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Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Jordan Smith] #14474471 01/16/20
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Originally Posted by Jordan Smith

Frame of reference matters, so to clarify, the angle between the bullet trajectory and the earth upon exiting the muzzle does not increase, and starts to decrease the moment the bullet leaves the gun. .


I never stated that the angle between bullet trajectory and the earth upon exiting the muzzle increased, and agree that in almost every case it starts to decrease the moment the bullet leaves the gun.

As you said, frame of reference matters - Your average Joe describing ballistics isn't visualizing a line of bore - they're using the only reference they have, line of sight.

Originally Posted by Jordan Smith

The projectile does not counteract the force of gravity, and begins falling the instant it is no longer being supported by the barrel.


A projectile launched at a horizontal target with an upward trajectory from the line of bore relative to the line of sight absolutely does NOT begin falling the instant it leaves the bore. It begins falling the instant it leaves barrel relative to the line of bore (again frame of reference). But relative to the only things that you can actually see when squeeze the trigger - the line of sight and terra firma, the bullet does indeed rise for a brief period of time.

Being subject to the force of gravity does not equal falling. I am subject to the force of gravity right now - I'm not falling. When I get in my plane next week for my scheduled trip, I will be subject to the force of gravity, but I will in fact rise as I begin that journey (for a significant period of time I hope!) Because the bullet is no longer accelerating when it leaves the muzzle, the upward force counteracting gravity is quickly overwhelmed, and the force of gravity accelerates it's return to earth. But the energy of the bullet imparted to an upward trajectory does in fact counteract the force of gravity for a brief period of time.

Pretend I've got a target paper (no backstop) placed at 1000 yards exactly 4 feet off the ground. I fire a rifle perfectly sighted in to be dead on at 1000 yards from that rifle that has the end of the muzzle precisely 4 feet off the ground while simultaneously dropping a bullet from exactly 4' in elevation immediately next to the muzzle - which bullet hits the ground first?

Please don't tell me that you believe they will hit at the exact same time, because they won't. The bullet dropped from the hand will hit significantly sooner than the bullet fired at the 1000 yard target. The difference in impact times can only be because a force (a portion of the bullet's initial velocity) was utilized to counteract the force of gravity (briefly).


If it helps you or anyone else purely to describe ballistics from the reference point of line of bore, I have no issue with it. To jump up and down and tell someone that bullets don't rise when people stating such are clearly referencing light of sight is silly and disingenuous.

David

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Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Canazes9] #14474658 01/16/20
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Originally Posted by Canazes9

A projectile launched at a horizontal target with an upward trajectory from the line of bore relative to the line of sight absolutely does NOT begin falling the instant it leaves the bore. It begins falling the instant it leaves barrel relative to the line of bore (again frame of reference). But relative to the only things that you can actually see when squeeze the trigger - the line of sight and terra firma, the bullet does indeed rise for a brief period of time.

No problem here. I should have said "accelerating downward". The statement I made is true in the case of a horizontal barrel.

Originally Posted by Canazes9

Being subject to the force of gravity does not equal falling. I am subject to the force of gravity right now - I'm not falling. When I get in my plane next week for my scheduled trip, I will be subject to the force of gravity, but I will in fact rise as I begin that journey (for a significant period of time I hope!) Because the bullet is no longer accelerating when it leaves the muzzle, the upward force counteracting gravity is quickly overwhelmed, and the force of gravity accelerates it's return to earth. But the energy of the bullet imparted to an upward trajectory does in fact counteract the force of gravity for a brief period of time.

Pretend I've got a target paper (no backstop) placed at 1000 yards exactly 4 feet off the ground. I fire a rifle perfectly sighted in to be dead on at 1000 yards from that rifle that has the end of the muzzle precisely 4 feet off the ground while simultaneously dropping a bullet from exactly 4' in elevation immediately next to the muzzle - which bullet hits the ground first?

Please don't tell me that you believe they will hit at the exact same time, because they won't. The bullet dropped from the hand will hit significantly sooner than the bullet fired at the 1000 yard target. The difference in impact times can only be because a force (a portion of the bullet's initial velocity) was utilized to counteract the force of gravity (briefly).

David,

With all due respect, that's not how physics works. The direction of acceleration is a function of the net force on an object, not just a single, specific force. I specifically said that the bullet begins falling once it is no longer supported by the barrel because at that point gravity is the only force acting on the bullet. The reason that you are not falling right now is that gravity is not the only force acting on your body. The normal force from the chair or the floor is exactly equal and opposite the force of gravity. But I can see why the word "falling" causes confusion. As you said, the bullet does begin falling (ie, it has a downward velocity vector) from the axis of the bore the moment it leaves the barrel, but not necessarily relative to the axis of the LOS. Perhaps a more precise statement would be that the bullet begins accelerating downward towards the earth the moment the force of gravity is the only force acting on it along the vertical axis (aka when it leaves the muzzle). This is true relative to both the LOS and the axis of the bore.

