Paintball Spin Physics - Getting to the final Answer

[hitech]

BJJB,

Thanks. I was thinking of using angles, but I'll rethink that. I found another site that had many formulas that were over my head. I think I used the yahoo search engine, advanced search for all the words: vortex shedding sphere force. I'll see if I can find it again.

[ben_JD]

Originally posted by ben_JD
i.e. do vortices shed by prior paintballs further disrupt subsequent paintballs in a long string of shots (indepedent of the subsequent paintball's own vortex shedding issues)?

This is an exaggerated model, due to the objects only being 7 diameters apart, but it is a reference point.

Animation

[hitech]

Okay, it appears that a sphere does NOT perform like a cylinder. We have been basing out arguments on cylinder experimentation, which is inaccurate. The problem is that good experimental data for sphere does not appear to exist. Here is what I found:

Based on these studies it is known that vortex shedding in the sphere wake occurs for Reynolds numbers greater than about300. As the Reynolds number is increased beyond this value the vortex shedding process goes through a series of bifurcations which successively increase its complexity.

One characteristic feature of vortex shedding from cylinders is that every shedding cycle involves the formation of two counter-rotating vortices. As a result of this, the lift oscillates at the shedding frequency whereas the drag oscillates at twice the shedding frequency. Thus the cylinder wake exhibits a strong superharmonic component. In contrast vortex shedding from a sphere at low Reynolds numbers involves the formation of one vortex loop per shedding cycle and thus a significant superharmonic component does not exist. This difference between the two wakes is expected to result in a markedly different response to flow perturbations.

All of this was taken from the following link. BJJB, you might want to check out that link, there is much more there including some formulas.

http://project.seas.gwu.edu/~fsagmae/AIAA-99-3806.pdf

[hitech]

Originally posted by ben_JD
This is an exaggerated model, due to the objects only being 7 diameters apart, but it is a reference point.

Pretty interesting. It seems reasonable that at high rates of fire the turbulent wake of the previous paintball has a damping effect on the wake of the following paintball.

[joeyjoe367]

..so the faster you shoot, the more accuracy you will have?

[ben_JD]

Attempting to post a spreadsheet for you guys with Reynolds calculations.....nevermind, cannot post a .xls file. Look at this:

Reynolds Number = 92,634.533 (Re = U*(D/v))

U = Velocity (m/s) 91.44
D = Diameter (m) 0.017526
v = Viscosity 1.73E-05


Check my work, please...but it looks like a paintball (.69 inches in diameter) traveling through air at sea level at 300fps will have a Reynolds number of 92,634? Seems high to me, but I am not sure. Most of the articles and theses have been focused on <800 Re.

Also, there are 3.28084 feet in every meter for those of you who (like me) often forget to convert to metric before calculating.

[edited to insert URL for viscosity]

[joeyjoe367]

Take a Screenshot, if possible of the spreadsheet and post it up. If you don't know how to do that, PM me.

[AGD]

There is some historical, anecdotal evidence for fast firing rates to produce tighter groups. In the early 90's Paintball Consumer Reports International clamped a minimag in a vise with an auto response trigger and fired full speed at a target.

The results stated that the configuration was the most accurate they had ever tested. This is just a commentary without any verification but I thought it was interesting.

I personally think there might be a chance that a drafting paintball might stay in line better. (thats why I always shoot fast )

Lots of good stuff here guys, this is really facinating!!

AGD

[bjjb99]

Originally posted by ben_JD

Check my work, please...but it looks like a paintball (.69 inches in diameter) traveling through air at sea level at 300fps will have a Reynolds number of 92,634? Seems high to me, but I am not sure. Most of the articles and theses have been focused on <800 Re.

Yep, that's a reasonable value. The formula I used includes fluid density in the numerator (air is around 1.23 kg/m^3), so my numbers will be a bit higher than yours.

A lot of the articles deal with slow moving objects (and thus low Re values) because they are much, much eaiser to collect data from.

For those following the entire thread, I did finally conclude that the variable corresponding to the paintball's dimensions was supposed to be its diameter, not its radius. Thus my 50,000 or therabouts Reynolds number value should actually be 100,000 or therabouts.

BJJB

[bjjb99]

Originally posted by hitech
Okay, it appears that a sphere does NOT perform like a cylinder. We have been basing out arguments on cylinder experimentation, which is inaccurate. The problem is that good experimental data for sphere does not appear to exist. Here is what I found:

[Snip of quoted material]

All of this was taken from the following link. BJJB, you might want to check out that link, there is much more there including some formulas.

http://project.seas.gwu.edu/~fsagmae/AIAA-99-3806.pdf

That's some good stuff there, though it is a bit beyond my understanding in places. I never did learn much about fluid dynamics. However, the 3-d plots at the end of the document were worth a zillion words towards convincing me that at high Re values the vortex shedding around a sphere becomes chaotic. It actually makes sense on a less technical level if you think of it in terms of "where on the surface can vortices be shed?"

On a cylindrical surface, you basically have two choices... a vortex can shed from the top or the bottom semicircle. If the first one sheds off the top, then it is likely that the next one will be off the bottom due to the imbalances created in the airflow. Thus, you end up with a periodic top-bottom-top-bottom... behavior. Slight vortex asymmetries along the axis parallel to the cylinder's axis still result in the vortex shedding from the same side of that cylinder. There's no way for chaotic behavior to build up.

With a spherical surface, you have a whole mess of choices as to where the vortex will shed. When one vortex sheds, any asymmetry in the way it comes off the sphere will influence where the next vortex will shed. A slight shift one side and you're no longer symmetric about the axis of flight, and the next vortex will come off at some goofy angle, and so on, leading to increasingly chaotic behavior. I see it as a physical manifestation of the "butterfly wingbeats ultimately causing a hurricane halfway around the world" concept often tossed about in studies of chaos.

I'll buy that the orientations of vortices shed from spheres at high Re values are chaotic in nature.

Now let's find a good reference for the force imparted by one of these vortices. The last one, while interesting, didn't really help me much since I don't know what I need to multiply their lateral force coefficients by to get a usable force (read "something with Newtons as its unit").

Nice work, Hitech.

BJJB

[AGD]

If you can't find a good factor to calc the lateral force due to asymetric flow then it might be possible to ball park it. If you take a snap shot of the shedding vortex as it's pealing away you will see that it has somewhat of a wing shape.

There should be laminar flow around the outside of this turbulent area causing the area encompased by the ball and vortex to flow air like a wing. In other words, the air flowing around one side of the ball travels farther than air flowing around the other side. This is how a wing creates lift.

If you can find a wing lift program that we can plug the shape into then it should give us some idea of the lifting force. If you can find a real calculation then forget this.

AGD

[AGD]

uppity

[hitech]

Originally posted by AGD
...If you can find a wing lift program that we can plug the shape into then it should give us some idea of the lifting force. If you can find a real calculation then forget this.

Is it safe to assume that you do not have the formula? I didn't have much trouble finding what vortex shedding is, and even that sphere's are different. I haven't been able to find anything on how much force is generated.

[AGD]

Deep Blue guys lets keep posting in the Deep Blue forum and leave this one for general discussion.

thanks

AGD

[Cypres0099]

On the topic of drafting at a high ROF... Have you found at what ROF it has any significant effect on accuracy?