The Four (Actually Five) Questions Every Long Range Rifle Instructor Gets Asked Eventually

By John Simpson on June 6, 2016

Snipers are trained to see patterns and since I became a sniper instructor in 1985, I’ve noticed that when it comes to the long-range rifle marksmanship part of the instruction that there are 5 questions I keep getting from students over the years. And this includes both military and police students.

They are:

  1. Does the near or far wind have a greater effect on my bullet?
  2. What effect does the rotation of the Earth have on my bullet?
  3. What effect does spindrift have on my bullet?
  4. What effect does hitting a raindrop have on my shot?
  5. CENSORED

The one for #5 is just so stupid I’d rather not bring it up myself if you don’t mind.

I know that every instructor (no matter what the subject) has it beaten into them, “That The Only Stupid Question Is The One Not Asked.” My response to that cliché is that anyone who believes there’s no such thing as a stupid question hasn’t spent that much time teaching because I know I’ve heard some and I’m sure you have too.

Anyway, we’ll concentrate on the first four because that will provide us with some useful information.

As before, I like to start out by defining some of the terms that we’ll be using.

Accuracy: Is a measure of the distance between our Point of Aim (POA) and the Point of Impact (POI) of our shot. When we fire a shot group, we can mathematically determine the “center of gravity” of all shots fired and find the location of the Mean Point of Impact (MPI). So each shot is a Point of Impact, and multiple shots fired under similar circumstances have a Mean Point of Impact. When we adjust our sights we’re moving the Mean Point of Impact based on the shot group we just fired to apply to a shot group we haven’t fired yet. Heavy, huh?

You’ll find that most of today’s subjects affect where a shot lands or Accuracy.

Minute of Angle: This is an angular unit of measurement which works out to 1/60th of a degree or approximately 1” for every 100 yards. DON’T get into the lazy habit of just saying it’s “an inch” and leaving it at that. The reason for this is that whole angular unit of measure thing. Permit me an example first of what isn’t an angular unit of measurement.

Imagine a yardstick being coincidentally 36 inches long as you hold it in your hands. You hand it off to an able assistant and have him walk 100 yards away from you. He’ll tell you that 100 yards away from you the yardstick is still 36 inches long.

Now I can feel your scorn but now I can say that the minute of angle doesn’t share this same consistency with linear units of measurement like yards, meters, miles and such.

A minute of angle (or MOA) 100 yards away from you is one inch across. That same minute of angle however, is a half an inch across at 50 yards and 2 inches across at 200 yards.

One of Simpson’s Laws is If I Can’t Show You The Math It’s Just My Opinion so I’ll show you how to come up with the size of a minute of angle for any distance mathematically.

1. There are 360 degrees in a circle and each of those degrees is divided up into 60 minutes (and yes each minute is divided by 60 seconds but we’ll stop there) meaning that any circle contains 21,600 MOA.

2. Now you should use a calculator with a π (pi) key for the next part. If we want to do a MOA at 100 yards we enter 100.

3. Since we want our answer in inches we multiply 100 yards by 36 to get 3600 inches.

4. Since we want the diameter of our circle we double our last result and get 7200 inches.

5. 7200 inches times pi or 3.14159 changes the diameter of 7200 inches to a circumference of 22,619.147 inches.

6. We then divide the circumference of 22,619.147 inches by the 21,600 MOA in a circle and get at 100 yards a MOA equals 1.04719422119 inches we can mercifully round off to one inch.

By the way, I don’t have any patience for various instructors that insist their guys use a MOA out to like 3 or 4 decimal places instead of rounding it off. A rounded off MOA at 1000 yards comes to 10 inches. Going to two decimal places makes it 10.47 inches. I’ll state for the record I’ve never shot well enough to see the difference and although we probably haven’t met I’m going to go out on a limb here and say that you don’t either.

