The Flight of a Nerf Vortex Disk

I told you I would return to complete my training. The last time I looked at the Nerf Vortex gun, I just examined the launch speed.

Well, what else is there too look at? Clearly, there is plenty.

Air Resistance

How big of a factor is the air resistance on the flying disk? First, I need some data. Step one, get a child to shoot the gun and record the video. Step two, use Tracker Video to get position-time data for a Nerf shot.

And now for the analysis. Here is the same force diagram I used before for a disk after it was shot.

So, does the disk have the typical velocity-squared dependent air resistance force on it? This is the air resistance model I will test:

I feel like I have said this a million times, but maybe you didn't read those previous posts. The ρ is the density of air. A is the cross sectional area and C is the drag coefficient. One problem is that the air resistance would change if the orientation of the disk changes with respect to the direction the disk is moving. Let me just assume that the disk stays essentially horizontal. Actually, let me look at this problem by just considering the horizontal motion. In fact, this isn't quite correct since as the disk starts to move down, there will also be an upward air resistance force.

Here is a plot of the horizontal position of the disk.

Here I have a linear fit to the data (which isn't a a terrible fit). This gives a slope (x-velocity) of 11.8 m/s. Strange. This doesn't quite agree with my previous finding of a launch speed around 15 m/s. Maybe there was some wind. Well, if I can get one data point for velocity I can get a whole bunch. Here is a histogram of the different speeds. Oh, this is just the x-velocity. I assumed all of the shots were level even though that is not likely true.

This distribution has an average of 10.99 m/s with a standard deviation of 0.51 m/s. Seems to be launching at a fairly consistent speed (except for that one shot). Not sure why my indoor speed is so much different. Alas, I must have made a mistake. I doubt the difference is due to wind because this seems it would cause a larger variation in the speed.

Back to air resistance. Since the horizontal speed doesn't change too much, the air resistance force should be fairly constant. Let me fit a quadratic function to one of these x-position plots as though the disk had a constant horizontal velocity. What the heck, I will do this for all of the above shots.

So, this is an average horizontal acceleration of -2.87 m/s² with a standard deviation of 0.925 m/s². Ok, not perfect data but I am going with it anyway. If I assume that the air resistance is the only force in the x-direction and that the velocity is mostly constant over this short time frame (mostly true), then I can write:

I just so happen to know that the mass of a disk is about 2.47 grams. Also, I measured the dimensions.

The disk seems to have a width of about 3.94 cm and a height of 0.94 cm. Now, I can solve for the one thing I don't really know - the drag coefficient for this shape.

I guess I need to estimate the area. The cross sectional area is almost diameter times height, but not quite since it is rounded a bit. Let me just go with a value of 3.5 x 10^-4 m². I will also use a density of air of 1.2 kg/m². Plugging these in, I get a drag coefficient of 0.279.

But wait! There's more! Why stop there? Why settle for a coefficient like that? What I need is the uncertainty in this drag coefficient. Let me assume that both the velocities and the horizontal accelerations are normal distributions. I can then use the Monte Carlo method for error propagation to determine the uncertainty in the drag coefficient. Basically, I will randomly generate 1000 velocities (normally distributed with a standard deviation of 0.51 m/s) and 1000 horizontal accelerations. Then I will calculate the the drag coefficients 1000 times and look at the mean and standard deviation of this data. Oh, when I say "I will calculate" I really mean "I will make my computer calculate".

Here is a plot of the 1000 drag coefficients that I (my computer) calculated:

This gives an average drag coefficient of 0.281 with a standard deviation of 0.095 (no units on the drag coefficient). But is this a reasonable value? I guess so. It seems in the range of other values. In particular, a sphere shaped object would have a drag coefficient of about 0.47. Mostly, I am happy.

Lift

Oh, you thought I would end with the drag force? No. I must go on. So, what about this lift force? Really, this is what makes the vortex Nerf gun rather cool. If you have just a normal Nerf dart gun, the only way to increase the range is to shoot the dart with a higher speed. Of course higher speed means the dart will be a little bit more dangerous. Also, higher speeds on the darts means a higher drag.

With the vortex disks, the range of the launched disks can be increased by making them "fly" in essentially the same way a frisbee flies. I am not too sure of the best way to model this lift force, but I would assume it would depend on the speed of the disk. Unfortunately, the speed of the disks doesn't change too much in the data that I collect. Well, the best I can do is to calculate this lift force for the above shots.

If I assume the lift force is only in the vertical direction (probably not a terrible assumption), then I can write the following force equation:

I already know the mass. I already know the gravitational constant (g) and I can measure the vertical acceleration from the videos. Here is the distribution of accelerations from all the shots.

So, an average of -2.73 +/- 0.37 m/s² (where the plus-minus part is the standard deviation). Using the mass from above, this gives a lift force of 0.0174 Newtons. I know it would be cool to have a better model for the lift force - you know, see how it changes with speed. Oh well. I guess I will have to come back to this thing once again. Maybe I can throw some frisbees and look at their motion.