Down hill shot question

[308Nut]

Who is Chuck Norris?

[308Nut]

This of course assumes that the bullet leaves the gun 40 degrees below the horizontal. If the target was 40 degrees below the horizontal the bullet would actually leave the gun at slightly less than 40 degrees since hunting rifles are sighted for a 100 or 200 yd zero.

 

In theory, this would be negated by the fact that the trajectory is parabolic and would arrive at the target at the same angle it left the rifle. Or 180 degrees depending on how you choose to look at it.

[rclouse79]

In theory, this would be negated by the fact that the trajectory is parabolic and would arrive at the target at the same angle it left the rifle. Or 180 degrees depending on how you choose to look at it.

 

I may once again have to beg to differ. Saying the bullet angle does not change (“arrive at the target at the same angle”) is another way of saying the bullet travels in a straight line path which is not true. Below I set up a theoretical situation to illustrate that.

 

Lets say you shoot at a paper target 100 yds away on flat ground at the same height as the barrel and hit the bullseye. Lets just say the initial angle of the bullet was 1 degree above the horizontal. The angle of the bullet would be 1 degree below the horizontal when it hits the target at 100 yds (not the same). Now imagine the first target was on the beginning of a hill that perfectly matched the trajectory of the bullet. A target is placed every 100 yds from 100 yds to infinity. The same bullet continues to punch a hole through the center of each target every 100 yds. Imagine that the hill perfectly matching the bullets path is infinitely long. Eventually the horizontal component of the bullets velocity would be next to zero because it has been decelerating due to air resistance. By this time the bullet would have reached its terminal velocity in the vertical direction (the weight of the bullet is equal and opposite to the air resistance resulting in zero net force and zero acceleration in the vertical direction. a.k.a. maximum speed.) At this point the bullet would be traveling close to 90 degrees below the horizontal still punching a hole through every target every 100 yds. The angle of the bullets travel is not equal to its initial angle, but it still hits the target every 100 yds. Along the way the bullet hit every single angle from 1 degree above the horizontal to just about 90 degrees below the horizontal.

[308Nut]

In theory, this would be negated by the fact that the trajectory is parabolic and would arrive at the target at the same angle it left the rifle. Or 180 degrees depending on how you choose to look at it.

 

I may once again have to beg to differ. Saying the bullet angle does not change (“arrive at the target at the same angle”) is another way of saying the bullet travels in a straight line path which is not true. Below I set up a theoretical situation to illustrate that.

 

Lets say you shoot at a paper target 100 yds away on flat ground at the same height as the barrel and hit the bullseye. Lets just say the initial angle of the bullet was 1 degree above the horizontal. The angle of the bullet would be 1 degree below the horizontal when it hits the target at 100 yds (not the same). Now imagine the first target was on the beginning of a hill that perfectly matched the trajectory of the bullet. A target is placed every 100 yds from 100 yds to infinity. The same bullet continues to punch a hole through the center of each target every 100 yds. Imagine that the hill perfectly matching the bullets path is infinitely long. Eventually the horizontal component of the bullets velocity would be next to zero because it has been decelerating due to air resistance. By this time the bullet would have reached its terminal velocity in the vertical direction (the weight of the bullet is equal and opposite to the air resistance resulting in zero net force and zero acceleration in the vertical direction. a.k.a. maximum speed.) At this point the bullet would be traveling close to 90 degrees below the horizontal still punching a hole through every target every 100 yds. The angle of the bullets travel is not equal to its initial angle, but it still hits the target every 100 yds. Along the way the bullet hit every single angle from 1 degree above the horizontal to just about 90 degrees below the horizontal.

 

No worries. I understand you have a lot of practical shooting knowledge. I understand the part of your post I discussed was not relevant to most of the readers. Most people were more interested only in the information on how to compensate for different shots which you provided.

 

Now for the fun part!

 

If I knew the velocity of the gun I would multiply by the cos(40) to find the horizontal component of the velocity, and by the sin(40) to find the initial negative (downward) vertical velocity of the gun.

Since there is no air resistance there are no horizontal forces which means the horizontal component of the bullet will be constant until it hits the target. I would use the equation t=d/v where d is the distance to the target and v is the horizontal component of velocity. This will tell you time it takes for the bullet to reach the target.

During the time it takes the bullet to reach the target it is accelerating downward at (negative) 9.81 m/ss. Its downward displacement would be given according to the equation d = 1/2gtt + vt where g is acceleration due to gravity, t is the time to reach the target and v is the initial vertical component of the velocity.

 

After you know how far it will drop you could compare that to the drop distance of your scope.

 

This of course assumes that the bullet leaves the gun 40 degrees below the horizontal. If the target was 40 degrees below the horizontal the bullet would actually leave the gun at slightly less than 40 degrees since hunting rifles are sighted for a 100 or 200 yd zero.

 

In the post I was refering to you said 'Since there is no air resistance' (highlighted in bold) and now you shoot me down by using an example involving air resistance by saying 'due to air resistance' (highlighted in bold). So.....Which is it?