In my own experiments about when to round off I’ve come up with the Rule of 22. If you’re dealing with a distance of 2,200 yards or adjusting 22 MOA up or down use 1.047” instead of rounding off.

For those of you using the metric system we can substitute the number of centimeters in a meter, do the math and see that at 100 meters a MOA equals approximately 2.908 centimeters, which we can also mercifully round off to 3 centimeters. This is why riflescopes for the European market are calibrated in 1/3 MOA clicks so that at 100 meters a click equals one centimeter. But I digress.

The MOA is important today due to the fact that when I start addressing the questions we’ll see that there’s an effect but in most cases it’s beyond our efforts to correct for it.

More often than not if you read about correcting for long-range wind in a long range shooting book the author shows the wind as being the same direction and velocity all the way from the muzzle to the target.

In real life though I’ve shot on rifle ranges where wind would enter into various places through gaps in the terrain or the surrounding woods.

The classic is a wind blowing one way at the shooter’s position and an opposite wind blowing at the targets. What do you do?

Some folks say that the wind closest to the target because the farther downrange the bullet flies the slower it goes and thus is affected by the wind to a greater degree.

Fewer folks say the wind at the shooter’s position because that’s where an initial change in the bullet’s direction is amplified as it travels downrange.

A lot of folks say that you should use the mid-range wind or “average together” the near and far winds.

Rather than hold you in suspense much longer the wind closest to the muzzle has a greater influence than the wind closes to the target.

Based on the pushback I’ve received since I first wrote about this in the 1990’s I hasten to add that I didn’t “discover” this by any stretch of the imagination but at one time used to be common knowledge among shooters.

In fact, I learned about this from an excellent book titled Canadian Bisley Shooting: An Art and Science by Desmond Burke. My only innovation was figuring out how to calculate the influence of the wind in 100-yard/meter segments as well as for the bullets we use today.

Keeping in mind Simpson’s Law mentioned previously I’ll show you how to figure it out for yourself, or you can just take my word for it and skip ahead.

There’s a real wind formula out there that’s in the 1929 British Textbook of Small Arms and it’s NOT that R times V nonsense everyone loves republishing.

This one is based off of the “lag time” or difference between the time of flight of a bullet in air and through a vacuum.
D=(x-x’)W

Where D is the deflection in feet, x is the Time of Flight in air, x’ is the Vacuum Time of Flight and W is the wind speed in Feet Per Second

So x we can get by using ballistic software:

Distance in Yards      TOF in Seconds
0-500                        .706075
0-1000                      1.781697

We have to calculate x’ by using the muzzle velocity or the velocity along the trajectory that we want to examine.

Range in Yards        Velocity in FPS
0                               2600
500                           1742

So to find the vacuum time of flight for the first 500 yards we multiply by 3 to convert 500 yards into 1500 feet. If we fired in a vacuum with no air to slow the bullet down it would cover that 1500 feet at 2600 feet per second or .576923077 seconds.

So TOF minus the VTOF is .706075 minus .576923077 or .129151923 seconds for 0-500 yards.

If we do that for the second 500 yards we subtract our 1000-yard TOF from our 0-500 yard TOF to get a 500-1000 yard TOF of 1.075622 seconds.

The VTOF or x’ for that same distance is 1500 divided by 1742 or .8610792 seconds.

So, keeping all this handy we start by calculating a uniform 10 mph wind from 0-1000 yards.

We want W in feet per second so we multiply 10 MPH by 1.467 to convert it to 14.67 FPS.

For 0-1000 yards the TOF is 1.781697 seconds and now we need to 1000 yards VTOF. 1000 yards is 3000 feet and the muzzle velocity is 2600 FPS. So the Vacuum Time of Flight is 1.153846 seconds.

We now have the following numbers to plug in to our Wind Formula.

x=1.781697 seconds
x’= 1.153846 seconds
x-x’= 0.627851 seconds

W = 14.67 Feet Per Second

D= 0.627851 times 14.67

D=9.21057417 feet or approximately 110.5 inches

This is shown graphically in Figure A as looking at a 1000 yard from directly overhead.

