Angle Shooting Simplified..

[Jim K]

Deleted

[JLA]

deleted?

[JLA]

"The cosine has no connection with gravity. All your formula gives you is a number, which may or may not relate to the holdover distance, the cubic centimeters of my car's engine, or the length of the king's foot. Or something."

Ill try to explain this more clearly..

When you shoot over an angle, up or down hill is irrelevant, you have a line of sight to the target.. Laser that range and you have a number, that number is the Hypotenuse of the triangle.. Multiply that number by the cosine of the angle your are shooting over, again up or down hill is irrelevant, and it will give you the length of the adjacent, or in this case the distance the bullet will be affected by gravity.. plug that distance into your range calculations instead of your 'hypotenuse' range and your bullets will hit the target instead of flying over it. This is about as simple as I can explain it. It makes absolute perfect sense in my head.

In response to the quote.. The cosine of the angle youre shooting over, when multiplied by the ranged line of sight gives you the range you should calculate your trajectory for, in other words IT IS THE HOLDOVER DISTANCE. simple as that. It has absolutely nothingh to do with the size of an engine or the length of a kings foot. That is just silly..

[JLA]

Forgot one more thing...

No the cosine does not have any connection with gravity whatsoever. Not once in any of my posts in this thread does it say that it does. I have proofread ALL of them and corrected mistakes (save for some spelling) and re-worded things to make them more easily grasped because I know sometimes, something will make absolutely perfect sense in my head but when I try to explain it nobody understands. Thats been the story of my life my entire life.

Again, to re-itterate, the cosine of the angle you are shooting over, when multiplied by the measured line of sight to the target will give you the holdover distance correct for calculating the proper ballistics to achieve a hit on target and nothing more. Its just mathematics, I wont call it simple math, but then again i have looked at some advanced formulas and equations and said YIKES! WTF, wheres the door...

[Jim K]

(Goofed on the delete and wanted to keep the order right so I just left it there, sorry.)

Now, you write

"it doesnt. it simply defines the distance the bullet is affected by gravity... multiplying the line of sight by the cosine of the angle only gives you the hold over distance."

How does the cosine incorporate gravity? As for "the distance the bullet is affected by gravity", when, pray tell, is the bullet NOT affected by gravity between the end of the barrel and the time it strikes something? The figure you arrive at is a number; it could equal the holdover distance, or the number of CC's in my car's engine, or the number of beans in a can, or something.

You give the whole thing away by saying that, "the trajectory of a given round calculation is still made using the ballistics of the cartridge you are shooting." Bullet drop figures are included in most balistic tables and can be calculated by some very complex formulae, but not the way you have tried to do it.

Your calculation produces nothing.

Jim

[Country101]

EEEKK! MATH!

[JLA]

Youre not understanding Jim..

The cosine of the angle does not incorporate gravity. the distance youre shooting over incorporates gravity. The cosine is used to figure the distance the bullet is affected by gravity and nothing more, because when you shoot over an angle, the line of sight distance is longer than the distance the bullet is affected by gravity. Which is why you aim low when shooting at something up or down hill instead of holding center like you normally would if you were shooting a level line of sight.

Here I just spent the last 15 minutes making a drawing in paint so maybe I can help you to understand..

[JLA]

Refer to the drawing above...

The line of sight is the distance between the hunter and the deer, figure c (Hypotenuse of triangle) which is 300 yds...

The distance the bullet is affected by gravity is the adjacent, figure b...

You calculate this figure using what you know.. The lasered distance to the target (figure c) and the angle your shooting over (either estimated or measured with a shooting square) to arrive at the angles cosine using the chart in the link in my first post.

Figure b, the distance the bullet is affected by gravity would be unchanged if the deer and hunter swapped places. the bullet must travel that distance no matter the direction it shot in (up or down hill) that is why this calculation is relevant to shoting over angles.

The average MV of modern cartridges is roughly 3000 fps. A bullet is going to travel 300 yds (900 feet) in 1/3 of a second which is the same time it takes the bullet to travel the 245.73 yds of the adjacent at the angle you shoot it at...

I really cant make it any plainer. I really hope this helps you understand..

