From the moment a pellet leaves the muzzle the twist rate will "increases" the further it travels?
Twist rate question true or false?
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If this is a guessing contest i would say no. The reason being the pellet is never going faster than right before it leaves the barrel. So I would ASS-UME the twist rate would not get faster either. But as my wife says I'm usually wrong. lol
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Maybe you are talking about the "Barrel Twist Rate". Twist rate is normally referred to in Barrels, not like the rotation speed of a pellet.
OK, that's kind of out of my league, but I know enough to be dangerous. The tighter twist rates (correct me if I'm wrong) like 1 in 16 spin the pellet faster than the slower twist rates like 1 in 20. (1 twist in 16 inches sort of thing). I've heard but you'll need to validate that the tighter twist works better for heavier pellets with greater mass. Just backwards of what I thought it should do.
I believe Critahitta is correct, that once the pellet leaves the barrel, the rotation will actually decrease due to friction. Twist rate is usually refers to the rifling in the barrel. Can you be a bit more clear on whether you mean riflings of the barrel or projectile rotation?
OK, that's kind of out of my league, but I know enough to be dangerous. The tighter twist rates (correct me if I'm wrong) like 1 in 16 spin the pellet faster than the slower twist rates like 1 in 20. (1 twist in 16 inches sort of thing). I've heard but you'll need to validate that the tighter twist works better for heavier pellets with greater mass. Just backwards of what I thought it should do.
I believe Critahitta is correct, that once the pellet leaves the barrel, the rotation will actually decrease due to friction. Twist rate is usually refers to the rifling in the barrel. Can you be a bit more clear on whether you mean riflings of the barrel or projectile rotation?
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Twist rate is fixed by the barrel manufacturer and cannot be altered without re-barreling. The rate of rotation and the velocity both will drop as the projectile travels farther from the barrel.
As stated above the "Twist" rate of a barrel is fixed and constant. Twist rate simply refers to the number of inches of barrel length required for one complete rotation of the projectile.
The longer a projectile the "faster" (lower number ... shorter length for one rotation) the rate of spin needs to be for stabilization. Diabolo pellets are less sensitive to twist rate since the skirt "drag" stabilizes them to an extent much like the vanes on an arrow.
Thurmond
As stated above the "Twist" rate of a barrel is fixed and constant. Twist rate simply refers to the number of inches of barrel length required for one complete rotation of the projectile.
The longer a projectile the "faster" (lower number ... shorter length for one rotation) the rate of spin needs to be for stabilization. Diabolo pellets are less sensitive to twist rate since the skirt "drag" stabilizes them to an extent much like the vanes on an arrow.
Thurmond
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Yes ok I didn't word that very well sorry. I'll try to be clearer. The pellet leaves the rifle spinning and traveling forward.
In a certain distance it would spin on it's axis however many times (rotational rate)
Just say in 20in it spins once.
If we could measure that again at 100 yards would that pellet be spinning more than once in 20in or less than once in 20in?
In a certain distance it would spin on it's axis however many times (rotational rate)
Just say in 20in it spins once.
If we could measure that again at 100 yards would that pellet be spinning more than once in 20in or less than once in 20in?
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This is just something I read and thought would be a good brain teaser.
Also must be relevant to keeping pellets stable.
Just a bit of fun.
Also must be relevant to keeping pellets stable.
Just a bit of fun.
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The spin rate should decay slightly. Once the impulse of energy has been applied to the pellet in the gun, there is no more increase in energy possible.
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Jim yes spin rate will decay but what about velocity?
What happens to the ratio spin rate to distance travelled as velocity drops?
It starts at 1:20 what happens to that ratio as the pellet travels further and velocity drops.
The pellet doesn't need more energy to rotate more times in a given distance....
What happens to the ratio spin rate to distance travelled as velocity drops?
It starts at 1:20 what happens to that ratio as the pellet travels further and velocity drops.
The pellet doesn't need more energy to rotate more times in a given distance....
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G
Guest
Guest
Ok, The projectile leaves the muzzle at some velocity, say 1000 fps (as this is a convenient number). The twist rate of the rifling has spun the projectile up to 1 in 12 (because this is also a convenient number), you rifle would naturally vary. When the projectile leaves the muzzle the rate of spin expressed in revolutions per second is 1000 RPS. The rate can also be expressed in revolutions per minute (RPM) and is 60000 RPM for this example. That's about four or five times as fast as your Dremel tool.
As the pellet travels down range it encounters an enormous amount of drag because it is traveling so fast. That 1000 fps pellet is leaving the barrel at 681 MPH. So there is a lot of drag slowing the forward progress of that pellet down. It quickly falls off. If it were an H&N FTT out of one of my .22s it would be slowed down to 500 fps at about 145 yards. It would have been in the air only about 0.636 seconds. So it looses half it's velocity in just over half a second.