Blowing in the Wind Graphics FIG A

So now, let’s imagine building a 500 yard long wall from the 500 yard line to the target 1000 yards distant like this:

Blowing in the Wind Graphics Fig B
We see that the bullet is deflected by the same 10 mph wind until it crosses the wall and just continues on the path started under the first 500 yards of influence showing 72.76 inches of deflection caused by the first 500 yards of wind.

But now we laboriously tear down the wall and rebuild it to block the wind for the first 500 yards and see the result:

Blowing in the Wind Graphics FIG C
So we see that even though the bullet is indeed flying slower, the influence of the wind has less time to act so the deflection by a 10 mph wind working from 500-1000 yards is only 37.7 inches.
The first thing we see wrong with the “averaging” school is that if we were to see a 10 mph wind blowing from the left at 10 mph for the first 500 yards and a 10 mph wind from the right for the last 500 yards, we would average the two winds together and cleverly come up with zero wind effect which would cause us to miss.

Rather, we can see from the graphics that the wind deflection caused by the first 500 yards of 72.76 inches right would only countered by the wind deflection of the second 500 yards of 37.7 inches for a total wind deflection of 35.06 inches right!

When I mentioned pushback on this earlier I wasn’t kidding! I’ve gotten people basically asking if I built a 500 yard wall on a 1000 yard range all the way to one pinhead who tried impressing me that he was a retired Gunners Mate in the Navy and thought that in Figure B above when the bullet passed by the wall and out of the influence of the wind it would resume flying straight again. DON’T BE THAT GUY!

As a matter of interest, since I first published this I’ve since confirmed this with live fire as documented in my second book, Snipercraft: Laying the Groundwork for a Career as a Sniper.

And finally as covered in my first book, The Snipers Notebook, if we were to have a situation on a 1000 yard range with a 10 mph wind blowing one direction from 0-100 yards, that would deflect our bullet about 15 inches. In comparison it would take a wind at the 900-1000 yard blowing in the opposite direction at around 80 mph to push the bullet the same 15 inches.

This is one of those ones that turns up when people want to impress you with how much thought they put into a shot.

We’re going to be discussing the Coriolis effect and how the rotation of the earth influences the strike of our bullet.

First be aware that there are two different effects that the Earth’s rotation has on our bullet with the Coriolis Effect causing deflection and the Eötvös Effect causing elevation changes.

Second, this is all because the Earth is a sphere rotating at over 1000 mph at the Equator.

Third, the Coriolis Effect is least visible at the Equator and most visible closer to the North or South poles.

Fourth, no matter which way you shoot the Coriolis Effect causes your shot to deflect to the right in the Northern Hemisphere and to the left in the Southern Hemisphere.

Fifth, the Eötvös Effect causes a bullet fired due East to land higher and a bullet fired due West to land lower. The extent depends on what azimuth your shooting along with shots due North and due South having no vertical effect.

We’ll have to define some additional terms here:

Azimuth: This is the direction that we’re pointing our rifle in relation to North when we fire with 0 (or 360) degrees pointing North, 90 degrees pointing East, 270 degrees pointing West and 180 degrees pointing South like so:

azimuth 5921_47954_i1
Latitude: These are the imaginary lines that go around the Earth from East to West and allow us to locate how high above or below the Equator some place is.

This is shown graphically below with an illustration from the 1941 Army Advanced Map Reading manual (because I’m all about the Old School).

1941 lat long

 

So for example here in Kennesaw I would use 34 degrees North latitude for my calculations.

Coriolis Effect

To demonstrate this graphically think in terms of what your point of view is.

Let’s say that you’re standing on the ground and hovering above you is a helicopter. Although that helicopter is staying in the same spot as far as you are concerned, it’s because you both are spinning on the Earth at the same speed.

