Aerodynamic Jump with 22lr.

[orkan]

You better know about this if you want to shoot small with a 22lr in the wind!

 

[Near miss]

Thanks for testing. In my coutry and the landscape we have this test would be impossible.

Over time I have started changing my holds depending on the wind without even much noticing it. Up/dowm about 1/3 or 1/4 of the wind variable.

What is the specific relation you have found?
I use AB mainly but Strelok lets you input manually the relation.
So for every Wind component it calculates AJ component in relation that is given.

 

[orkan]

What is the specific relation you have found?

As I mention in the video, its essentially on tenth of a mil, per 10mph wind, full value. Another relative conversion that I haven't proofed much is essentially 1/5th of the wind solution, will be your jump offset.

 

[Near miss]

Thanks, I heard the 10mph estimation but thought that you would have numbers too and that 1/5 is pretty close what I have felt to be.

 

[barronian]

here is an old but still valid report

https://apps.dtic.mil/dtic/tr/fulltext/u2/a229713.pdf

 

[Merlin O]

Its really quite simple. This is what I use...
A simple math problem which can be done at the line with any cell phone calculator feature/application.



(mN∗)=∑n=1+∞Ps(mn)=∑n=1+∞1ζ(s)(mn)s=1ms P s ( m N ∗ ) = ∑ n = 1 + ∞ P s ( m n ) = ∑ n = 1 + ∞ 1 ζ ( s ) ( m n ) s = 1 m s

.Now since mN∗∩m′N∗=lcm(m,m′)N∗ m N ∗ ∩ m ′ N ∗ = lcm ( m , m ′ ) N ∗ , we have

Ps(mN∗∩m′N∗)=Ps(mN∗)Ps(m′N∗)⟺gcd(m,m′)=1 P s ( m N ∗ ∩ m ′ N ∗ ) = P s ( m N ∗ ) P s ( m ′ N ∗ ) ⟺ gcd ( m , m ′ ) = 1​

Let (pn)n∈N∗ ( p n ) n ∈ N ∗ be the strictly increasing sequence shake movement both on the lateral plane and vertically. From what we just proved we can deduce that (pkN∗)k∈N∗ ( p k N ∗ ) k ∈ N ∗ is a familly of mutually independant events (this is the key argument which is equivalent to regrouping by p-adic valuations in a very subtle way!).

Therefore the 7th lunar phase coupled with tidal flow events are also mutually independent, thus:

Your product

∏p≤y ∏ p : p ≤ y specifies p≤y p ≤ y , so let's try y=5 y = 5 , so that p∈{2,3,5}. p ∈ { 2 , 3 , 5 } . Your sum ∑n∣n, p≤y ∑ n : p ∣ n , p ≤ y requires n n to be divisible only by those primes, so where looking at integers divisible only by 2 2 , 3 3 , or 5 5 and by no other primes.

××=(1+12s+14s+18s+⋯)(1+13s+19s+127s+⋯)(1+15s+125s+1125s+⋯)a sum in which, for example, 17200s appears because7200=25×33×5 has no primefactors besides 2, 3, and 5;thus 17200s=125s×132s×15s.(a) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) × ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) × ( 1 + 1 5 s + 1 25 s + 1 125 s + ⋯ ) = a sum in which, for example, 1 7200 s appears because 7200 = 2 5 × 3 3 × 5 has no prime factors besides 2, 3, and 5; (a) thus 1 7200 s = 1 2 5 s × 1 3 2 s × 1 5 s .​

Within the expression (1+12s+14s+18s+⋯) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) one finds the term 1/25s 1 / 2 5 s ; specifically, it is the sixth term in that sum. Likewise, within the sum (1+13s+19s+127s+⋯) ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) one finds 1/32s 1 / 3 2 s ;

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=∏k=1+∞(1−1psk) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = ∏ k = 1 + ∞ ( 1 − 1 p k s )​

Finally, add Cornelius effect. Therefore

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=Ps({1})=1ζ(s) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = P s ( { 1 } ) = 1 ζ ( s )​

Which states that

ζ(s)=∏k=1+∞(1−1psk)−1=∏k=1+∞∑i=0+∞1psik​

Concluding with - hold a little Kentucky windage in less than ideal conditions.

​

 

[Arc Light]

Its really quite simple. This is what I use...
A simple math problem which can be done at the line with any cell phone calculator feature/application.



(mN∗)=∑n=1+∞Ps(mn)=∑n=1+∞1ζ(s)(mn)s=1ms P s ( m N ∗ ) = ∑ n = 1 + ∞ P s ( m n ) = ∑ n = 1 + ∞ 1 ζ ( s ) ( m n ) s = 1 m s

.Now since mN∗∩m′N∗=lcm(m,m′)N∗ m N ∗ ∩ m ′ N ∗ = lcm ( m , m ′ ) N ∗ , we have

Ps(mN∗∩m′N∗)=Ps(mN∗)Ps(m′N∗)⟺gcd(m,m′)=1 P s ( m N ∗ ∩ m ′ N ∗ ) = P s ( m N ∗ ) P s ( m ′ N ∗ ) ⟺ gcd ( m , m ′ ) = 1​

Let (pn)n∈N∗ ( p n ) n ∈ N ∗ be the strictly increasing sequence shake movement both on the lateral plane and vertically. From what we just proved we can deduce that (pkN∗)k∈N∗ ( p k N ∗ ) k ∈ N ∗ is a familly of mutually independant events (this is the key argument which is equivalent to regrouping by p-adic valuations in a very subtle way!).

