Do bullets rise?

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ScottsGT

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OK, I'm sure lot's of discussions have been had about this subject. I did a search and found nothing, but here goes again.
When the bullet leaves the end of a barrel, does it rise? I've read about a .30-.30 having a greater arch of rise than say a .308 due to the blunt nose. Or is it just aiming the end of the barrel upwards (to compensate for drop) to hit a target at a given distance giving the perceived effect of rise? I've heard about aerodynamics (shape of the bullet) effecting the ammount of rise, etc....
Now I'm not wanting to get into the shooting uphill or downhill theory, that will be a later discussion! But one of my sighting targets tells you to zero one inch low at 25 yards to be on center at 100 yds. This tells me that the bullet rises.
If there is any published data from bullet mfgrs. or anyone else credible, could you please post the source since my brother is having a heated debate with a coworker of his and I was asked to get the skinny on the truth. :confused:
 
Like Steve said, gravity goes to work immediately. The perceived rise is the shooters line of view crossing the bullet path feet or yards from the muzzle (depending on how high your eye is above the bore) and further downrange. I can picture it but I can't explain it too well.
 
Yep, the "rise" is simply the elevation of the barrel reguired to hit a target at a specific distance. At an intermediate distance the bullet will most likely be slightly above the line of sight to the target. A 30-30 will have a higher intermediate rise due to the lower velocity it has compared to a .308.

Someone on here (or TFL) said he believed a bullet does indeed rise due to some aerodynamic phenomenon, but that has yet to be proven satisfactorily with standard bullets. The bullet is very symmetrical and spinning rapidly. There is nothing to cause the bullet to rise assuming calm air. I suppose some precession could cause an effect of rise, but it would be a cyclical effect around the CoG so that couldn't be called true rise.
 
To be precise, there is no upward force on the bullet once it leaves the barrel. It may start off with some upward velocity however.

Since gravity acts as a force pulling the bullet towards the earth immediately, to have the rifle "zeroed" such that it hits at point-of-aim at anything further than point-blank range one must arrange the sights and barrel such that the barrel is pointing "up" relative to the line-of-sight through the sights. Since the sights are above the bore axis, the bullet starts off below the point-of-aim, passes it at the "near zero", will impact above the point-of-aim for some range, and then will hit at the point-of-aim at the "far zero." From there onward, the point of impact will be below the point of aim.

-z
 
Someone on here (or TFL) said he believed a bullet does indeed rise due to some aerodynamic phenomenon, but that has yet to be proven satisfactorily with standard bullets. The bullet is very symmetrical and spinning rapidly. There is nothing to cause the bullet to rise assuming calm air.
The effect of rise will occur when an object moving through air has a greater surface area on top than on bottom, an airfoil being the ideal cross-section. The air is stretched on top, pressure is reduced and the resultant pressure differential causes the object to rise. Whether or not this occurs on a bullet with a positive angle of attack or whether this effect would be disrupted by the rotation of the bullet, I have no idea, but I would say that it might be possible. I can't imagine that it would be significant though.
 
Rock jock, if the bullet is not stable and is not spinning precisely around it's longitudinal axis, it is possible for it to have an attitude slightly different from a straight on attitude. However, that variation might cause drop just as easily as rise in the aerodynamic sense of those words. As far as I know there is nothing to cause a consistent state of rise, and only rise, in a symmetrical bullet. Variations of all kinds can cause all manner of variations of flight, but we must deal with the theoretical state of a perfect bullet fired from a perfect gun to truly answer the question asked.
 
I won't bet my life on this, but I'm pretty sure that: (1) relative to the barrel itself, or the vector of the barrel, the bullet does not rise, but (2) relative to your sighting system (scope, iron sights, red dots, whatever) the bullet will seem to have risen for the first 99 yards if you've sighted the gun at 100 yards or 200 yards or whatever. With sights, the bullet will "seem" higher at, say, 60 yards than it will at 99 yards, assuming a 100-year sight-in.
 
The bullet only "rises" in relationship to your line of sight because your line of sight is at an angle to the axis of the bore. If your line of sight is perfectly parallel to the bore axis the bullet drop would be more apparent. Under anything aproaching normal conditions the bullet begins to fall away from the axis of the bore after leaving the muzzle.


