Is it possible for a rifle to be more accurate at distance than 100 yards?

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Bart B. said:
Then please explain why round cannon balls drift with wind even though they may be spinning on some axis other than that of their trajectory. Or golf, base, ping pong, basket, bearing or croquet balls, for that matter, when propelled into moving air. They all get sucked into the lower air pressure side. Same as wings on airplanes.

If the bullet's spin is so critical to wind drift, why is rifling twist not an input to all ballistic software calulating wind drift with the bullet's rpm rate? It slows down much less than velocity does.

Anyone ever calculate the speed bullets move at right angles to the line of sight and compare a given bullet at different velocities it has at different down range points?
Click to expand...

Bart, you can "prove" just about anything conceptually. As you point out at the end of your post, it is important to do the math.

Assume a cross wind of 10 MPH, and a 55 grain .224" bullet launched at 3000 FPS, with a typical BC for that type of bullet.

Time of flight to 500 yards is .611 seconds, and horizontal deflection is close to 19.5" according to my ballistics software.

The wind cannot horizontally accelerate a bullet to a speed faster than the air flow in the wind. So the maximum possible horizontal speed imparted to the bullet by the wind blowing against it is 10 MPH. In reality, the horizontal speed imparted will be significantly less than this.

10 MPH = 14.66 FPS = 1.22"/S.

1.22"/S for .611 seconds (the time of flight) yields an upper limit on horizontal deflection of .746". Yet the real answer is 19.5". So the effect of wind blowing against the side of the bullet cannot account for the deflection we actually see.

It is as I have said: The bullet slightly noses into the wind so that the drag vector points from tip to center of base. This gives the drag vector a horizontal component, and this horizontal component of the drag vector is much more influential than the effect of the wind blowing against the side of the bullet.

Once that is understood, it is possible to have a meaningful discussion about the bullet very slightly corkscrewing its way to the target, which it does for the first part of its flight.

If you are somehow confounding spin with the horizontal component of the drag vector, then you are getting that from someone else's posts.
 
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The speed bullets deflect in crosswinds can be calculated with the time of flight across a range band and how far the bullet drifts in that range band. For example, the 20 yard band centered at 100, 300, 600 and 1000 yards. A rate - time - distance problem.
 
Bart B. said:
The speed bullets deflect in crosswinds can be calculated with the time of flight across a range band and how far the bullet drifts in that range band. For example, the 20 yard band centered at 100, 300, 600 and 1000 yards. A rate - time - distance problem.
Click to expand...

A much better analysis can be done. The force of the wind against the side of the bullet can be calculated, and a spot of integral calculus will give you horizontal deflection and speed for all distances in the trajectory.

The upper limit calculation I have shown is much simpler, and very adequately answers the question at hand: Can the force of wind against the side of the bullet account for the horizontal drift we observe? The answer is, it cannot.
 
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Wind drift on a projectile is the effect of a projectile blown off course and calculable within a good degree of accuracy for POI using External ballistics calculations. A product is something produced by an action or set of actions, or a quantity obtained by multiplying quantities together, or from an analogous algebraic operation. http://www.dictionary.com/browse/product Both apply!

As far as barrel flex affecting accuracy at certain distances it can and it has if you have been following along. A slower bullet leaving at a higher elevation due to flex will cross at some point in flight with a faster projectile that leaves a barrel on a flatter plane during descent. But as it has already been pointed out it the group opens back up after the cross point as the slower bullet drops. Its all in velocity at the arc or angle the projectile leaves as to when this occurs. So accuracy to only one point at distance and much worse at further distances is not real accuracy because accuracy should follow along the path of flight in equal MOA to the distance capability of the caliber.

However and once again the bullet itself does not become more accurate at distance. The effect of the barrel flex can make a more accurate group at a certain distance based on the aforementioned variables in angle of bullet and velocity but the bullet itself cannot produce an improved group consistently in same MOA past 100 yards out to distance in different intervals with the same constants in load rifle.
 
The calculation of wind drift must account for when or where the the wind is having the effect during flight for an accurate result. The slower the projectile the greater the effect and the faster the projectile the lesser the effect. So flight path is affected more on the end of bullet flight than in the beginning. To get really good at using the drift factor is to get good at reading wind speed and distance to the effect.

