Bullet twist question.....
- Thread starter Winny Fan
- Start date
Varminterror
New member
The INITIAL spin rate will be 1:9", but as the bullet slows down (air speed), it's rotation will NOT proportionately slow down, so as Cat and others mentioned, it will actually make each revolution in SHORTER lateral travel. In other words, it rolls over in about the same TIME regardless of range, but as the bullet slows, it will travel less distance in that same period (length of time).
Example:
At the muzzle, my 24" AR is 3400fps with a 1:9" twist, (3400ft/sec) * (1 rev 9") * (12"/ft) * (60sec/min) = 272000RPM. Take out the 60sec factor and you get 4533.333rev/sec.
INVERSELY, that means it takes 0.0002206sec to make each revolution (sec/rev)
At 100yrds, my bullet has slowed down to 2950fps.
(2950ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 7.8"/rev, for a 1:7.8" twist.
At 400yrds, my bullet has slowed to 1875fps:
(1875ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 4.9"/rev, for a 1:4.9" twist.
No, nothing is accelerating the spin rate, but YES, the lateral speed is decreasing, while the rotational speed is barely effected at all, so the turns per foot traveled WILL increase.
Example:
At the muzzle, my 24" AR is 3400fps with a 1:9" twist, (3400ft/sec) * (1 rev 9") * (12"/ft) * (60sec/min) = 272000RPM. Take out the 60sec factor and you get 4533.333rev/sec.
INVERSELY, that means it takes 0.0002206sec to make each revolution (sec/rev)
At 100yrds, my bullet has slowed down to 2950fps.
(2950ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 7.8"/rev, for a 1:7.8" twist.
At 400yrds, my bullet has slowed to 1875fps:
(1875ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 4.9"/rev, for a 1:4.9" twist.
No, nothing is accelerating the spin rate, but YES, the lateral speed is decreasing, while the rotational speed is barely effected at all, so the turns per foot traveled WILL increase.
Winny Fan
New member
Originally Posted By: VarminterrorThe INITIAL spin rate will be 1:9", but as the bullet slows down (air speed), it's rotation will NOT proportionately slow down, so as Cat and others mentioned, it will actually make each revolution in SHORTER lateral travel. In other words, it rolls over in about the same TIME regardless of range, but as the bullet slows, it will travel less distance in that same period (length of time).
Example:
At the muzzle, my 24" AR is 3400fps with a 1:9" twist, (3400ft/sec) * (1 rev 9") * (12"/ft) * (60sec/min) = 272000RPM. Take out the 60sec factor and you get 4533.333rev/sec.
INVERSELY, that means it takes 0.0002206sec to make each revolution (sec/rev)
At 100yrds, my bullet has slowed down to 2950fps.
(2950ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 7.8"/rev, for a 1:7.8" twist.
At 400yrds, my bullet has slowed to 1875fps:
(1875ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 4.9"/rev, for a 1:4.9" twist.
No, nothing is accelerating the spin rate, but YES, the lateral speed is decreasing, while the rotational speed is barely effected at all, so the turns per foot traveled WILL increase.
Good post. Your illustration assumes no loss at all in rotational speed with the predictable and expected decline in lateral momentum, but it describes what happens very well.
Example:
At the muzzle, my 24" AR is 3400fps with a 1:9" twist, (3400ft/sec) * (1 rev 9") * (12"/ft) * (60sec/min) = 272000RPM. Take out the 60sec factor and you get 4533.333rev/sec.
INVERSELY, that means it takes 0.0002206sec to make each revolution (sec/rev)
At 100yrds, my bullet has slowed down to 2950fps.
(2950ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 7.8"/rev, for a 1:7.8" twist.
At 400yrds, my bullet has slowed to 1875fps:
(1875ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 4.9"/rev, for a 1:4.9" twist.
No, nothing is accelerating the spin rate, but YES, the lateral speed is decreasing, while the rotational speed is barely effected at all, so the turns per foot traveled WILL increase.
Good post. Your illustration assumes no loss at all in rotational speed with the predictable and expected decline in lateral momentum, but it describes what happens very well.