Therefore the 7th lunar phase coupled with tidal flow events are also mutually independent, thus:

Your product

∏p≤y ∏ p : p ≤ y specifies p≤y p ≤ y , so let's try y=5 y = 5 , so that p∈{2,3,5}. p ∈ { 2 , 3 , 5 } . Your sum ∑n∣n, p≤y ∑ n : p ∣ n , p ≤ y requires n n to be divisible only by those primes, so where looking at integers divisible only by 2 2 , 3 3 , or 5 5 and by no other primes.

××=(1+12s+14s+18s+⋯)(1+13s+19s+127s+⋯)(1+15s+125s+1125s+⋯)a sum in which, for example, 17200s appears because7200=25×33×5 has no primefactors besides 2, 3, and 5;thus 17200s=125s×132s×15s.(a) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) × ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) × ( 1 + 1 5 s + 1 25 s + 1 125 s + ⋯ ) = a sum in which, for example, 1 7200 s appears because 7200 = 2 5 × 3 3 × 5 has no prime factors besides 2, 3, and 5; (a) thus 1 7200 s = 1 2 5 s × 1 3 2 s × 1 5 s .​

Within the expression (1+12s+14s+18s+⋯) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) one finds the term 1/25s 1 / 2 5 s ; specifically, it is the sixth term in that sum. Likewise, within the sum (1+13s+19s+127s+⋯) ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) one finds 1/32s 1 / 3 2 s ;

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=∏k=1+∞(1−1psk) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = ∏ k = 1 + ∞ ( 1 − 1 p k s )​

Finally, add Cornelius effect. Therefore

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=Ps({1})=1ζ(s) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = P s ( { 1 } ) = 1 ζ ( s )​

Which states that

ζ(s)=∏k=1+∞(1−1psk)−1=∏k=1+∞∑i=0+∞1psik​

Concluding with - hold a little Kentucky windage in less than ideal conditions.

​

I find it's closer to 1/64s, but otherwise spot on!

 

[Kisssofdeath]

Its really quite simple. This is what I use...
A simple math problem which can be done at the line with any cell phone calculator feature/application.



(mN∗)=∑n=1+∞Ps(mn)=∑n=1+∞1ζ(s)(mn)s=1ms P s ( m N ∗ ) = ∑ n = 1 + ∞ P s ( m n ) = ∑ n = 1 + ∞ 1 ζ ( s ) ( m n ) s = 1 m s

.Now since mN∗∩m′N∗=lcm(m,m′)N∗ m N ∗ ∩ m ′ N ∗ = lcm ( m , m ′ ) N ∗ , we have

Ps(mN∗∩m′N∗)=Ps(mN∗)Ps(m′N∗)⟺gcd(m,m′)=1 P s ( m N ∗ ∩ m ′ N ∗ ) = P s ( m N ∗ ) P s ( m ′ N ∗ ) ⟺ gcd ( m , m ′ ) = 1​

Let (pn)n∈N∗ ( p n ) n ∈ N ∗ be the strictly increasing sequence shake movement both on the lateral plane and vertically. From what we just proved we can deduce that (pkN∗)k∈N∗ ( p k N ∗ ) k ∈ N ∗ is a familly of mutually independant events (this is the key argument which is equivalent to regrouping by p-adic valuations in a very subtle way!).

Therefore the 7th lunar phase coupled with tidal flow events are also mutually independent, thus:

Your product

∏p≤y ∏ p : p ≤ y specifies p≤y p ≤ y , so let's try y=5 y = 5 , so that p∈{2,3,5}. p ∈ { 2 , 3 , 5 } . Your sum ∑n∣n, p≤y ∑ n : p ∣ n , p ≤ y requires n n to be divisible only by those primes, so where looking at integers divisible only by 2 2 , 3 3 , or 5 5 and by no other primes.

××=(1+12s+14s+18s+⋯)(1+13s+19s+127s+⋯)(1+15s+125s+1125s+⋯)a sum in which, for example, 17200s appears because7200=25×33×5 has no primefactors besides 2, 3, and 5;thus 17200s=125s×132s×15s.(a) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) × ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) × ( 1 + 1 5 s + 1 25 s + 1 125 s + ⋯ ) = a sum in which, for example, 1 7200 s appears because 7200 = 2 5 × 3 3 × 5 has no prime factors besides 2, 3, and 5; (a) thus 1 7200 s = 1 2 5 s × 1 3 2 s × 1 5 s .​

Within the expression (1+12s+14s+18s+⋯) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) one finds the term 1/25s 1 / 2 5 s ; specifically, it is the sixth term in that sum. Likewise, within the sum (1+13s+19s+127s+⋯) ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) one finds 1/32s 1 / 3 2 s ;

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=∏k=1+∞(1−1psk) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = ∏ k = 1 + ∞ ( 1 − 1 p k s )​

Finally, add Cornelius effect. Therefore

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=Ps({1})=1ζ(s) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = P s ( { 1 } ) = 1 ζ ( s )​

Which states that

ζ(s)=∏k=1+∞(1−1psk)−1=∏k=1+∞∑i=0+∞1psik​

Concluding with - hold a little Kentucky windage in less than ideal conditions.