One way is mirage. By sighting in at different intervals and checking the mirage angle helps to guess at what speed and what direction the wind is blowing and by checking different distances you have a better idea how to compensate for effect.

If a bullet leaves a barrel on a 1000 yard flight and encounters a 10 mph wind from left to right for the first 500 yards and a 10 mph crosswind from right to left during the second half of flight the bullet will impact left to the POA.

Its easier to imagine that in time of flight and broken down at distance. For the first 500 yards the time of flight or TOF is shorter so the wind will be blowing on the bullet for less time than in the second half of flight because the bullet will be moving slower.

In many instances wind will have different speeds at different angles throughout flight. Even more often with hills and valleys.

 
If you still think it is the effect of the wind blowing against the side of the bullet, consider this: In a 10 MPH cross wind, using 150 grain 30 cal bullets launched at 2700 FPS, at 500 yards the Hornady FMJ BT will deflect 12.5". The Hornady RN will deflect 32". Same weight bullet, same general silhouette, same initial speed, but very different deflections. Why?

It's because the RN has a much greater drag vector, so when the bullet noses into the wind, the horizontal component of drag is also much greater.

A model based on the wind blowing against the side of the bullet cannot explain the real-world results that we see.

You have to understand the drag vector thing before you can understand why the bullet follows a slightly corkscrew path during the first part of its trajectory.
 
denton said:
If you still think it is the effect of the wind blowing against the side of the bullet, consider this: In a 10 MPH cross wind, using 150 grain 30 cal bullets launched at 2700 FPS, at 500 yards the Hornady FMJ BT will deflect 12.5". The Hornady RN will deflect 32". Same weight bullet, same general silhouette, same initial speed, but very different deflections. Why?

It's because the RN has a much greater drag vector, so when the bullet noses into the wind, the horizontal component of drag is also much greater.

A model based on the wind blowing against the side of the bullet cannot explain the real-world results that we see.

You have to understand the drag vector thing before you can understand why the bullet follows a slightly corkscrew path during the first part of its trajectory.
Click to expand...
My calculations get a 25.8" drift with a 150gr FMJ @2700fps with a 10mph crosswind not 12.5" and 28.1" with a 150grrn @2700fps with a 10mph crosswind.



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Our ballistics calculator includes both basic and advanced features, allowing you to customize the shooting conditions to replicate your environment. Choose standard or metric, and enter the appropriate data to view your favorite load.

Standard values are provided in the sight height based on the firearm type, but may be changed to match your setup. Default values are entered in the temperature and pressure spaces but can be changed. Once calculated, we offer a printable "cheat sheet" to tape to your firearm for quick reference.











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Description:
Ballistic Coefficient (0.01 - 1.2 )
Velocity (100 - 4600 ft/s)
Weight (5 - 1000 grains)
Maximum Range (100 - 2000 yds)
Interval ( yds )
Zero Range (0 - 2000 yds)
Sight Height (0.1 - 4 inches)
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Shooting Angle (-90 - 90 degrees)
Wind Speed (0 - 70 mph)
Wind Angle (0 - 359 degrees)
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Your Input Variables
Ballistic Coefficient 0.186 Velocity (ft/s) 2700 Weight (grains) 150
Maximum Range (yds) 500 Interval (yds) 100 Drag Function G7
Sight Height (inches) 1.5 Shooting Angle (degrees) 0 Zero Range (yds) 100
Wind Speed (mph) 10 Wind Angle (degrees) 90 Altitude (ft) 0
Pressure (hg) 29.53 Temperature (F) 59 Humidity (%) 0.78
Ballistics Results
Range (yards) Velocity (fps) Energy (ft.-lb.) Trajectory (in) Come UP in MOA Come UP in Mils Wind Drift (in) Wind Drift in MOA Wind Drift in Mils
Muzzle 2700 2428 -1.5 0 0 0 0 0
100 2463 2021 0 0 0 1 0.9 0.3
200 2239 1669 -4.3 2 0.6 3.9 1.9 0.5
300 2025 1366 -15.5 4.9 1.4 9.1 2.9 0.8
400 1823 1107 -35.2 8.4 2.4 17.1 4.1 1.2
500 1632 887 -65.5 12.5 3.6 28.1 5.4 1.6

The following table is provided as a "cheat sheet" that you can tape to your gun.
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Your Round 150 gr. 0.186 B.C.
Range Muzzle 100 200 300 400 500
Trajectory -1.5 0 -4.3 -15.5 -35.2 -65.5
Come UP in MOA 0 0 2 4.9 8.4 12.5
Come UP in Mils 0 0 0.6 1.4 2.4 3.6
Wind Drift 0 1 3.9 9.1 17.1 28.1
Wind Drift in MOA 0 0.9 1.9 2.9 4.1 5.4
Wind Drift in Mils 0 0.3 0.5 0.8 1.2 1.6
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That's an interesting discrepancy.