To someone orbiting the Earth they see you and the helicopter spinning around and back as fast as the Earth is spinning.

So let’s set up that you’re on the North Pole and shooting at a target some distance away.

Graphically that looks like this:

coriolis 001

The arrow of course shows the direction of spin. We take aim (accounting for the wind naturally) and fire.

coriolis 003
From the point of view of someone in space looking down the path of the bullet appears straight while you and the target continue to spin with the Earth. Essentially the bullet is traveling straight while the Earth rotates underneath it. The bullet is landing where the target was not where it wound up.

coriolis 004
Down on the ground however with you and the target the bullet follows a curved path to the right as far as the shooter is concerned. If we set this up with an opposite spin for the South Pole and the Southern Hemisphere we would see a curve to the left from the shooter’s point of view.

The Horizontal Deflection is calculated using this formula

0.00007292 times X times SIN (Lat) times TOF

With

0.00007292 is a constant representing the spin of the Earth
X = range to the target in feet
SIN (Lat) being the sine of the Latitude you’re located with a + for North of the Equator and – for South of the equator. Hopefully your calculator has a SIN key.
TOF= time of flight of our shot obtained from a ballistics program.

And the answer will be in feet.

So from our previous chapter let’s look at a 1000-yard TOF of 1.781697 seconds for 34 degrees of Latitude:

0.00007292 times 3000 feet times the sine of 34 or .5591929 equals .218 feet or 2.6 inches.

Having mastered this and MOA we see that at 1000 yards this 2.6 inches comes to 0.26 or about a quarter a minute of angle.

Hardly seems worth it, you think?

Eötvös Effect

This one is a bit more complicated and you need to know a lot more before you can calculate it so I’m only going to show it for one example.

But first, this influence is seen to the greatest effect shooting at the Equator either due East or due West.  Shots fired in an easterly direction will land higher than normal while shots to westerly directions will be lower.

Technically if you want to zero your rifle for elevation and eliminate the rotation of the Earth you would have to point your muzzle either straight North or straight South.

The formula is:

2016-06-09_12-56-59

Where
MV = muzzle velocity in feet per second
COS (Lat) is the cosine of the Latitude
SIN (Az) is the sine of the direction of fire expressed as an azimuth
32.2 is the gravitational constant

The Correction factor is then multiplied by the bullet’s drop at the desired range to see what the difference in drop is.

The bullet I’ll use is a 168-grain Hornady TAP round with a 1000-yard drop of 466.75 inches and our Latitude is still 34 degrees but our Azimuth is due East or 90 degrees.

2 times the constant times the MV is 0.379184 which divided by 32.2 is 0.0117759006.

When that gets multiplied by the cosine of 34 and the sine of 90 we get 0.00976271329, which we subtract from 1 to get a correction factor of .99024.

We multiply our 1000-yard drop of 466.75 inches by this factor and get a corrected bullet drop for the rotation of the Earth of about for a whopping four and a half inch difference which comes to less than half a minute of angle at 1000 yards!

So the next time you hear someone in a movie say that they have to take account of the rotation of the Earth and they don’t whip out a compass, just chalk it up to Ignorance in Action (aka Stupidity).

And by the way, if you’re going to all this effort you need to use True North instead of Magnetic North that can be way apart depending on where you are.

In real life, when we fire sighting shots at a known distance range we’re actually compensating for this and the next one we’ll cover.

So to answer the question, the rotation of the Earth doesn’t have that much practical effect.

This one was really a tough one to quantify with shooters usually having to rely on published data until some of the best work on nailing this one was published by ballistician Bryan Litz in Applied Ballistics for Long-Range Shooting.

Spin drift is caused because rifle bullets by definition of being fired in rifles barrels are spinning as they travel downrange.

If you’ve seen a toy gyroscope while it’s spinning, if you apply force in one direction to tip it over it goes off in a direction to the side.

