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125 foot accuracy test

#61 User is offline   Leftystrikesback 

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Posted 26 March 2009 - 02:01 PM

View PostJack Wood, on Mar 26 2009, 03:55 AM, said:

That's what I was trying to say in the (one of man) discussion we had on this subject in another thread. Was it the Hammerhead thread? Preferential shedding.

So, how do we try and get a look to see if this really happens? I feel that even a very slow rotation that is near-perpendicular to the axis of flight could lead to small degrees of preferential shedding. Some college geek somewhere must have written a paper on the influence on vortex shedding in rotating spheres in a flow.........


That's a great idea, I just found a ton of articles on Compendex having to do with Vortex shedding in spheres, luckily I'm still a student and can download most of the articles for free. I'll spend some time looking at them.

View Postcockerpunk, on Mar 26 2009, 07:20 AM, said:

if you haven't already, you gotta check out the first video showing vortex shedding.

this guy is using lazes to illuminate micro hydrogen bubbles to see the flow. its freakin sweet.


Where's the link to this? It sounds sweet!
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#62 User is offline   Spitlebug 

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Posted 26 March 2009 - 02:42 PM

Youtube on cockerpunks channel.

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#63 User is offline   Jack Wood 

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Posted 27 March 2009 - 05:04 AM

Can't find it.

http://www.youtube.c...h?v=_AJgEa2dbJU

You talking about this one?
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#64 User is offline   Jack Wood 

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Posted 27 March 2009 - 05:15 AM

View PostLeftystrikesback, on Mar 26 2009, 07:01 PM, said:

View PostJack Wood, on Mar 26 2009, 03:55 AM, said:

That's what I was trying to say in the (one of man) discussion we had on this subject in another thread. Was it the Hammerhead thread? Preferential shedding.

So, how do we try and get a look to see if this really happens? I feel that even a very slow rotation that is near-perpendicular to the axis of flight could lead to small degrees of preferential shedding. Some college geek somewhere must have written a paper on the influence on vortex shedding in rotating spheres in a flow.........


That's a great idea, I just found a ton of articles on Compendex having to do with Vortex shedding in spheres, luckily I'm still a student and can download most of the articles for free. I'll spend some time looking at them.

View Postcockerpunk, on Mar 26 2009, 07:20 AM, said:

if you haven't already, you gotta check out the first video showing vortex shedding.

this guy is using lazes to illuminate micro hydrogen bubbles to see the flow. its freakin sweet.


Where's the link to this? It sounds sweet!


I found what looked like a good number of articles on the subject. Most are at lower Reynolds numbers though. The problem was that they are not the cheapest thing to pick up, and you don't know what you're going to get until you pay for it.

So if you have free access to most of them, then get them downloaded and give us an insight into stability of vortex shedding due to rotation of a sphere :)
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#65 User is offline   Snipez4664 

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Posted 27 March 2009 - 07:38 AM

I'm not sure I would bother with anything below the transition regime for reynolds numbers - the consensus is that you can 'freeze' the vortex shedding with tranverse rotation at a few harmonic frequencies. The article by KW Poon has some amazing images guaranteed to set you thinking...though the highest Re tested was 250. http://espace.library.uq.edu.au/eserv/UQ:1..._afmc_16_07.pdf

A more exhaustive treatment can be found at: http://dtl.unimelb.edu.au/view/singleviewe...2.jsp?frameId=1

Again, similar findings. Note that OVERSPIN actually creates WORSE oscillatory vortex shedding (should be a wider gaussian, number of steps is higher?).

There is also an article suggesting that streamwise (I think we tend to think of it as rifled) rotation delays the onset of vortex shedding by tilting the vortex loops back into the steam wake for a while (remember talking about the surface point of view? It does hurt me to be so insightful, yes. :-/ ). http://www.flair.monash.edu.au/publication...ur_mgfc2001.pdf

The other thing that falls out of almost all of these vortex shedding studies at low Re is that the wake looks like it does walk around the back of the ball. Unfortunately, none of this has a practical application, since there are a couple of transitional regimes between here and there.