CatShooter
New member
Originally Posted By: VarminterrorThe INITIAL spin rate will be 1:9", but as the bullet slows down (air speed), it's rotation will NOT proportionately slow down, so as Cat and others mentioned, it will actually make each revolution in SHORTER lateral travel. In other words, it rolls over in about the same TIME regardless of range, but as the bullet slows, it will travel less distance in that same period (length of time).
Example:
At the muzzle, my 24" AR is 3400fps with a 1:9" twist, (3400ft/sec) * (1 rev 9") * (12"/ft) * (60sec/min) = 272000RPM. Take out the 60sec factor and you get 4533.333rev/sec.
INVERSELY, that means it takes 0.0002206sec to make each revolution (sec/rev)
At 100yrds, my bullet has slowed down to 2950fps.
(2950ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 7.8"/rev, for a 1:7.8" twist.
At 400yrds, my bullet has slowed to 1875fps:
(1875ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 4.9"/rev, for a 1:4.9" twist.
No, nothing is accelerating the spin rate, but YES, the lateral speed is decreasing, while the rotational speed is barely effected at all, so the turns per foot traveled WILL increase.
The parts in bold are not true.
It must be interesting to watch you make an omelet.
As a minor point, you cannot make suppositions, and then carry them to four (and seven) places.
It just takes up space...
... but it looks impressive to the kids
.
Example:
At the muzzle, my 24" AR is 3400fps with a 1:9" twist, (3400ft/sec) * (1 rev 9") * (12"/ft) * (60sec/min) = 272000RPM. Take out the 60sec factor and you get 4533.333rev/sec.
INVERSELY, that means it takes 0.0002206sec to make each revolution (sec/rev)
At 100yrds, my bullet has slowed down to 2950fps.
(2950ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 7.8"/rev, for a 1:7.8" twist.
At 400yrds, my bullet has slowed to 1875fps:
(1875ft/sec) * (0.0002206 sec/rev) * (12"/ft) = 4.9"/rev, for a 1:4.9" twist.
No, nothing is accelerating the spin rate, but YES, the lateral speed is decreasing, while the rotational speed is barely effected at all, so the turns per foot traveled WILL increase.
The parts in bold are not true.
It must be interesting to watch you make an omelet.
As a minor point, you cannot make suppositions, and then carry them to four (and seven) places.
It just takes up space...
... but it looks impressive to the kids
.
Varminterror
New member
Originally Posted By: CatShooterAs a minor point, you cannot make suppositions, and then carry them to four (and seven) places.
It just takes up space...
... but it looks impressive to the kids
Disagree.
Read about "significant figures". I used 4 significant figures in 0.0002206sec, just like I used 4 significant figures in my 3400fps, 1875fps, etc etc. Sure, I should have made them "100.0yrds" and "400.0yrds", but I'm confident that the voting public will forgive me for assuming they'd know I wasn't talking about 101.7yrds when I said 100.
It just takes up space...
... but it looks impressive to the kids
Disagree.
Read about "significant figures". I used 4 significant figures in 0.0002206sec, just like I used 4 significant figures in my 3400fps, 1875fps, etc etc. Sure, I should have made them "100.0yrds" and "400.0yrds", but I'm confident that the voting public will forgive me for assuming they'd know I wasn't talking about 101.7yrds when I said 100.
Varminterror
New member
Originally Posted By: CatShooterThe parts in bold are not true.
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available), I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available), I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
Last edited:
Winny Fan
New member
Just for the record, my comment about a constant rotational spin was not meant as a criticism. I simply pointed it out for those who might not fully understand.
Again, your post illustrates very well what happens. Adding a sheer drag factor for rotational velocity to your illustration won't affect the overall result of how you clearly explained what happens to a bullet down range.
Overall, it's one of the better and more concise explanations I've seen. Some explanations I've ran across concerning the same thing have been voluminous. Even with no comments concerning sheer drag or boundary layer included.
Again, your post illustrates very well what happens. Adding a sheer drag factor for rotational velocity to your illustration won't affect the overall result of how you clearly explained what happens to a bullet down range.