​

Freaking A for effort.

 

[eldritch]

Great, informative video... thanks for posting!

 

[1badDart]

Its really quite simple. This is what I use...
A simple math problem which can be done at the line with any cell phone calculator feature/application.



(mN∗)=∑n=1+∞Ps(mn)=∑n=1+∞1ζ(s)(mn)s=1ms P s ( m N ∗ ) = ∑ n = 1 + ∞ P s ( m n ) = ∑ n = 1 + ∞ 1 ζ ( s ) ( m n ) s = 1 m s

.Now since mN∗∩m′N∗=lcm(m,m′)N∗ m N ∗ ∩ m ′ N ∗ = lcm ( m , m ′ ) N ∗ , we have

Ps(mN∗∩m′N∗)=Ps(mN∗)Ps(m′N∗)⟺gcd(m,m′)=1 P s ( m N ∗ ∩ m ′ N ∗ ) = P s ( m N ∗ ) P s ( m ′ N ∗ ) ⟺ gcd ( m , m ′ ) = 1​

Let (pn)n∈N∗ ( p n ) n ∈ N ∗ be the strictly increasing sequence shake movement both on the lateral plane and vertically. From what we just proved we can deduce that (pkN∗)k∈N∗ ( p k N ∗ ) k ∈ N ∗ is a familly of mutually independant events (this is the key argument which is equivalent to regrouping by p-adic valuations in a very subtle way!).

Therefore the 7th lunar phase coupled with tidal flow events are also mutually independent, thus:

Your product

∏p≤y ∏ p : p ≤ y specifies p≤y p ≤ y , so let's try y=5 y = 5 , so that p∈{2,3,5}. p ∈ { 2 , 3 , 5 } . Your sum ∑n∣n, p≤y ∑ n : p ∣ n , p ≤ y requires n n to be divisible only by those primes, so where looking at integers divisible only by 2 2 , 3 3 , or 5 5 and by no other primes.

××=(1+12s+14s+18s+⋯)(1+13s+19s+127s+⋯)(1+15s+125s+1125s+⋯)a sum in which, for example, 17200s appears because7200=25×33×5 has no primefactors besides 2, 3, and 5;thus 17200s=125s×132s×15s.(a) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) × ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) × ( 1 + 1 5 s + 1 25 s + 1 125 s + ⋯ ) = a sum in which, for example, 1 7200 s appears because 7200 = 2 5 × 3 3 × 5 has no prime factors besides 2, 3, and 5; (a) thus 1 7200 s = 1 2 5 s × 1 3 2 s × 1 5 s .​

Within the expression (1+12s+14s+18s+⋯) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) one finds the term 1/25s 1 / 2 5 s ; specifically, it is the sixth term in that sum. Likewise, within the sum (1+13s+19s+127s+⋯) ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) one finds 1/32s 1 / 3 2 s ;

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=∏k=1+∞(1−1psk) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = ∏ k = 1 + ∞ ( 1 − 1 p k s )​

Finally, add Cornelius effect. Therefore

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=Ps({1})=1ζ(s) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = P s ( { 1 } ) = 1 ζ ( s )​

Which states that

ζ(s)=∏k=1+∞(1−1psk)−1=∏k=1+∞∑i=0+∞1psik​

Concluding with - hold a little Kentucky windage in less than ideal conditions.

​

Merlin posted the simple method for figuring wind holds.

 

[rick137]

The projectile, i.e. ballistics speak for bullet in this case, is a spinning cylindrically symmetrical solid body in a gas, i.e. the atmosphere. Assume the crosswind flow is laminar, i.e. the streamlines/flowlines are parallel. Then over half the surface area of the projectile the rotation has a component in the same sense as the direction of flow. Over the other half the rotation has a component opposite the direction of the flow. This creates a pressure difference which results in vertical motion, either up or down depending on direction of the crosswind and the sense of rotation. Bernoulli's effect.

 

[usafa77]

Thanks for the video, I learned a lot. I have been zeroing out my W1 in the Kestrel- just like i do for centerfire. I will but it back in- and hopefully correct some of those first shot .1- .2 errors. I quess it's time to go back to the Kestrel forum and figure out why I'm doing that again. I don't want to drift the thread- thanks again.

 

[orkan]

Kestrel with AB typically accounts for aerodynamic jump. It's a big deal. Someone recommended you shut that off?

I haven't used the kestrel for 22lr, but I've used the AB phone app, and it does a great job.

 

[Dthomas3523]

Kestrel with AB typically accounts for aerodynamic jump. It's a big deal. Someone recommended you shut that off?

I haven't used the kestrel for 22lr, but I've used the AB phone app, and it does a great job.

With centerfire, most shut it off because past a few hundred yards, unless you are on 100% flat land (most matches aren’t), you cannot distinguish if the vertical component is cause by either

Vertical wind from terrain features
Or
Aerodynamic jump

Not to mention reticle wobble in general be it off a prop or bipod/bag (F class supports are needed to truly take out all wobble, at least in the sense of aerodynamic jump amounts)

Basically, it gets lost in the weeds and just isn’t needed.

300yd and closer benchrest is going to be where most people will need to worry about AJ. This would translate into the 50/100 yd Rimfire benchrest area as well.

While most things increase in importance at longer ranges, AJ is a more important consideration at closer ranges. And typically only in a benchrest setting.