When the bullet leaves the muzzle, there is no upward force at play (assuming no aerodynamic jump, etc) to counteract gravity. There is also no energy imparted to an upward trajectory. I think what you are trying to describe is the difference between velocity and acceleration. Assuming a positive angle between the axis of the bore and the surface of the earth, as you've described, when the bullet leaves the muzzle it has an initial velocity. We can break this velocity into two vector components, a vertical component and a horizontal component. We are only concerned with the vertical component in this discussion. The bullet has an initial velocity with a positive/upward vertical component. The net force on the bullet is simply the force of gravity, which results in an immediate downward acceleration as soon as the bullet leave the muzzle. This downward acceleration immediately begins to decrease the magnitude of the upward velocity, and eventually the magnitude of the vertical velocity vector reaches zero and then becomes a downward velocity vector. This is the exact same principle as if you were to throw a ball straight up in the air, and then catch it again. The ball starts with an upward velocity, but begins accelerating downward the moment it leaves your hand. It eventually has a velocity magnitude of zero (the turning point) and starts to fall back down toward your hand.

In your example above, the bullet fired at the 1000 yard target hits the ground after the bullet dropped from the hand, not because the bullet fired from the rifle has some force acting on it as a part of its initial velocity, but because it is given an initial velocity with a positive vertical component, while the bullet dropped from the hand is not. It's just like comparing the drop times of a ball that you throw straight up and allow to hit the ground, versus one that you drop straight down from your hand and allow to hit the floor. It's obvious that the ball thrown upward will hit the ground after the ball dropped from your hand directly.

Originally Posted by Canazes9

If it helps you or anyone else purely to describe ballistics from the reference point of line of bore, I have no issue with it. To jump up and down and tell someone that bullets don't rise when people stating such are clearly referencing light of sight is silly and disingenuous.

David


I can readily describe the ballistics of a bullet from whichever reference frame you like, so it doesn't matter to me, but the reason that guys are quick to correct the statement that "bullets rise when they leave the bore" is that many people are confused by such statements, and believe that the bullet rises relative to the axis of the bore after leaving the muzzle. This is a myth that has been perpetuated for decades. Clearly you are not under that misconception, but many are.

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Jordan Smith] #14474803 01/16/20
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Originally Posted by Jordan Smith

David,

With all due respect, that's not how physics works. The direction of acceleration is a function of the net force on an object, not just a single, specific force. I specifically said that the bullet begins falling once it is no longer supported by the barrel because at that point gravity is the only force acting on the bullet. The reason that you are not falling right now is that gravity is not the only force acting on your body. The normal force from the chair or the floor is exactly equal and opposite the force of gravity. But I can see why the word "falling" causes confusion. As you said, the bullet does begin falling (ie, it has a downward velocity vector) from the axis of the bore the moment it leaves the barrel, but not necessarily relative to the axis of the LOS. Perhaps a more precise statement would be that the bullet begins accelerating downward towards the earth the moment the force of gravity is the only force acting on it along the vertical axis (aka when it leaves the muzzle). This is true relative to both the LOS and the axis of the bore.

.


Jordan,

With all due respect, that is precisely how physics works. I am not confused by the word falling - to descend freely by the force of gravity.

If a body is not descending it is not falling (it is still subject to the force of gravity).

Your statement that the bullet is falling because it is no longer supported by the barrel and gravity is the only force working on it is incorrect - it's not falling if it's not descending.

I agree, a more precise statement would be that the bullet begins accelerating downward towards the earth the moment the force of gravity is the only force acting on it. The bullet is still rising in that instant, not falling.

When I have discussed ballistics with fellow hunters I haven't experienced any that believe there is some mystery force making the bullet rise - I have experienced quite a few that are annoyed with people attempting to condescendingly tell them bullets don't rise (I'm not referencing you) when they clearly do.

David

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14474831 01/16/20
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Damn...you two gotta stop.

Headache!

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14474902 01/16/20
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Confuscious once said "only a rising bullet has been fired upwards".

Last edited by Musicianized; 01/16/20.

You can lead a horse...
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Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Canazes9] #14474907 01/16/20
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Originally Posted by Canazes9
Originally Posted by Jordan Smith

David,

With all due respect, that's not how physics works. The direction of acceleration is a function of the net force on an object, not just a single, specific force. I specifically said that the bullet begins falling once it is no longer supported by the barrel because at that point gravity is the only force acting on the bullet. The reason that you are not falling right now is that gravity is not the only force acting on your body. The normal force from the chair or the floor is exactly equal and opposite the force of gravity. But I can see why the word "falling" causes confusion. As you said, the bullet does begin falling (ie, it has a downward velocity vector) from the axis of the bore the moment it leaves the barrel, but not necessarily relative to the axis of the LOS. Perhaps a more precise statement would be that the bullet begins accelerating downward towards the earth the moment the force of gravity is the only force acting on it along the vertical axis (aka when it leaves the muzzle). This is true relative to both the LOS and the axis of the bore.

.


Jordan,

With all due respect, that is precisely how physics works.


I'm sorry, but it's not. There is no force counteracting gravity when the bullet leaves the muzzle (again, neglecting more advanced aerodynamic factors like aerodynamic jump, etc). There is no energy imparted in the upward trajectory that counteracts gravity. Energy is a scalar quantity without direction, and it certainly can't be added to forces to obtain a net effect using vector addition. The results you're describing are correct, but the physical concepts you're using are not.