The same is not true of the rotation. The rotational velocity falls off MUCH more slowly. This is because there is less drag affecting the rotation of the pellet than there is affecting the forward progress of it. This means that at 145 yards the pellet is still rotating at about the same speed as it was when it left the barrel. I don't know of any studies of this information so I'll just do what people do when they don't have a clue, pick a number from somewhere the sun don't shine... I'd be willing to bet that the pellet still had 3/4 of it's original spin going at the half velocity point. That means that the pellet would be spinning more in less distance. With those numbers it would be spinning at something like 45000 RPM which is 750 revolutions per second; however, it would still be traveling forward at 500 fps so it would be doing 1.5 turns in 12" (1 turn in 8).
Now somebody has obviously told you how this is a big deal at long ranges because the bullet gets "over stabilized". With bullets this can be a problem because the bullet tends to fly tail down at extreme ranges when it is over stabilized. This happens for the simple reason that the bullet is initially aimed at a point above the intended target. At longer ranges the trajectory is high and the bullet "wants" to stay pointed in the direction it left the barrel. Pellets are not so bad because they have that lovely parachute hanging on their backside and it tends to correct for "pitch".
Now for a real example (these numbers are from chairgun using settings specific to my D460):
D460 H&N FTT 1:16" Twist 744 fps at muzzle.
(744*12)/16 = 558 RPS; 558*60 = 33480 RPM at the muzzle.
At 150 yards remaining velocity is 338 fps, flight time was .863 seconds.
Rotation/Travel assuming 90% retention of rotation rate at launch:
338/558 = 0.606 feet per revolution; This is 1 turn in 7.25 inches.
Remaining energy is 4.6 fpe.
As the pellet travels down range it encounters an enormous amount of drag because it is traveling so fast. That 1000 fps pellet is leaving the barrel at 681 MPH. So there is a lot of drag slowing the forward progress of that pellet down. It quickly falls off. If it were an H&N FTT out of one of my .22s it would be slowed down to 500 fps at about 145 yards. It would have been in the air only about 0.636 seconds. So it looses half it's velocity in just over half a second.
The same is not true of the rotation. The rotational velocity falls off MUCH more slowly. This is because there is less drag affecting the rotation of the pellet than there is affecting the forward progress of it. This means that at 145 yards the pellet is still rotating at about the same speed as it was when it left the barrel. I don't know of any studies of this information so I'll just do what people do when they don't have a clue, pick a number from somewhere the sun don't shine... I'd be willing to bet that the pellet still had 3/4 of it's original spin going at the half velocity point. That means that the pellet would be spinning more in less distance. With those numbers it would be spinning at something like 45000 RPM which is 750 revolutions per second; however, it would still be traveling forward at 500 fps so it would be doing 1.5 turns in 12" (1 turn in 8).
Now somebody has obviously told you how this is a big deal at long ranges because the bullet gets "over stabilized". With bullets this can be a problem because the bullet tends to fly tail down at extreme ranges when it is over stabilized. This happens for the simple reason that the bullet is initially aimed at a point above the intended target. At longer ranges the trajectory is high and the bullet "wants" to stay pointed in the direction it left the barrel. Pellets are not so bad because they have that lovely parachute hanging on their backside and it tends to correct for "pitch".
Now for a real example (these numbers are from chairgun using settings specific to my D460):
D460 H&N FTT 1:16" Twist 744 fps at muzzle.
(744*12)/16 = 558 RPS; 558*60 = 33480 RPM at the muzzle.
At 150 yards remaining velocity is 338 fps, flight time was .863 seconds.
Rotation/Travel assuming 90% retention of rotation rate at launch:
338/558 = 0.606 feet per revolution; This is 1 turn in 7.25 inches.
Remaining energy is 4.6 fpe.
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Hi Oldspook. Your answer typifies why I like being on AGN. I know it must sometimes be frustrating to trot out facts and figures on concepts that are so well formed in your head but I and I'm sure many others appreciate it.
I read one of Harry's posts where he explained as above that velocity decays more quickly ect ect and to me it was one of those moments that made me think. It is a little counter intuitive.
Thats the only reason I posed the question. Because I thought it was thought provoking.
I didn't do such a good job explaining what I was after in ballistic lingo.
I have a question. How would you describe an overly stabilised pellet in comparison to an overly stabilise bullet?
I read one of Harry's posts where he explained as above that velocity decays more quickly ect ect and to me it was one of those moments that made me think. It is a little counter intuitive.
Thats the only reason I posed the question. Because I thought it was thought provoking.
I didn't do such a good job explaining what I was after in ballistic lingo.
I have a question. How would you describe an overly stabilised pellet in comparison to an overly stabilise bullet?
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Forgive me for not reading your mind. Your original question was a bit vague. There should be a tiny amount of "spin-drift" toward the end of the flight due to rotational friction. How to quantify the amount is beyond my meager skills.
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Jim you sound like my wife!
Yes I know sorry for the confusion. I did try to explain a little better further down the tread.
Yes I know sorry for the confusion. I did try to explain a little better further down the tread.