David
 
I was only half right.

The definitive reference on exterior ballistics is Modern Exterior Ballistics: The Launch and Flight Dynamics of Symmetric Projectiles by Robert L. McCoy.

In the general case of a long spinning projectile flying through the air, there is a lift force if the nose is pointed "above" the instantaneous trajectory. In particular, this force is proportional to the "sine" of the difference in angle, and acts normal to the trajectory. Note that the rotational inertia of the projectile is what keeps it pointed in a constant direction. This is from McCoy pp 34-35.

However, this lift force only has noticable magnitude when the angle is large -- remember it scales with the "sine" of the angle.

For military small arms and sporting arms trajectory calculations, a "flat fire" approximation is used universally. This notes that the angle will be miniscule for the majority of all rifle bullet trajectories. (McCoy pg 88).

In summary: There is a lift effect, but it does not normally come in to play in rifle trajectories.

-z
 
OK here goes qas best as I can recall from Physics class long ago. There are four forces tat act upon an object in flight, Lift, Drag, thrust and Gravity.

As relates to a projectile fired from a gun, thrust is provided by the powder charge ignition. It is of short duration, and tgereafter the bullet becomes in effect an unpowered mass. Drag on the bullet begins immediately in the form of wind resistance. With no thrust being applied once the projectile leaves the barrel, the forth force, Gravity, comes into play and causes the bullet to return to Earth.

The reason bullets appear to rse s the result of havng to "hold over" in order to compensate for the forces of drag and Gravty. barrng a freak occurnce such as the bullet being deflected by hitting a twig in flihjy, there is no lift so the downward path begins immrdiatelyb
 
The line of sight is angled to the bore so the projo cuts the line of signt twice in its flight; once near the muzzle and the second time at the range it's sighted for. That's also known as the point blank range - not the typical news reporter's conception of shooting a guy at "point blank range" means jamming the gun into his guts and pulling the trigger. :rolleyes:
 
If the axis of the bullet is not coincident with its velocity vector then there may be some lift created by the spinning motion of the bullet. However it is small and depending on the angle in question it may also pull the bullet down not up.
 
The above post ( Big G ) is the definition that I subscribe to. If you look at ballistics tables for a particular cartridge and load you often see figures like:

@ 50 yd. + 1.5"
@ 100yd. 0"
@ 150yd. - 3.25"
@ 200yd. - 7.5"

Where 100 yards is the sighted range being "dead on" @ 100 yd. is "point blank range" at least that's how I interpreted it. I may be corrected in a later post. If I'm incorrect, another THR member please clarify.
 
Here's a related query. I saw a reference somewhere to rifling producing a LATERAL force on the bullet. The direction, left or right, depends on the twist direction. This is an effect that is quite separate from coriolis and can be used to provide some correction for that effect.

Can anyone provide more information?
 
Yes, bullets do rise, at least in certain guns.

For instance, the bullets (in the brass cartridge) rise in the magazine when the top cartridge is stripped off by the bolt or slide, and the others are pushed upward by the magazine spring and follower.

:D
 
Depends. If you've got the gun pointed up a little (or a lot), it rises, at least for a while. If you've got the gun pointed dead-level or down, it falls.

Now, if you're talking about departure from the line in which the axis of the bore lies, or from the line of sight, that's a whole nother thing. There is explanation of this in the previous posts.

BTW, axis of bore and line of sight are usually adjusted so that a six-foot-tall man aiming at the eye of a six-foot-tall man two hundred yards away on level ground is pointing his gun up a little.

Hope this helps.
 
In the general case of a long spinning projectile flying through the air, there is a lift force if the nose is pointed "above" the instantaneous trajectory.
Remember that because the bullet is spinning rapidly, gyroscopic effects (specifically precession and nutation) will come into play that will prevent the spin axis from staying consistently above (or below, or to the side) of the line of sight. Most likely, if the bullet started out with a spin axis slightly offset from the line of flight, the spin axis would continuously precess, so that the lift vector also precesses. So the path of the bullet would become slightly helical around its parabolic trajectory.
 
The McCoy book contains a full derivation and expression of all the forces acting on projectiles using vector calculus. If one wants the "right" answer one need only pick up this book.
 
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