From: sfaber@ihlpb.att.com (Steven R Faber) Subject: Re: Rifling Twist Angles - stability Organization: AT&T Came across the book "Methods of Exterior Ballistics" written just after WW1 that is a detailed mathematical treatment of the subject, and has the answer we've been looking for. Instead of using the angular momentum over the transverse moment as a measure of stability, we should be looking at the rate of change of the yaw angle and considering the moment of inertia around the transverse axis. rate of change in yaw angle (y) = torque around trasverse axis/ It where "It" is transverse moment of inertia. Then the assumption is made that all the force acting on the projectile is due to the drag on the front cross sectional area and that is independent of yaw angle assuming small yaw angles. Then scaling a given projectile by radius "r" including length. y*It* sin(angle) = const (c) * dens of air (p) * area (r^2) * moment arm (r) sin(angle) (It) is proportional to r^5 (1/2*m*r^2) so: y = c*p*v^2/r^2 Then after an involved derivation he finds that a projectile is stable when y < 1/4*w^2*(Il/It)^2 where Il is the longitudinal moment of inertia and w is the spin rate. so the stability ratio is: substituting w=2*pi*v*twist(in^-1) (v*tw)^2*(Il/It)^2*r^2/(p*v^2) = r^2*(Il/It)^2/p Note that velocity drops out! Of course this assumes all the velocity is obtained from the gun. now substitute rifling angle (r.a) = 2*pi*r*twist so the stab ratio = (r.a.)^2*(Il/It)^2/p --------------------- So stability is independent of velocity or the caliber! Don't you love it when everything drops out:) Guess Bart's post really summed things up. The above ratio implies that for greater stablity you want to concentrate the mass in the center of the bullet, so a longer projectile gives poorer stability, and denser air gives poorer stability. Stability is independent of the density of the projectile since it has the ratio of moments of inertia along the different axes. The book mentioned that you can be over stabilized since the bullet has to follow an arcing trajectory, which could then lead to forces that could disintegrate the bullet if it is overstable. If we increase the length of a projectile by factor "L" and assume a ogival projectile which has approximately the same moments of inertia as a cylinder: the Il will be proportional to L and the It will be proportional to L^3 so to keep the same stability we must increase the rifling angle by L^2. Steve

From: sfaber@ihlpb.att.com (Steven R Faber) Subject: Re: Rifling Twist Angles - stability Organization: AT&T Just noticed a couple dropped variables in that post so here are the corrections. The results are the same except there was a mistake in the scaling of length by "L" part - the twist should be increased by L^3/2 instead of L^2. ----------------------------------------------------- # Came across the book "Methods of Exterior Ballistics" written # just after WW1 that is a detailed mathematical treatment of # the subject, and has the answer we've been looking for. Author: Forest Ray Moulton, Dover 1962 ed. reprinted from 1926 ed. # # Instead of using the angular momentum over the transverse moment # as a measure of stability, we should be looking at the rate of # change of the yaw angle and considering the moment of inertia around # the transverse axis. # # rate of change in yaw angle (y) = torque around trasverse axis/ It actually angular acceleration (y) # # where "It" is transverse moment of inertia. # Longitudinal is along its rotating axis and transverse is the "bad spin" axis perpendicular to the direction of travel. # Then the assumption is made that all the force acting on the projectile # is due to the drag on the front cross sectional area and that is # independent of yaw angle assuming small yaw angles. # # Then scaling a given projectile by radius "r" including length. # y*It* sin(angle) = const (c) * dens of air (p) * area (r^2) *v^2* moment arm (r) sin(angle) # # (It) is proportional to r^5 (1/2*m*r^2) # # so: y = c*p*v^2/r^2 # # # Then after an involved derivation he finds that a projectile is stable # when # # y < 1/4*w^2*(Il/It)^2 # # where Il is the longitudinal moment of inertia and w is the spin rate. # # so the stability ratio is: substituting w=2*pi*v*twist(in^-1) # # (v*tw)^2*(Il/It)^2*r^2/(p*v^2) = # # # r^2*(Il/It)^2/p # (r*tw)^2*(Il/It)^2/p # # Note that velocity drops out! # Of course this assumes all the velocity is obtained from the gun. # # now substitute rifling angle (r.a) = 2*pi*r*twist so the stab ratio = # # (r.a.)