 

[Near miss]

The projectile, i.e. ballistics speak for bullet in this case, is a spinning cylindrically symmetrical solid body in a gas, i.e. the atmosphere. Assume the crosswind flow is laminar, i.e. the streamlines/flowlines are parallel. Then over half the surface area of the projectile the rotation has a component in the same sense as the direction of flow. Over the other half the rotation has a component opposite the direction of the flow. This creates a pressure difference which results in vertical motion, either up or down depending on direction of the crosswind and the sense of rotation. Bernoulli's effect.

Yoar trying to science talk us down with fancy words?
Well how well did that Bernoulli fella did with rimfire??

 

[Cuzz]

Merlin posted the simple method for figuring wind holds.

too much simple for me!

 

[Longshot231]

Its really quite simple. This is what I use...
A simple math problem which can be done at the line with any cell phone calculator feature/application.



(mN∗)=∑n=1+∞Ps(mn)=∑n=1+∞1ζ(s)(mn)s=1ms P s ( m N ∗ ) = ∑ n = 1 + ∞ P s ( m n ) = ∑ n = 1 + ∞ 1 ζ ( s ) ( m n ) s = 1 m s

.Now since mN∗∩m′N∗=lcm(m,m′)N∗ m N ∗ ∩ m ′ N ∗ = lcm ( m , m ′ ) N ∗ , we have

Ps(mN∗∩m′N∗)=Ps(mN∗)Ps(m′N∗)⟺gcd(m,m′)=1 P s ( m N ∗ ∩ m ′ N ∗ ) = P s ( m N ∗ ) P s ( m ′ N ∗ ) ⟺ gcd ( m , m ′ ) = 1​

Let (pn)n∈N∗ ( p n ) n ∈ N ∗ be the strictly increasing sequence shake movement both on the lateral plane and vertically. From what we just proved we can deduce that (pkN∗)k∈N∗ ( p k N ∗ ) k ∈ N ∗ is a familly of mutually independant events (this is the key argument which is equivalent to regrouping by p-adic valuations in a very subtle way!).

Therefore the 7th lunar phase coupled with tidal flow events are also mutually independent, thus:

Your product

∏p≤y ∏ p : p ≤ y specifies p≤y p ≤ y , so let's try y=5 y = 5 , so that p∈{2,3,5}. p ∈ { 2 , 3 , 5 } . Your sum ∑n∣n, p≤y ∑ n : p ∣ n , p ≤ y requires n n to be divisible only by those primes, so where looking at integers divisible only by 2 2 , 3 3 , or 5 5 and by no other primes.

××=(1+12s+14s+18s+⋯)(1+13s+19s+127s+⋯)(1+15s+125s+1125s+⋯)a sum in which, for example, 17200s appears because7200=25×33×5 has no primefactors besides 2, 3, and 5;thus 17200s=125s×132s×15s.(a) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) × ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) × ( 1 + 1 5 s + 1 25 s + 1 125 s + ⋯ ) = a sum in which, for example, 1 7200 s appears because 7200 = 2 5 × 3 3 × 5 has no prime factors besides 2, 3, and 5; (a) thus 1 7200 s = 1 2 5 s × 1 3 2 s × 1 5 s .​

Within the expression (1+12s+14s+18s+⋯) ( 1 + 1 2 s + 1 4 s + 1 8 s + ⋯ ) one finds the term 1/25s 1 / 2 5 s ; specifically, it is the sixth term in that sum. Likewise, within the sum (1+13s+19s+127s+⋯) ( 1 + 1 3 s + 1 9 s + 1 27 s + ⋯ ) one finds 1/32s 1 / 3 2 s ;

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=∏k=1+∞(1−1psk) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = ∏ k = 1 + ∞ ( 1 − 1 p k s )​

Finally, add Cornelius effect. Therefore

Ps(⋂k=1+∞pkN∗¯¯¯¯¯¯¯¯¯¯)=Ps({1})=1ζ(s) P s ( ⋂ k = 1 + ∞ p k N ∗ ¯ ) = P s ( { 1 } ) = 1 ζ ( s )​

Which states that

ζ(s)=∏k=1+∞(1−1psk)−1=∏k=1+∞∑i=0+∞1psik​

Concluding with - hold a little Kentucky windage in less than ideal conditions.

​

Excellent explanation. However, you should have mentioned the Chandler effect or Chandler wobble.

This is a small deviation in the Earth's axis of rotation relative to the solid earth, which was discovered by American astronomer Seth Carlo Chandler in 1891. It amounts to change of about 9 metres (30 ft) in the point at which the axis intersects the Earth's surface and has a period of 433 days. This wobble, which is a nutation, combines with another wobble with a period of one year, so that the total polar motion varies with a period of about 7 years.

The Chandler wobble is an example of the kind of motion that can occur for a spinning object that is not a sphere; this is called a free nutation. Somewhat confusingly, the direction of the Earth's spin axis relative to the stars also varies with different periods, and these motions—caused by the tidal forces of the Moon and Sun—are also called nutations, except for the slowest, which are precessions of the equinoxes.

On the other hand, I can understand why you would not factor in the Chandler effect as it would be an impossibility to mathematically model any mitigating factors that would lessen the Cornelius effect. The processional factors are almost an impossibility to quantify and so inconsequential that most ballisticians either ignoring it or assigning an arbitrary fractional multiplier after reducing the equation by 1 in the last line of your solution so that it reads as...