[H-D]

Finally!!!! a picture thanks the stick man and deer helped too .I finally got what you were saying, the bullet travels the full 300 yards but since elevation has no difference on the gravity , the bullet drops the same amount (due to gravity) that it would in only 243.75 yds because that is the lateral distance from you to the target..whew

[JLA]

EXACTLY. stickman always works..

I wish theyd put stick figure drawings in all advanced math books..

[bluesea112]

I thought these calculations became a mute point with the invention of the laser range finder.

[JLA]

No, the laser range finder provides a quick, simple and accurate means of measuring the line of sight to the target.

[JLA]

I was thinking about this thread today while driving back from the deer lease and it hit me. People that would use this kind of math to calculate proper holdover would be bowhunters in a treestand.

We all know a bow and arrow has a rainbow trajectory over a relatively short distance, and a miscalculation of as little as 5 yards can result in a miss. Sometimes a big miss if the range is longer than average. A bowhunter 30 feet or so up in a treestand with a deer in his sights will need to know precise holdover so he doesnt watch his arrow graze the top of the deer. This formula of multiplying the line of sight by the cosine of the angle to determine correct holdover would help bowhunters most I think.

[Jim K]

Nice drawing. But the flight path of the bullet is always GREATER than the line of sight to the point of impact, because the bullet travels in a parabolic curve, no matter whether one is firing on the level, or up hill, or down hill. Since the bullet is "affected by gravity" from the time it leaves the muzzle until it strikes something that stops it, it is "affected by gravity" for its entire flight path.

Jim

[JLA]

cant disagree with any of that Jim. It is all correct. But the math posted above for the last 2 pages is still accurate and still applies to shooting firearms, bows, slingshots, etc. etc. etc over any angle.

Just because you dont understand why something works or how something works doesnt mean it doesnt work. Perhaps it would be easier for you to think of it as an unexplainable shooting phenomenon and leave it at that. Because I really cant simplify what im trying to explain anymore than I already have.

Just trust that if you ever find yourself shooting downhill at a 300 yd target over a 35* angle, I promise you if you shoot at the target like its 246 yds away instead of 300 your bullet will hit it where the crosshairs lay. Shoot for the full 300 yd distance and your bullet will fly high.

[Jim K]

I understand better where you are coming from, calculating b from c cos (35 degrees). That works fine and does give you the length of line b (the calculation assumes the angle ba is a right angle). But your basic falacy is in believing that the bullet is "affected by gravity" only for the horizontal distance b. That is simply not true, depending in part on the velocity and caliber. The bullet is affected by gravity (falling) at every point on its path between the muzzle of the hunter's rifle and the deer. And that path is a parabolic curve, which means it is longer than either b or c. To see this, draw a curve upward from the line of sight (a plain arc will do for illustrative purposes). Now measure ALONG THAT CURVE. That is the length of the bullet's path and the distance (and time) for which the bullet is "affected by gravity." And your calculations based on straight lines go out the window.

Now let me be clear. I am not saying there is no need to aim high/low in cases where there is a vertical distance between shooter and target. I think if you draw a true parabolic curve on a strip of paper, pin it at the shooter and swing it around, you will be able to see what happens, not exactly, but close enough.

What I am saying that your method of calculating the amount of holdover/holdunder is not valid. Can it be an approximation? That depends on the length of the bullet's path, which depends on its trajectory - a .45-70 will have a longer path (sharper curve) than a .30-'06. And of course its velocity, which you consider a constant, will be dropping all the time. All those things have been calculated by smarter folks than I, but your method is simply not valid.

Jim

[sting75ray]

I am just glad it is flat out here in western Ok and don't have to worry about all of the angles while shooting my 45/70 downrange. Great info though. I remember reading something along those lines in a Shooting Times magazine a few years back.

[goofy]

Jim K sory The Numbers that JLA is saying IS corect I spent 4 hours with a cornell profesor last night talking about it.The bullet is efected by gravety the second it leaves the barrel WE MAKE IT ARCH if we don't it will hit the ground quick. So we adjust our scopes so it will not.The round arches because we are shooting it up in the air and setting our scopes so that it hits the target.When a projectile is lunched it immediately starts to drop WE create the arch so it can go farther.And gravity effects it differantly shooting up and down.When shooting up gravity pulls it back(slowing it down)and when shooting down it slows down slower(keeping a more constent speed).The arches that are created are different.The arch when shooting up is greater then shooting down.This is due to the diference of the track due to the effects of gravity. Gravity remains constent at the same angle wether you are shooting up,down,leavel. It pulls straight to the center of the earth.Now there are ALOT of variables weight of the projectile,speed,humidity,wind and angle.But the numbers that JLA gave are corect(simplifyed) but correct.And this goes for ANY projectile bullet,arrow,rocket,bottle,stone it does not matter.

