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1/25/2018 7:38:29 AM
Posted: 11/13/2003 6:11:30 PM EST
It at least tumbles, right?
Link Posted: 11/13/2003 7:46:21 PM EST
[Last Edit: 11/13/2003 7:49:30 PM EST by Troy]
Link Posted: 11/13/2003 7:48:46 PM EST
If I had to pick, I would take the first one in the gut.
Link Posted: 11/14/2003 5:02:05 AM EST
In my limited adventures in testing slow varmint ammo, I have noticed that V-Maxes that are going slow (1300-1400 FPS tested) tend to mushroom a little when striking the target. It is not a pretty mushroom, and the core of the bullet does seperate, but the lead core is flattened in front and the jacket is pulled all the way to the rear and missing some pieces. Weight retained was better than 60%. It almost reminds me of some 180 grain Golden Saber bullets I found in the backstop after firing from a 10mm. There were core seperations on these too, and the jacket looked like a brass daisy. I'll have to look around the reloading room and see if I can find an example.
Test load: Bullet: 35 gr. V-Max Powder: 4 gr. Blue Dot Primer: CCI 400 Case: Winchester 223 COL: 2.1"
This is my yard pest load. It is hotter than .22 LR, but not by much.
Link Posted: 11/14/2003 7:37:50 AM EST
What troy said. Yes, they will.
Link Posted: 11/15/2003 7:31:02 AM EST
Of course bullete tumble! Almost all of them do! Tumble means "to fall or roll end over end." To "flip end to end" is to tumble. [i]All projecticles with lengths longer than their diameters will tumble[/i] when they go unstable such as when striking dense media like animal flesh or water. Pointed projectiles will typically only [i]tumble[/i] 180° and stabilize in a base first configuration. Some projectiles like the 5.45mm M74 will tumble several rotations and never statilize in animal tissue unless you're shooting an elephant, there isn't enough tissue depth. -- Chuck
Link Posted: 11/15/2003 12:20:44 PM EST
Link Posted: 11/15/2003 1:03:58 PM EST
All projecticles with lengths longer than their diameters will tumble
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It has nothing to do with the ratio between length and width, it has everything to do with the position of center of gravity. It will be most stable when the center of gravity is most foreward. Thus, even a sphere would rotate so that the heaviest part for foremost. A bullet could easily be designed that was long and thin and had it's center of gravity towards the front. They just don't do that, because a bullet with that design would be less aerodynamic, thus it would decellerate faster. That is why most bullets have the center of gravity in the rear, and twist is used to keep them from rotating in flight.
Link Posted: 11/16/2003 6:30:04 AM EST
If you define tumble as continual rotation end over end then few bullets tumble in dense media. This [i]isn't[/i] the defination of tumble, however. Tumble is a simple rotation end for end. Even a half rotation. But moving on.... [;)] Tendency to [i]tumble[/i] is a factor of projectile [i]length[/i] not Center of Gravity. Remember the required rifling twist to achieve bullet stability is based on bullet [i]length[/i], not weight (although with identical construction length and weight will be in a constant ratio). Long bullets are much more unstable and require faster twists to stabilize them in air (or any media). The denser the media the more spin required, hence bullets stable in air will go unstable in animal tissue or water. Unstable bullets will tumble. Pure cylinder shaped wad cutters with their Center of Gravity exactly in the center will tumble when they strike dense media. Clearly an aft CG doesn't cause tumbling. What an aft CG does do, however, is allow the bullet to stabilize after the bullet has tumbled to a base first position -- moving the GC to the front. Wad cutters may tumble several times in enough depth. You'll have a hard time making a .45ACP bullet tumble, even 180° because it's length is barely longer than it's diameter. Military rifle ball ammo projectile length is typically 3x - 4x the diameter and has an aft CG. Because of their pointed shape and aft CG rifle FMJ bullets typically [i]tumble[/i] 180° to a stable, base first position and with enough momentum will exit the body base first. Unless they fragment. -- Chuck
Link Posted: 11/16/2003 9:20:49 AM EST
[Last Edit: 11/16/2003 9:24:20 AM EST by stuh505]
Tendency to tumble is a factor of projectile length not Center of Gravity. Remember the required rifling twist to achieve bullet stability is based on bullet length, not weight (although with identical construction length and weight will be in a constant ratio). Long bullets are much more unstable and require faster twists to stabilize them in air (or any media). The denser the media the more spin required, hence bullets stable in air will go unstable in animal tissue or water. Unstable bullets will tumble.