ζ(s)=∏k=1+∞(1−1psk)−1(__/9m)=∏k=1+∞∑i=0+∞1psik

The numerator, naturally, is up to the discretion of the ballistician who wishes to make his solution fit the actual results. More experienced ballisticians recognize this inverse multiplier as an exercise in futility because of the glacial speed of the Chandler effect on anything going faster than a baseball. Therefore it is completely ignored as in your solution.

It is interesting to note that NASA has stated that: "Because of the precession of the poles over 26,000 years, all the stars, and other celestial objects, appear to shift west to east at the rate of .014 degree each year (360 degrees in 26,000 years). This apparent motion is the main reason for astronomers as well as spacecraft operators to refer to a common epoch such as J2000.0."

One esteemed colleague advocated for a standard value to used for factoring in the Chandler Effect based on NASA's modeling. "To do otherwise," he contended, "would be a disservice rimfire perfectionists everywhere." The value he uses is .014/26,000 or 5.384615384615385e-7‬. For the purposes of obscurity this number is commonly rounded off to 5e-7.

ζ(s)=∏k=1+∞(1−1psk)−1(5e-7/9m)=∏k=1+∞∑i=0+∞1psik

It is sad to say that the National Bureau of Standards isn't as enlightened as our colleague and hasn't adopted his suggestion as the standard.

 

[Near miss]

I know this effect and thus just hold my reticle 0.00001 mil to the right. So far it seems to do the trick!

 

[Longshot231]

I know this effect and thus just hold my reticle 0.00001 mil to the right. So far it seems to do the trick!

If you go to Australia you will have to hold to the left.

 

[1badDart]

I know this effect and thus just hold my reticle 0.00001 mil to the right. So far it seems to do the trick!

If the hash marks in your scope are that coarse you need a better scope. The hash marks are so fine (15 points to the right of the decimal) that I have to have a magnifying glass to see them.

I wet my thumb to check wind speed and direction, then figure in my head using Merlins short method for my hold offs. The whole process can be done pretty quickly, in fact I use it hunting for hold offs on running deer at extremely long range. Easy preasy with just a little practice.

What's the old saying, "the first one don't stand a chance". LOL

 

[Cuzz]

I know this effect and thus just hold my reticle 0.00001 mil to the right. So far it seems to do the trick!

Whodathunkit, that’s exactly what i do!

 

[rick137]

I know this effect and thus just hold my reticle 0.00001 mil to the right. So far it seems to do the trick!

You have an amazing sighting system. Holding the rifle for windage also adjusts the elevation to compensate for aerodynamic jump. Bet it cost a few bucks.

 

[Near miss]

You have an amazing sighting system. Holding the rifle for windage also adjusts the elevation to compensate for aerodynamic jump. Bet it cost a few bucks.

Yeah this sighting system I have is a one of a kind.
Usually it works like wonder but often it can be tricky to get it properly calibrated.

Sometimes during day-long range sessions the sighting system crashes completely but it is restored and back in order with snacks and water.

 

[rick137]

Although the equations presented in previous posts have nothing to do with the situation they do suggest the complexity of the physics. The attached is from a BRL report, AD441598, by Lieske and McCoy, "Equations of Motion of a Rigid Projectile" which contain the actual equations. Quoting, "The basic vector equations of motion of a rigid body are described, and an aerodynamic force-moment system is developed. The vector forces and moments due to gravity and rotation of the earth are added to the system. ..... the resulting vector differential equations of motion of a symmetric rigid projectile are derived."

(5.1) is the equation of motion for the center of mass of the projectile:

(5.2) is the equation of motion for the instantaneous rate of rotation of the projectile:

 

[Longshot231]

Although the equations presented in previous posts have nothing to do with the situation they do suggest the complexity of the physics. The attached is from a BRL report, AD441598, by Lieske and McCoy, "Equations of Motion of a Rigid Projectile" which contain the actual equations. Quoting, "The basic vector equations of motion of a rigid body are described, and an aerodynamic force-moment system is developed. The vector forces and moments due to gravity and rotation of the earth are added to the system. ..... the resulting vector differential equations of motion of a symmetric rigid projectile are derived."

(5.1) is the equation of motion for the center of mass of the projectile:

View attachment 7288352

(5.2) is the equation of motion for the instantaneous rate of rotation of the projectile:
View attachment 7288353

You are communicating with marksmen on this forum that have a superior intellect, insight and skills far above ordinary plinkers. Do you really feel the need to discuss this topic with the same rudimentary and simplistic vernacular used in a plebeian first grade classroom?

 

[usafa77]

Cabin fever has taken control.

 

[Sheldon N]

As I mention in the video, its essentially on tenth of a mil, per 10mph wind, full value. Another relative conversion that I haven't proofed much is essentially 1/5th of the wind solution, will be your jump offset.

Interestingly enough, if you play with a Kestrel that's exactly what it gives for a wind jump solution. Exactly 0.1 mil per 10mph (pure AJ, regardless of distance). I wonder if that is based on some complex math or just a rule of thumb applied to the ballistic calculator.

I will use the AJ in my kestrel, but it's seemed a bit aggressive to me so I typically will put about 1/2 the wind speed value in Wind 1 field for a more conservative AJ figure. My gut says that 0.2 mils of AJ in a 20mph wind is a bit much for a centerfire rifle, but I wish I could test more.

