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# Left @ 300 yds - Right @ 1000 yds? WTH?

#127
12-23-2008, 08:09 PM
 Platinum Member Join Date: Jan 2003 Location: The rifle range, or archery range or behind the computer in Alaska Posts: 3,740
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

Quote:
 Originally Posted by royinidaho The impact on the trajectory based on bullet impact of shadows could be conducted. BTW, does a shadow have mass or is it a lack of mass? Hmmmmm
And I thought my brain was busy!
__________________
__________________
Long range shooting is a process that ends with a result. Once you start to focus on the result (how bad your last shot was, how big the group is going to be, what your buck will score, what your match score is, what place you are in...) then you loose the capacity to focus on the process.
#128
01-11-2009, 10:24 PM
 Platinum Member Join Date: Feb 2005 Location: USA Posts: 2,598
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

Deflections and Drift of a Bullet in a Crosswind
William T. McDonald
On July 8, 2003 we received a question in the Forum section of Sierra’s Exterior Ballistics Website from Mr. David Hollister. This equation is repeated verbatim here:
“Your explanation of the turning of a bullet in a crosswind to produce vertical deflection is the opposite to that experienced by shooters. A wind from the right sends a bullet to 10 o’clock and a wind from the left to 4 o’clock. (A right wind produces a little more vertical deflection than a left wind.) This has always been my experience shooting smallbore (22 rimfire) and 300 m ISSF with both your 155, 168 and 180gr MK and 6mmBR 107MK. Even a 5mph wind produces this phenomenon, but I must say that that as the crosswind increases the vertical component does not continue to increase relative to the horizontal.
What is the cause of this deflection?”
This was an excellent question and a difficult one to answer. This writer replied a few days later in the Forum that he needed more time to study this phenomenon and promised an answer. The writer has performed this study, mainly using Modern Exterior Ballistics by Robert L. McCoy, but with the aid of other references also. Mr. Hollister’s observations are correct. Section 4.3 of the Exterior Ballistics chapter of the Sierra Rifle and Handgun Reloading Manual, Edition V, is not correct. This author will eat some humble pie and correct that section at the next printing of the Sierra Manual. This article has been prepared to answer Mr. Hollister’s question and to serve as a reference on crosswind deflections until the error in the Sierra Manual has been corrected and published. Regrettably perhaps, some mathematics are necessary to answer the question, but the writer will do his best to explain in English what the mathematics are telling us about the motions of the bullet.
In this article we will assume a level fire shooting situation, that is, where the firing point and the target are at the same (or nearly the same) elevation above sea level. This is usually true for target shooting. We also will describe the effects of a crosswind only, because the analysis is simplified a great deal, vertical air currents (vertical winds) are usually very small on level ranges, and headwinds or tailwinds are known to cause much smaller deflections of a bullet on a target than crosswinds. We also will assume that the bullet has a right hand spin, but we will consider the effects of crosswinds in both directions. Right hand spin is produced by rifles with right hand rifling twist, which is the usual case.
When a bullet flies through the air, the velocity of the bullet relative to the ground is affected by the wind. The following vector equation is the beginning point for every analysis of wind deflections:
Vbullet relative to the ground = Vbullet relative to the air + Vair relative to the ground (1)
where Vbullet relative to the ground is the bullet velocity relative to the ground
Vbullet relative to the air is the bullet velocity relative to the air mass through which
it flies
Vair relative to the ground is the velocity of the air mass relative to the ground (i.e.,
the wind velocity)
The boldface letter (V) in each symbol in equation (1) denotes that each quantity is a vector. Velocity is a vector quantity, that is, a quantity which has both a magnitude and a direction. The magnitude of any velocity is speed, but the direction of each velocity must be taken into account when using equation (1). All aerodynamic forces and torques on the bullet as it flies through the air are caused by the velocity of the bullet relative to the air, that is, Vbullet relative to the air.
When a bullet exits the muzzle of a gun, it immediately begins some angular pitching and yawing motions which have several possible causes including the crosswind. These angular motions are small, cyclical, and transient. They typically start out with amplitudes of a degree or so, and damp out, or at least damp to some very small residual values, after the bullet travels a relatively short distance. From our experience measuring ballistic coefficients, these motions damp within 100 yards or less of bullet travel downrange.
Throughout the trajectory, including the initial transient period, the bullet has an “average” angular orientation which aligns the longitudinal axis almost exactly with Vbullet relative to the air. In this orientation the principal aerodynamic force on the bullet is drag, which acts in a direction opposite to Vbullet relative to the air. The side forces on the bullet are essentially nulled in this “average” angular orientation. Only a tiny lift force and a tiny side force are maintained to generate moments of torque which cause the nose of the bullet to turn. We will describe these tiny effects a little later. First, let us consider the “average” angular orientation.
