force. Or does it? Lets make a simple check.

Let’s assume the line rider is a sphere with a diameter of 0.75 meters (since his sled is 1 meter long, it is probably not as wide). When an object falls in the presence of an air resistance force, we can draw a diagram representing the forces (we like to call this a free body diagram).

![freebody for linerider](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/freebody-for-linerider.jpg)

This is a complicated situation to analyze because the air resistance force depends on the velocity (which depends on the forces acting on the line rider). Perhaps the easiest method to examine the motion of the rider (with air resistance) is using numerical methods. In a numerical solution, the problem will be broken into many small time intervals. During each time interval, the forces won’t change too much so we can assume they are constant. The only problem with this approach is that there will be many, many small problems to solve. To solve these many little problems, we could employ a 4th grader to do all these tedious calculations, or we could use a computer. I vote to use a computer as our labor force. Might as well use them now before they take over the world (you know, like in the movie Terminator or the Matrix).

Here is the basic recipe that will be used to look at the velocity of a falling object with air resistance:

**1./ Calculate the forces on the rider (this will be gravity and the air resistance force). The gravitational force near the surface of the Earth is simply proportional to the mass of the rider (more on this later). The air resistance force will be proportional to the cross sectional area of the rider as well as the velocity squared.**

**2./ Update the momentum using the momentum principle: (I will write it for just the y-direction so that it is a scalar equation)**

![momentum](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/momentum.jpg)

Where momentum (p) is:

![py](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/py.jpg)

**3./ Update the position. This can be accomplished by rearranging the expression for the y-velocity:**

![delta y](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/delta-y.jpg)

Again, this assumes that the time interval is small

**4./ Rinse, add conditioner and repeat.**

I know this looks like it is cheating, but it works.

Air resistance force
For the magnitude of the air resistance force, we can use the following model

![air resistance model](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/air-resistance-model.jpg)

Where rho is the destiny (I am your density, I mean…your destiny) of the fluid (in this case, air has a density of approximately 1 kg/m3
A is the cross sectional area of the object
C is the drag coefficient. The drag coefficient for a sphere is 0.1
v is the magnitude of the velocity.
The direction of this force is in the opposite direction to the velocity.

For this comparison, A will be approximated as a rectangle 1 meter by 0.4 meters (I completely made that up – well, not the 1 meter)
The drag coefficient is more complicated to guess. According to the ultimate source of truth ([wikipedia](http://en.wikipedia.org/wiki/Drag_coefficient#Cd_in_other_shapes)), a smooth brick has a coefficient of 2.1. For this calculation, a coefficient of 1 was used.

The mass of the falling object is also required. According to that child growth chart, a 5 year old is about 19 kgs. Add the sled and the mass can be approximated as 24 kg (again, a made up number).

Here is the program to calculate the position as a function of time for both an object with air resistance and without. The program was written in python using the [VPython modules](http://www.vpython.org).

Here are the results:

![comparing data](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/comparing-data.jpg)

Note that the numerical model without air resistance and the line rider data a little off. This is likely due to dropped frames in the movie of the line rider.

Another method to test for air resistance is to look at the horizontal motion. The line rider starts with some initial velocity in the horizontal direction. If there is no air resistance, this velocity should remain constant (because there are no forces acting in the horizontal direction). Below is a graph of the x-position from both the line rider data and the numerical model with air resistance.

![position time](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/position-time.jpg)

It appears as though the line rider data shows a mostly constant horizontal velocity. There is a transition in horizontal velocity (from 0.73 m/s to 1.52 m/s) that occurs a little before 1 second. The only thing that I can think of that this would relate to is when the video transitions from showing the line rider moving to having the background move.

The point is: clearly the line rider data is more straight that curved like the numerical data.

I claim there is no air resistance in the line rider game. To further test this, one would need to let the rider fall for a much longer time, but I was too impatient to do that.

Comments

  1. #1 andrew
    January 14, 2009

    Fantastic

  2. #2 andrew
    January 14, 2009

    Fantastic. I really needed to know this for line rider…

    Doing A-Level Physics so I understand everything here!

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