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you over complicate it guys, imagine if you have a 18 inch barrel and the twist rate is 1 in 18.
each barrel legth distance the pellet will do one revolution approx.
from the muzzle to until you hit a target there is a relatively low number of revolutions can be worked out pretty good
so roughly how many barrel lenghts you can fit between the muzzle and the target.
not so many crazy high number as one would think! Even Teds videos show pellets spiralling "slowly" from the ST barrel
use common sense.
maybe the x barel will twist faster but it will be still slower than a conventional barrel. there is no real grab just skidding action.
each barrel legth distance the pellet will do one revolution approx.
from the muzzle to until you hit a target there is a relatively low number of revolutions can be worked out pretty good
so roughly how many barrel lenghts you can fit between the muzzle and the target.
not so many crazy high number as one would think! Even Teds videos show pellets spiralling "slowly" from the ST barrel
use common sense.
maybe the x barel will twist faster but it will be still slower than a conventional barrel. there is no real grab just skidding action.
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Hi Sirk. Your analogy would be true if the pellet maintained exactly the same velocity and spin rate all the way to the target. The point of the post was to point out from the moment the pellet leaves the barrel it no longer spins once in 18in because it immediately slows down very quickly. The spin rate slows but less so.
I can't explain it any better that oldspooks explanation above. Take some time to go through it and look at the example he gives.
It is a constantly changing ratio of revolutions or spin rate / unit of distance travelled.
The further it goes the greater the number of revolutions / unit of distance travelled.
The new smooth x barrel does fully engage the pellet. No more skidding through like the smooth twist
barrels do.
I can't explain it any better that oldspooks explanation above. Take some time to go through it and look at the example he gives.
It is a constantly changing ratio of revolutions or spin rate / unit of distance travelled.
The further it goes the greater the number of revolutions / unit of distance travelled.
The new smooth x barrel does fully engage the pellet. No more skidding through like the smooth twist
barrels do.
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back to the original question:
if the pellet is fully engaged in the barrel then the twistrate will be what you can measure just by pushing a pellet through a barrel and counting the rotations. and measuring the barrel length. there are videos on how to mark a pellet and do it. so no theories pure exact measurement
if the pelllet skids through, then the twistrate is BS. the value will be lot lower to that what it is published.
i still have not seen a pellet skirt picture from a x barrel.
if the pellet is fully engaged in the barrel then the twistrate will be what you can measure just by pushing a pellet through a barrel and counting the rotations. and measuring the barrel length. there are videos on how to mark a pellet and do it. so no theories pure exact measurement
if the pelllet skids through, then the twistrate is BS. the value will be lot lower to that what it is published.
i still have not seen a pellet skirt picture from a x barrel.
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Hi I probable should have thought a little more about the way I posed the question.
A twist rate is a ratio of how far a pellet travels to complete a revolution within the barrel. Once the pellet leaves the barrel there will still be a relationship between how far the pellet travels to complete a revolution. That ratio once the pellet leaves the barrel is what I was getting at.
A twist rate is a ratio of how far a pellet travels to complete a revolution within the barrel. Once the pellet leaves the barrel there will still be a relationship between how far the pellet travels to complete a revolution. That ratio once the pellet leaves the barrel is what I was getting at.
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Yeah, this RPM stuff cracks me up since the pellet is normally on target in well under a second at normal ranges (not minutes) and turns 1 time for every 18inch of travel which equals 2 turns per 1 yard. Made it easy for you. 50 yards = 100 turns. Time of flight depends on velocity and target distance. Stability is projectile weight, BC (shape and drag), twist and velocity. This is why you sometimes see slow spiralling thru the scope.
Regards,
DT
Regards,
DT
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It could be that one spiral does not translate to one rotation of the pellet. That is to say you see 1 spiral but the pellet revolves around its axis many times in that spiral.
Even at slower pellet speeds your getting at least 500rev per second.
Do you think your seeing 500 spirals per second?
Michael
Even at slower pellet speeds your getting at least 500rev per second.
Do you think your seeing 500 spirals per second?
Michael
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G
Guest
Guest
I have to disagree with this. The pellet is "spun up" inside the barrel. When it leaves the barrel it starts to lose forward velocity faster than it loses "spin". This happens because the drag slowing the forward velocity of the pellet is greater than the drag of the pellet spinning against the air. At the end of it's travel is is indeed spinning slower than when it started out BUT is it spinning more turns in less distance as well.
For your example, in the beginning the pellet is traveling at some velocity. Lets use 1000 fps because it is convenient. When it leaves the muzzle it is traveling 1 foot every millisecond. It immediately starts to slow down. Likewise when it leaves the muzzle it is rotating 1 full turn every 18" of travel. It loses forward velocity faster than it loses spin because of the way drag works. The end result is that as it slows down the distance that it takes to make one full turn becomes less and less. By the time that it reaches your 100 yard mark it is probably traveling half as fast as when it left the barrel but it is likely still rotating at 70 or 80 percent of the rotation it had when it left the muzzle. The end result is that the pellet, while it IS rotating more slowly than it was when it left the muzzle, is STILL rotating more turns in much less distance.
If we had a barrel that was a hundred yards long then the pellet would always be making one revolution every 18" but once it leaves the muzzle it is free to make one turn in whatever distance it takes to make that turn. Surely you have seen on TV the guys that shoot bullets into the ice on the Great Lakes in winter then go locate the bullet still spinning on the ice? That bullet is still spinning but no longer has ANY forward movement.
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