When the bullet exits the muzzle, its nose turns upwind. This is the necessary direction to align the longitudinal axis with Vbullet relative to the air. If the bullet did not turn in this direction, the crosswind would cause a strong side force on the bullet which would ultimately destabilize it. The bullet turns because of its gyroscopic stabilization. When the bullet exits the muzzle, there is an initial misalignment between the bullet axis and Vbullet relative to the air. This misalignment disappears quickly due to the gyroscopic stabilization, and the bullet takes the “average” angular orientation. Looking at Figure 2 below, at the muzzle the “average” angle the bullet must turn into the wind is almost exactly given by:
θmuzzle = arctan [Vcrosswind / Vmuzzle] (2)
where Vcrosswind is the speed of the crosswind (the V symbol without boldface denotes
the magnitude of the V vector velocity)
Vmuzzle is the bullet speed at the muzzle (“muzzle velocity”)
arctan means “the angle whose tangent is”
This angle is small. For example, for a rifle firing the 308 Winchester cartridge with Sierra’s 168 grain MatchKing bullet at 2650 fps muzzle speed in a 25 mph crosswind, θmuzzle = 0.793 degree = 47.6 minutes of angle. For lower wind speeds higher muzzle velocities values of θmuzzle are even smaller.
Figures 1, 2, and 3 have been prepared to support the rest of this discussion. Figure 1 is a side view of the trajectory as would be seen from a position to the right of the trajectory. The X-axis is downrange from the shooter toward the target; the Y-axis is vertically upward at the firing point; and the Z-axis is horizontal and toward the shooter’s right at the firing point. The side view then looks in the direction backward along the Z-axis. The black circle marked “Z” is meant to indicate that the viewer “sees” the Z-axis arrowhead.
The trajectory is then viewed as if it were projected on the X-Y vertical plane. On this plane the elevation angle of the trajectory, called φ, is the angle between the horizontal direction (the X-axis) and the velocity vector projected onto the X-Y plane. The projected velocity vector, Vprojected, is the projection of both Vbullet relative to the air and Vbullet relative to the ground, because the wind velocity Vair relative to the ground is parallel to the Z-axis and points at the viewer.
t every point on the trajectory the trajectory elevation angle φ is the angle between the
l
is
and
XYVmuzzleVprojectedZФat muzzleΦat downrange pointTarget Figure 1 Side View of Trajectory
A
horizontal direction (parallel to the X-axis) and the projected velocity vector. As shown in Figure 1, φ changes as the bullet flies. For the level fire situation, at the muzzle φ is the superelevation of the bore required to put the bullet on the target, and this is a smalpositive elevation angle (positive because the trajectory starts upward). For the 308 Winchester example used above, the superelevation angle for a target at 1000 yards 0.739 degrees = 44.3 minutes of angle. As the bullet flies the trajectory curves downward, and φ decreases, going through zero at the summit of the trajectory,
then going negative and increasing in magnitude as the trajectory steepens. The bulnoses downward, and this is one of the small turning motions caused by a moment of torque on the bullet. let
igure 2 shows a top view of the trajectory as would be seen from a point above the
is
ote that all the velocity vectors in Figure 2 are projections on the horizontal X-Z plane.
nly Vcrosswind is always parallel to the horizontal plane.
a crosswind blowing from the
hooter’s left. The trajectory then curves to the shooter’s right to follow the crosswind.
osswind [1.0 – (Vdownrange / Vmuzzle)] (3)
een defined under equation (2). Now Vdownrange = Vmuzzle at the muzzle, but
ero
nrange.
F
shooter. The top view then looks in the direction downward (backward) along the Y-axis. The black circle marked “Y” is meant to indicate that the viewer “sees” the Y-axarrowhead. The trajectory relative to the ground is then seen as if it were projected on the X-Z horizontal plane.
XYZθprojat muzzleprojpointbulletrelative to air at muzzleVbulletrelative to ground at muzzleVbulletrelative to groundVbulletrelative to airVcrosswindVcrosswindTrajectoryWindFigure 2 Top View of Trajectory, Left to Right Crosswind
V θ at downrange
N
O
Figure 2 is drawn for a bullet with right hand spin, and for
s
As the trajectory curves the bullet gains a component of velocity relative to the ground inthe crossrange direction. It is moving crossrange, so it must have a component of velocity in that direction. The direction of this velocity component is parallel to the Z-axis, and its speed is: Vcrossrange = Vcr
where Vdownrange is the downrange speed of the bullet, and the other speeds have
b
Vdownrange decreases as the bullet flies. So equation (3) tells us that Vcrossrange is zat the muzzle and then grows toward the value Vcrosswind as the bullet flies dowFigure 2 attempts to show what this means. As the bullet flies downrange the velocity
vector Vbullet relative to the air turns gradually to the right, approaching an orientation where it would lie in a vertical plane parallel to the X-Y plane.
In the formal analysis of bullet motion the angle θ is defined mathematically. This angle
of
on
θ = arctan {[(Vcrosswind – Vcrossrange) / Vdownrange] (cos φ)} (4)
here all parameters in this equation have been defined previously. Equation (4) is
s
there were no crosswind, equation (1) tells us that Vbullet relative to the air would be
full
ll
t
the crosswind blows from the shooter’s right to the left, the top view of the trajectory
ir
ns
is important to understand that the trajectories shown in Figures 2 and 3 are
is a measure of how far the velocity vector Vbullet relative to the air is turned away from a vertical plane parallel to the X-Y plane at each point on the trajectory. This angle is complicated to describe, and even more complicated to sketch, without lots of vectoralgebra. However, it is necessary to the analysis, and Figure 2 shows the projection this angle on the horizontal X-Z plane, denoted by the symbol θproj. For a level fire shooting situation, the projected angle θproj is almost equal to the true angle θ at all points between the muzzle and the target. So, Figure 2 is a pretty good representatiof how the true angle changes as the bullet flies from the gun to the target. The true angle is given by the equation:
