there is such a thing as an over stabilised bullet

Snap.

So what happens to a bullet when is it over stabilised, does it become unstabilised?

John

hey mate i'm not looking to start anything just giving my opinion...

And when a bullet becomes overstabilisied, once it reaches the highest point of the flight curve the tip of the bullet should be slighty nose down....

now in an over stabalisied bullet, at this point it tends to be tip facing slightly skyward, and the air flow becomes the problem as the air is flowing faster over one surface than the other,

this causes the bullet to loose its stated BC. value, the trajectory changes, and the bullet impacts slightly heigher than it should...

hence some of the top marksmen in the world figured this out...

** ****The force of gravity **- <LI class=feature>
**e **j - Unit vector, opposite to the direction of the acceleration of gravity **F **G - *The force of gravity *

Explanation: The force of gravity is proportional to the mass of the projectile and the local acceleration of gravity. The force is directed towards the center of the earth and attacks at the CG. The force of gravity is responsible for the bending of the trajectory.

**The centrifugal force **

Explanation: The figure above shows a cut through the globe. The formula gives the components of the centrifugal force in an xyz - reference frame, the y -axis being antiparallel to the force of gravity. The y - component of the centrifugal force can be regarded as a correction of the force of gravity, the other components are generally neglected in ballistics because of their smallness.

**The Coriolis Force ***F c - *Coriolis force*v - *Velocity vector with respect to xyz - coordinate system*w - *Vector of the angular velocity of the earth´s rotation with respect to xyz - coordinate system.

**Explanation **The magnitude of the

**Coriolis force **is so small that it is usually completely neglected.

The

**Coriolis effect **can be demonstrated by the following experiment.

Imagine a disk which first does not rotate. A mass m which moves with constant velocity v from the center of the disk towards the rim moves on a straight line. No force affects the movement of the mass.

Now let the disk rotate with a constant angular velocity w (see figure above). Still no external force affects the mass. An observer who rotates with the disk - and has no indication of this rotation - sees the mass moving along a bended path. This observer would state that the mass must have been affected by a "force", which acts perpendicular to the direction of movement.

This fictitious "force" is called the Coriolis force

**F **c

The real problems in this type of calculation is that we are shooting on a sphere and the results will be dependant on where you are on the earth at the time of the shot and what direction you shoot in. It is more straight forward if you shoot only perfectly north or south (according to the earths axis of rotation) and on a flat surface. If you shoot any other direction other than perfectly north or south it gets much more complicated very quickly. So as you can see, knowing where you are on the surface of the earth is extremely important. You would need to know where you are to a greater degree of accuracy than a standard civilian GPS unit could place you. The calculation gets more complicated when you consider the fact that lines of longitude are “Great Circles” and represent differing amounts of linear measure depending on how far you are from the equator.

Also you must take into consideration any east/west component of the bullets path. Then so-called “centripetal acceleration” must also be factored into the equation with the ultimate end of this being the projectile being put into orbit. Spherical geometry is not my best subject but it is an important one when dealing with this subject.

Simple exterior ballistics (more integral calculus) cannot be excluded either because they determine how long the bullet will stay in the air before returning to the surface. Keep in mind this is not a flat surface but one that is curving away from us thus the “centripetal acceleration” factor in our calculations.

That said, here's a brief explanation of the basics. I am sure that others reading this with more understanding will find faults with my logic, math and explanations so have at me guys.

The Coriolis effect can be looked at from one of two perspectives as all things in physics. All things are relative. The simplest way to look at it is from the projectiles perspective. It is expressed by the following equation:

F c = -2m(wv i )

Where F c = Coriolis effect

m = mass of the object

w = Angular velocity of rotating surface

v i = Velocity of rotating surface

* For the earth the angular velocity is 1 rev per day or about 7x10 -5 radians/24 hours

-Sort of the same radians as in our mil-dot reticles but that's a different issue-

*v i will change depending on where latitudinally you are on the earth. The v i of 1100mph applies only at the equator. At the poles it is 0mph

**Note that this is only a linear equation so the rate that the projectile loses its velocity is not considered here

The other consideration is called the

Magnus Force. This is the same force that causes a curve-ball to curve. It mainly depends on bullet geometry, spin rate, velocity and angle of yaw. Modern projectiles yaw very little but it does happen.

This is a long and lengthy discussion that barely scratches the surface. These effects appear to be negligible when compared to other more significant factors such as the wind and velocity of the target.

