Angle Firing

[camies]

Hi.

REgarding angle firing, shooting at an angle it is easy to calculate the real range by using Cos(angle) X laser range.
The question I'm looking for an answer to is how does the angle affect the power.
For example
You are shooting at 1 o'clock up a hill. Laser range says 300 yards, actual range = 150 but does that half or double the ft lb delivered?

 

[RS14]

Naively, I would expect it to simply decrease the projectile energy by the amount required to raise the projectile to a given height, which is weight*height. Height should be Sin(angle)x[LaserRange]. So with e.g. a 150 grain bullet (~0.02lbs) , and the above example, you would lose about 17 ft-lbf.

 

[guntech59]

I think the best answer is.....Not enough to be concerned about.

Energy does not kill animals......destruction of vital organs kills them.

In the scenario you describe, the actual length of the shot is about 100 yards (maybe less, I approximated it with two pencils of the same length).

 

[PT1911]

I am having a difficult time fathoming this idea as 300 yards uphill is still 300 yards... not 150... if I walk 2700 feet, no matter how steep the hill, I just walked 300 yards. perhaps laser range finders work on a principle I am not familiar with but from here to there is as far as from here to there regardless of the angle you took to get there.


I guess I dont think in enough dimentions.

I can see that fireing a bullet at a high angle, then at its farthest point it did not go as far (horizontally) as the bullet actually traveled (vertically and horizontally) through the air, but does the bullet really know the difference? I imagine the energy at your 300 yard (100 yard actual) would be the same as that bullets performance at 300 yards as that is the distance it actually traveled or something like that.

 

[Mr_Rogers]

Forget the energy, any change is only theoretical.

Range is always important to allow for bullet drop but your chosen sighting-in range should compensate, to some extent, for this.

At an angle, choosing your point of aim on the target is the most important factor because you want to hit a point inside the target. The point on the outside of the target that you must hit so that the bullet penetrates to the correct location on the inside of the target is critical.

 

[benEzra]

You are shooting at 1 o'clock up a hill. Laser range says 300 yards, actual range = 150 but does that half or double the ft lb delivered?

The range is still 300 yards, and the energy delivered will be essentially the same (within a few ft-lb). The only difference is how much drop you have to allow for. Shooting uphill or downhill reduces the amount of drop slightly, but it does not change the range.

 

[JohnBT]

Well, 300 yards straight up is fighting gravity all the way and 300 yards straight down is going with the flow, so I don't know about all that angle cosine stuff. I agree it doesn't much matter except theoretically. Gravity isn't all that strong over a short distance like 300 yards.


"You are shooting at 1 o'clock up a hill."

Is that Eastern Standard Time or Eastern Daylight Savings?

John

 

[Wheeler44]

The correct answer is simple really............Anything that you currently own is obviously insufficient for the task.....You now must purchase something adequate...

You are welcome

W44

 

[Uncle Mike]

Energy will be like at 300y.
Trajectory will be like at 150y.
Velocity will be like at 300y.

This is of course during daylight savings time....hehehehehe

 

[45crittergitter]

Gravity isn't all that strong over a short distance like 300 yards.

Gravity is pretty much the same anywhere on the Earth's surface. Approximately 32.2 feet per second per second. That's acceleration, not strength.

 

[ucrt]

Here is how I have explained it over the years (But I’ve never done explained it without a sketch?) -
When you sight a rifle in you are supposed to do so on a level plane. If you sight a rifle in at 200-yards, the bullet will (approx) cross the line of sight at 25-yards going upward, then gradually come back down across the line of sight at 200-yards and gradually keep dropping until it strikes the earth.

With a 200-yard sighted-in rifle and you shoot 45 degrees uphill at a 200-yard target, your bullet will only be exposed to (about) 140-yards of gravity. So, the bullet will still be above the line of sight, close to where it would be if you had actually shot at a level target only a 140 yards away.

When the angle changes, the affect of gravity will change accordingly. The steeper the angle, (whether up or down) the less effect gravity has on it. Gravity pulls down perpendicular to a earth's level surface, so, when you shoot uphill or downhill, you are actually exposing the bullet to less gravity.