w
accurate for all values of the trajectory elevation angle φ. For the level fire situation φalways has a very small value everywhere between the muzzle and the target so that (cos φ) has a value very nearly equal to 1.0 in this segment of the trajectory. Under thicondition equation (4) tells us that θ begins at the muzzle with the value given by equation (2), and then grows smaller as the bullet flies downrange.
If
equal to Vbullet relative to the ground. Figure 2 for this special case would show that thetrajectory of the bullet would lie in the vertical X-Y plane. (This is not quite true as we will see later; because the yaw of repose would cause a drift to the right.) Under this condition equation (4) tells us that the angle θ is zero, which means the same thing. However, the trajectory shown in Figure 1 would not change. The trajectory would sticurve downward from the muzzle, and Vbullet relative to the air, which is the same as Vbullerelative to the ground, would turn downward as the bullet flies.
If
would be as shown in Figure 3. This figure is again drawn for a bullet with right hand spin, but the crosswind direction is reversed. The velocity of the bullet relative to the ais upwind as shown, and the bullet initially turns to align itself with Vrelative to the air. As the bullet flies downrange it gains a component of crossrange velocity relative to the ground. The angle θ decreases in magnitude, and the vector Vbullet relative to the air turgradually toward a vertical plane which is parallel to the X-Y plane. The magnitudes of the angles given by equations (2) through (4) are the same. And Figure 1 still applies tothis case; the bullet pitches downward as the trajectory curves downward, and from Figure 3 Vbullet relative to the air turns gradually to the left as the bullet flies.
It
deflections of the bullet as it tries to “catch up” with the crosswind. In order for the
velocity vectors to turn, torques must be applied to the bullet, and the forces which
create those torques cause additional small drifts of the bullet. We will now describethose drifts. At this point w
e have established graphically and verified analytically that the bullet
elocity vectors turn in both the vertical direction (Figure 1) and the horizontal direction
his
o the air.
lt of the bullet nose in any
irection with respect to Vbullet relative to the air. This is known as “aeroballistic yaw.”
se
it flies,
ter 9 of Modern Exterior Ballistics
XYZθprojat muzzleθprojat downrange pointVbulletrelative to air at muzzleVbulletrelative to ground at muzzleVbulletrelative to groundVbulletrelative to airVcrosswindVcrosswindWindFigure 3 Top View of Trajectory, Right to Left Crosswind
Trajectory
v
(Figure 2 or 3) as the bullet flies in a crosswind. The bullet itself stays aligned with tvelocity vector on the “average,” and so it also turns in both of these directions. We know from the laws of physics that when a gyroscopically stabilized bullet turns, torquesmust be applied to cause this to happen. These torques can come only from aerodynamic forces. To generate such aerodynamic forces, the bullet must have small aerodynamic yaw and pitch angles relative to the velocity vector Vbullet relative tThese angles are small deviations from the “average” angular orientation of the bullet, and they are known collectively as the “yaw of repose.” Ballisticians have long referred to “yaw” as an angle of ti
d
“Aeroballistic yaw” is composed of an aerodynamic pitch angle (nose up or down) followed by an aerodynamic yaw angle (nose right or left). The yaw of repose is an aeroballistic yaw, and it can have both an aerodynamic pitch component and an aerodynamic yaw component. The yaw of repose is just the angle of tilt the bullet nomust have relative to Vbullet relative to the air to cause the bullet to remain stable asand turn as it moves along the trajectory. We will follow McCoy’s presentation in Chap . The yaw
f repose of a bullet in a crosswind has two components. The first is an angular tilt of
(5)
o
the nose of the bullet to the right of Vbullet relative to the air (for a bullet with right hand spin). This tilt is the same for both directions of crosswind and has the value: β = αRo (cos θ)
Because the
erodynamic side force to the right is generated. This force causes a moment of torque
then θ would be 0, (cos θ) = 1.0, and β = αRo. So, regardless of
hether there is wind or not, the bullet nose is tilted slightly to the right of Vbullet relative to
f
he
t
ose would be tilted to the left, and the aerodynamic force would be directed to the left.
ynamic and static
haracteristics of the bullet and on environmental parameters:
(6)
t
p is the bullet spin rate
r through which the bullet is flying
of the bullet, where d is the bullet diameter
In equ d easily, except for
. This aerodynamic characteristic of a bullet is measured in firing tests in spark
ts.
ding
e causes of bullet deflections observed in the field. It also varies with the speed of the
nose of the bullet is tilted to the right with respect to Vbullet relative to air, an
a
(another vector) directed vertically downward. This moment causes the bullet to pitch downward as it flies. If there were no wind,
w
the air. This generates the side force on the bullet pointed to the right and the moment otorque directed vertically downward. The rotational equations of motion of a spin stabilized bullet tell us that the angular momentum vector (pointed out the nose of the bullet for right hand spin) turns toward the moment of torque vector. This causes tnose of the bullet to pitch downward as it flies, keeping the longitudinal axis of the bulleessentially tangent to the trajectory. At the same time, the side force causes an acceleration of the bullet to the right, which in turn causes the bullet to drift to the right. This description is for a bullet with right hand spin. If the bullet had left hand spin, the
n
The aerodynamic torque moment would be directed upward. But since the angular momentum vector would be pointed out the tail of the bullet, it would still pitch downward as it flies and remain tangent to the trajectory. The angle term αRo in equation (5) depends on both aerod