Overall, it's one of the better and more concise explanations I've seen. Some explanations I've ran across concerning the same thing have been voluminous. Even with no comments concerning sheer drag or boundary layer included.
Ridgeline17
New member
Originally Posted By: VarminterrorOriginally Posted By: CatShooterThe parts in bold are not true.
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available), I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
Ahh ok
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available), I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
Ahh ok
Last edited:
CatShooter
New member
Originally Posted By: Varminterror
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available), I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
The parts in bold are not true.
.
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available), I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
The parts in bold are not true.
.
Seems like the "old" drag being pushed of the back of the bullet would be replaced by the new drag encountered by the front of the bullet. Kinda like running in the rain... the drops you out run (that fall behind you) are replaced by the drops you run into (that would have fallen in front of you) had you not been running. Not a expert but hard to believe you can outrun rotational drag or the rain.
Jeff
Jeff
Last edited:
Varminterror
New member
Originally Posted By: SoTexSeems like the "old" drag being pushed of the back of the bullet would be replaced by the new drag encountered by the front of the bullet. Kinda like running in the rain... the drops you out run (that fall behind you) are replaced by the drops you run into (that would have fallen in front of you) had you not been running. Not a expert but hard to believe you can outrun rotational drag or the rain.
Jeff
Yes, "new drag" does occur, but the overall effect is that the boundary layer isn't able to establish nearly as dramatically as it would say if the bullet were not moving.
So the net effect is much much lower. It's called boundary layer shedding. I can write you a chapter out of a physics book, but I think that sums it up.
Jeff
Yes, "new drag" does occur, but the overall effect is that the boundary layer isn't able to establish nearly as dramatically as it would say if the bullet were not moving.
So the net effect is much much lower. It's called boundary layer shedding. I can write you a chapter out of a physics book, but I think that sums it up.
Varminterror
New member
Originally Posted By: CatShooterOriginally Posted By: Varminterror
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available),[/i] I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
The parts in bold are not true.
.
The last bold section was a question, not a statement. I'd be incredibly surprised to learn the roughness factor of copper tubing were significantly different than that of a bullet jacket, but if you have proof one way or the other as to what the roughness factor for jackets are, that's the only info I need.
As to the truth of the 1st several bold sections, I guess we'll have to agree to disagree. When I was contracting with the gov't (naval 120mm cannon projectiles), that was my job. Agreeably, no, it isn't perfect, but yes, it is sound science. The form drag significantly out shadows the sheer drag due to rotation, so the bullet will slow much faster that its spin rate.
If you have a function for the retard rate (rather the drag coefficient across a parallel plane) of the angular velocity of a bullet in flight, I'd be more than happy to incorporate it into my above analysis.
The reason I said "barely affected at all" is because relative to the lateral speed, the decrease in rotational speed is null. The resistance to the angular speed is the drag of air against the rotation of the bullet, which is minimal because it is a SHEER drag, like wind across western Kansas. Additionally, the boundary layer (the static air that induces the sheer drag against the surface of the bullet) is actually reduced by the lateral velocity of the bullet. In essence, the lateral velocity pushes the boundary layer (the sheer drag resisting the spin, NOT the form drag) off of the back of the bullet, minimizing its effect.
So yeah, sure, you're correct, the spin WILL slow down, and yes, I simplified the function by neglecting the sheer drag induced against the spin, but in relative terms, the sheer drag resisting the spin is REASONABLY negligible compared to the FORM drag resisting the flight of the bullet.
Just for grits and shins, if you'd agree that copper tubing has a relatively similar roughness-factor to a copper jacket (because reference materials for tubing roughness factors are readily available),[/i] I can probably produce a formula tomorrow that incorporates the sheer drag, to illustrate how little it effects the spin rate relative to the form drag?
The parts in bold are not true.
.
The last bold section was a question, not a statement. I'd be incredibly surprised to learn the roughness factor of copper tubing were significantly different than that of a bullet jacket, but if you have proof one way or the other as to what the roughness factor for jackets are, that's the only info I need.
