Question about the Twist Rate formula
- Thread starter tpcollins
- Start date
Art Eatman
Staff in Memoriam
Odds are that the polymer tip is an inconsequential part of the bullet's weight.
I'd calculate both ways, just from curiosity, but I'd favor the 0.91 for decision.
I'd calculate both ways, just from curiosity, but I'd favor the 0.91 for decision.
nicknitro71
New member
The tip must be included in the calculation.
Also for utterly precision, your formula must be multiplied by the square root of the specific gravity of the bullet/10.9.
Also for utterly precision, your formula must be multiplied by the square root of the specific gravity of the bullet/10.9.
Actually I think Art is right you subtract the length of the plastic tip. If you use JBM's Bullet Stability Calculator they have you input the length of the plastic tip and the OAL of the bullet. According to their calculator if I put in the total length of a 70 grain Nosler BT of .910 the bullet is marginally stable in my 6X47 with a 1:12 twist with a stability factor of 1.247. Once I subtract the tip .1" I get a stability factor of 1.55 which puts me in the green for a stable bullet, and shooting them in my X47 proves this to be true.
Art Eatman
Staff in Memoriam
I don't see how there is enough mass in the plastic tip to affect the stability of the bullet. I'm not being arbitrary, but I'd want some sort of established knowledge to show how my assumption is incorrect.
Start by removing a tip and weighing it, to learn the percentage of the total weight.
Start by removing a tip and weighing it, to learn the percentage of the total weight.
Art Eatman
Staff in Memoriam
I understand about length, but it would seem that uniformity of mass plays a part.
An obviously extreme example would to have an inch of toothpick protruding from the front. I doubt that it would affect the stability.
An obviously extreme example would to have an inch of toothpick protruding from the front. I doubt that it would affect the stability.
Brian Pfleuger
Moderator Emeritus
Center of mass is the issue, I believe, and the plastic tip adds considerable length with virtually no change in CoM. It's essentially "phantom" length, in terms of balance.
The calculator wants to know that part of the bullet is plastic so it can ignore it.
The calculator wants to know that part of the bullet is plastic so it can ignore it.
Thats part , this video explains why the rifle bullet needs to be spun
http://www.youtube.com/watch?v=kThMXchUJ3A
More detailed explanation
http://www.youtube.com/watch?v=Z3h0pMQAFUU
I found these to be interesting to watch . If what he is saying is true . I'd think the center of pressure would change with the polymer tip on or off . That would in turn change the distance between the center of gravity and the center of pressure and that will effect how much spin is needed .
My understanding is that the length is important because it relates to the "leverage" (for lack of a better term) that the atmospheric drag can exert on the bullet to cause it to yaw/tumble. If that's the case, the formula would need the total length even though the extra length added by the ballistic tip is very light compared to the rest of the bullet.
Art Eatman
Staff in Memoriam
Heh. It would be interesting to compare answers to this question from several different bullet manufacturers.
Oh: And a couple or three Physics profs.
Oh: And a couple or three Physics profs.
Warning--stream of consciousness follows... 
I don't have a good enough handle on gyroscopic concepts to be able to intuitively think through the spin-stabilized case, but I can think through the aerodynamically stabilized case. I'm going to start from there and then see if that gives me any insight into the spin-stabilized case.
In the aerodynamically stabilized case, it's largely a matter of where the center of mass and the center of pressure (due to drag) are located on/in the object. If they are co-located (or close to being co-located) on/in the object, the object will tend to tumble, if they are significantly separated, the object will tend to want to fly with the center of mass forward and the center of pressure rearward.
If that concept is difficult to visualize, imagine dropping a straw that has a paperclip attached to one end. The paperclip end will consistently hit the ground first no matter how you orient the straw before you let go of it. The center of mass is near the paperclip end but the center of pressure (focus of drag) is somewhere in the middle of the length of the straw. So the drag exerts force on the straw and that means the center of mass ends up on the "front" end of our projectile.
Now let's change things up by increasing the length significantly without significantly affecting the center of mass. That would change the center of pressure and could potentially affect stability depending on where the length is added. If it's added toward the rear, it would be expected to improve stability, if it's added toward the front, it would be expected to make the object less stable.