c
αRo = (2Ixpg) / (ρSdV3CMα)
where Ix is the polar momen of inertia of the bullet
g is the acceleration due to gravity
ρ is the density of the ai
S = πd2/4 is the cross sectional area
V is the magnitude of Vbullet relative to the air
CM is the overturning moment coefficient for the bullet
α
ation (6) all the parameters are known or can be measure
C
Mαphotography ballistic ranges, usually not available to manufacturers of sporting bulleCMα is measured on military bullets, but is unknown for almost all sporting bullets. The angle αRo is small but unknown for sporting bullets, but it is useful for understan
th
bullet, because the bullet speed appears directly in equation (6) and CMα changes with Mach number.
Now consider thpward. This is
e other component of the yaw of repose, the angular tilt downward or
either a negative or positive angle of attack:
downrange] (cos θ) (sin φ) (cos φ) (7)
downrange] (cos θ) (sin φ) (cos φ) (8)
ullet sp (cos s always positive, and (cos φ) is always positive. However, (sin φ)
llet
e
crosswind d Vcrossrange are both equal to zero. Equation (4) then tells us that the
(9)
ir as th
of attack is
zero, that is, the nose of the bullet is not tilted either upward or downward with respect
tilt
the crossrange drift of the
es of the bullet which are simply unknown for sporting
ullets and unmeasurable in a manufacturer’s laboratory. So, a calculation of this drift
is not possible. A target shooter therefore must determine the drift experimentally for
u
For a crosswind blowing from left to right:
αattack = - αRo [(Vcrosswind - Vcrossrange) / V
For a crosswind blowing from right to left:
αattack = + αRo [(Vcrosswind - Vcrossrange) / V
In equations (7) and (8) the crosswind speed is always greater than the crossrange
b
eed, θ) i
is initially positive on the ascending part of the trajectory (see Figure 1) before the bureaches the summit, but it is negative on the descending part of the trajectory beyond the summit. Now, let us interpret these equations. First, consider that there is no wind. In this cas an
V
angle θ = 0 for no wind. This means that the bullet trajectory lies almost completely in the X-Y plane. There will be a small drift out of this plane due to the yaw of repose. In equation (5) for no wind the term (cos θ) = 1.0, and then:
βno wind = αRo
which means that the nose of the bullet is tilted slightly to the right of Vbullet relative to the e bullet flie
a
s.
Equations (7) and (8) then tell us that, for the case of no wind, the angle
to Vbullet relative to the air. So, the yaw of repose for the no wind case is simply a noseto the right (away from the X-Y plane) as the bullet flies. With a slight nose tilt to the right, the aerodynamic force to the right will accelerate the ullet to the right (out of the X-Y plane). This in results in
b
bullet observed by target shooters. This drift is to the right of the target if the bullet hasright hand spin. If the bullet has left hand spin, the drift is to the left of the target. If thespin rates are the same (barrel twist rates are the same), the drift to the left will be the same as the drift to the right. There is an equation for the force caused by the yaw of repose. However, this equationvolves aerodynamic properti
inb
each cartridge and each bullet, muzzle velocity, and range distance to the target, and apply appropriate windage corrections for different range distances. It is known that tcrossrange drift is relatively small, being a few inches at 1000 yards range distance for most cartridges. Now, consider the case when a crosswind is present. The bullet follows the crosswind, so that the bullet crossrange speed Vcrossrange approaches the crosswind speed Vcrosswind. So, the bullet.deflects in the direct
his
ion of the crosswind, and we know
at this is quite a large deflection. This deflection is calculated by the Sierra Infinity
pose.
e equation
)] and then decreases toward zero as the crossrange speed of the bullet grows toward
nd there will be a
ose tilt of the same magnitude and in the same direction as there is with no wind. This
and spin) superimposed on the deflection caused by the bullet following the wind.
(11)
the muzzle and the summit of the
ero at the summit, and then is negative between the summit and the
th
software. In addition to this deflection there will be drift caused by the yaw of reThis drift will be in both the horizontal and vertical directions. We cannot calculate thisdrift, but the equations will tell us the directions and relative magnitudes. In equation (4) the term (cos φ) is almost exactly equal to 1.0 because for level fire φ is always small over the bullet trajectory between the muzzle and the target. So, equation(4) tells us that the angle θ begins at the muzzle with the value θmuzzle [se
(2
the crosswind speed. So, the (cos θ) terms in equations (5), (7), and (8) can be set to 1.0, as can the term (cos φ). For this condition equation (5) becomes: βcrosswind = αRo (10) Equation (10) is the same as equation (9), telling us that with a crosswi
n
means that for a crosswind in either direction there will be a drift to the right (for right
h
However, when the crosswind is left to right, the deflection is to the right and the drift adds to the deflection. When the crosswind is right to left, the deflection is to the left and the drift subtracts from the deflection. So the total horizontal movement of a bullet on the target should be a little larger for a crosswind blowing from left to right than it would be from a crosswind of the same speed but blowing from right to left. This effect reverses if the bullet has left hand spin. Now consider the angles of attack. Equations (7) and (8) become: For a crosswind blowing from left to right:
αattack = - αRo [(Vcrosswind - Vcrossrange) / Vdownrange] (sin φ)
For a crosswind blowing from right to left:
αattack = + αRo [(Vcrosswind - Vcrossrange) / Vdownrange] (sin φ) (12)
In these equations note that φ is positive between
trajectory, is z
target. Equation (11) tells us that for a crosswind blowing from left to right the angle of