**The Drag Force** - <LI class=feature>
*c D *- Drag coefficient *F D *- Drag force

Explanation: The drag force

**F **D is the component of the force

**F **W in the direction opposite to that of the motion of the centre of gravity. The force

**F **W results from pressure differences at the bullet's surface, caused by the air, streaming against the moving body. In the case of the absence of yaw, the drag

**F **D is the only component of the force

**F **W .

The drag force is the most important aerodynamic force. Given the atmosphere conditions

*p,T,h *, the reference area

*A *and the momentary velocity

*v w *, the drag force is completely determined by the the drag coefficient

*c D *.

**The Drag Coefficient **The drag coefficient

*c D *is the most important aerodynamic coefficient and generally depends on - bullet geometry (symbolic variable

*B *),

- <LI class=feature>Mach number
*Ma, *<LI class=feature>Reynolds number *Re* - The angle of yaw d

The following asssumptions and simplifications are usually made in ballistics:

1.

*Re *neglection

It can be shown, that with the exception of very low velocities, the

*Re *dependency of

*c D *can be neglected.

2. d dependency

Depending on the physical ballistic model applied, an angle of yaw is either completely neglected ( d =0) or only small angles of yaw are considered. Large angles of yaw are an indication of instability.

For small angles of yaw the following approximation is usually made:

a) *c D *( *B,Ma, *d ) = *c Do *( *B,Ma *) + *c D d *( *B,Ma *) * d 2 /2

Another theory which accounts for arbitrary angles of yaw is called the "crossflow analogy prediction method". A discussion of this method is far beyond the scope of this article, however the general type of equation for the drag coefficient is as follows:

*c D *( *B,Ma, *d ) = *c Do *( *B,Ma *) + *F *( *B,Ma,Re, *d )

3. Determination of the zero-yaw drag coefficient

The zero-yaw drag coefficient as a function of the Mach number

*Ma *is generally determined experimentally either by wind tunnel tests or from Doppler Radar measurements.

Fig.: Zero-yaw drag coefficient for two military bullets

M80 (cal. 7.62 x 51 Nato)

SS109 (cal. 5.56 x 45)

There is also software available which estimates the zero-yaw drag coefficient as a function of the Mach number from bullet geometry. The latter method is mainly applied in the development phase of a new projectile.

4. Standard drag functions

Generally each bullet geometry has its own zero-yaw drag coefficient as a function of the Mach number. This means, that specific - time-consuming and expensive - measurements would be required for each bullet geometry. A widely used simplification makes use of a "standard drag function"

*c Do standard *which depends on the Mach number alone and a form factor

*i D *which depends on the bullet geometry alone according to:

*c Do *(

*B,Ma *) =

*i D *(

*B *) *

*c Do standard *(

*Ma *)

If this simplification is applicable, the determination of the drag coefficient of a bullet as a function of the Mach number is reduced to the determination of a suitable form factor alone. It will be shown that the concept of the ballistic coefficient , widely used in the US for small arms projectiles follows this idea.

- <LI class=feature>
*c D - Drag coefficient; **c D (B,Ma,Re, *d ) <LI class=feature>*c Do standard *- Zero-yaw standard drag function *i D *- Form factor

**The Ballistic Coefficient ( ***bc *) The 'ballistic coefficient' or

*bc *is a measure for the drag experienced by a bullet moving through the atmosphere, which is widely used by manufacturers of reloading components, mainly in the US. Although, from a modern point of view,

*bc *s are a remainder of the pioneer times of exterior ballistics, ballistic coefficients have been determined experimentally for so many handgun bullets, that no treatise on exterior ballistics would be allowed to neglect it..

The bc of a test bullet

*bc test *moving at velocity

*v *is a real number and defined as the deceleration due to drag of a "standard" bullet devided by the deceleration due to drag of the test bullet.

The standard bullet is said to have a mass of 1 lb (0.4536 kg) and a diameter of 1 in (25.4 mm). The drag coefficients of the standard bullet can be derived from the G1-function given in literature and will be named

*c Do G1 *(

*Ma *) .

Using

*c Do test *(

*B,Ma *) =

*i D test *(

*B *) *

*c Do G1 *(

*Ma *)

one finds for the

*bc *(assuming "standard" atmosphere conditions)

*bc test =1 / i D test ( * m test / d 2 test *This formula also shows that the

*bc *and the form factor

*i D *of a "test" bullet are two aspects of the same principal simplification: the substitution of the (unknown) particular drag function of a bullet by the (given) "standard" drag function of the standard bullet.

