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Theory of barrel twist rate and bullet mass and diameter?

Discussion in 'Rifle Country' started by Bill_Rights, Nov 15, 2012.

  1. Vern Humphrey

    Vern Humphrey Well-Known Member

    Go and look at the target. If the bullets are not stabilized, not only will few of them hit the target, those that do will leave elongated holes, indicating that they were more or less sideways when they hit. That's called "keyholing" and when you see that you know you have stability problems.
    It isn't weight or mass that determines twist rate, it's bullet length.

    The confusion comes in because to make a bullet heavier, you usually have to make it longer. But you could make a 100 grain .22 bullet of depleted uranium that would be no longer than a lead-core 55 grain bullet and shoot it in a 1 in 12 twist very nicely.
  2. Bill_Rights

    Bill_Rights Well-Known Member

    Thanks for the replies - we're getting there....

    helotaxi, OK, the Miller formula is more what I need for longer, slender bullets. What is the Miller formula? (Or just give a link to it, please. Thanks!)

    DanTheFarmer, Excellent that you actually did the experiment! My hat is off to you.

    1858, Excellent point about the "moment" of force that is trying to flip a bullet.
    I take it that this moment of force is the actual force mltiplied by the lever arm length, and a) the actual force is due to drag forces from the air resistance and b) the moment arm length (lever arm) is longer for a longer bullet. Excellent understanding!

    Zoogster, Yes, as you say:
    I agree. I assert that any amount of spin, excessive or not, detracts from muzzle velocity. Of course, some spin is necessary, as we've been discussing. I will calculate the bullet energy consumed by spin and post it later...

    pseudonymity, I think you for your interpretation of the Greenhill Formula:
    What I understand you to say is (correct me if I get it wrong), in the Greenhill Formula, the density (specific gravity) of lead is defined to be "1", therefore its units-of-measure [grams/cc, pounds/cu. ft., etc.] don't matter. The effect of the measurement units is built into the numeric factor "150". Only if you deviate from lead material for the bullet do you need to factor in density.

    Vern, You hammer home the point well!
  3. Bill_Rights

    Bill_Rights Well-Known Member

    gyroscopic stabization

    OK, so we, or at least some of us, are saying that twist rate imparts spin rpm, and this spin stabilizes the bullet by gyroscope forces, exactly like a spinning top. Similarly, I used to take a bicycle wheel and have somebody spin it while I held both ends of the axle in my two hands, I could barely change the angle of the axle when the wheel was spinning very fast. So this gyroscope force would keep a bullet from tumbling when the center of air resistance force is forward of the center of mass. Cool :cool:

    Maybe I need to go and look up the theory/formula of gyroscopes, but intuitively I "know" that the smaller the diameter of the bullet (or bicycle wheel), the faster it has to be spinning in order to exhibit the same gyroscopic force resisting tumbling (or tilting the bicycle wheel axle). None of you posters, nor the Greenhill Formula, seem to capture that trend. Yet I do see it in bullets and barrels, as we have them. Big bore guns have lazy twist rates and small caliber guns have fast twist rates. To quote my own OP:
    So in which formula is this gyroscopic force correlated with bullet diameter?
  4. MrBorland

    MrBorland Moderator

    Well, I'm neither an engineer nor a physicist, but here's my understanding:

    A bullet can rotate/spin 2 ways - along its longitudinal axis (i.e spinning as it does while in flight) and/or along its transverse axis (i.e. yaw). Each of these rotations is associated with its own gyroscopic force, so for max stability, you want the longitudinal one as high as possible, relative to the transverse one.

    So, here's the rub: Each gyroscopic force is directly proportional to the moment of inertia and spin rate along that axis, and because of its relatively long length, a bullet has a relatively high moment of inertia through its transverse axis. If the longitudinal gyroscopic force is to dominate, then, it can't do it through its small moment of inertia, and must do it through its spin, so it spins fast. The longer the bullet relative to diameter (i.e. caliber), the greater the potential for yaw, so the spin needs to be even faster for the longitudinal gyroscopic force to completely dominate & prevent yaw.
    Last edited: Nov 17, 2012
  5. MCMXI

    MCMXI Well-Known Member

    Bill_Rights, I put together a Miller bullet stability calculator in Excel if you're interested. It has SG corrections for velocity and atmospheric conditions. I used the formulas shown in Bryan Litz's book and confirmed that the calculator is correct by entering the values used in his example. The current values in the calculator are for a .308 175gr SMK bullet and a 1:10 twist barrel.