When I squirrel hunt with a 22, the rodent may be in a very tall pine tree, (let’s say 25-yards tall). The easiest way to calculate drop is to find the base (bottom) of the tree that the squirrel is in, and estimate how far away you are from the base of the tree. If the base of the tree is 50 yards away, even though the squirrel may actually be 65-70 yards away from you, gravity will only have 50 yards to pull the bullet down. So, aim at the squirrel as if he was at the base of the tree.

If you aimed at this squirrel like the 65-70 yards away he actually was, you would have missed. If you were shooting 45 degrees at a Mule Deer that was 300-yards away, as if he was 300 yards away, your bullet would have hit him like he was 210-yards away or depending on the cartridge, possibly 6" high?

Well I hope all of this wasn't too long-winded and makes a little sense.

 

[Mt Shooter]

urct, welcome and about time you post something.

Assuming they are shooting at something larger than a squirrel, say a Deer. This is based on a 30.06. As you stated, 25 yards is the same as shooting 100 yds. at that distance impact should be about an 1 1/2 high. At 200 yds impact should be dead on, at 300 yds, impact should be about 2 inches low of zero. This is on a flat plain with zero wind effect.
So we go from 1 1/2 high to 2 inches low from 25 yards all the way to 300. A vital shot on lets say a deer as an area of roughly 8 inches. Is 1 inch either way really going go make a difference?

 

[MinnMooney]

In your example of shooting at the extreme up-hill angle of 1:00 and the actual distance to your target is 300 yards then the "ballistic distance" or "bullet-drop distance " would be : Sine 30deg. = BDD/300yds OR
0.5=BDD/300yds OR Bullet drop distance is 150 yds as you said.

It would be the same as shooting at an extreme down-hill angle of 5:00.

The difference is that your bullet would slow at a much greater rate going uphill than going downhill therefore the bullet would have much greater energy hitting the downhill target.

 

[hadmanysons]

Well I hope all of this wasn't too long-winded and makes a little sense.

That always did kind of bug me, thanks for clearing that up

 

[ucrt]

urct, welcome and about time you post something.

Assuming they are shooting at something larger than a squirrel, say a Deer. This is based on a 30.06. As you stated, 25 yards is the same as shooting 100 yds. at that distance impact should be about an 1 1/2 high. At 200 yds impact should be dead on, at 300 yds, impact should be about 2 inches low of zero. This is on a flat plain with zero wind effect.
So we go from 1 1/2 high to 2 inches low from 25 yards all the way to 300. A vital shot on lets say a deer as an area of roughly 8 inches. Is 1 inch either way really going go make a difference?

Thanks, I just have been soaking stuff up and not commenting a whole lot.
(Even a fool seems wise if he keeps his mouth shut)

I'm not sure what you meant with "As you stated, 25 yards is the same as shooting 100 yds. at that distance impact should be about an 1 1/2 high." but I'll try to explain some more.

Generally, (and I mean generally) a high powered rifle sighted in at 200 yards will be -1.5" at the muzzle, dead-on at 25 yards, +1.5" at a 100 yards, dead-on at 200 yards, -2.5" at 250 yards and -7" at 300 yards.

Most high-powered rifles have a point-blank trajectory out to about 225 yards. 225 yards and less, gives you about 1-1/2" high/low if you aim dead-on.

If you were taking a vital area shot at 45 degrees at 300 yards and did not compensate for a 45 degree angle but did shoot 7" high to compensate for 300 yard trajectory, your shot would be about 6" high and out of the 8" vital area. A neck shot might even be a miss.

I keep using "45 degree angle" because that is the only angle I know how to figure mathmetically. To calculate the "distance exposed to gravity": take the distance of the 45 degree shot to the target and divide it by 1.41, that number will be the actual trajectory yardage of the bullet. 300/1.41=212.

Take it easy.

 

[Mt Shooter]

If you were taking a vital area shot at 45 degrees at 300 yards and did not compensate for a 45 degree angle but did shoot 7" high to compensate for 300 yard trajectory, your shot would be about 6" high and out of the 8" vital area. A neck shot might even be a miss.