attack is negative (nose down) between the muzzle and the summit of the trajectory,
tor Vbullet
rce directed downward
goes through a zero value at the summit, and then is positive (nose up) between the summit and the target. This is for a bullet with right hand spin; for left hand spin thenose is up between muzzle and summit, and then down between summit and target. If the crosswind is in the opposite direction, equation (12) tells us that these conditions simply reverse. Both equations show us that the angles of attack decrease in magnitude as the crossrange speed of the bullet grows toward the crosswind speed. Equation (11) tells us that for the portion of the trajectory between the muzzle and thesummit of the trajectory the bullet nose turns downward with respect to the vecrelative to the air for a crosswind from left to right. This means that a small aerodynamic
fo is applied to the bullet before it reaches the summit. This
downward force accelerates the bullet downward. This force goes to zero at the summit. After the bullet passes the summit the angle of attack becomes positive. The bullet nose then turns upward and the aerodynamic force becomes directed upward between the summit and the target. Using Infinity to search a number of level fire trajectories for the summit, we have foundthat the summit occurs at about 58 % of the range distance between muzzle
et.
ard before it reaches the summit is larger than the
pward acceleration after it passes the summit. The downward acceleration propagates
ation is downward, reducing the upward drift of the bullet. This
ends the bullet imprint on the target to 10 o’clock. Mr. Hollister’s observation (b)
equations above do
ot directly reveal the cause of this effect. However, it was mentioned above that the
t
left-to-right crosswind of a certain speed produces a horizontal bullet displacement at
and targThe acceleration of the bullet downw
u
into a downward ddrift of the bullet. The upward acceleration after the bullet passes the summit is smaller in magnitude and has less time to act on the bullet before it reaches the target. But, it tends to reduce the downward drift of the bullet. This is the reason for the observations in Mr. Hollister’s question that (a) the total bullet drift in a crosswind blowing from left to right is toward 4 o’clock, and (b) as the crosswind speed increases, the vertical drift of the bullet does not increase as rapidly as thecrossrange drift. If the crosswind blows from right to left, equation (12) tells us that the stronger acceleration of the bullet is upward as it deflects to the left to follow the crosswind, and the weaker acceler
s
above also happens in this case because of the same effect. There remains one more part of Mr. Hollister’s question to answer. This is his observation that a crosswind blowing from right to left produces a little more vertical drift than a crosswind blowing from left to right. The
n
horizontal drift produced by the yaw of repose adds to the horizontal deflection of a bullet following a left-to-right crosswind, and subtracts from the deflection of a bullefollowing a right-to-left crosswind. The vertical drifts, however, have the same magnitudes, provided the crosswind speeds are the same, but different directions. This writer believes that the reason for Mr. Hollister’s observation is the following. A
the target equal to the deflection caused by the crosswind plus the horizontal dto the yaw of repose. Suppose the horizontal deflection is 10 inches and the horizo
rift due
ntal
rift is 4 inches. Further, suppose that the vertical drift caused by the yaw of repose is
-
er,
nt. Other aerodynamic effects (Magnus force and moment,
itch damping force and moment, and others) are smaller and assumed negligible.
.
erodynamic characteristics of sporting bullets are unknown and unmeasurable without
e
e
ion
d
minus 2 inches. So, the bullet hole on the target should appear 14 inches to the rightand 2 inches down, relative to the bullseye. Now, a right-to-left crosswind of the samespeed produces a horizontal bullet displacement at the target equal to the crosswind deflection minus the horizontal drift due to the yaw of repose. The vertical drift caused by the yaw of repose is plus 2 inches for the right-to-left crosswind. Then, the bullet hole on the target for this crosswind would appear 6 inches to the left and 2 inches up,compared to the bullseye. This would make it appear to the shooter as though the rightto-left crosswind produced vertical drifts greater than a left-to-right crosswind. Howevthis would be an illusion. If this is not true, then the reason for this observation by Mr. Hollister could possibly be in the assumptions that McCoy lists leading to the derivation of the equation for the yaw of repose. These assumptions make the yaw of repose dependent only on the lift force and the overturning mome
p
If a bullet has left hand spin, the bullet still deflects to follow the wind. However, the yaw of repose reverses direction, and the aerodynamic forces associated with the yaw of repose reverse direction. The displacement of the bullet on the target then is toward 2 o’clock for a left-to-right crosswind, and toward 8 o’clock for a right-to-left crosswind
The bullet deflections by a crosswind depend on the ballistic coefficients of the bullet, which can be measured in a manufacturer’s shooting laboratory. The deflections therefore can be calculated, and this is done in Infinity. However, the horizontal and vertical drifts due to the yaw of repose cannot be calculated because the necessary
a
very expensive laboratory instrumentation. Luckily, the drifts are small compared to thdeflections. But, the drifts can be observed by skilled target shooters. So, it may bimportant to measure them experimentally on the shooting range and use both elevatand windage corrections when adjusting the sights on the gun for targets at various range distances.
__________________
range it,check the wind, dial in correction, aim and only one shot
#129
01-12-2009, 07:54 PM
 Platinum Member Join Date: Feb 2004 Location: Pennsyltucky Posts: 2,630
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