    Miller Bullet Stability Calculator
  6. Bill_Rights

    Bill_Rights Well-Known Member

    Yes, thanks 1858. I will see if I can download the Excel spreadsheet and understand it. I am interested!

    BTW, what is "SG" again? (I know I knew it before, but old-age CRS syndrome is creeping in.....)
  7. Bill_Rights

    Bill_Rights Well-Known Member

    Miller Formula calculator


    I was able to download the Excel spreadsheet. Thanks! (I didn't know we could do this.)

    Yeh, so... looking at it, I see that you don't ask for the caliber anywhere, in actual dimensions, such as inches. But the twist and the bullet length are expressed in units of calibers. OK, so I get it. It does not matter what the caliber measurement is, you just make sure to express the other two parameters (twist rate and projectile length) in multiples of caliber diameters.

    Cool. :cool:
  8. MCMXI

    MCMXI Well-Known Member

    Bullet diameter (in.) is the caliber. You enter bullet diameter and length in inches and the spreadsheet calculates those values in calibers and then uses those values to calculate SG.

    SG is a unitless value to indicate gyroscopic stability.
  9. Bill_Rights

    Bill_Rights Well-Known Member

    Oh! Excuse me. I was wrong. You corrected me by saying
    For some reason, I did not see the data entry field for bullet diameter in inches. In addition to CRS, I must be coming down with CSS (can't see sh!#) :mad:.
  10. Bill_Rights

    Bill_Rights Well-Known Member

    Bullet spinning energy - is it significant?

    I said earlier
    So, here it is. The needed formulas tying the spin energy to bullet mass, diameter and spin speed rely on the moment of inertia of the bullet. The moment of inertia about the spin axis is a property of the shape of the bullet and the density of the material. I approximated the shape as a cylinder and took the material to be solid lead (density is 11340 kg/m^3 = 11.34 g/cc = 185.83 g/cu.in. = 2867.7 gr/cu. in.). Here are the formulas, in which "W" is the energy or work just to spin up the bullet:
    In the formulas, "M" is the mass of the bullet, "r" is the radius of the bullet (half the caliber diameter) and omega is the spin speed, as explained. This all has to be worked out in consistent units. I have a spreadsheet for that, if anybody wants me to upload it.

    It turns out that the spin energy carried by the bullet to the target is small, less than 1 %, compared to the linear speed energy, which is technically called the translational energy. Here are three examples sort of covering some extremes.

    A 180 grain .308 inch diameter lead bullet travelling at 2620 fps out of a barrel with a twist of 12 inches/turn has:
    Bullet translational energy of 3719 Joules = 2743 ft*lbf
    Bullet rotational energy of 12.1 Joules = 8.92 ft*lbf.
    So the ratio of rotational to translational energy is 0.325 %.
    Bullet length, effective (cylinder approximation): 0.842 inches

    A 300 grain .452 inch diameter lead bullet travelling at 1500 fps out of a barrel with a twist of 30 inches/turn has:
    Bullet translational energy of 2032 Joules = 1499 ft*lbf
    Bullet rotational energy of 2.28 Joules = 1.68 ft*lbf.
    So the ratio of rotational to translational energy is 0.112 %.
    Bullet length, effective (cylinder approximation): 0.652 inches

    A 62 grain .223 inch (5.56 mm) diameter lead bullet travelling at 3002 fps out of a barrel with a twist of 7 inches/turn has:
    Bullet translational energy of 1682 Joules = 1240 ft*lbf
    Bullet rotational energy of 8.42 Joules = 6.21 ft*lbf.
    So the ratio of rotational to translational energy is 0.501 %.
    Bullet length, effective (cylinder approximation): 0.554 inches

    These energies carried by the spin of a bullet do not include the extra friction of the faster spinning bullet lost in the barrel or to the air. Zoogster pointed out these effects earlier. This friction energy is lost as heat, not delivered to the target.
    Last edited: Nov 27, 2012
  11. helotaxi

    helotaxi Well-Known Member

    Those energies have been shown to be somewhere around negligible when compared to the same energies from a straight rifled barrel. Essentially you lose the tiny fraction of a percent used to get the bullet spinning and otherwise the difference is nil.
  12. Bill_Rights