Too much thinking, by the time the calculations are made it will be gone.
How about just moving closer.

 

[GRIZ22]

You are shooting at 1 o'clock up a hill. Laser range says 300 yards, actual range = 150 but does that half or double the ft lb delivered?

You are misguided on bullet energy and range. Go here:

http://www.ipgeneral.com/~renegade/rifle/rifle1.shtml

You will see energy is not cut in half by range.

Any influence of gravity reducing the velocity at the ranges you're talking about is so little it is of no consideration.

 

[45crittergitter]

With a 200-yard sighted-in rifle and you shoot 45 degrees uphill at a 200-yard target, your bullet will only be exposed to (about) 140-yards of gravity. So, the bullet will still be above the line of sight, close to where it would be if you had actually shot at a level target only a 140 yards away.

When the angle changes, the affect of gravity will change accordingly. The steeper the angle, (whether up or down) the less effect gravity has on it. Gravity pulls down perpendicular to a earth's level surface, so, when you shoot uphill or downhill, you are actually exposing the bullet to less gravity.

False. For all practical purposes, if you shoot 200 yards, the time of flight is the same regardless of angle, and therefore the actual drop due to gravity is the same regardless of angle, since the acceleration of gravity (feet per second per second, NOT feet per foot) acts on the bullet over TIME, not DISTANCE. However, since gravity is pulling an angled-shot bullet vertically down, which is not perpendicular to the line of flight of a steep angle shot, what APPEARS to you to be the drop is less.

 

[MT GUNNY]

PT1911 ; you are right If you were to walk the distance. However the Dynamics of a shot from a firearm is different. Yes the bullet still travels the same distance you would have walked, however ballisticly speaking the bullet travels a shorter distance.

Imagine a Tree 20ft in front of you, the tree is 100ft tall. You shoot at the tip of the tree, the bullet actual distance traveled is about 105ft. Ballisticly it Only traveled 20Ft.

 

[Mt Shooter]

dont forget to figure in air density and elevation....

 

[45crittergitter]

however ballisticly speaking the bullet travels a shorter distance.

Imagine a Tree 20ft in front of you, the tree is 100ft tall. You shoot at the tip of the tree, the bullet actual distance traveled is about 105ft. Ballisticly it Only traveled 20Ft.

Still false. See my last post.

 

[kingmt]

PT1911 ; you are right If you were to walk the distance. However the Dynamics of a shot from a firearm is different. Yes the bullet still travels the same distance you would have walked, however ballisticly speaking the bullet travels a shorter distance.

Imagine a Tree 20ft in front of you, the tree is 100ft tall. You shoot at the tip of the tree, the bullet actual distance traveled is about 105ft. Ballisticly it Only traveled 20Ft.

This is the closest thing to the right answer I have seen. Look at it this way: If you shoot 50 yards 2' off the ground then 6' off the ground then both shots will hit where you aim. The bullet doesn't drop because it has to go 4' higher. If the target is up on the hill you don't have the add the angle just aim at it but the most truth comes from this post:

Forget the energy, any change is only theoretical.

At an angle, choosing your point of aim on the target is the most important factor because you want to hit a point inside the target. The point on the outside of the target that you must hit so that the bullet penetrates to the correct location on the inside of the target is critical.

I can't add anything to this because it is said perfectly.

This is the only time I have seen someone else understand this.

 

[NMGonzo]

If you are on a threadmill, will that affect the results?

 

[MinnMooney]

by guntech59 - post #3 :

In the scenario you describe, the actual length of the shot is about 100 yards (maybe less, I approximated it with two pencils of the same length).

Sorry guntech59, your "two pencil" routine leaves something to be desired. Maybe you should brush up on your math before telling someone (who is correct, BTW) that he's wrong.

The 'ballistic distance' is, as the OP says, 150 yards.

 

[MinnMooney]

to : crittergitter :

I'm thinking you'd better get a better understanding of geometry or leave the calculating to those who have gone thru the classes. You are so very wrong in calling the other posters incorrect. They ARE correct!

Still false. See my last post.

Sorry...... you'are wrong. The 100' tree that's 20' away example by is correct in every way.