 

[orkan]

My gut says that 0.2 mils of AJ in a 20mph wind is a bit much for a centerfire rifle, but I wish I could test more.

That seems right to me... but I shoot in flat prairie, and I think the main disconnect on aerodynamic jump comes from people not actually seeing the full wind call often times. So if the calculator is calling for it, but the bullet doesn't land per the firing solution... usually it is human error in not properly seeing downrange conditions.

 

[rick137]

You are communicating with marksmen on this forum that have a superior intellect, insight and skills far above ordinary plinkers. Do you really feel the need to discuss this topic with the same rudimentary and simplistic vernacular used in a plebeian first grade classroom?

View attachment 7288401

And we don't need no stinkin' formulas.

 

[Merlin O]

That seems right to me... but I shoot in flat prairie, and I think the main disconnect on aerodynamic jump comes from people not actually seeing the full wind call often times. So if the calculator is calling for it, but the bullet doesn't land per the firing solution... usually it is human error in not properly seeing downrange conditions.

I understand the concept and found your vid thought provoking and will now be conscious/aware of the possibilities when I see my bullets landing somewhere other than where I think they should be landing. Hard for me to test this out for myself as the ranges I am privy to would most definitely frown upon the opposite direction shooting....
Looking back 60 years I can remember my BBs spinning to the right and curving away from line of sight but sure don't remember the wind values... LOL

 

[orkan]

Hard for me to test this out for myself as the ranges I am privy to would most definitely frown upon the opposite direction shooting....

Just keep a log. One day the wind will be 3:00, sometime in the future it will be 9:00. (unless you don't shoot often) It will be easy to "prove" the existence of the effect if you can shoot small enough. Lots of folks can't legitimately shoot a tenth mil... and that's why it just isn't discussed often, especially with centerfire.

With rimfire though... lots of people can shoot tenth of a mil @ 50yds. That's why I used the rimfire section to demonstrate it. Lots of people shooting precision 22lr these days, and it's increasing in popularity. Super easy to see it with 22lr up close, even if you can only shoot 1/2" groups. You'll still see the trend.

 

[1813Benny]

Orkin
Your video provide a good base for further explanation of the exterior ballistics of a rimfire projectile. Will you be following up with demonstration of the full reversed flattened Z pattern that results in wind from not only full value (i.e. 3 and 9 o'clock), but also "half" and "zero" value?

Regards,
Ken
ASSA Secretary/Treasurer

 

[orkan]

Orkin
Your video provide a good base for further explanation of the exterior ballistics of a rimfire projectile. Will you be following up with demonstration of the full reversed flattened Z pattern that results in wind from not only full value (i.e. 3 and 9 o'clock), but also "half" and "zero" value?

Regards,
Ken
ASSA Secretary/Treasurer

This is the only video I had planned on aerodynamic jump. Once introduced to the topic, it's not hard to run theoretical scenarios with AB to fill in any blanks someone has.

 

[1813Benny]

Orkin,
Thx again.

For other’s reference, the pattern is simple and easy to learn and apply without any need for calculators, etc. If you are interested, there is a good thread with the pattern and depicted hold off points on RimFire Accuracy.

As a point of order: IMHO, it is somewhat of a disservice by calling it “aerodynamic jump” since what is occurring is only part of .22 rimfire exterior ballistics and it’s result is not always a higher point of impact. There is always a vertical and horizontal impact by wind - even at the 12 and 6 “no value” directions.

I guess it is what it is.

 

[barronian]

i'm assuming that nobody objects to me cross-posting ?
Here is a nice simple visualisation of the effects of wind on POI and POA for .22 lr (someone elses thread on another forum)

Rimfire Accuracy Forum

 

[orkan]

Orkin,
Thx again.

For other’s reference, the pattern is simple and easy to learn and apply without any need for calculators, etc. If you are interested, there is a good thread with the pattern and depicted hold off points on RimFire Accuracy.

As a point of order: IMHO, it is somewhat of a disservice by calling it “aerodynamic jump” since what is occurring is only part of .22 rimfire exterior ballistics and it’s result is not always a higher point of impact. There is always a vertical and horizontal impact by wind - even at the 12 and 6 “no value” directions.

I guess it is what it is.

This is primarily due to the fact that there is almost never a true 6:00 or true 12:00... there is always an off axis element, it's just difficult to see... which is why they are very tough to shoot in. I personally have not seen a true 12 or 6 condition that produces vertical. If the wind is above 5mph, then it is almost NEVER true 6 or 12. A lazy, consistent, 6 or 12 under 5mph... I hold center bull and that's where the shots land. Above that speed... I'm looking for those tiny little twitches in mirage and grass to tell me whether it's off axis.

I thought it self evident that it doesn't take a full value wind to create the effect to a lesser degree.

 

[HuskerP7M8]

Here's another example of "Aerodynamic Jump" that I did back in the winter of 2008.

Two targets were placed 25 yds from the bench and were aligned with the wind perpendicular to the course of fire (+/- 15 deg's).

25 shots fired at 279 deg's followed by a second 25 shots fired at 99 deg's.

4 electronic wind sensors located along the course of fire recorded the wind. Target #1 had an average crosswind of 6.9mph with one recorded maximum gust of 10.6 mph. Target #2 had an average crosswind of 6.1mph with one recorded maximum gust of 8.9mph.

