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Posted: 3/13/2003 9:24:32 AM EDT
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Always seems to be question on what "weight bullet" for what "barrel twist rate", but the "length" of the bullet is never questioned. I just found a small article with some very interesting info and thought some other may find it interesting. A gentleman named Sir George Greenhill ( a Brit, if his name was something like Pierre Du Squat`, I would have taken this info as being false.) developed a hypothesis that the twist rate T multiplied by bullet length in calibers L is equal to 150. So the equation is as follows straight from the article, so bear in mind a writer spelled out the equation not a mathetician: (T)x(L)=150 We are looking for (T) so now the equation looks like this: (T)=150/(L) You with me? Now it get a bit confusing. To find the length of the bullet in calibers L you have to divide the bullet's length in inches by it caliber width. (L)= bullet length in inches/bullet caliber Here is an example: A .224 cal 75 gr. Hornady BTHP length is .97", so the length is calibers = .97"/.224 or 4.33. So (L)=4.33. The equation looks like this: (T)=150/4.33 or (T)=34.642. You're probably thinking that is an alfully slow twist rate, "my god, my muzzleloader is 1x28". True, but right now (T) is equal to calibers, not inches. To find (T) in inches, multiply (T) by the bullet's caliber, as such: Twist in inches = 34.642 x .224 or 7.76 So by using this equation a Hornady .224 cal 75 grain bullet is best suited for a barrel with a 1x7.76, which they do not make. It falls closer to a 1x8 twist, so a 1x7 twist barrel would not be optimal for this bullet. Big deal huh? Everyone knows to use a 1x8 for bullets 75 grains and up, right? What about those 80 grain Sierra (one the ones that wont fitted in a standard mag) or Hornady's 75 grain A-max bullets in 1x8 twist barrels. Follow this equation and you will see a significant difference. I promise to repost a much easier equation to work with soon. |
| I've been Googling for some dimensions on the 80 gr. Sierra. Closest I've come is [url=http://216.239.53.100/search?q=cache:gO086GOovu8C:no-treason.com/laissezfirearm/tracer.htm+sierra+.223+80+grain&hl=en&ie=UTF-8]here[/url], where it's said to be 1/32" shorter than the M856, which would equal 1 1/16" in length. |
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Article was is the January '03 edition of the [u]American Rifleman[/u]. The article never discussed weight as a variable in the equation. With this equation, the 80 grain Sierra would be best suited for 1X7 twist (7.09 was the figure). I know a lot of HP shooters using 80 grain bullets in 1x8 barrels. Need some bullet lengths on lighter weigth bullets from comparison. Any help? |
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The Greenhill formula is not real precise and assumes solid lead bullets. When things like copper jacktes and partial steel cores get tossed into the mix it doesn't work as precisely as the formula implies. With consistent bullet construction a longer bullet is, not surprisingly, heavier and vv. -- Chuck |
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From: [email protected] Newsgroups: rec.guns Subject: Barrel twist, Part 1 Message-ID: Date: 27 Aug 93 15:01:36 GMT Sender: [email protected] Organization: AT&T Lines: 96 Approved: [email protected] 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 |
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From: [email protected] Newsgroups: rec.guns Subject: Barrel twist : Part 2 Message-ID: Date: 31 Aug 93 00:10:50 GMT Sender: [email protected] Organization: AT&T Lines: 93 Approved: [email protected] 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 |
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You lost me right here: 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 What's the decimal number in the [s]first[/s] second column? -- Chuck |
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Interesting. GOtta find some time and read it all, remininsent of transport phenomena in Grad School. Someone was looking for the length of the Sierra 80 gr HPBT MK (Item# 9390) it is 1.090" in the QuickLoad database Also from QuickLoad, the following: "The author provides these formulas only as a source of comparative and historical information. The user may decide himself about the usefulness of such formulas.) Described by Julian S. Hatcher in : Hatcher's Notebook, 1962, Formula: (150 * Squareroot(10.9/Specific gravity of bullet material) / length of bullet in calibers = twist length in calibers)." We use a proprietary formula that takes into account bullet velocity as well .... tends to give good results with subsonic rounds ... Marty |
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I love it when people discover that length plays more of a role in determining twist rate than weight. Best example I use when trying to explain it to them is the ordinary 62grain SS109 bullet which can stabilize in a 1:9 twist and then compare it to the comparable tracer round which has all the length of an 80grain match bullet and requires a 1:7 twist to stabilize. Neat thing would be going with something like the corbin bullet press where you can make use of powdered metals for bullet cores with tungsten being one of them. Having a specific density that is higher than lead, it would be possible to make a bullet with a tungsten core having a lower volume than one of equal weight using lead. The resulting tungsten core bullet should have a much reduced over all length which could open up other twist rate possibilities to the shooter. I often wonder what a 100grain powdered tungsten core bullet would be like in over all length, close to a 70grain bullet perhaps lower? |
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Quoted: I often wonder what a 100grain powdered tungsten core bullet would be like in over all length, close to a 70grain bullet perhaps lower? Powell 87 gr BT Tungsten 223 bullet = 1.000" Berger 80 gr VLD 224 bullet = 1.087" Hornady 75 gr AMax 224 bullet = 1.070" Hornady 100 gr FBHP 224 bullet = 1.145" Marty |
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