this is the other reason i lean my crosshairs CCW. i just didn't know it!
#130
01-12-2009, 08:16 PM
 Bronze Member Join Date: Feb 2008 Location: Starkville Mississippi Posts: 40
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

I am going to have to agree squirrelduster on this one. A friend of mine has a 700 Rem. 7mm mag. and it was doing the same thing. I used a set of levels from Wheeler called Level,Level,Level and set the scope and the rifle level to each other and it followed a straight vertical path up and down. We also put a bubble level on the rear of the rail mount so when he got down on the rifle he could be sure it was all level just by looking at his bubble level. It seemed to fix all his problem with that rifle. Hope this helps in some way!!!!!!!!!

Jim Pruitt
#131
01-14-2009, 11:32 PM
 Platinum Member Join Date: Apr 2005 Location: Alaska Posts: 4,618
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

I used levels on the receivers of both of my rifles to ensure the barreled receivers were leveled. I hung a plumb bob about 50 yds away and then aligned the vertical crosshair of my scopes with the plumb bob string. In addition, I had to true up the scope bases on my 7mm Rem Mag because the scope was mounted pointing slightly to the right. Then I purchased and installed bubble levels on both rifles to minimize any canting. Lastly, a 1 mph crosswind can produce as much drift at 1000 yds as spindrift. So you have to shoot in absolutely windless conditions. When I shot there was a light snow falling vertical - straight down. I checked it multiple times at multiple locations. No observable wind whatsoever. After everything was trued up and plumbed level with the world, both of my rifles produced the spin drift I identified in my prior posts. And my total left to right drift experienced at 990 yds was a very close match to the sum of the typical spindrift experienced with VLD match bullets, plus the LoadBase 2.0 calculated coriolis drift. Which is why you can count me in the believer's camp. I had a little more rightward drift than could be explained by spindrift alone. When I calculated the Coriolis drift and added it to spindrift, I had virtually a perfect match with my measured rightward drift..

