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 December 5, 2013, 11:59 AM #1 tpcollins Senior Member   Join Date: February 18, 2009 Location: SE Michigan Posts: 443 Question about the Twist Rate formula I understand the Twist Rate formula of: TR = C x D squared / L but concerning bullet length, I have a bullet that's 1.102" including the polymer tip, .910" not including the polymer tip. Which measurement should I use? Thanks. __________________ What direction did that last shot at Kennedy come from?
 December 5, 2013, 12:20 PM #2 Art Eatman Staff Lead   Join Date: November 13, 1998 Location: Terlingua, TX, USA Posts: 21,891 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. __________________ You're from BATFE? Come right in! I use all your fine products!
 December 5, 2013, 12:43 PM #3 nicknitro71 Senior Member   Join Date: February 5, 2008 Location: DeLand, FL Posts: 110 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. __________________ http://nitrorigging.com
 December 5, 2013, 01:01 PM #4 taylorce1 Senior Member   Join Date: November 18, 2005 Location: Colorado Springs, CO Posts: 5,218 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. __________________ NRA Life Member The Truth About Guns
 December 5, 2013, 06:36 PM #5 Art Eatman Staff Lead   Join Date: November 13, 1998 Location: Terlingua, TX, USA Posts: 21,891 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. __________________ You're from BATFE? Come right in! I use all your fine products!
 December 6, 2013, 11:40 AM #6 kraigwy Senior Member   Join Date: June 16, 2008 Location: Wyoming Posts: 9,131 Bullet length is used to determine twist, not weight. In the case of the poly. tip, use the whole length. __________________ Kraig Stuart CPT USAR Ret USAMU Sniper School Oct '78 Distinguished Rifle Badge 1071
 December 6, 2013, 10:34 PM #7 Art Eatman Staff Lead   Join Date: November 13, 1998 Location: Terlingua, TX, USA Posts: 21,891 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. __________________ You're from BATFE? Come right in! I use all your fine products!
 December 6, 2013, 10:51 PM #8 Brian Pfleuger Staff   Join Date: June 25, 2008 Location: Central, Southern NY, USA Posts: 17,111 Question about the Twist Rate formula 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.
December 8, 2013, 01:59 AM   #9
Metal god
Senior Member

Join Date: April 10, 2012
Location: San Diego CA
Posts: 1,574
Quote:
 Center of mass is the issue,
Thats part , this video explains why the rifle bullet needs to be spun

More detailed explanation

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 .
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 December 8, 2013, 04:38 AM #10 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 17,829 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. __________________ Did you know that there is a TEXAS State Rifle Association?
 December 8, 2013, 09:24 AM #11 Art Eatman Staff Lead   Join Date: November 13, 1998 Location: Terlingua, TX, USA Posts: 21,891 Heh. It would be interesting to compare answers to this question from several different bullet manufacturers. Oh: And a couple or three Physics profs. __________________ You're from BATFE? Come right in! I use all your fine products!
 December 8, 2013, 11:23 AM #12 Cary Senior Member   Join Date: November 18, 2010 Location: Boise, ID Posts: 103 Yes this is good stuff to discuss when it is too cold to go out and shoot. -5 F here according to the National Weather Service. Cary __________________ Shooters. We are a community. United we stand divided we fall.
 December 8, 2013, 03:28 PM #13 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 17,829 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. __________________ Did you know that there is a TEXAS State Rifle Association?
 December 8, 2013, 03:54 PM #14 Metal god Senior Member   Join Date: April 10, 2012 Location: San Diego CA Posts: 1,574 lol yeah what I said , just with 7 more paragraphs and no videos for those of us that have bad reading comprehension . __________________ If you have a problem with everybody . It's not everybody that has the problem
 December 8, 2013, 04:10 PM #15 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 17,829 Different people assimilate and/or dispense information in different ways. Some might find watching 30 or 40 minutes of video preferable to spending 4 or 5 minutes reading, others might be just the reverse. __________________ Did you know that there is a TEXAS State Rifle Association?
 December 8, 2013, 05:00 PM #16 Jimro Senior Member   Join Date: October 18, 2006 Posts: 5,166 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 __________________ "Gorsh" said Goofy as secondary explosions racked the beaten zone, "Did I do that?" http://randomthoughtsandguns.blogspot.com/

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