Think of taking a longer straw and attaching the paperclip so that the same amount of straw trails it as in the first example but now there's some straw extending forward of the paper clip. Assuming that there's enough added forward length, the straw will now flutter or tumble when it's dropped. The center of pressure has been moved forward somewhat relative to its initial position with respect to the center of mass. That's because the paperclip is still significantly heaver than the straw and therefore the center of mass will still be very close to where the paperclip is. But now the straw extends forward of the paper clip, so even though the center of pressure (focus of drag) is still near the middle of the length of the straw, that center of pressure is now farther forward with respect to our center of mass.
So, let's try to extend that to the spin-stabilized case. We undertand that the spin-stabilization is there to compensate for the fact that the projectile isn't inherently stable (either the center of mass and center of pressure are too close together or the center of mass is actually rearward of the center of pressure).
Earlier, with our straw experiment, we established the idea that extending the length of the projectile forward without changing the mass of the projectile significantly would have a tendency to move the center of pressure forward while leaving the center of mass nearly unchanged. The effect of the change was that it made the projectile less stable. We would expect a less stable projectile to be harder to stabilize and therefore we would logically expect that we might need more spin to achieve the stability we want.
Getting back to our specific question, extending the length of the bullet in the forward direction without significantly changing the weight seems like it would make the projectile less stable and that suggests that we would want our formula to reflect that change so we can calculate an accurate twist to stabilize it.
I don't have a good enough handle on gyroscopic concepts to be able to intuitively think through the spin-stabilized case, but I can think through the aerodynamically stabilized case. I'm going to start from there and then see if that gives me any insight into the spin-stabilized case.
In the aerodynamically stabilized case, it's largely a matter of where the center of mass and the center of pressure (due to drag) are located on/in the object. If they are co-located (or close to being co-located) on/in the object, the object will tend to tumble, if they are significantly separated, the object will tend to want to fly with the center of mass forward and the center of pressure rearward.
If that concept is difficult to visualize, imagine dropping a straw that has a paperclip attached to one end. The paperclip end will consistently hit the ground first no matter how you orient the straw before you let go of it. The center of mass is near the paperclip end but the center of pressure (focus of drag) is somewhere in the middle of the length of the straw. So the drag exerts force on the straw and that means the center of mass ends up on the "front" end of our projectile.
Now let's change things up by increasing the length significantly without significantly affecting the center of mass. That would change the center of pressure and could potentially affect stability depending on where the length is added. If it's added toward the rear, it would be expected to improve stability, if it's added toward the front, it would be expected to make the object less stable.
Think of taking a longer straw and attaching the paperclip so that the same amount of straw trails it as in the first example but now there's some straw extending forward of the paper clip. Assuming that there's enough added forward length, the straw will now flutter or tumble when it's dropped. The center of pressure has been moved forward somewhat relative to its initial position with respect to the center of mass. That's because the paperclip is still significantly heaver than the straw and therefore the center of mass will still be very close to where the paperclip is. But now the straw extends forward of the paper clip, so even though the center of pressure (focus of drag) is still near the middle of the length of the straw, that center of pressure is now farther forward with respect to our center of mass.
So, let's try to extend that to the spin-stabilized case. We undertand that the spin-stabilization is there to compensate for the fact that the projectile isn't inherently stable (either the center of mass and center of pressure are too close together or the center of mass is actually rearward of the center of pressure).
Earlier, with our straw experiment, we established the idea that extending the length of the projectile forward without changing the mass of the projectile significantly would have a tendency to move the center of pressure forward while leaving the center of mass nearly unchanged. The effect of the change was that it made the projectile less stable. We would expect a less stable projectile to be harder to stabilize and therefore we would logically expect that we might need more spin to achieve the stability we want.
Getting back to our specific question, extending the length of the bullet in the forward direction without significantly changing the weight seems like it would make the projectile less stable and that suggests that we would want our formula to reflect that change so we can calculate an accurate twist to stabilize it.
Twist rate formulas are approximations anyways, at least as far as stability is concerned.
I wouldn't worry about the polymer tips of bullets, the amount of bearing surface on the bullet seems to be a better indicator of needing a faster twist than anything else. For example two bullets of the same mass but different construction (one being solid gilding metal, the other cup and core construction) the monolithic bullet will have a longer bearing surface (if the nose and base profiles are the same).
Jimro
I wouldn't worry about the polymer tips of bullets, the amount of bearing surface on the bullet seems to be a better indicator of needing a faster twist than anything else. For example two bullets of the same mass but different construction (one being solid gilding metal, the other cup and core construction) the monolithic bullet will have a longer bearing surface (if the nose and base profiles are the same).
Jimro