As to the truth of the 1st several bold sections, I guess we'll have to agree to disagree. When I was contracting with the gov't (naval 120mm cannon projectiles), that was my job. Agreeably, no, it isn't perfect, but yes, it is sound science. The form drag significantly out shadows the sheer drag due to rotation, so the bullet will slow much faster that its spin rate.
Last edited:
CatShooter
New member
Originally Posted By: Varminterror
"... so the bullet will slow much faster that [sic] its spin rate."
That was never in question... though "much" is an iffy number.
.
"... so the bullet will slow much faster that [sic] its spin rate."
That was never in question... though "much" is an iffy number.
.
OldTurtle
Moderator - Deceased
Back to the original Question...Quote:So, if you shoot the same bullet in the same rifle, but the velocity is increased to 3200 FPS, does the bullet spin faster than one revolution in 9" of traveled distance like it did at 2800 FPS?...IMHO, NO...and I'm not an engineer by any means...9" is 9"...
doggin coyotes
Well-known member
Quote:So, if you shoot the same bullet in the same rifle, but the velocity is increased to 3200 FPS, does the bullet spin faster than one revolution in 9" of traveled distance like it did at 2800 FPS?
Recovering that bullet, reloading it and shooting it again is gonna be a big PIA. Can we just shoot one that is identical to it? One outta the same box should be close enough. No?
We got a new knife sub forum. How about a engineer sub forum? I'd likely stay outta there.
Recovering that bullet, reloading it and shooting it again is gonna be a big PIA. Can we just shoot one that is identical to it? One outta the same box should be close enough. No?
We got a new knife sub forum. How about a engineer sub forum? I'd likely stay outta there.
Winny Fan
New member
Originally Posted By: doggin coyotesQuote:So, if you shoot the same bullet in the same rifle, but the velocity is increased to 3200 FPS, does the bullet spin faster than one revolution in 9" of traveled distance like it did at 2800 FPS?
Recovering that bullet, reloading it and shooting it again is gonna be a big PIA. Can we just shoot one that is identical to it? One outta the same box should be close enough. No?
We got a new knife sub forum. How about a engineer sub forum? I'd likely stay outta there.
Dang, doggin'.... I never thought of that....!
'Supppose shooting an identical bullet and not the same one will move where the aspirin is located?
As many have stated, a 9" twist barrel will only result in one revolution or turn of the bullet in 9" of lateral movement of the bullet once it exits the barrel. As has also been stated, as distance from the bore increases, lateral movement of the bullet (FPS velocity) deteriorates much faster than does the spin rate of the bullet so that at various points down range the bullet might make one complete revolution in 6" of lateral bullet movement, and so on, based on the retained speed of each variable. And it can all be calculated very precisely if that's your thing.
I posted the question simply because you see this explained in many different ways on the internet. Some of them indeed result in people looking for an aspirin, and some, like several here, are spot on from the perspective of an engineer or a physicist. Who, by the way, need love too....
I like the knife forum....
Could we post pictures in the new engineer's forum? I'll go first.
How to engineer, design, and build a swing:
Recovering that bullet, reloading it and shooting it again is gonna be a big PIA. Can we just shoot one that is identical to it? One outta the same box should be close enough. No?
We got a new knife sub forum. How about a engineer sub forum? I'd likely stay outta there.
Dang, doggin'.... I never thought of that....!
As many have stated, a 9" twist barrel will only result in one revolution or turn of the bullet in 9" of lateral movement of the bullet once it exits the barrel. As has also been stated, as distance from the bore increases, lateral movement of the bullet (FPS velocity) deteriorates much faster than does the spin rate of the bullet so that at various points down range the bullet might make one complete revolution in 6" of lateral bullet movement, and so on, based on the retained speed of each variable. And it can all be calculated very precisely if that's your thing.
I posted the question simply because you see this explained in many different ways on the internet. Some of them indeed result in people looking for an aspirin, and some, like several here, are spot on from the perspective of an engineer or a physicist. Who, by the way, need love too....
I like the knife forum....
How to engineer, design, and build a swing: