# Ballistic trajectory of a projectile (ballistic loophole)

• Gmtom
In summary, a ballistic loophole is a small hole in a wall or structure that can be used to shoot through it without being close to the wall. The height from the bore to the center of the optic must be 2.5 inches, the distance to the target must be 300 yards, and the projectile must have a weight of 62 grams and a velocity of 2980 fps.
Gmtom
I am trying to find the formula for the so called "ballistic loophole" which is a technique where one will cut a small hole in a wall or structure (no bigger than 3 inches wide or tall) and shoot through it at a target without placing himself to close to the wall.

I know I need the height from bore to center of optic which is 2.5inches
Distance to target 300yards
100yards zero so its 3"low at muzzle to 10yards, 1"low at 50, 0"low from 70yards to 130yards, 0.3"low at 150yards, 2"low at 200yards, 5" low 250yards,and 10"low at 300 yards.
weight of projectile 62grains (4.017grams) velocity-2980fps at the muzzle

I'm just going by eye and I can shoot through a hole 1.5inches wide and 2 inches tall and hit targets 470yard away but I would like an actual formula. Thank you for the help. I apologize if this isn't the place to ask this question

My grandson is a sniper in the US Army. He told me that the answer to the OP's question is highly secret.

Of course basic ballistics physics is well known and hardly secret; a rule-of-thumb formula that people in the field can execute in their heads is something else. The suggestion is that would-be terrorists would like to know the answer to that question.

I guess the key is that the hole must be both on the line of sight to the target and on the ballistic path of the bullet. The maths can't be too hard but since a lot of web posts claim it's secret I don't plan on doing the calculations.

No, its not highly secret that's what's going around and it just simply isn't so. Terrorists don't want to "snipe" you that's a myth circulated by movies they want to cause as much damage as possible as quickly as possible. There are commercial pda devices available that will calculate the formula for you. Besides anyone with 30minds some cardboard and a rifle can do it well enough without the formula. I'm asking for a simple math formula so I can understand in depth what's actually going on.

EDIT-I've found the formula but man oh man am I in over my head. I'd really appreciate some help

BD=Bullet diameter (.224inches)
SH=sight height (2.5inches)
LD=distance to loophole (120inches)
LH=Loophole Height (3inches
NZ=near zero (50yards)
FZ=far zero (200 yards)
HA=Holdover angle (5.3 degree)

H0=BD/2
H1=SH(NZ-LD)/NZ for LD<=NZ
H2=-LD*Tan(HA)~=-LD*Sin(HA)
LH=H0+the greater of (H1+H2orH0)~=H0+H1+H2~=H1+H2

H1 is the simple triangular interpolation of the sight radius onto the Loophole (we can consider the trajectory linear at such short distances). H2 is an amount you can shrink the loophole at the top for holdover (you don't need the whole scope field of view to see) or the amount you have to add to the keyhole height at the top to be able to see the target in a holdunder situation. H1 is 0 at the near zero and far zeroCan someone spell this out for me? I guess I'm going to have to go to the local library because a lot of this is over my head. I will certainly be glad once I start college classes.

Last edited:
To put people at ease

Look its a simple technique if you want to just get the raw jist of it. If it makes you feel better I'm not asking you to make anything public that's not available in classes or from ballistics software. I'll even show you. http://longshotsoftware.com/eshop.php?view=productListPage&category=3 pc software to calculate it (I have a version similar to this) and here http://www.horusvision.com/pda.php either one of these 3 will do the math for you and much much more.

I have the formula I just need help understanding it.

I would start by putting all that data onto a drawing.

It's quite possible you know more about the problem than we do. Many people here aren't shooters but will understand the ballistics side. If you have to take into account things like the field of view of telescopic sights it sounds like it's going to be pretty complicated.

## What is the ballistic trajectory of a projectile?

The ballistic trajectory of a projectile refers to the path that a projectile takes when it is launched into the air and affected by gravity. It is a curved path that can be described by mathematical equations.

## How is the ballistic trajectory of a projectile affected by distance?

The ballistic trajectory of a projectile is affected by distance in terms of its shape and the amount of time it takes to reach its target. As the distance increases, the trajectory becomes longer and flatter, and the projectile takes longer to reach its target.

## What factors influence the ballistic trajectory of a projectile?

The ballistic trajectory of a projectile is influenced by several factors, including the angle of launch, initial velocity, air resistance, and gravity. These factors can be manipulated to achieve a desired trajectory for a given projectile.

## What is the "ballistic loophole" and how does it relate to the trajectory of a projectile?

The "ballistic loophole" is a term used to describe the method of achieving a flatter, more efficient trajectory for a projectile by exploiting the laws of physics. This can involve adjusting the angle of launch, speed, and other factors to achieve a desired trajectory with minimal drag and air resistance.

## What are some applications of understanding ballistic trajectories of projectiles?

Understanding the ballistic trajectories of projectiles has many practical applications, such as in sports like golf, baseball, and archery where precision and distance are important. It is also crucial in military and law enforcement operations, where knowledge of projectile trajectories can help in aiming and predicting the movement of a target. Additionally, understanding ballistic trajectories is important in the design and testing of weapons and other objects that are launched into the air.

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