I use the RSI Shooting Lab, and normally just take the default G1 BC. Reading through your results, I see that you're using the G7 coefficient, so I switched to that to see what would happen.

With the G7 model, using .186 for the BC for the RN bullet, and .398 for the BC for the FMJ BT, I get 30.65" and 9.65" deflection respectively, rather than 32" and 12.5".

Using Strelok, an Android app, I get 74.25" and 25.5" respectively.

Good grief!
 
I used Sierra's G1 data at their default atmosphere conditions:

Drift Bullet Data.jpg

Here's the standard Range Card for the bullet's ballistics data I used through 1100 yards:

Range Card.jpg
 
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LOL.... I just bought the Nosler app ($1.99), and it gives 63.87" and 23.43".

Bart, could you run the Sierra app with 30 cal 150 grain RN with BC .186, and 150 grain FMJ-BT with BC .386?

I'm using my local elevation, 4550' so our answers might be a little different.
 
I tried the Hornady calculator, and get different numbers yet. (Using my elevation)

For G1, I get 56.2" and 21". For G7, I get 22.7" and 9.5".

Well, in any case, RSI, Hornady, and Strelok all give very different deflections for low and high BC bullets. I'm not sure why Wreck-n-Crew and I are getting such different results with the Hornady software. That's a puzzle to be solved.

If it were simply a case of wind blowing against the side of the bullet, we wouldn't get results that different for high vs. low BC.

I just wish I knew which, if any, of these calculators I can believe.
 
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You have to use G1 for flat base RN and G7 for the boat tail for the Hornady Ballistics calculator or the numbers will be off.
 
Bart B:

I probably understand why you think barrel vibrations have nothing to do with smaller groups at longer distances in a particular rifle. Lots of people think the same. All bullets do not leave ar exactly the same LOF angle relative to the LOS angle.
Click to expand...
Nothing about that explains why the groups would be smaller at a longer distance than they are at 100 yards.

That's addressing overall accuracy and how to get the smallest groups at all distances.

Instead of continuing to repeat the same things, why not copy and paste it specific portion that proves a rifle can be more accurate at long distance than it is at a shorter distance.
 
Snyper said:
Nothing about that explains why the groups would be smaller at a longer distance than they are at 100 yards.
Click to expand...
It explains why groups would subtend smaller MOA values. That's the point I and others have made. They can also be smaller in extreme spread as the two trajectories shown in post #61 shows at mid range in their trajectories where the vertical spread is more than at target ranged where it's zero.

If a bullet drops 40 inches more at 1000 yards leaving 100 fps slower than another, the barrel has to point 4 MOA higher to positively compensate for that. 4 MOA ain't very much for the last several inches of the barrel vibrating 3 to 4 times that of the barrel's resonant frequency. If it's 7 inches long, that's about .008 inch; what 4 MOA subtends over 7 inches.
 
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denton said:
I'm not sure why Wreck-n-Crew and I are getting such different results with the Hornady software
Click to expand...

I don't know if this plays into the differences you are seeing, but I think there is an anomaly in the way the advanced Hornady ballistics calculator works. I have discussed it with the Hornady folks, and they DON'T believe it to be an anomaly, so perhaps it's only my thinking that is the anomaly. Anyway...

The advanced calculator on their site has entries for both elevation and local pressure. If you know them both but enter only one or the other and leave the other to the default value, the trajectory calculated will be accurate. However, if you know them both and you enter them both the trajectory will be incorrect as the difference from sea level will be accounted for twice. Now, if you are interested in ranges under 400 yards or so, the difference won't be that great to the average shooter. But if you are interested in ranges from, say, 600 to 1000 yards or beyond, then there might be a problem.