[JLA]

And to add, once you figure the holdover and adjust your scope for THAT range, you have just set the arc required for the round you are shooting at the holdover distance..

Bullets travel too fast to take into account the distance being up or down hill, they only meet atmospheric resistence and gravity, one of which is constant and the other of which relates directly to the bullets Ballsitic co-efficient of the bullet.

[JLA]

Originally Posted by Jim K

I understand better where you are coming from, calculating b from c cos (35 degrees). That works fine and does give you the length of line b (the calculation assumes the angle ba is a right angle). But your basic falacy is in believing that the bullet is "affected by gravity" only for the horizontal distance b. That is simply not true, depending in part on the velocity and caliber. The bullet is affected by gravity (falling) at every point on its path between the muzzle of the hunter's rifle and the deer. And that path is a parabolic curve, which means it is longer than either b or c. To see this, draw a curve upward from the line of sight (a plain arc will do for illustrative purposes). Now measure ALONG THAT CURVE. That is the length of the bullet's path and the distance (and time) for which the bullet is "affected by gravity." And your calculations based on straight lines go out the window.

Now let me be clear. I am not saying there is no need to aim high/low in cases where there is a vertical distance between shooter and target. I think if you draw a true parabolic curve on a strip of paper, pin it at the shooter and swing it around, you will be able to see what happens, not exactly, but close enough.

What I am saying that your method of calculating the amount of holdover/holdunder is not valid. Can it be an approximation? That depends on the length of the bullet's path, which depends on its trajectory - a .45-70 will have a longer path (sharper curve) than a .30-'06. And of course its velocity, which you consider a constant, will be dropping all the time. All those things have been calculated by smarter folks than I, but your method is simply not valid.

Jim

IT ALWAYS WILL BE A RIGHT ANGLE. It can never not be.

[Jim K]

"Bullets travel too fast to take into account the distance being up or down hill..."

That is not quite true either, but I refuse to get in any deeper. And pardon me, but didn't you just negate your whole argument? You claimed to have found a precise number that will give a holdover distance, apparently without regard to velocity, bullet shape, air resistance or any of those other nasty little details. Now you say that they do count, which means your precise figure is still invalid.

Hi, Goofy,

Most of that is perfectly correct. But how does it jibe with JLA's contention that gravity only affects the bullet while it is travelling the distance of line b and not at any other point in its own (curved) path?

Jim

P.S. With that, I give up. That darned pig just won't sing.

JK

[goofy]

Hi Jim K Well on your last post about your question. I if you are saying that gravity efects the bullet in the whole arch of it's flight you are right. And if JLA is saying that gravity only efects the bullet for the distance of B he is also right.You both are saying the same thing.The bullet may travel a total distance of say 300 feet in a arch up and away(I did not write down JLA numbers) But only travels 200 feet across the earth then the bullet is effected by gravity for all of it's 300 foot arch(up and away) with in the 200 lateral feet traveled.If you were to draw stright lines up from line B that would represent gravity even thow the bullet travels 300 feet(and is effected by gravity the whole time) up and away. It is only traveling 200 lateral feet across the earth.(effected by gravity the whole time)So you are both right about that.

[cutter]

That Info is very useful!! Thanks very much! It made sense to me, it give the true shooting distance and thats all. You need to know what the bullet drop is for your specific rd. Thanks again.

[JLA]

Originally Posted by cutter

That Info is very useful!! Thanks very much! It made sense to me, it give the true shooting distance and thats all. You need to know what the bullet drop is for your specific rd. Thanks again.

Excellent. Someone else that perfectly understands how to use this math..

[wyoredot]

I'm heading out to the fabled land 'o plenty dumdogs. My trusty .257 Rbts and I intend to do crowd control. I will apply these things I have lernt when I try the 250 and 300 yd shots. I think it will be a blast. For me anyway.