Simply put, because a rifle bullet is following a curved path to the target the nose points to the right (for right twist barrels) at an angle from the direction of travel.

spin drift

Under 1000 yards spindrift can be ignored. In this Army ballistics tables for the M16A1 rifle firing M193 Ball ammo you can see in Column 5 labeled DRIFT that the spin drift is measured in another angular unit of measurement called Mils. I won’t go into Mils here but one mil equals 3.375 MOA so at 1500 meters where we see 1.8 Mils of Drift that’s over six MOA of spin drift to correct for.

Pages from firingTables-5.56mm

For ranges beyond 1000 yards I can’t help but recommend Bryan’s book as being one of the best I’ve read on the subject of rifle ballistics as well as spin drift.

The short answer is basically nothing. However, a friend of mine in the business just couldn’t shake this student who kept pestering him to prove that it doesn’t have an effect.

Ignoring of course that the burden of proof is on the person wanting to show that there is an effect I’ll press on.

This is a pretty old one and extends back just past WW2. Antiaircraft Artillery was becoming a hot career field in the Army due to the growing threat of Soviet bombers. Because more and more aircraft had thin aluminum skins, projectile fuzes were made more sensitive so they would detonate on first impact and not after exiting the other side of the aircraft.

During live fire tests of some 20mm shells in the rain, most would perform satisfactorily while some would travel part way to the target before going off. It was determined that impact with raindrops was making the shells go “BOOM”.

In the July 1978 American Rifleman magazine this very same question was asked and the late W.C. Davis, a ballistician I still hold in the highest regard answered the letter. His answer is still relevant today:

“It would be very difficult to test the effect of rain on the flight of a rifle bullet because the statistical probability that a bullet will hit one or more raindrops is quite small during ordinary rainfall, and you could not be sure which bullets you observed actually hit raindrops and which had not.

I participated in tests some years ago to determine the effect of raindrops on 20 mm projectiles fired from aircraft cannon. The point-detonating fuzes of the high explosive ammunition were very sensitive, as was required to assure that they would function on the thin aluminum skin of hostile aircraft.

Some of the projectiles occasionally burst prematurely in mid-air when fired during heavy rainfall. A test was devised using spray nozzles on fire hoses which were supplied with water by several truck pumpers and adjusted so that water fell along the line of fire of the aircraft gun on its test stand. It was established that raindrops could cause functioning of the sensitive fuzes, which were then modified and retested until the premature functioning had been eliminated.

In the course of this investigation, some calculations were made to estimate the statistical probability that any particular projectile would encounter a raindrop in a flight of 2000 yds or so. The probability was relatively small under any usual conditions of rainfall. and the calculation was thus consistent with the fact that only a small percentage of projectiles fired in actual or simulated rainfall were observed to burst prematurely.
I’m not able to answer your question about the consequences of a bullet striking a raindrop, but the above establishes that the probability of such an event occurring is small.

However, some of the best smallbore rifle scores have been shot in the rain, so the overall effect of rain on accuracy is probably not very great. During a light rain, wind is often diminished and the effect of wind on accuracy is probably far more significant”

This led to the development of what are called Rain Insensitive Fuzes and a lot of time and money went into developing methodologies to test these fuzes under realistic conditions.

Now, I’m going to address the two subtopics of this question, specifically:

1. What are the chances of a bullet hitting a raindrop?

2. What happens when a bullet hits a raindrop?

What Are The Chances?

The first thing we have to do is really look at rainfall. The Army has already done this for us and created a table comparing different rainfall statistics for different parts of the world. I’ll just show two of them for comparison:

2016-06-06_15-25-06

In this case, just remember that Median is the number marking the point where 50% of all the raindrops are smaller than that number and 50% are larger.

Now don’t be too impressed with the last column above because that includes even the very small raindrops down to 0.5 millimeters or about 0.2 inches in diameter. Now the last thing we need to do is to look at how many raindrops of each diameter we can get for a rainfall of so many inches per minute.