*c Do test - Zero-yaw drag coefficient of test bullet **c Do G1 - Zero-yaw G1 standard drag coefficient **i D test* - Form factor of test bullet*bc test *- Ballistic coefficient of test bullet*m test - Mass of test bullet in lb **d test - Diameter of test bullet in inches *

**The Lift Force ***c L - Lift coefficient; **c L (B,Ma.Re, *d ) *e L- *Unit vector*F L - **Lift force *

Explanation: The lift force

**F L **(also called

**cross-wind force **) is the component of the wind force

**F W **in the direction perpendicular to that of the motion of the center of gravity in the plane of the yaw angle. The lift force vanishes in the absence of yaw and is the reason for the drift of a spinning projectile even in the absence of wind.

**The Spin Damping Moment **** **Explanation: Skin friction at the bullet's surface retards its spinning motion. The spin damping moment (also: roll damping moment) is given by the above formula. The spin damping coefficient depends on bullet geometry and the flow type (laminar or turbulent).

**The Magnus Force **F m = P (p x T x h)/2 x A x c Mag (B x M a x R e (w x d/v w ) x (w x d/v w ) x S) x

(w x d/v w ) x v w 2 x e M

Where

- e M = 1/sin
*S *(e c x e t ) - M a = v w /@(p . T . h)
- R e = P(p x Tx H) x
*l *x v w /u x T - C mag = Magnus force coefficient
- e M = Unit Vector
- F m = Magnus Force
- A = Bullet cross section area
- a = Velocity of sount in air
- B = Symbolic Variable, indicating bullet geometry
- d = bullet diameter
- e c = Unit vector into the direction of the bullet's longitudinal axis
- e t = Unit vector into the direction of hte tangent ot the trajectory
- g = Acceleration due to gravity
- h = Relative humidity of air
*l *= bullet length- m = bullet mass
- M a = Mach number
- p = air pressure
- R e = Reynolds number
- T = Absolute air temperature
- u = absolute viscosity of air
- v w = bullet velocity with respect to wind system
- @ = Azimuth angle
- S = Yaw angle
- P = Air density
- w = spin rate of bullet (angular velocity)

Explanation: The Magnus force

**F **M arises from an asymmetry in the flow field, while the air stream against a rotating and yawing body interacts with its boundary layer and applies at the CPM. Depending on the flow field, the CPM may be located ahead or behind the CG. The Magnus force vanishes in the absence of rotation and in the absence of a yaw angle.

The Magnus force is usually very small and mainly depends on bullet geometry, spin rate, velocity and the angle of yaw. In exterior ballistics, the above expression is used for the Magnus force.

For the whole bullet, the

**Magnus effect **(which arises from the boundary layer interaction of the inclined and rotating body with the flowfield) results in the

**Magnus force F M **which applies at its centre of pressure CPM. The location of the CPM varies as a function of the flowfield conditions and can be located either behind or ahead of the CG.

The figure above assumes that the CPM is located behind the CG. Experiments have shown that this comes true for a 7.62 x 51 FMJ standard Nato bullet at least close to the muzzle in the high supersonic velocity regime.

**The Overturning Moment **- <LI class=feature>
*c M - Overturning moment coefficient, **c *M ( *B, Ma, Re, *d ) <LI class=feature>**e **W - Unit vector **M **W - Overturning moment

Explanation: The point of the longitudinal axis, at which the resulting wind force

**F **1 appears to attack is called the centre of pressure CPW of the wind force, which, for spin-stabilized bullets is located ahead of the CG. As the flow field varies, the location of the CPW varies as a function of the Mach number. Due to the non-coincidence of the CG and the CPW, a moment is associated with the wind force. This moment

**M **W is called

**overturning moment **or

**yawing moment**. For spin-stabilized projectiles

**M **W tends to increase the yaw angle and destabilizes the bullet. In the absence of spin, the moment would cause the bullet to tumble.

The forces

**F **1 and

**F **2 form a free couple, which is said to be the

**aerodynamic moment of the wind force **or simply

**overturning moment M **w. This moment tries to rotate the bullet about an axis through the CG, perpendicular to the axis of symmetry of the bullet. The overturning moment tends to increase the angle of yaw d .

