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| Sept
1999
1999
by Art Pejsa There
is a great deal of confusion in the minds of many precision
shooters, who are otherwise quite savvy, about the effect of
wind on projectiles. I have found
that this is mainly due to a lot of false information circulated
by self-styled experts who do not have a clear understanding
of the basic physics that is learned in Dynamics 101. I
first began teaching dynamics more than half a century ago
as a young lad of twenty-five, the youngest professor on the
faculty at the US Naval Academy in Annapolis, and have been
doing so in various capacities ever since. First,
results of correct analysis are FACTS and are not matters of
opinion. The basic laws of nature governing
such motion were first put forth more than 300 years ago by Isaac
Newton, the greatest scientist who ever lived. Correct
analysis enables us to predict the free flight of a 6000 mile
ICBM to within a fraction of a mile or the precise position
of Mars a hundred years in the future. Analyzing the
effect of wind on a bullet is utterly trivial by comparison.
To understand clearly what is going on we must return to Newton’s
basic laws of motion. The first law states: “A body
at rest remains at rest or if in motion, retains that motion (velocity)
until acted upon by an outside force.” The second law
says: “The change in motion (acceleration)
of a body is proportional to the outside force acting on it and is
in the direction of that force.” These laws, though seemingly
simple, are very profound; they govern all of ballistics! Cross-Wind Analysis To
keep things simple, let us ignore gravity for now and consider
only the force of air drag acting on the bullet. To keep
the math as simple as possible, assume a bullet is fired horizontally
at 3000 fps at a target 3000 feet away. Without air drag,
flight time would of course be one second (sec). With
air drag, we find that it takes 1.50 sec for the same bullet
if it has a ballistic coefficient of BC = .498. The key
fact is that because of drag, it takes 1.50-1.00 or 1/2 sec
longer. Now, without
any forces, in 1.5 sec it would go 4500 feet, or 1500 feet farther
than with air drag. We
now have a precise measure of the effect of drag; it moves
the bullet backward 1500 feet, relative to where it would be
if there had been NO force acting. This 1500 feet is,
of course, the product of the 1/2 sec time DELAY caused by
drag and the 3000 fps initial velocity. A key point in
thoroughly understanding this is that, according to the above
Laws, even if the bullet had originally been motionless, that
same force over the same period of time would have moved it
backward the same 1500 feet! Let
us now introduce a continuous crosswind of 1 fps. The
relative wind causing the drag force now has a tiny added perpendicular
component. The resulting relative wind is represented
by the diagonal of a rectangle whose sides in this case are
3000 fps and 1 fps, as shown in Figure 1. From
the standpoint of the bullet, the resultant relative wind is
now about 3000.0002 fps, at an angle from head-on of 1/3000 radian
or about 1/52 of one degree! The
magnitude of the relative wind and the resulting force on the
bullet has not changed perceptibly but is now at a tiny angle
of 1/3000 radian. Now, since
by the 2nd Law the motion is in the direction of the applied force,
the above 1500 feet backward displacement is now at an angle
of 1/3000 radian from directly backward, as shown in Figure
2. The
resulting lateral displacement is: 1/3000 X 1500 = 1/2 ft. We
have just calculated wind deflection per fps of cross-wind
for a bullet with muzzle Vm = 3000 fps and BC = .498.
To show that it depends on the time DELAY, and not on muzzle
velocity or distance, I’ve constructed a similar problem. I
find that a bullet with Vm = 2000 fps and BC = .219 takes 1.25 sec
to go 500 yards or 1500 feet/ Now, without drag, it would take
1500/2000 = .75 sec to go 500 yards. Drag,
as above, causes a: 1.25
- 0.75 = 1/2 sec time DELAY. The bullet is in this case, thus
displaced 2000/2 = 1000 feet behind where it would be if there were
no drag.
With
a 1 mph crosswind, our relative wind rectangle is now: 2000
by 1. The resultant
relative wind is about 2000.0002 fps at an angle of 1/2000 radian
from head-on. The resulting 1000 feet displacement
is now at an angle of 1/2000 radian from directly backward so that
the lateral displacement rate, again, is: 1/2000 X 1000
= 1/2 ft/fps.
Thus, although this bullet’s muzzle Vm is 2/3 as great and
it goes half as far, its crosswind deflection rate is the same because
the time DELAY is the same. Hence, in a crosswind Vw fps and
a time DELAY of dt sec, deflection: W
= Vw x dt feet.
We rarely know flight time and don’t need it.
I’ve derived an accurate formula for crosswind deflection
W. For wind velocity Vw
in mph, muzzle velocity Vm in fps, and retardation coefficient Fo
in ft, at range R yards: W = 79.2 x R x Vw inches.
(Fo/R - 1.5) Vm
As
an example, for: Vm = 2800,
Fo = 3000 and Vw = 10 mph, at R = 300 W = 79.2 x 300
x 10 = 10.0 inches. 3000/300
- 1.5 2800
For
a wind at an angle 2 from cross-range, we of course use its cross-range
component, Vwxcos (2). F can be found either from velocity
loss or from the BC. As shown in previous issues of PS,
for velocity loss Vd fps over a distance r feet, at average velocity
Va: F = r x Va/Vd. Given a BC based on the G1 drag curve: F
= 165 VxBC (approx.). The
Effect of a Wind Gust at Various Ranges To
illustrate this case, I like to use my bowling ball example. Let
a ball roll down the center of an alley 60 feet long and 3
feet wide at 20 fps so that it takes 3 sec to go down the alley. During
its first inch of travel, let it be hit with a crosswind gust
enough to import 1/2 fps of cross-alley velocity. Since
its forward velocity is still 20 fps it takes 3 sec to go down
the alley, during which time it goes 3 x 1/2 = 1 1/2 feet laterally
and into the gutter!
Again, by Newton’s 1st law, it retains that velocity during
the rest of the time down the alley. A wind gust during the
last inch would have virtually NO effect because there is no time
left over which to accumulate a displacement. The
key point here is that a FORCE does NOT impart a displacement
or even a velocity, it imparts an acceleration which in a
period of time accumulates a change in velocity and which
in that same time accumulates a displacement change. Because
of this fact, the effect of a wind gust on a bullet depends
on the time to go to the target. Hence a gust near the
shooter has far more effect than one near the target. To
illustrate this, I’ve calculated the effects of a 10mph
crosswind during the various 100 yard portions of a typical
400 yard flight, for Vm = 3000 fps and BC = .425. Note
for example, Figure 3, with no wind in the last 100 yards total
deflection is reduced only 1 inch, (from 13 to 12 inches)! Headwind
and Tail-Wind Effects Headwinds
or tailwinds are very easy to analyze correctly.
From the bullet’s standpoint, in a 10 fps headwind, drag
is the same as if its velocity is increased 10 fps, from, say, 3000
fps to 3010 fps, or 1/3 of 1%. We can show that for each 1%
change in V, drag typically changes 1.5%. Hence the 10 fps
headwind causes an increase in drag of 1.5 x 1/3 = 1/2%. This
can be simulated by a 1/2% DECREASE in BC (or F), from, say, .400
to .398. With Vm = 3000 fps, this causes an
increase in bullet drop at 500 yards of 0.12 inch. The deflection due to a 10 fps crosswind is found to be 15.72 inches, 130 times that of an equal headwind or tailwind! The effects of head or tailwinds are utterly trivial by comparison; so ignore them! In fact, the vertical change in impact point from a crosswind due to “gyroscopic coupling” is far greater than that due to a head or tailwind!
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