If you experience no rightward bullet drift at 1000 yds, then you should consider yourself amongst the blessed. Because I believe if you've got everything set up plumb and proper, and you don't cant your rifle, you should be experiencing enough rightward drift at that distance to be able to visually confirm it on the range.

If your not getting any rightward drift out in the field with your rifle, then you certainly shouldn't factor any right-drift dope into your long range shots.
#132
05-16-2010, 12:02 AM
 Gold Member Join Date: Jan 2003 Location: North Louisiana Posts: 747
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

Old posting but just the same....needs to be brought up that this is a common law of physics that occurs!

Go here:

I'm not surprised that this isn't common knowledge by most shooters!
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Monte Walsh
#133
05-16-2010, 01:31 AM
 Gold Member Join Date: Apr 2009 Location: Carrollton, Ohio Posts: 637
Re: Left @ 300 yds - Right @ 1000 yds? WTH?

I first encountered spin drift when I was changing bullets from a 220gr RN to a 180gr SP at 400 yards. The 180 hit 4 inches left of my POI than with the 220gr, I have a right hand twist.

I was confused at first but after some testing and reading I concluded it was spin drift. I figure I get around 3 inches of spin drift at 400 yards with my 30-06 using 190gr SPBTs. I get about a 7" drift using the 220gr RN, it drops about twice as much.

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