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the reason that long bullets require more twist is not because they are longer, but because their center of gravity is further back. this correlation exists because bullets are given an ogival or rounded or otherwise aerodynamic front end which causes the front end of the bullet to be less dense than the back end. if you fire a long bullet backwards it won't yaw because it's center of gravity is already forwards. EDIT - perhaps you will be skeptical of anything i say, so I just found a quote to support me from the ammo faq:
All FMJ bullets with tapered noses will tumble in flesh with enough velocity, because their center of gravity is aft of their length center--causing them to want to travel "tail first" in denser mediums (like water and tissue).
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Link Posted: 11/16/2003 2:23:22 PM EST
The fact that bullets normally have pointed noses and their CG aft isn't the reason they're unstable. Fire a bullet base first thru a non rifled barrel and you'll hear it whirr on the way down range as it tumbles continually. So much for forward CG. (If it was this easy we'd just put something heavy in the nose of bullets and forget the rifling.) Likewise, launch a cylindical wadcutter down a smooth bore and it'll start tumbling just as will a pointed bullet fired either end first. The longer a bullet gets, the more unstable it gets. That's why longer bullets need to be spun faster to stabilize them. Run these bullets down a rifled barrel with a fast enough twist and they'll run downrange just fine, but will tumble as soon as they strike dense media. It's currently physically impossible to spin a rifle bullet fast enough to be stable in water or animal tissue, hence they tumble. If their CG is aft they will eventually stabilize base first as forward CG is more stable than aft and this shows up more graphically in dense media. When General Hatcher participated in the [i]bullets from the sky[/i] experiment after WW1 they found the bullets returned from max ordinate continually tumbling most of the time and base first the other times. Bullets rarely tumble more than 180° in dense media, but in air they can go end for end several times if not constantly. Velocity of these bullets from the sky was very low because of the drag induced by their constant tumbling and it's doubtful they'd break the skin. Rifled muskets, using round balls use almost no twist in their rifling, or at least not much as we're used to seeing it. A round ball in the size of about .60 caliber only needs a rifling twist of one turn in 7-1/2 feet or so to be stable. You gotta spin M856 about 11 times faster or it will -- wait for it -- tumble in air. -- Chuck
Link Posted: 11/16/2003 4:06:40 PM EST
[Last Edit: 11/16/2003 4:09:12 PM EST by stuh505]
Chuck, we have been repeating ourselves a couple times now. I know why I am not agreeing with you yet, and that is because what you are saying seems to defy not only logic, but also personal experience, and also quotes from Martin Fackler. Let me perform a little mental experiment. It is so simple that I don't even need to actually do the experiment because I think the results are obvious. If I make a paper airplane 6 inches long with a 2 inch wingspan, and glue a lead weight to the tail of the plane, and throw it, the plane will of course tumble once until the weight is in the front and then it will continue. If I throw this airplane base first, it will just continue base first, because that is the heavy side. In this case the difference in center of mass is more pronounced in a bullet, but it is the same principle. According to your logic, the paper airplane would flip around once regardless of if I threw it point first or base first. That just doesn't make any sense.
Link Posted: 11/17/2003 12:54:55 AM EST
No he is saying the paper airplane would flip end over end multiple times if there were no weights on it. The problem is that evenly ballanced bullets seldom "tumble" and even unbalanced bullets can fail to yaw past 180 degrees before leaving a human body.
Link Posted: 11/17/2003 1:35:11 AM EST
Link Posted: 11/17/2003 2:35:46 AM EST
If you were to reverse the construction of M855 Ball and put the steel in the aft half and lead in the front do you think you could shoot this accurately from an unrifled barrel? I wonder why those ballistic engineers didn't think of this long ago and save all the nonsense and expense of rifling? I will readily agree that Center of Gravity forward is more stable, but that's a relative term. Real world cylindrical projectiles are unstable in air regardless of of where their CG happens to be. You can't duplicate your paper airplane example in a bullet. If your massively aft weighted paper airplane is launched nose first it will tumble into "stable" CG forward flight. It will indeed tumble 180°. However if you spin it fast enough it won't as pointed bullets don't. Or if you launch it in a vacuum. The media thru which the projectile is moving has great effect on this. Projectiles are less stable in dense media -- as we see when bullets strike dense media. If you want to shoot your Glock underwater put the bullets in the cases backwards so they don't have to tumble 180°. -- Chuck
Link Posted: 11/17/2003 2:37:39 AM EST
There's a couple of smileys [;)] missing above.... This is a discussion, not an arguement. It's governed by both the laws of physics (fluid dymanics) and what happens in the real world. -- Chuck
Link Posted: 11/17/2003 3:26:41 AM EST
If you were to reverse the construction of M855 Ball and put the steel in the aft half and lead in the front do you think you could shoot this accurately from an unrifled barrel? I wonder why those ballistic engineers didn't think of this long ago and save all the nonsense and expense of rifling?
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I think that it would be stable, but it is not done because that bullet would lose velocity extremely quickly.
Link Posted: 11/17/2003 3:51:04 PM EST
You can stabilize projectiles in two ways: spin stabilize, or fin stabilize them. Forward CG is not a reliable stabilization means. Certainly not in light media like air at high velocities. I suggest shooting an arrow without any flecthes to see the stability of a forward center of gravity in a long projectile. Didn't we all try this as boys? Pretty dramatic. Arrow goes out a few feet, tumbles and stops like it hit a wall. Even short crossbow bolts need fin stabilization. -- Chuck
Link Posted: 11/17/2003 5:51:11 PM EST
[Last Edit: 11/18/2003 9:45:17 AM EST by stuh505]
Chuck, it looks like I am forced to break out the physics. It's all about torque and the drag force. 1) Let me represent a bullet with a vector for simplicity. It goes from (0,0) to (0,10) (so that means its vertical pointing down). It has a center of gravity at (0,2). 2) This bullet is travelling to the left (X approaches negative infinity) at some velocity which produces a drag force which is uniform over the entire length of the bullet (assume that the bullet is simply a uniform cone). We cannot calculate the drag force, but since it is constant, we can give it an arbitrary value of 2. 3) We can calculate the torque about both sides of the radius which is the center of gravity. 4) Torque = RxF (r [i]cross[/i] f, this is a cross product, not a multiplication). This is equivalent to |r||f|sin(theta), where theta is the angle between the direction of the radius and the direction of force being applied. Since the radius is going vertically and the force is being applied from the left, the theta is 90 degrees. Sin90 = 1, so our equation can be further simplified: T = |r||F| 5) Torque on the lower half would then be the integral of |r||f|dr from r = 0 to r = 2. f we already said was a constant of 2. This constant can be moved out of the integral. So we have T = 2 integral(rdr)from r=0 to r =2. This very simple integration is obviously 4 (2^2-0^2). 6) Torque on the upper half is 2 integral(rdr)from r = 2 to 8, so that comes out to 60 (8^2-2^2). 7) Thus, since the torque on the upper half is greater, the bullet will rotate clockwise with the upper half coming down until it is parallel. Since it rotated about the center of gravity, the bullet is now in the position (-2,2) to (8,2) 8) At this point, the torque about the upper and lower half are both 0, so it will not rotate any more. This is the stable position. 9) At first, it might seem that this same logic would mean that it would rotate due to torque of gravity. This is false, because remember that force due to gravity = m*g, and since torque is about the CM (center of mass) point, that means that force due to gravity will be zero on both sides of the CM.
Link Posted: 11/18/2003 12:40:51 PM EST
[Last Edit: 11/18/2003 1:31:44 PM EST by stuh505]
Continuation of the math... 10) a diagram can be constructed to visually represent what the equations in my last post were showing. [img]http://ardan-nights.org/members/downloads/nwn/other/t1.JPG[/img] 11) In figure 1, you see a bullet represented by a line. It does not have evenly distributed mass; the CM (Center of mass) is denoted by the black dot which is on the dotted CM line. The drag force is even along each point. 12) Given this distribution of drag force, the torque equation can be shown to produce greater torque around the upper part of the bullet because the CM is at the back. Torque is an object's tendency to rotate about the CM. 13) Figure 3 shows potential results of this. This is how it would yaw in any medium. The drag force is greater in tissue so that's why we see yawing in tissue. I am ignoring spin for now. 14) Upon striking the medium, the bullet is tipped slightly upwards. This is because the bullet was fired slightly upwards from the horizontal of the ground because the shooter was compensating for bullet drop. Step 9 explains why the bullet stays at this angle and is not affected by gravity to rotate. It will rotate a very miniscule amount due to downward drag forces but since the aerodynamics are so uniform I will ignore this. 15) In Fig 3, torque is actually a vector which comes out of the page. I am trying to represent it with a curved line like in Fig2 to show how it would rotate, but due to technical difficulties, you will have to just imagine that it is curved. The effect is to show you the magnitude of the torque, and how it will cause it to rotate. 16) As you can see once it is base first there is no more torque causing it to rotate due to horizontal drag forces. There is no gravitational torque. Any vertical drag force can be ignored. The bullet still has some rotational momentum. This does not cause it to continue to rotate because the drag force through tissue is much larger. 17) If the bullet had originally been fired base first, you can see the effects would cause it to remain base first after colliding with tissue. EDIT: 18) So how does spin factor come into play? If the bullet is spinning fast enough then it will have equal torque telling it to rotate in counteracting directions which will cancel out and cause it to not rotate. Faster spin rates are required to cancel out larger torques due to larger drag forces. Thus the spin required to prevent rotation(or yaw) in tissue is much higher than to prevent yaw in a less dense medium such as air. Hope this helps you to understand Chuck.
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