 

 

The theory that I posed was based on your earlier previous post where you used a theoretical senario using NO air resistance. Without that, the trajectory would be a pure parabolic. In your last post you use air resistance. That said, in a vacume, on level ground the end of the trajectory will be at the same angle as the begining only in reverse. In our enviornment (air resistance), the angle wil be much different. In THAT context, you are right.

[rclouse79]

Same thing would end up happening without air resistance at a different rate. You could use the same analogy but the shape of the hill would change because the trajectory would change. The horizontal velocity would stay constant but the vertical velocity would increase by 32 ft/s every second. The instantaneous angle could be found by taking the inverse tangent of the vertical velocity over the horizontal velocity. As the vertical velocity approaches infinity the angle approaches 90 degrees below the horizontal.

 

After 100 seconds the vertical velocity would be about 3200 ft/s. If the bullet were shot out of the gun at 3200 ft/s it would be traveling at 45 degrees below the horizontal at that moment.

after 1000 seconds the vertical velocity would be around 32000 ft/s and the horizontal velocity would still be 3200 ft/s. Inverse tangent of 32000/3200 is about 84 degrees below the horizontal.

 

I could insert the word roughly in front of the 1 degree below the horizontal in my last post. It would be pretty close.

[Broken Wheels]

Wow! This is way too heavy for me and a definite Makers Mark and Coke discussion around the campfire though I wonder how the old timers used to figure it out "back in the day"? I will leave this one to the experts for sure... Broken Wheels

[Ernesto C]

You know what guys........next time you get a down hill shot..... easy no complications....just call me; I'll do it for you.

 

Ernesto C

[308Nut]

Assuming that we are talking about he same thing.......The fact still remains that a bullets trajectory in a vacume is still a parabolic.

 

In a vacume, the vertical velocity (due to gravity) would decrease from the original velocity to zero on the ascending branch of the trajectory and increase from zero to the original velocity on the descending branch. Zero vertical velocity occurs at the highest point in the trajectory.

 

Fact: The summit or peak of a trajectory in a vacume is half way between the begining and the end.

 

Fact: The angle of the fall in a vacume is equal to the angle of elevation.

 

x = vt cos θ

y = vt sin θ - ½ g t²

 

v is the velocity

 

θ is the angle you launch it at

 

g is the downwards accelaration caused by gravity = 9.81 m/s²

 

t is time in seconds

[lionhunter]

Again thanks for all the input, but you guys are giving me a headache! I (like most others on this sight) am a hunter; not a mathematician. I missed what I consider an easy shot, one that I have made a hundred times, and am probably just looking for an excuse as to why I missed at 380 yards. At any rate, if I ever need to take a computer hunting with me to figure out my shot, I will probably stick to my bow.

Thanks again, great debate..........

Whitey

whoever said get the new range finder with the angle compensation hit it on the head...

 

 

[rclouse79]

In theory, this would be negated by the fact that the trajectory is parabolic and would arrive at the target at the same angle it left the rifle. Or 180 degrees depending on how you choose to look at it.

 

I have found the root of our new disagreement. I am talking about the angle relative to a set line, the horizontal, which does not change. You are talking about the angle above the straight line path from your position to the target. This line will change depending on where your target is located. If that stipulation is added I think your statement would be correct. You would also have to say it would arrive at the same angle below that straight line to the target which is technically not the same angle, just the same magnitude. If that is not added I would still disagree because the bullet is constantly changing direction and will only be going a given direction at one instant in time.

[308Nut]

Just in case we are just misunderstanding eachother, the picture below illustrates what I am getting at. This illustrates a trajectory in a vacume. The summit is 1/2 way between the start and the end and the angle of departure is the same as the angle of arrival.

[rclouse79]

Exactly. You are talking about just connecting the shooter to the target with the straight line and measuring your angles from that straight line. Without a point of reference an angle does not mean much. If I tell you to walk at 32 degrees it is worthless. If I say 32 degrees east of north you could do it.

[Ernesto C]

Honey??.......please bring me another bud light!!

[rclouse79]

THIS IS KILLING ME! I am supposed to be packing to leave for a HUNTING trip tomorrow and I am sitting on my butt doing this. I think I need counseling. It has been fun, thanks for the friendly discussion 308nut. I will check back after my hunt to see if there is anything else for us to ponder.

[Coach]

Not to throw gasoline on the fire or anything...BUT, this guy came on here asking whether to aim low or high at a steep angle, and all kinds of stuff entered into the mix, including the vast, gravity-less, airless expanses of the universe, plus some higher end trigonometry and physics discussion. Not that that's bad - It makes for entertaining reading, but, it's way overkill IMO.

 

We're not talking about the perfect 1.5 mile shot calculating altitude, air density, Coriolis Effect...The guy is making a 358 yard shot at an angle, in field conditions, and wasn't sure whether to hold over or under. This is a very realistic hunting situation, and doesn't take a Ph.D. to give a simple answer to.

 

Lionhunter, you hold *under* at an angle, either up or downhill. You hold for the actual horizontal distance from muzzle to target. You might be a little high or a little low, but if you hold under correctly to compensate for the angle, you'll hit the deer.

 

Not bashing 308Nut or orclouse79 - they have their own discussion going on, and no doubt they know what they are talking about, but I would argue that it really isn't germane in the context of the question you asked.