^2*(Il/It)^2/p # --------------------- # # So stability is independent of velocity or the caliber! # Don't you love it when everything drops out:) # Guess Bart's post really summed things up. # # The above ratio implies that for greater stablity you want to # concentrate the mass in the center of the bullet, so a longer projectile # gives poorer stability, and denser air gives poorer stability. # Stability is independent of the density of the projectile since it # has the ratio of moments of inertia along the different axes. # # The book mentioned that you can be over stabilized since the bullet # has to follow an arcing trajectory, which could then lead to forces # that could disintegrate the bullet if it is overstable. # # # If we increase the length of a projectile by factor "L" # and assume a ogival projectile which has approximately the same # moments of inertia as a cylinder: # # the Il will be proportional to L and # the It will be proportional to L^3 # Unfortunately we can't just substitute this since we have to go back to the beginning and separate "r" from length "L" when we scaled things so the result is the rifling angle or twist must be increased by a factor of L if all the force is on the front of the bullet or L^3/2 if it all on the side of the bullet. # so to keep the same stability we must increase the rifling angle # by L^2. (wrong) # # # Steve

From: sfaber@ihlpb.att.com (Steven R Faber) Subject: Re: Rifling Twist Angles - stability Organization: AT&T #From article <1992Sep30.151911.24878@u.washington.edu>, by toby@carson.u.washington.edu (Toby Bradshaw): # In article <30SEP199208590643@ariel.lerc.nasa.gov> smpod@ariel.lerc.nasa.gov (Stefan) writes: # #toby@milton.u.washington.edu (Toby Bradshaw) writes... # ##... # ##About the only way to get a higher BC without going to a longer # ##bullet (assuming the form factors are already pretty good) is # ##to go to a larger caliber. Maybe that's why some hunter class # ##shooters are using those Speedy Gonzales 110gr match bullets # ##in their short .308 cases...with S-L-O-W twists (like 1:16). # # # #I thought that BC is proportional to mass/drag and going to larger but # #relatively light bullets would increase drag faster than mass and therefore # #decrease BC. # # # # # BC is some function of sectional density (squared, I think)*form # factor. I'm mostly thinking of bullets that require approximately # equal twist rates for minimum stability in benchrest, like 1:14 # for 52gr .224 as well as 125gr .308. The 125gr .308 will have # a BC about 0.100 higher than the .224, which makes a difference # in wind-bucking :) and recoil :( # For a fixed length the sectional density should stay the same when you increase diameter, maybe there are some other effects on BC. You would get a stability advantage with increasing radius which is consistent with the slow twists as seen below. Going back to the equation and substituting for r and L (length) in the moments of inertia for cylindrical bullets and simplifying I get : (rifling angle)*r/(l*sqrt(dens of air)) for a stability ratio under the same initial assumptions. The author switched assumptions to force on the side of the bullet when he came up with the rifling angle increase proportional to L^3/2 and I don't understand why, otherwise it would be proportional to L as above. As to Toby's other comment on how hotter loads seemed to overcome marginal stablity. It was pointed out to me that the theory assumed a constant coefficient of drag. It also doesn't account for slowing of the bullet spin, so maybe some of these effects change things at long range. Oh yeah, I forgot to mention the results assume the bullet is supersonic. Steve