The force

**F **W , which applies at the CG can be split into a force, opposite to the direction of the movement of the CG (the direction of the velocity vector

**v **), which is called the

**drag force F **D or simply

**drag **and a force, perpendicular to this direction, which is called the

**lift force F **L or simply

**lift **.

**The Magnus Moment***c Mp *- Magnus moment coefficient; *c Mp (B,Ma,Re, *w , d )*e MM *- Unit vector*M M - Magnus moment *

Explanation: As the Magnus force applies at the CPM, which does not necessarily coincide with the CG, a Magnus moment

**M M **is associated with that force. The location of the centre of pressure of the Magnus force depends on the flow field and can be located ahead or behind the CG. The Magnus moment turns out to be very important for the dynamic stability of spin-stabilized bullets. For the Magnus moment, the above expression is used in exterior ballistics.

The Magnus force, which applies at its centre of pressure CPM can be substituted by a force of the same magnitude and direction, which applies at the CG, plus a moment, which is said to be the

**Magnus moment M M **This moment tries to rotate the bullet about an axis, perpendicular to the longitudinal axis of the bullet.

<P class=feature>However, the gyroscopic effect also applies for the Magnus moment and the bullet´s axis will be shifted into the direction of the moment. Thus, as far as the conditions of the figure above are valid, the Magnus moment will have a

**stabilizing **effect as it tends to

**decrease **the angle of yaw d . It can be easily shown that this is only true, if the centre of pressure of the Magnus force CPM is located

**behind **the CG. The Magnus force

**destabilizes **the bullet and

**increases **the angle of yaw, if its centre of pressure is located

**ahead **of the CG, which may come true in a specific velocity regime.

**The Gyroscopic Stability Condition **- <LI class=feature>
*c M a *- Overturning moment coefficient derivative; *c M a (B,Ma *) *s g* - Gyroscopic stability factor

Explanation: A spin-stabilized projectile is said to be

**gyroscopically stable **, if, in the presence of a yaw angle d , it responds to an external wind force

**F **1 with the general motion of nutation and precession. In this case the longitudinal axis of the bullet moves into a direction perpendicular to the direction of the wind force.

<P class=feature>It can be shown by a mathematical treatment that this condition is fulfilled, if the gyroscopic stability factor

*s g *exceeds unity. This demand is called the

**gyroscopic stability condition. **A bullet can be made gyroscopically stable by sufficiently spinning it (by increasing w! ). <P class=feature>As the spin rate

*w *decreases more slowly than the velocity

*v w *, the gyroscopic stability factor

*s g *, at least close to the muzzle, continuously increases. <P class=feature>Thus, if a bullet is gyroscopically stable at the muzzle, it will be gyroscopically stable for the rest of its flight. The quantity

*s g *also depends on the air density r and this is the reason, why special attention has to be paid to guarantee gyroscopic stability at extreme cold weather conditions. <P class=feature>Bullet and gun designers usually prefer

*s g *> 1.2...1.5, but it is also possible to introduce too much stabilization. This is called over-stabilization. <P class=feature>The gyroscopic (also called static) stability factor depends on only one aerodynamic coefficient (the overturning moment coefficient derivative

*c M a *) and thus is much easier to determine than the dynamic stability factor. This may be the reason, why some ballistic publications only consider static stability if it comes to stability considerations. However, the gyroscopic stability condition only is a

**necessary condition **to guarantee a stable flight, but is by no means sufficient. Two other conditions - the conditions of dynamic stability and the tractability condition must be fulfilled.

This figure shows the

**gyroscopic stability factor **of the 7.62 x 51 Nato bullet M80, fired at an angle of departure of 32°, a muzzle velocity of 870 m/s and a rifling pitch at the muzzle of 12 inches. The M80 bullet shows static stability over the whole flight path as the static stability condition

*s g *>1 is fulfilled everywhere. The value of

*s g *adopts a minimum of 1.35 at the muzzle.

<P class=feature>Generally it can be stated that if a bullet is statically stable at the muzzle, it will be statically stable for the rest of its flight. This can be easily understood from the fact, that the static stability factor is proportional to the ratio of the bullet´s rotational and transversal velocity (see formula ). As the the rotational velocity is much less damped than the transversal velocity (which is damped due to the action of the drag), the static stability factor increases, at least for the major part of the trajectory. Bullet and gun designers usually prefer

*s g *> 1.2 ..1.5 at the muzzle, however it has been observed that many handgun bullet show

**excessive **static stability.

**The Dynamic Stability Condition **- <LI class=feature>
*c D *- Drag coefficient *c L a *- Lift coefficient derivative<LI class=feature>*c Mp a *- Magnus moment coefficient derivative <LI class=feature>*c mq +c m a *- Pitch damping moment derivative <LI class=feature>*s g *- Gyroscopic (static) stability factor *s d *- Dynamic stability factor

Explanation: A projectile is said to be dynamically stable , if its yawing motion of nutation and precession is damped out with time, which means that an angle of yaw induced at the muzzle (the initial yaw) decreases.

