revolutions per minute debate

Brown Dog
You are right about the revs per distance traveled changing as velocity drops with distance from the barrel. I read GG's question several times but never quite grasped what he was asking.

Good job!
 
I'm not going to disagree with you. I was trying to illustrate that the rifling twist ratio is only part of the picture regarding the stability of a bullets rotation. This ratio must be coupled with MV to equal some RPM. We get used to saying that this or that bullet fired from this or that cartridge will either stabilize or not. What we really mean is that X length bullet of Y weight and Z caliber needs to spin at some approximate RPM. I wrote my illustration with no regard to an actual example and used numbers that had been used before in the discussion.

I also used "somewhat" and "almost" in paretheses for exactly the reason you mentioned. There is the transsonic issue but a bullets stability in crossing it (I believe and I may be wrong)is more a function of RPM because the speed of the bullets crossing it is always about the same. Some remain stable and some not as much. Yes there are other factors that contribute to this transition as well.

Yes, a bullet spinning like a top on a table must spin faster than a bullet in supersonic flight to stay stable. That spin rate is measured in RPM for both bullets. It is not correctly measured as a twist ratio. That's all I meant.
 
yes but a bullet leaving rifle at 3000' in a one in 8'twist matching the rifiling as bullet looses speed the rate of twist stays the same but the distance traveled has been extended same twist but longer distance to make the same twist. I think
 
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4ked, that is not right. The rotational angular velocity (spin) required for a particular stability factor is dependant on the drag forces being experienced. Obviously this then depends on the translational velocity. As others have stated, spin decays more slowly than translational velocity, <font color="brown"> increasing stability downrange. </font> .. <font color="red"> This may still not be enough to remain stable during the transonic transition. </font>

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Slight *apparent* contradiction there. Yes, bullet is over-stabilized as it slows down <font color="purple"> UNTIL </font> it hits transonic where there is huge transient turbulence increase.

Most bullets would require such fast twist rates to stabilize thru the transonic boundary:
[*]You couldn't make a barrel with that fast a twist ( 1:4) (guessing here - help) [*]Over stabaliztion causes increased spin drift problems (this is true) - that's why the BR crowd goes for the slowest twist that properly stabilizes a bullet
 
Longtooth

U misunderstand. This is not worry, it is boredom waiting on the weekend and it finally gets here and the turkeys are gobbling and the shad and stripers are in but it is now scheduled to rain all Saturday and Sunday. What else is there to do except replace leaky faucets and change the cat litter?
 
There are a lot of interesting concepts here, but the answer to GG's question is simple. The relative rate of spin starting at 1:8 is increasing downrange. Whether or not that is optimal re spin drift, mass imbalances etc, or whether it will remain stable through the transonic region is another issue.
 
Boredom That I understand. It has led me since semi retirering to do some very strange things. Take this weekend I will be seeing how large a spray pattern I can get on ground squirrels with a 17 rem and a 6X284 Maybe I'll study it at several different ranges, I may even try to figure out the revolutions of the bullet, but then maybe not.
 
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The relative rate of spin starting at 1:8 is increasing downrange.

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But the RPMs are decreasing. /ubbthreads/images/graemlins/wink.gif
 
Ok, here is my thinking and it is probably flawed but here goes. Using the 3000'per sec. with the 1 in 8 twist rate. 3000'x12 = 36000" then divided by 8" = 4500 revolutions per second or 4500 revolutions at the 3000' mark.
Now the 4000' MV 4000'x12 = 48000 then divided by 8" = 6000 revolutions per second or 6000 revolutions at the 4000'mark. Now how many revolutions did the 4000'MV make at the 3000'mark? 3000 is 75% of 4000 so 6000 x .75 =4500 revolutions. They made the same amount of revolutions. This is all assuming the MV remained constant and drag or other forces are not acting on the spin of the projectile. Is there a professor in the house?
 
Okiehunter,
All you are saying is that a 1:8 twist is a 1:8 twist regardless of velocity.
GG was asking what happens (not a vacuum trajectory) downrange, for example the "786th yard".
The answer is that the bullet will make MORE than 4 1/2 complete revolutions during that yard.
This is because the translational velocity decays much more quickly than angular. As 4ked pointed out, the actual RPM of spinning is decreasing, its just that the velocity is decreasing much faster. So the relative rate of spinning (eg 1:8 initial) is increasing down range.
 
I'm far from obtaining my rocket science degree, so much of this sounds foriegn to me. The problem wasn't time it's the dam tuition cost. But I think I have some real world experience with this. Last Sunday I shot 16 porcipines, unfortunatly 3 were gut shot. This produces a spectacularly gruesome site. Porcipines are 60% guts and as they climb and hang on their guts are particularily suseptible to gravity. Well my friend was intrested in seeing the carnige up close. All shots were with a 22-250, ranges from 100-365 yards. And I'll be dammed if all those intestinese and tape worms didn't have about a 1 in 12 twist imparted on them! Call me crazy but I think I solved it! /ubbthreads/images/graemlins/grin.gif /ubbthreads/images/graemlins/grin.gif /ubbthreads/images/graemlins/grin.gif
 
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