Most interesting thing I found:

"In contrast, there was little evidence of vortex shedding in the supercritical range (ReD> 4 x 105), consistent with many earlier observations in the literature; however, flow visualization studies in the near-wake clearly showed the existence of a three-dimensional vortex-like structure exhibiting random rotations about the streamwise axis. In this range of Reynolds numbers, surface flow visualization studies indicated the existence of a laminar separation bubble which was followed by a transitional/turbulent reattachment and an ultimate separation around 0S = 145°."


That's pretty interesting. The cool thing is a lot of this has been done since TK left...
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#66 User is offline   Spitlebug 

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Posted 27 March 2009 - 05:08 PM

Mmmmmm spermatozoa...

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#67 User is offline   Jack Wood 

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Posted 30 March 2009 - 07:38 AM

View PostSnipez4664, on Mar 27 2009, 01:38 PM, said:

Most interesting thing I found:

"In contrast, there was little evidence of vortex shedding in the supercritical range (ReD> 4 x 105), consistent with many earlier observations in the literature; however, flow visualization studies in the near-wake clearly showed the existence of a three-dimensional vortex-like structure exhibiting random rotations about the streamwise axis. In this range of Reynolds numbers, surface flow visualization studies indicated the existence of a laminar separation bubble which was followed by a transitional/turbulent reattachment and an ultimate separation around 0S = 145°."


That's pretty interesting. The cool thing is a lot of this has been done since TK left...


I'm struggling to vizualize what they mean. Are they saying there is something akin to a spiral vortex behind the sphere? And what does a laminar seperation bubble look like?

And are they saying that at high Reynolds numbers, there is not the type of Vortex shedding that we have been discussing?
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#68 User is offline   Leftystrikesback 

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Posted 30 March 2009 - 07:30 PM

I looked through some research articles and I found a few things worth noting:

I think we're looking at what's called "subcritical flow" where the Re is high, but not high enough for the drop off in drag associated with very high Re and a fully turbulent boundary layer.

First off, the near wake mechanism for subcritical spheres from "PIV analysis of near-wake behind a sphere at a subcritical Reynolds number", Young Il Jang, Sang Joon Lee

Quote

The flow around a sphere shows steady axi-symmetric flow in the range of Reynolds number 20–210. The axisymmetry
is then broken, and planar-symmetric flow appears until Re = 280. From Re = 280, unsteadiness
starts to occur in the planar-symmetric flow, and hairpin vortices are periodically shed. In the range of Reynolds
number 420–800, asymmetric flow is observed, and unsteadiness continues (Taneda 1956; Nakamura 1976;
Wu and Faeth 1993; Johnson and Patel 1990; Leweke et al. 1999). Many experimental and numerical research
works on sphere wake have been carried out in this Reynolds number range to study laminar flow separation
and laminar wake. On the other hand, from Re = 800, the large-scale low-frequency vortex shedding and small-scale
high-frequency shear layer instabilities become prominent flow phenomena. A large-scale vortex is shed with wavy
shape, and turbulence occurs in the far field. At the critical Reynolds number of Re = 3.7 9 105, the drag
coefficient is rapidly reduced. This results from sequential laminar separation, reattachment, and turbulent separation
from the boundary layer of the sphere. The sphere wake becomes fully turbulent beyond this critical Reynolds
number. In the subcritical Reynolds numbers from 800 to 3.7 x 10^5, the drag coefficient of a sphere has almost
constant values, and the flow separates laminarily from the sphere. In addition, Kelvin–Helmhortz instability
occurs in the separating shear layers, and the wake becomes turbulent.


So the wakes are different on paintballs than they would be at Re ~ 100

I found this article, you'll need to purchase it if you want to read it, I could post figures and quotes but I'm not sure if I can get in trouble for doing that. It Covers the wake mechanism of a sphere at subcritical and critical Re but not spinning spheres. if you are a student ask your library and they can get you a free copy (usually):

"Numerical investigations of flow over a sphere in the subcritical and supercritical regimes", Physics of fluids [1070-6631] Constantinescu, George (2004) volume: 16 issue: 5 page: 1449 -1466

Some figures:
Sub critical drag coefficient fluctuations: Cz and Cy are basically lift coefficients
Posted Image
compared to the supercritical equivalent figures:
Posted Image

This shows the sideways forces on the ball that contribute to "random walk".