<P class=feature>A dynamic stability factor

*s d *can be defined from the linearized theory of gyroscopes (assuming only a small angle of yaw) and the above

**dynamic stability condition **can be formulated. An alternate formulation of this condition leads to the illustrative

**stability triangle **.

*s d *however depends on five aerodynamic coefficients. Because these coefficients are hard to determine, it can become very complicated to calculate the dynamic stability factor, which varies as a function of the momentary bullet velocity.

**The Stability Triangle ***s g *- Gyroscopic stability factor*s d* - Dynamic stability factor

Explanations: The dynamic stability condition can be expressed in an alternate way. leading to a very illustrative interpretation of bullet stability. In using a quantity s , according to the above definition, the dynamic stability condition takes a very simple form (see above formula). This means that for a bullet to be gyroscopically

**and **dynamically stable, a plot of s vs.

*s d *has to remain completely within the stability triangle (green area in the figure below).

**The Tractability Condition**- <LI class=feature>
*f *- Tractability factor <LI class=feature>*f l *- Low limit tractability factor; *fl *» 5.7 <LI class=feature>*s g *- Gyroscopic stability factor *d *p - Yaw of repose vector

Explanation: The

**tractability factor ***f *characterizes the ability of the projectile's longitudinal axis to follow the bending trajectory. The quantity

*f *can simply be defined as the inverse of the yaw of repose. It can be shown that the tractability factor

*f *is proportional to the inverse of the gyroscopic stability factor.

Over-stabilized bullet

This figure schematically shows an

**over-stabilized bullet **on a high-angle trajectory.

An over-stabilized bullet rotates too fast and its axis tends to keep its orientation in space. The bullet´s longitudional axis becomes uncapable to follow the bending path of the trajectory. Over-stabilization is said to occur, if the angle enclosed between the bullet´s axis of form and the tangent to the trajectory (the yaw of repose) exceeds a value of approximately 10°.

<P class=feature>Over-stabilization of a bullet is most probable, if a bullet has

**excessive static stability **(a high value of s g and a low value for the tractability factor ) and is fired at a high angle of departure, especially when fired vertically. An over-stabilized bullet on a high-angle trajectory lands base first. However, when firing bullets from handguns, over-stabilization is of minor importance in normal shooting situation, but

**must **be considered when firing at high angles of elevation.

**The Yaw of Repose <A name=yawrepose>**- <LI class=feature>
*c M a *- Overturning moment coefficient derivative coefficient **d **p - Yaw of repose vector

Explanation: The repose angle of yaw (or

**yaw of repose **, also called

**equilibrium yaw **) is the angle, by which the momentary axis of precession deviates from the direction of flight. As soon as the transient yaw induced at the muzzle has been damped out for a stable bullet, the yaw angle d equals the yaw of repose.

<P class=feature>The magnitude of the yaw of repose angle is typically only fractions of a degree close to the muzzle, but may take considerable values close to the summit especially for high-elevation angles. <P class=feature>The occurrence of the yaw of repose is responsible for the side drift of spin-stabilized projectiles even in the absence of wind. The spin-dependent side drift is also called

**derivation **. <P class=feature>It can be shown that for right-hand twist, the yaw of repose lies to the right of the trajectory. Thus the bullet nose rosettes with an average off-set to the right, leading to a side drift to the right. The above formula for the yaw of repose vector is an approximation for stable bullet flight.

If a bullet flies stable (gyroscopically

**and **dynamically!) and the transient yaw has been damped out, usually after a travelling distance of a few thousands of calibres, the bullet´s axis of symmetry and the tangent to the trajectory deviate by a small angle, which is said to be the yaw of repose .

For bullets fired with right-handed twist, the longitudinal axis points to the right and a little bit upward with respect to the direction of flight, leading to a side drift to the right. The yaw of repose, although normally measuring only fractions of a degree, is the reason for the side deviation of spin-stabilized bullets.

As i said i'm not looking for an argument.... just the opposite... i seek true enlightenment.... lol

Snap.