Originally Posted by Canazes9

I am not confused by the word falling - to descend freely by the force of gravity.

If a body is not descending it is not falling (it is still subject to the force of gravity).

Your statement that the bullet is falling because it is no longer supported by the barrel and gravity is the only force working on it is incorrect - it's not falling if it's not descending.


As I said, my statement about the bullet falling the moment it leaves the barrel is true in the case of a horizontal barrel; a specific case. I used that case to illustrate the point that the bullet does not somehow rise relative to the bore because it is going so fast that is skips up on the air like a water-ski skipping up out of the water when the boat starts pulling (I have come across this misunderstanding multiple times, typically with older hunters). The statement about the bullet accelerating downward applies to the general case, regardless of which direction the barrel is pointing.

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: joshf303] #14474922 01/16/20
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Originally Posted by joshf303
Damn...you two gotta stop.

Headache!

Sorry, buddy. I'm a physicist. This is what I do all day. grin

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Jordan Smith] #14475002 01/16/20
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Originally Posted by Jordan Smith

I'm sorry, but it's not. There is no force counteracting gravity when the bullet leaves the muzzle (again, neglecting more advanced aerodynamic factors like aerodynamic jump, etc). There is no energy imparted in the upward trajectory that counteracts gravity. Energy is a scalar quantity without direction, and it certainly can't be added to forces to obtain a net effect using vector addition. The results you're describing are correct, but the physical concepts you're using are not.



Jordan

I had to go back through my earlier statements and realized I did state the bullet has an upward force, which is incorrect. The bullet has momentum, which is a vectored quantity possessing both mass and direction.

With regards to your statement about accelerating downward applying to the general case, regardless of which the muzzle is pointed, I agree. I was referring to my example.

David

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14475015 01/16/20
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Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14475026 01/16/20
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Does this mean you guys are going to have each others babies now?

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Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: huntsman22] #14475029 01/16/20
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Originally Posted by huntsman22
Does this mean you guys are going to have each others babies now?


I'm too old for Jordan.

David

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14475081 01/16/20
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The bullet rising above the bore upon firing reminds me of a buddy of mine that tried to tell me his arrow is faster 20 yards downrange than it is right out of the bow due to archers paradox... lol

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: huntsman22] #14475111 01/16/20
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Originally Posted by huntsman22
Does this mean you guys are going to have each others babies now?

I've already got all I can handle. grin

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: huntsman22] #14475132 01/16/20
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Originally Posted by huntsman22
Does this mean you guys are going to have each others babies now?


That's gross.



A wise man is frequently humbled.

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14475320 01/17/20
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Jordan,


You are a patient individual. And quite often helpful, just as Form. For a while there I didn't know if there was a prayer of this ending well. I am not a genius, however I do engage my brain before my mouth. And don't suffer fools very well.




Take care, Willie


Cry to the heavens and let slip the dogs of war. For they must feed on the bones of tyranny. In order for men to have freedom and liberty.
Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: wdenike] #14475617 01/17/20
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Originally Posted by wdenike
Jordan,


You are a patient individual. And quite often helpful, just as Form. For a while there I didn't know if there was a prayer of this ending well. I am not a genius, however I do engage my brain before my mouth. And don't suffer fools very well.




Take care, Willie













You think Jordan said you were right?

Bless your heart.


David

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14475634 01/17/20
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Did y'all just become best friends?.........

"So much room for activities"

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Formidilosus] #14475847 01/17/20
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I think there's a good lesson here. Between Jordan and David's discussion, I have a little better understanding of how to explain trajectory to someone who doesn't understand what their bullet is doing when it leaves the muzzle.

When someone uses the word "rise", they may not have in mind what I do. I'll better most people understand the curve of the trajectory better than they do the relationship between barrel and scope.

But then again, there's a few guys I know who have 7mags that shoot "flat out to 500 yards", grin.

Re: How to use that new SWFA scope- reticle, turrets and proper usage. [Re: Canazes9] #14476505 01/17/20
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Originally Posted by Canazes9
Originally Posted by Jordan Smith

I'm sorry, but it's not. There is no force counteracting gravity when the bullet leaves the muzzle (again, neglecting more advanced aerodynamic factors like aerodynamic jump, etc). There is no energy imparted in the upward trajectory that counteracts gravity. Energy is a scalar quantity without direction, and it certainly can't be added to forces to obtain a net effect using vector addition. The results you're describing are correct, but the physical concepts you're using are not.



Jordan

I had to go back through my earlier statements and realized I did state the bullet has an upward force, which is incorrect. The bullet has momentum, which is a vectored quantity possessing both mass and direction.

With regards to your statement about accelerating downward applying to the general case, regardless of which the muzzle is pointed, I agree. I was referring to my example.

David

One more thing to clarify, just so nobody is confused, a vector quantity has both magnitude and direction, not mass and direction. Momentum is a function of mass and velocity, but that functional relationship is not necessarily related to momentum being a vector quantity.

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