I honestly can't find anything about streamwise rotating spheres at high reynolds numbers. :( The article Snipez posted though is pretty sweet.
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#69 User is offline   brycelarson 

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Posted 30 March 2009 - 07:34 PM

ok, so all of that aside - what Reynolds number are we looking at at standard paintball velocities for a projectile the size of a paintball and mostly spherical?

#70 User is offline   Leftystrikesback 

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Posted 30 March 2009 - 09:33 PM

we calculated it a couple times in this thread and discussed implications: http://www.techpb.com/forum/index.php?show...20&start=20
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#71 User is offline   HAZE_243 

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Posted 02 April 2009 - 01:53 PM

Question: What is the maximum velocity for a paintball? Does vortex shedding effect the paintball at that speed? If so, would it be possible to build a vertical wind tunnel to achieve maximum velocity (make the paintball fall continuiosly in one place) and measure the vortex shedding. I suppose to see it you may have to use a colored air/smoke agent. I am considering building this and posting a video, but just curious to see if its worth trying or even possible.

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#72 User is offline   lazylink 

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Posted 11 June 2010 - 08:41 PM

Hi I'm new here, but I have a quick question.

How did you calculate the "vector" number? I know its sqrt(x^2 + y^2) but the number in the data sheet next to "vector" is clearly an average of some kind. I just wanted to know how exactly you calculated that average. I have calculated the "vector" for every data point and then averaged them but I don't get the same values as in the data sheet (for example in the two piece .682 / 12" barrel I get 14.08, while the data sheet says 11.23), did you use the sample mean instead of the average? did you set the "zero" as the mean x and mean y and computed the vector as the distance from that instead of from the point (0,0)? I'm a curious person. Thanks!

#73 User is offline   Steephill 

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Posted 11 June 2010 - 08:45 PM

View Postlazylink, on 11 June 2010 - 08:41 PM, said:

Hi I'm new here, but I have a quick question.

How did you calculate the "vector" number? I know its sqrt(x^2 + y^2) but the number in the data sheet next to "vector" is clearly an average of some kind. I just wanted to know how exactly you calculated that average. I have calculated the "vector" for every data point and then averaged them but I don't get the same values as in the data sheet (for example in the two piece .682 / 12" barrel I get 14.08, while the data sheet says 11.23), did you use the sample mean instead of the average? did you set the "zero" as the mean x and mean y and computed the vector as the distance from that instead of from the point (0,0)? I'm a curious person. Thanks!

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`kook, on 01 February 2012 - 06:06 PM, said:

The gold standard isn't what is most popular.By that logic, Justin Beiber is the gold standard of music.

#74 User is offline   BEASTY 

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Posted 11 June 2010 - 08:51 PM

holy shit...

you went back a fucking year to post?? what the tits?

View Postthed3monvisionz, on 29 September 2011 - 04:09 AM, said:

What you NEED to do is mark your territory, Pee on/in the WHOLE school. I mean EVERYWHERE just SOAK IT.

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#75 User is offline   PacosTacos88 

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Posted 11 June 2010 - 08:52 PM

View Postlazylink, on 11 June 2010 - 08:41 PM, said:

Hi I'm new here, but I have a quick question.

How did you calculate the "vector" number? I know its sqrt(x^2 + y^2) but the number in the data sheet next to "vector" is clearly an average of some kind. I just wanted to know how exactly you calculated that average. I have calculated the "vector" for every data point and then averaged them but I don't get the same values as in the data sheet (for example in the two piece .682 / 12" barrel I get 14.08, while the data sheet says 11.23), did you use the sample mean instead of the average? did you set the "zero" as the mean x and mean y and computed the vector as the distance from that instead of from the point (0,0)? I'm a curious person. Thanks!



You can just start a new thread if you want...

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