The results:

The "slope" you expect to see, but steeper than the equation used to predict it. You would expect to see the 9:30 to 3:30 slope that McCoy's work produced, but my testing resulted in a steeper slope of 36 deg's from the horizontal or a 10:00 to 4:00 slope. Even with the range of statistical uncertainties factored in, it appeared my results were statistically different than McCoy's.

Upon farther research and testing I became aware that it's likely this "slope" changes based on the distance to the target. This is due to both the scaling and the angular measurements involved with increased distance. If I had done my testing at the distances involved in McCoy's project, I believe the slope I produced would shallow out to match McCoy's results.

I only mention this because a lot of you guys are shooting at much longer distances and there will probably be a point where "Jump" may get lost in the statistical noise except under laboratory conditions or at the theoretical level.

It might be very interesting if some of you try doing what the OP and myself have done at other than the 50 yd or 25 yd distance. 200 yds would be my choice for gathering some interesting data, but you'll have to be very careful to match the test methodology to the conditions or you'll only get a bunch of useless numbers that aren't conducive to analysis.

I see while I was composing this post that "barronian" gives a link to a thread I started on Rimfire Accuracy a while back. Thanks for that and I'd like to think it's worth a read for those interested.

Landy

 

[orkan]

Landy, I don't suppose you've done any analysis with a left twist barrel?

 

[Dthomas3523]

While physically a 12 o’clock wind will result in a reduced velocity on target (and cause a lower POI) and a 6 o’clock will result in an increased velocity on target (and cause a higher POI), I have yet to see or hear of anyone who can shoot the actual difference in the POI to factor it in.

It is generally the ADJ from not being a perfect 12 or 6 wind.

 

[Merlin O]

Here's another example of "Aerodynamic Jump" that I did back in the winter of 2008.

Two targets were placed 25 yds from the bench and were aligned with the wind perpendicular to the course of fire (+/- 15 deg's).

25 shots fired at 279 deg's followed by a second 25 shots fired at 99 deg's.

4 electronic wind sensors located along the course of fire recorded the wind. Target #1 had an average crosswind of 6.9mph with one recorded maximum gust of 10.6 mph. Target #2 had an average crosswind of 6.1mph with one recorded maximum gust of 8.9mph.

The results:

The "slope" you expect to see, but steeper than the equation used to predict it. You would expect to see the 9:30 to 3:30 slope that McCoy's work produced, but my testing resulted in a steeper slope of 36 deg's from the horizontal or a 10:00 to 4:00 slope. Even with the range of statistical uncertainties factored in, it appeared my results were statistically different than McCoy's.

Upon farther research and testing I became aware that it's likely this "slope" changes based on the distance to the target. This is due to both the scaling and the angular measurements involved with increased distance. If I had done my testing at the distances involved in McCoy's project, I believe the slope I produced would shallow out to match McCoy's results.

I only mention this because a lot of you guys are shooting at much longer distances and there will probably be a point where "Jump" may get lost in the statistical noise except under laboratory conditions or at the theoretical level.

It might be very interesting if some of you try doing what the OP and myself have done at other than the 50 yd or 25 yd distance. 200 yds would be my choice for gathering some interesting data, but you'll have to be very careful to match the test methodology to the conditions or you'll only get a bunch of useless numbers that aren't conducive to analysis.

I see while I was composing this post that "barronian" gives a link to a thread I started on Rimfire Accuracy a while back. Thanks for that and I'd like to think it's worth a read for those interested.

Landy

I am definitely riding the short bus here. Left wind low right impacts? Right wind high left impacts...?

 

[orkan]

I am definitely riding the short bus here. Left wind low right impacts? Right wind high left impacts...?

With a right twist barrel... yes.

 

[HuskerP7M8]

Landy, I don't suppose you've done any analysis with a left twist barrel?

Hi orkan,

Unless I were to explore some of the more esoteric and/or smaller aerodynamic forces in play such as say "Spin Drift" for example, the science involved is well proven and backed up by empirical data for predicting projectile behavior in both left and right twist barrels. For all practical purposes, unless I'm missing something, analyzing the flight of a projectile in a left twist barrel would only be a duplication of testing already completed.

You'll notice I left myself an "out" by stating "unless I'm missing something". Fact is, it's easy to miss something because there are so few serious studies or research into transonic/imbalanced/malleable projectiles. Most true ballisticians want nothing to do with the inherent problems associated with this type of projectile and the hoops you have to jump through for acquiring quality data.

Landy

 

[orkan]

I agree... but you know how it is... curiosity prevails, and I am always interested in first hand accounts.

 

[Shawn1492]

Very cool. Thanks for posting this. Using eyeball measurements it seems to align with what 4DOF predicts.

 

[rick137]

Hi orkan,

Unless I were to explore some of the more esoteric and/or smaller aerodynamic forces in play such as say "Spin Drift" for example, the science involved is well proven and backed up by empirical data for predicting projectile behavior in both left and right twist barrels. For all practical purposes, unless I'm missing something, analyzing the flight of a projectile in a left twist barrel would only be a duplication of testing already completed.

You'll notice I left myself an "out" by stating "unless I'm missing something". Fact is, it's easy to miss something because there are so few serious studies or research into transonic/imbalanced/malleable projectiles. Most true ballisticians want nothing to do with the inherent problems associated with this type of projectile and the hoops you have to jump through for acquiring quality data.