From: sfaber@ihlpb.att.com (Steven R Faber) Subject: Rifling angle and stability update Organization: AT&T There are a couple interesting updates to the bullet stability ratio thread if you remember: Originally I claimed that the stability ratio was independent of density of the bullet since there was a ratio of moments of inertia in the equation, but this was an error since one of the moment of inertia terms was already substituted for assuming a constant shape. The net result if you substitute for density in all the moments of inertia etc and reduce, this is the stability ratio : (rifling angle)*r/l*sqrt(dens of projectile/dens of air) which is nice and unitless as you would want. So a denser bullet will be more stable. This is also what the author Moulton claimed (after reading some more). The formula given in the last posting is the same as above if you assume constant bullet density. Henry Schaffer also noted that the formula above is: # This is basically Greenhill's Formula without the constant. Quoting #from Hatcher's Notebook - pg. 446 "The twist required (in calibers) #equals 150 divided by the length of the bullet (in calibers)." #There are also corrections to this to account for different specific #weights of the bullet and different density of the medium or #flight. So the Greenhill's formula is the reciprocal of the above stability ratio. The above stability ratio assumed a simplistic formula for longitudinal moment of inertia. If you substitute the exact formula for "It" of a cylinder you get a more complicated expression: (r.a.)^2*d/((1+l^2/(3*r^2))*p) where p is dens. of air (1.29) and d is the dens. of the bullet. (11.35) Here is a table of some bullets and the ratio's: cal wt. twist tw*l(cal) (ra)*r/l*sqrt(d/p) (r.a.)^2*d/((1+l^2/(3*r^2))*p) .224 62gr 1/12 216 .0216 .00133 .308 220 1/12 189 .0246 .00177 .224 55 1/12 178 .0262 .00192 .224 62 1/9 162 .0287 .00239 .308 220 1/10 157 .0296 .00254 .308 173 1/10 135 .0345 .00341 .224 55 1/9 134 .0349 .00342 .308 147 1/10 118 .0400 .00453 The first entry I tried and it is definitely unstable. The next is a 220 grain from a 1/12 barrel which is marginally unstable according to the Lyman manual. The rest are stable in increasing order of stability. All of the ratios display the same trend, so the more complicated ratio probably does not buy us much. It would not be too difficult to find the exact moment of inertia for an ogive shaped bullet, but the formula gets pretty long and given these results is probably not worth the effort. The Greenhill formula seems to suggest stability under 180 at Lake Michigan air densities. Our simple ratio suggests stability over .025. Moulton also claimed that once a bullet leaves the gun and is stable it will stay stable as long as the velocity is above about 1600 fps. The coefficient of drag decreases a little from mach 3 to a bit above mach 1 so the stability will increase a bit more than v^2 as the velocity drops. He assumes that the spin will not decrease faster than that. Steve

From: sfaber@ihlpb.att.com Subject: Re: Disappearing bullets Organization: AT&T From article <C9qyHp.7v1@fc.hp.com>, by bartb@hpfcla.fc.hp.com (Bart Bobbitt): # Bullet spin rate is found by: # # Muzzle velocity times 720 # ------------------------- = RPM # Twist rate in inches # # Your 1:7 twist will spin a bullet with 3000 fps muzzle velocity at # 308,571 RPM. That's really fast for the 55-gr. 22 caliber bullets # you used. Those bullets are best spun at about 155,000 RPM when they # leave the muzzle. Considering the bullet's are spinning at about # twice what they shoot best at, no wonder the jackets came apart. # There's a lot of centrifugal force on bullet jackets. A lot of # folks have experienced bullet desintagration when they're fired # from faster than correct twists for the velocity used. An interesting question is why does the bullet disintegrate down range and not right away at the muzzle? The answer is that the most internal forces are generated when the over-stabilized bullet must follow an arcing trajectory, so it disintegrates when its trajectory starts requiring it to go where its not pointing. [[[ YGTBSM ]]] # If you use a powder type/weight that will give you about 2000 fps, # your 1:7 twist barrel will still spin 'em about 200,000 RPM. That's # probably OK and the bullets won't fly apart. But your accuracy may # not be as good as a 70 or 80 grain bullet fired at max velocity from # that twist. # # I've often asked Sierra Bullets to put some information in their # reloading manuals about bullet spin rates. The section that discusses # each bullet caliber, weight and style is a good place. Useful ranges # are mentioned in this manual, but the important thing for accuracy is # each rifle and