    Bill_Rights Well-Known Member

    Since the thread responses indicated the critical importance of bullet length, I added the calculated effective lengths of the bullets used for the sample calculations posted, two posts above. As explained there, for the sake of an easy, text-book calculation of bullet spin moment of inertia, I approximated the true bullet shape by a simple cylinder, with flat, square ends. The true bullet lengths would be longer than these approximate results, for the same weight of bullet.
  13. helotaxi

    helotaxi Well-Known Member

    But the cylinder would be the worst case scenario with regard to energy required to get up to speed. You also assumed a monolithic lead slug which would be worse than a jacketed bullet.
  14. Bill_Rights

    Bill_Rights Well-Known Member


    Yes, also true. But also the effect of the jacket (being lighter, less dense material) would be somewhere around negligible, unless the jacket was very thick.

    For my part, I was interested in terminal ballistics (at or inside the target). How much energy does bullet spin carry to the target and could this play any role on effectiveness at impact or penetration? Since the spin energy is well under 1% of the total bullet kinetic energy, my answer seems to be "no".

    You, on the other hand, seem more interested in firing chamber dynamics. If we are going to "dissect" that, we also should consider the energy required to engrave the rifling grooves into the bullet flanks. I am a materials scientist, so I could probably estimate the displacement or deformation energy of the bullet jacket, if I had some idea of the mass of the bullet material that gets moved and how far it gets moved. But this is probably also near to negligible, especially for soft materials such as dead-soft lead.

    We might be beating a dead flea on a dead horse :scrutiny:.
  15. Bill_Rights

    Bill_Rights Well-Known Member

  16. helotaxi

    helotaxi Well-Known Member

    Once you factor in the obduration of the bullet with the very high forces invloved, the engraving force is anything but negligible. The rate of onset of that force is one of the determining factors of the initial pressure curve of the cartridge. The leade and freebore are manipulated to modulate that onset rate and are the deterministic differences between the .223 and 5.56x45 chambers as well as the 6.8 SPC and 6.8 SPC II chambers. the reduction in the onset of the engraving force is such that a hotter round can be used while keeping the same max chamber pressure and the longer pressure curve of the hotter load results in higher velocity. In the case of the 6.8 SPC, the change from 1:10 to 1:11 rate of twist was the final adjustment used. Again a rate of onset issue. The peak in linear acceleration is also the peak in rotational acceleration and occurs at peak pressure. Reducing the end result RPM by 10% reduced the required rotational acceleration as well and the energy requirement for that acceleration.
  17. Bill_Rights

    Bill_Rights Well-Known Member


    Could you define "obduration" of the bullet? I looked it up, but it is not clear how the general definition applies:
    Regardless, I gather that it has to do with deformation of the bullet, which is resisted by the bullet material. It is not unknown to occur that the energy (and of course the force) required to deform a material increases with increasing rate of deformation. And the rate of deformation of a bullet will be FAST.

    Considering only the energy, energy is defined as a force acting through a distance (or power expended over a time duration). The distance over which the high engraving forces are active may be only the length of the flanks of the bullet, i.e., short. So the total energy expended in engraving may be small.

    Which brings the point (which I have known all along) that energy, itself, may not be the important parameter. In the case of terminal ballistics due to spin, angular momentum may be more important than spin energy, for example. As you point out, force can be critical. Force directly maps to pressure. [pressure] = [force per unit area]

    It bears thinking about which of the following quantities are most important to bullet kinematics (kinematics = motion dynamics). Here are the linear quantities, and each has it angular (spin) analog:

    position: .................. (self-explanatory)
    velocity (or speed): ... [rate of change of position]
    acceleration: ............ [rate of change of velocity]
    jerk: ........................ [rate of change of acceleration]
    momentum: ............... [mass mutiplied by velocity]
    force: ...................... [mass mutiplied by acceleration]
    pressure: .................. [force applied per unit area OR force divided by area]
    energy: .................... [force mutiplied by distance]
    power: ..................... [energy expended per unit time OR energy divided by time]
  18. helotaxi

    helotaxi Well-Known Member

    Basically when the pressure starts to move the bullet forward, the bullet gets "squished" for lack of a better term along its longitudinal axis and expands along its radius. This occurs at pretty much the exact time that the bullet is being squeezed into the rifling.

    The squishing of the bullet to fill/seal the bore is what is typically referred to as obduration in this context.
  19. WNTFW

    WNTFW Well-Known Member

  20. helotaxi

    helotaxi Well-Known Member


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