Landy

You may be correct on the number of studies but post #5 in this thread refers to a very serious study.

 

[HuskerP7M8]

You may be correct on the number of studies but post #5 in this thread refers to a very serious study.

Hi Rick,

Yes, the 1990 report by McCoy at the Aberdeen BRL is a serious study by perhaps one of the greatest ballisticians in modern times. However, as pointed out in the report, it was a "limited firing program".

The small sample sizes in the testing leads to higher statistical uncertainties and lower confidence levels in drawing conclusions. The report also doesn't delve into many of the variables that most likely have a large influence on the results. A few that come to mind would include bore and chamber dimensions along with the various styles of rifling.

Perhaps of most interest to the majority on this forum/thread, would be an analysis of flight behavior at ranges beyond 100 meters. We're quite literally in the dark when it comes to predicting flight at those extended ranges and I can't help much. A few years ago I considered extending my ballistic tunnel to 200 yds, but I couldn't justify the expense to satisfy my curiosity.

I can only dream about the things we would have learned if the Aberdeen facility and the Government had thrown their entire weight behind the project like they do with small arms fire using jacketed CF projectiles.

Landy

 

[rick137]

Hi Rick,

Yes, the 1990 report by McCoy at the Aberdeen BRL is a serious study by perhaps one of the greatest ballisticians in modern times. However, as pointed out in the report, it was a "limited firing program".

The small sample sizes in the testing leads to higher statistical uncertainties and lower confidence levels in drawing conclusions. The report also doesn't delve into many of the variables that most likely have a large influence on the results. A few that come to mind would include bore and chamber dimensions along with the various styles of rifling.

Perhaps of most interest to the majority on this forum/thread, would be an analysis of flight behavior at ranges beyond 100 meters. We're quite literally in the dark when it comes to predicting flight at those extended ranges and I can't help much. A few years ago I considered extending my ballistic tunnel to 200 yds, but I couldn't justify the expense to satisfy my curiosity.

I can only dream about the things we would have learned if the Aberdeen facility and the Government had thrown their entire weight behind the project like they do with small arms fire using jacketed CF projectiles.

Landy

Landy:

I hardily applaud someone with a 100 m ballistic tunnel. Do you have a 6DOF ballistic equation solver to complement it?

The bottoms-up approach is definitely interesting theoretically. However I suspect the top-down approach, as exemplified by WEZ analysis, is sufficient for most people, perhaps more than sufficient for most people since it predicts accuracy probabilities on basis of a few inputs.

Perhaps there will be a shooter with sufficient wealth to own or rent time on a supercomputer in order calculate the aerodynamic coefficients of small caliber rimfire projectiles from first principles.

Rick

 

[Seymour Fish]

Orkin
Your video provide a good base for further explanation of the exterior ballistics of a rimfire projectile. Will you be following up with demonstration of the full reversed flattened Z pattern that results in wind from not only full value (i.e. 3 and 9 o'clock), but also "half" and "zero" value?

Regards,
Ken
ASSA Secretary/Treasurer

Benny, Now were talking. Tony Boyers take on wind angles proves up. Seymour

 

[HuskerP7M8]

Landy:

I hardily applaud someone with a 100 m ballistic tunnel. Do you have a 6DOF ballistic equation solver to complement it?

The bottoms-up approach is definitely interesting theoretically. However I suspect the top-down approach, as exemplified by WEZ analysis, is sufficient for most people, perhaps more than sufficient for most people since it predicts accuracy probabilities on basis of a few inputs.

Perhaps there will be a shooter with sufficient wealth to own or rent time on a supercomputer in order calculate the aerodynamic coefficients of small caliber rimfire projectiles from first principles.

Rick

Hey Rick,

LOL....Don't applaud too hard cuz I'm self-taught in Ballistics and Statistics! Someone like say McCoy would describe my tunnel and methodology as a "Red Neck" solution for gathering data.

I've spent and incredible amount of time and money in an attempt to be as good as I can with what I have, but to approach the level of a true Ballistic Research Lab would require a few years of post graduate work and a pile of $100 bills that might reach the ceiling!

You don't really need a supercomputer for the bulk of work in ballistics, but you do need hardware/ instrumentation like Dopler Radar as well as other tools and the expertise to run them.
The same goes for the 6DOF software because it takes those aforementioned tools to know exactly what those inputs are for it. It's simply not possible for the amateurs like myself and others to determine these inputs through live fire.

To this date, my most notable accomplishment is probably the ability to measure drag coefficients and the resultant BC's through the use of a dual paired chronograph system I built. To the best of my knowledge, I may be the only amateur to achieve that and it was done a couple of years before Litz published "Modern Advancements In Long Range Shooting" (Volume II) where he did the same.

I was very gratified to find out my numbers were equal to his and in some cases better. That's no longer the case because he's advanced to a Doplar Radar unit that I'm nearly certain runs in the 5 figure range.

For the most part, I'm forced to theorize like the rest of you based on jacketed CF data/research, where I hope to find some commonalities with RF transonic ballistics.

Landy

 

[orkan]

I'm in the process of designing my 100yd tunnel. Landy, I wonder if I couldn't have you bend my ear one of these days. Would be great to learn what you'd do again and do differently if you were to do another one.