pistol bullet needs to be spun in a narrow RPM range # to be accurate. Once this RPM range is identified and the above formula # modified to give rifling twist rate for the muzzle velocity the bullet # will be fired at, getting best accuracy is easy. The RPM calculation is useful for determining when the bullet is being spun too fast, however I'm wondering how accurate a metric a range of RPM would be to determine best stability/accuracy for a given bullet. From a previous discussion on stability ratios I once claimed that the velocity of the bullet cancelled out and the stability was only determined by the twist rate (for given bullet and air density). You and others gave convincing evidence that the stability did depend on the velocity. Taking another look at the assumptions, it was assumed that the coeff. of drag (Cd) was constant and this is not true. The faster the bullet, the lower the Cd. This means the drag does not quite increase as v^2 so faster bullets will be more stable than predicted by the simple Greenhill formula which depends only on the twist. The point is that the stability should still be more sensitive to twist than velocity, so although an RPM range would be useful for practical guns and velocities used, there may be a more accurate way to give the lower stability limit in general. I'll try and come up with more quantitative results. Steve

From: sfaber@ihlpb.att.com Subject: Re: Reloading SS-109 in .22/250 Organization: AT&T From article <243q3e$4jt@news.u.washington.edu>, by toby@stein.u.washington.edu (Toby Bradshaw): # In article <241mmj$fu4@msuinfo.cl.msu.edu> elric@ap.cl.msu.edu (Tom # Wright) writes: # # # Whoah! We were talking SS109, which is ~61.5 grains, at least # #all the examples I've measured are. Where'd the 69 grain Sierras # #come from? :) # # My bust. But why on earth would one use a 1-9 twist with a 61.5gr # bullet? This weight of Berger bullets will stabilize well even # in a 1-12 out of a .223. I was just doing some calculations on twist required for different bullets and one of them I tried was the SS109. According to the calculation it required a twist of 9.75 in or faster ( for v = 3200 fps) You need to figure the moment of inertia around the long axis and the moment of inertia perpendicular at the base of the bullet ( this is best measured by making it into a type of pendulum and timing it). You also need the ballistic coeff. or form factor, which I guessed at, so if anyone knows a good value for this I could come up with a better estimate of the twist. The density of the SS109 is less than 9 g/ml due to the steel in it, so it ends up with a relatively long length. The Berger bullets are probably denser and shorter so they would require less spinning for their weight. Steve

From: sfaber@ihlpb.att.com Subject: Barrel twist, Part 1 Organization: AT&T This is the first part of a several part posting on the subject of determining the barrel twist rate needed to stabilize a given bullet. A spinning bullet behaves a lot like a spinning top. Imagine a top or toy gyroscope that is spinning and balanced on a post. If the spin is high enough it will remain balanced, and if it gets below a critical point the gyro will fall off. Now think of the spinning bullet as a top balanced on its base. Instead of the force of gravity, we have the air resistance acting on it from the head of the bullet, and the spin counteracting it. If the spin decreases enough, the bullet will fall or tumble, otherwise it will stay heading into the wind although there may be some precession evident. With the bullet problem, the spin is determined by the barrel twist and the bullet velocity. The drag is dependent on the air density, the cross sectional area of the bullet and the shape. The drag is expressed with the following equation: drag = Cd * r^2 * v^2 * p where Cd is the unitless coeff of drag (varies with velocity) r is the radius of the bullet v is the velocity of the bullet p is the air density. If Cd is constant, this says the drag is proportional to v^2. There is a table of drag force as a fuction of velocity for a standard projectile in Hatcher's Notebook, which one can graph and obtain Cd as a function of velocity. In computing Cd(v), r=.5in for the standard projectile and p = 1.221 g/liter, standard air density used for the table, 60 deg. F, 