I've compared the outputs with several other ballistics calculators and they agree with my findings that the affects are double counted on the Hornady calculator. But as I said, the Hornady folks don't agree.

So if anyone else wants to check that out and tell me I'm all wet, then great. Sometimes being dumb but open to correction is the best way to learn. (Provided being dumb doesn't get you killed before being corrected, of course!)
 
Assume a cross wind of 10 MPH, and a 55 grain .224" bullet launched at 3000 FPS, with a typical BC for that type of bullet.

Time of flight to 500 yards is .611 seconds, and horizontal deflection is close to 19.5" according to my ballistics software.

The wind cannot horizontally accelerate a bullet to a speed faster than the air flow in the wind. So the maximum possible horizontal speed imparted to the bullet by the wind blowing against it is 10 MPH. In reality, the horizontal speed imparted will be significantly less than this.

10 MPH = 14.66 FPS = 1.22"/S.

1.22"/S for .611 seconds (the time of flight) yields an upper limit on horizontal deflection of .746". Yet the real answer is 19.5". So the effect of wind blowing against the side of the bullet cannot account for the deflection we actually see.
Click to expand...

Good argument, but bad math! 14.66 FPS * 0.611 S = 8.95 Feet, way more than the predicted 19.5" from the calculator. Seems you divided 12 in/ft when you should have multiplied, so your result is 1.22 ft*ft/in*S which is a pretty hard to interpret number :)

The real deflection is much less than this for primarily two reasons. The bullet velocity is non-linear, slowing down with distance, and it takes some time for the wind to accelerate the bullet to the wind velocity. During this acceleration time the bullet is moving the fastest thus covering more ground with less drift greatly reducing the total drift over the entire flight path. Depending on the time of flight the bullet may never actually hit terminal velocity from the wind.

A bullet fired at a high enough angle might hit terminal velocity from both the wind and gravity and thus enter a period of linear trajectory near the end of its flight, but this "indirect fire" is normally of no use or interest to recreational shooters. Does long range artillery ever enter this aerodynamic regime?
 
Full disclosure, I have not read all the comments in this thread. However, earlier in the thread someone mentioned the spinning of a top and compared it to bullet spin. This seems to be a plausible explanation to me. As the top spin and slowed it would start making wider arcs, at times though, the top would spin back towards the center of where it started. Longer sweeping arcs, to shorter faster arcs, ( in appearance only) the top would actually be slowing.
 
Bullet rpm rate drops about 10% over a thousand yards of travel.

There's a 10% spread in twist rates most bullets will stabilize very nice in across that much spread in muzzle velocity.

How much does a top spin in rpm's degrade over 2 seconds of spin? That's the longest time of flight for most bullets shot at 1000 yard things.

An approximating formula Geoffery Kolbe (Border Barrels) published for spin decay is:

Where:
Nm = the spin rate the bullet had at the muzzle
N = the spin rate of the bullet after your time of flight to the range of interest
t = time of flight in seconds from the muzzle to the current bullet position
d = bullet diameter in inches
e = natural logarithm base, 2.71828…

N = Nm × (-0.035 × t / d)^e

Let's keep things plausible in our comparisons else we'll fall of the edge of the reality cliff.
 
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Bart B. said:
Bullet rpm rate drops about 10% over a thousand yards of travel.

There's a 10% spread in twist rates most bullets will stabilize very nice in across that much spread in muzzle velocity.

How much does a top spin in rpm's degrade over 2 seconds of spin? That's the time of flight for most bullet shot at 1000 yard things.

An approximating formula Geoffery Kolbe (Border Barrels) published for spin decay is:

Where:
Nm = the spin rate the bullet had at the muzzle
N = the spin rate of the bullet after your time of flight to the range of interest
t = time of flight in seconds from the muzzle to the current bullet position
d = bullet diameter in inches
e = natural logarithm base, 2.71828…

N = Nm × (-0.035 × t / d)^e

Let's keep things plausible in our comparisons else we'll fall of the edge of the reality cliff.
Click to expand...

Well, I am not going to say math was my best subject. The top analogy, in my mind seems plausible. But, what do I know, apparently I am just another Knuckle dragger who likes to shoot for the fun of it. Just offering an observation. Guess I live in an alternate reality. Although, on second thought, the bullet still slows, as you pointed out and we talking about going from 1.5moa to .75. might still be plausible.
 
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