2016-06-06_15-25-19

The last measurement we need is the diameter of the bullet tip (called the meplat). Rather than try to figure out where a ballistic tip begins and ends we’ll just use the flat tip of a military sniper round M118 Special Ball, which comes to .060 inches or .1524 centimeters.

The distance in meters ON AVERAGE before the bullet impacts with ONE raindrop is:

2016-06-06_15-26-47

I chose to cut it off at the 3mm diameter because the numbers got really ridiculous.

So we can see that our rifle bullet will travel an average distance of about 370.41 meters or 405 yards before hitting a half-millimeter diameter raindrop dead on. Our bullet would have to travel about 17 miles on average before hitting a 3-millimeter diameter raindrop under standard conditions.

Even when shooting in a Miami Continuous Rain, a bullet will likely travel 274 meters or about 300 yards before hitting a two millimeter raindrop. Keep in mind, however that visibility across distance decreases in a heavy rain so worry more about seeing that far in a rainstorm before worrying about shooting that far in a rainstorm.

Effects of Hitting a Raindrop

Okay, so hopefully we can see that the chances of your bullet directly hitting a raindrop are slim indeed. But let’s just say that our bullet impacts a raindrop dead center, now what?

Again the short answer is not much as evidenced by all the folks still hitting the target during the deluge that was Sniper Week West 2011.

The reason for this is that aside from the fact a very heavy (in comparison to a rain drop) bullet is traveling supersonically (at least until 600 yards), the raindrop doesn’t actually touch the bullet.

To see what I mean, here is a shadowgraph of a bullet traveling at just under Mach 2 or twice the speed of sound (you can figure that at 68 degrees Fahrenheit Mach 1 is 1126 feet per second so a typical .308 Winchester bullet leaves the muzzle at Mach 2.30).

shadowgraph_mach 1_99

The shock wave is trailing back from a bow wave just in front of the meplat. This bow wave consists of air piling up because it can’t get out of the bullets way fast enough. In other words, the raindrop is hit by the bow wave not the bullet itself. As the bullet slows down, the bow wave actually rides further in front of the bullet as we can see below on a bullet slowed to Mach 1.24:

shadowgraph_mach 1_24

Keep in mind that a 168 grain Hornady TAP round is still traveling at Mach 1.43 about 600 yards from the muzzle!

Conclusion

Technically speaking, there is an effect on the bullet’s energy. If physics teaches us anything it’s that there’s no free lunch. As a practical matter however, (and considering your chances of hitting a raindrop) worry more about where you place your trigger finger than whether you’ll hit a raindrop on the way to the target.
IN SUMMARY

As I keep telling people ballistics is a really useful science , unfortunately equations that were developed by launching artillery shells for several miles or launching ballistic missiles across the world can also be used for small arms by plugging in the appropriate numbers.

This causes people to worry about stuff that has little or no practical effect on their shooting (and it sounds good in the movies).

Hopefully you’ve already taken one of the best steps to improve your shooting performance and that’s buying and installing an MDT chassis on your rifle!

FURTHER READING

Snipercraft: Laying the Groundwork for a Career as a Sniper by John C. Simpson
Although written as an aid to student snipers this is more about position rifle shooting. More to the point, it contains the results of the live fire wind experiment that proved what I said. Available from Paladin Press
http://www.paladin-press.com/product/Snipercraft/Sniping

Sniper’s Notebook by John C. Simpson
My first book on sniping, it contains the detailed procedure for calculating the effects of Non-Uniform Winds. It’s only available from me for either Law Enforcement or Military snipers.

Modern Exterior Ballistics: The Launch and Flight Dynamics of Symmetric Projectiles; Robert McCoy

Understanding Firearms Ballistics; Robert Rinker. One of the few ballistics books that talks about shooting in the rain.

Applied Ballistics for Long-Range Shooting by Bryan Litz.
As I said previously, one of the best modern books on rifle ballistics available.
http://appliedballisticsllc.com