30" Hg, and 66% RH. This shows Cd rises from about 0.1 at 900 ft/sec just below the speed of sound to almost 1.0 around 1500 ft/sec, just above the speed of sound, and then drops off slowly from there reaching a value of 0.77 at 3600 ft/sec. The region between 1500 and 3600 ft/sec fits nicely to a line: Cd = 1.146 - 1.047E-4 * v Cd is 0.98 at 1500 ft/sec and decreases by 0.01 every 100 ft/sec. Since the drag of an arbitrary bullet may vary from that of the standard projectile, the form factor "i" is introduced to scale the observed to standard drag. This is related to the ballistic coefficient (BC). i = m/(d^2*BC) where m is the mass in lbs and d is the diameter of the bullet in inches. So the drag for an arbitrary bullet is: drag = Cd(v) * i * r^2 * v^2 * p Moulton's book on Exterior Ballistics gives the criterion for stability when transverse (tumbling) angular acceleration < 1/4 * w^2 * (Il/It)^2 where It is the transverse moment of inertia and Il is the longitudinal moment of inertia and w is the longitudinal angular velocity ( the spin ) the transv. ang. accel = drag * moment arm / It w depends on the barrel twist and initial velocity "v" w = 2*pi*v/tw where tw is the length of one turn of the barrel twist. Now we have everything we need to write the equation for barrel twist needed to stabilize the bullet when it just exits the barrel. Substituting for drag and using "Lm" for moment arm: tw = pi*Il/r * sqrt( 1/(Cd*p*Lm*It*i)) Note that velocity cancelled out, and the only velocity dependence is due to Cd(v). Velocity here was the initial (muzzle) velocity, but what happens down range? It would be safe to assume that the spin decays slower than the forward velocity, so with the decreased drag as the bullet slows down range, it should become more stable. As long as the bullet is supersonic above 1500 ft/sec, we would be safe to assume stability is determined at the muzzle exit point. To calculate the twist, we know Cd(v), p, r, and i can be determined from the ballistic coefficient (BC), by dividing the sectional density by the BC where sectional density is defined as the mass in lbs divided by the diameter of the bullet in inches. Il and It remain to be determined. Lm, is the length from the base of the bullet to the average point of the drag force. In the next chapter we will use the moments of inertia for a cylinder and derive the Greenhill equation. After that we can calculate moments for a boattail bullet with an ogive point and have a much more accurate method of determining the twist. Also since Cd(v) is now known, we know the velocity dependence. SRF

From: sfaber@ihlpb.att.com Subject: Barrel twist : Part 2 Organization: AT&T Last time we derived the equation for twist needed to stabilize a bullet: tw = pi*Il/r * sqrt( 1/(Cd*p*Lm*It*i)) eq (1) where: Cd is the unitless coeff of drag (varies with velocity) Cd = 1.146 - 1.047E-4 * v r is the radius of the bullet v is the velocity of the bullet p is the air density. i is the form factor i=m/((2*r)^2*BC)*in^2/lb Lm is the moment arm of the drag force It is the transverse moment of inertia around bullet base Il is the longitudinal moment of inertia Lets assume a cylindrical bullet so Il = 1/2*m*r^2 It = 1/3*m*L (approx) where L is length of the cylinder Lm = L m (mass) = d*pi*r^2*L #From the above equation we can write: tw/(2*r)*L/(2*r) = sqrt( (3*pi^3*d)/(4^3*Cd*i*p) ) eq (2) This is the Greenhill equation relating the twist length in calibers times the bullet length in calibers equal to a constant. According to Hatcher, the bullet used to derive the constant was a Krag 220 grain, L=1.35in., v=2000ft/sec, d=10.9 gm/ml, p=1.221gm/l For i=.61, the constant comes out to 150. The Greenhill equation gives a lot of insight into the problem, showing that the twist is mainly determined by the length to radius ratio and sqrt of the bullet density to air density ratio. Modern bullets have an i a bit less than .6 sometimes, and the velocities used are higher, so the Cd is a bit lower - affecting the results. Now we are in the position to write our own equation for the cylindrical bullet approximation where we can take into account those effects. We could also use the exact form of It = 1/3m*L^2 + 1/4*m*r^2 but the latter term is less than a 1% effect, so we will also ignore it, giving: tw = sqrt(3*pi^3*r^4*d/(4*Cd*i*p*L^2)) eq (3) Note that the L from substituting m in Il cancels Lm=L. If the bullet is pointed and not a cylinder, this equation is using L as if it were a cylinder given the density of the bullet. This gives a good approximation for Il. L then is a length to a point somewhere between the cylindrical part of the bullet and the tip which also gives a good approximation for Lm, these then cancel with good accuracy. The remaining L in eq. 3 is from "It". If we use the L calculated from the density here, it would underestimate It, since It is sensitive to the distance the mass is from the base of the bullet. It turns out that using the measured length is a much better approximation. The table below compares the results of the original Greenhill equation, our version that takes into account "i", and Cd, and equation 1 using accurately calculated values of "Il" and measured values of "It" (more on this later). 1. Greenhill : tw(calibers)*L(calibers)=150 2. equation 3 with appropriate "i" and calculated Cd, measured "L" 3. equation 3 with L determined from the density 4. equation 1 with accurate "Il" and measured "It". all bullets are FMJ boattail .30 caliber at 2500 ft/sec except the 55 and 62 grain (ss109) which are .224 caliber at 3200 ft/sec bullet "i" _1_ _2_ _3_ _4_ 150 gr .581 12.9 13.1 14.3 168 gr .533 12.0 12.8 12.7 174 gr .53? 11.1 11.8 11.6 220 gr .526 9.5 10.3 13.6 10.2 55 gr .626? 10.1 10.6 14.2 62 gr .61? 8.3 8.4 9.8 The results show that equation 3 with the "i" and calculated Cd (column 2) is a good improvement on Greenhill (column 1), and is in good agreement with equation 1 using exact moments of inertia (column 4) for the .30 caliber bullets that have a longer cylindrical section. One should use equation 1 with the bullets that are mostly point with little body, or that have a non-uniform density. Equation 1 requires more refined calculations of It, Il, and Lm. This is the topic of a subsequent installment. SRF

From: sfaber@intgp1.att.com Subject: Re: .308 Palma Bullet Loads Organization: AT&T From article <CCwysF.4LD@fc.hp.com>, by bartb@hpfcla.fc.hp.com (Bart Bobbitt): #This is what folks thought was the best in the early 1950s. In the 1960s, #when the military rifle teams finally began to build really accurate rifles, #and civilian highpower match rifle builders finally did the same thing, #it was found not to be true. Match M1 (and service) rifles had a 1:10 twist #barrel. The M14 had a 1:12 twist barrel. When the M14 started to be worked #over into a competition rifle, it was found that the slightly slower muzzle #velocity of the M2, 173-grain match bullet in the M14's 1:12 twist barrel #shot better than in an M1's longer barrel at higher velocity. This meant #that the 173-grain bullet was being spun too fast for best accuracy. But it #wasn't really well understood until the Navy started rebarreling their 30 #caliber M1 rifles with 7.62mm NATO barrels with 1:12 twists. As the match #conditioning success with M1 rifles was miles ahead of the M14 at the time, #folks took notice of the greatly improved accuracy of the M2 173-grain #military match bullet in the 7.62mm NATO M1 rifles. Then the M14s began to #be match conditioned to shoot as well as the M1s did. Their 1:12 twist #barrels were soon found to not quite spin the bullets as fast as they needed #to be in their 22-inch barrels. They went to 1:11 twists and the rest is #history. According to the above the 173 grain match bullets were used with 1:12 twist barrels. My best current calculation predicts a maximum 1:11.6 in. twist needed for that bullet. This would easily change to about 1:13 in. in the Rockys where the air is less dense, but the above suggests the 1:12 twist worked in general. The guy who sold me the 173 gr bullets also implied they would not work well in a 1:12 in gun (in Iowa.) Maybe I need to go back to the drawing board on this bullet or use a more accurate estimate of the B.C. if it turns out the 1:12 was good at standard air density. If anyone knows the B.C. for the 173 gr mil. match bullet I'd appreciate it. Steve

From: sfaber@intgp1.att.com Subject: Re: [RIFLE] nasty tumbling 243 bullets Organization: AT&T From article <9310190142.AA25299@hpfcnews.fc.hp.com>, by bartb@fc.hp.com (Bart Bobbitt): # Any given bullet needs to be spun within an RPM range to stablize it. # Whatever muzzle velocity and twist combination spins the bullet in that # range will work just fine. Here's the approximate spin rates .243 # spitzer/hollowpoint bullets need to stabilize: # # 65 - 70 grain 155,000 to 165,000 # # 71 - 80 grain 165,000 to 190,000 # # 81 - 90 grain 190,000 to 215,000 # # 91 -100 grain 215,000 to 245,000 # # 101 -110 grain 245,000 to 280,000 # # These are the spin rates competitive shooters use with 6mm bullets to # get the best accuracy. Hunting rifles have the same required spin rates # for their various bullet weights. These are probably good practical rules of thumb for determining the twists needed to stabilize the above bullets for the twists and velocities normally used, but I believe that specifying a given rpm needed to stablilize a given bullet will not work in general, since the velocity dependence of the stability is only determined by the variation of coefficient of drag with velocity, which is relatively small. From my previous derivation: tw = pi*Il/r * sqrt( 1/(Cd*p*Lm*It*i)) eq (1) where: Cd is the unitless coeff of drag (varies with velocity) Cd = 1.146 - 1.047E-4 * v for velocities between 1500 and 3600 ft/sec (could maybe extrapolate higher but not lower since the curve bends over at that point.) r is the radius of the bullet v is the velocity of the bullet p is the air density. i is the form factor i=m/((2*r)^2*BC) , r in (in.), m in (lb) Lm is the moment arm of the drag force It is the transverse moment of inertia around bullet base Il is the longitudinal moment of inertia Lets assume a cylindrical bullet so Il = 1/2*m*r^2 It = 1/3*m*L^2 (approx) where L is length of the cylinder Lm = L m (mass) = d*pi*r^2*L Substituting gives: tw = sqrt(3*pi^3*r^4*d/(4*Cd*i*p*L^2)) eq (2) I must confess that I don't have a good theoretical reason to use "It" as moment of inertia around the bullet base rather than around the center of mass of the bullet, and Lm the moment arm from the base instead of from the center of mass. Using the values from the base seems to give the best results consistent with Greenhill's formula (at the velocity he used) and experiment. This substitution does not affect my point though since it would not change the velocity dependence factor. For an example, here is a 168 grain Sierra: The table shows the maximum twist length needed to stabilize the bullet at the given velocity and the corresponding RPM as calculated: velocity twist RPM 1500 fps 12.1 in 89,260 2500 12.8 140,680 2800 13.0 154,700 It is clear that the calculated twist is not terribly velocity dependent, but you would be in trouble if you were to extrapolate the minimum RPM needed based on a result taken at a lower velocity. Steve

From: sfaber@intgp1.att.com (Steven R Faber +1 708 979 3147) Subject: Re: Bullet Length or Weight? From article <9405252354.AA24338@marlin.nosc.mil>, by njohnson@nosc.mil (Norman F. Johnson): # Joe, # # # I have always assumed (perhaps incorrectly) that since a bullet of a given # # caliber gets longer as it gets heavier that it's WEIGHT was the only reason # # it would not stabilize in a slow twist barrel. He claims that some of the # # solid brass/copper type bullets will not stabilize in slow twist barrels for # # their caliber due to their being too long and not too heavy. The reason the # # 40 gr. BT's are long for their weight is obviously because it has the long # # pointy polycarbonate tip. It adds quite a bit to length but very little to # # it's weight. # # May I introduce you to Greenhill: # # The Greenhill formula is an empirical equation that does a good # job of establishing the barrel twist necessary so that a bullet # of a given length will be adequately stabilized. # ... # Note that it is bullet LENGTH, not weight that is important. # Greenhill works well with all lead/lead-alloys commonly used for # bullets. The Greenhill estimate assumes a constant density bullet. The long pointy BT with the polycarbonate tip will not require as much spin to stabilize the bullet as a similar length bullet that was more dense at the ends. The actual formula depends on the moments of intertia around the transverse axis and the axial axis of symmetry. To get a more stable bullet, you want to get the transverse moment of inertia lower (less mass in the ends). Steve

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