bullet above bore's axis...?

Boyd,

Here are pictures of a sphere and a pointed bullet at supersonic speed. Do you believe wind blows on the surface of a supersonic projectile like your tin can?

Regarding others theories in past discussions, how could a correct theory not be accurate to describe the wind effect on both of these projectiles? By the pictures it should be obvious the same forces apply.
 

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I did not say that the same mechanism did not apply to spherical and elongated bullets, just that because the way that some have explained how the wind causes "drift" of elongated bullets would not seem to apply to a spherical bullet, and given that the latter is more influenced by the wind than the former, that the theory may be in error.
 
Boyd Allen said:
I did not say that the same mechanism did not apply to spherical and elongated bullets, just that because the way that some have explained how the wind causes "drift" of elongated bullets would not seem to apply to a spherical bullet, and given that the latter is more influenced by the wind than the former, that the theory may be in error.
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Boyd,

Am I correct in assuming your posts are more pointed towards previous discussions rather than specific things stated in this thread?
 
There are common elements in both. If you look at the paper that I posted a link to the search for, you will see that the whole question of how the wind couples to bullets in flight is not any part of the equation used to calculate wind drift, and I think that that has been overlooked in both discussions. In both discussions I was the only one to bring that information to the discussion, and to point out that the theories being discussed had serious flaws, because they did, and do not work for a spherical projectile, and therefore may be incorrect. Just because two things happen at the same time does not mean that one caused the other.
 
I did look over that paper Boyd. All 5 pages of information. I've attached it here to be sure it is the same one you linked. The description of spherical versus coned bullet flight is exactly what I was trying to say in my previous description of such flight. I have not read any technical descriptions of spherical bullet flight, specifically, so this paper reinforces my thought that I do understand the forces at play. So thank you for that.:D

Re your statement of: "you will see that the whole question of how the wind couples to bullets in flight is not any part of the equation used to calculate wind drift"

This paper certainly is not intended to discuss or describe the scientific details of aerodynamic flight. It appears to be a generalization of forces used to create a mathematical model to generally predic wind drift. Kinda like my tin can example of pushing on the side with your finger versus Keith's description using details from fundamental fluid mechanics. The origin of this thread revolved around the vertical component of drift seen on a target. (Ie. right wind blows a bullet left and up). This paper contains nothing that attempts to explain that. It doesn't address spin drift, aerodynamic jump or other real observed ballistic results.

If you still think I have said something that is contradicted by observed spherical bullet flight please point it out to me and I will attempt to clarify.
 

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Yes that is the paper. It is not so much that I was taking issue with anything that you wrote, but rather I was trying to filter the discussion a bit, and provide some context and focus. Comparing the two threads, I think that this one has a lot more to recommend it, and that what you have written is one of the main reasons.
 
I am a pilot, Private Airplane Instrument and CFI helicopter. So let us say we have a 10 mile an hour cross wind. That mass of air would move 10 miles in an hour or 52,800 feet, 880 feet in a minute, 14.66 feet in a second, and 1.46 feet in a tenth of a second which is about how long a bullet takes to travel a 100 yards. 1.46 feet is 17+ inches, we all know 10 mile an hour cross wind does not move it that far. So why is this so? Many years ago smooth bore muskets had a terrible problem of fouling, some one thought let us put groves in barrel so there will be somewhere for fouling to go. I believe 1 in 66 inch twist, not sure but damn if accuracy did not improve. Gyroscopic action. You all are making this way too complicated, the gyroscope resists movement off it's axis. I do not believe the wind pushes against the bullet. The bullet is in a mass of air only, it is resisting being moved off it's original axis. This force the bullet resists in the right to left direct cross wind is manifested 90 degrees which in right hand twist is bottom of bullet. Now then it resists, does not stop, so our little 68 grain boattail moves to the left some and it falls less than it would so it seems to go up. Now this is only my opinion, I do not think the nose of bullet is steering into wind and so on. Can not prove any of this but it is what I think, besides DMA!

Bill Greene
 
Boyd,
Lots of questions in your post. Responding to just a couple of them:
Boyd Allen said:
At the time, I brought up the round ball because it does not fit that model, and that therefore the model might be flawed.
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Would you accept a model with a variable distance between center of pressure and center of gravity, in which that distance can go to zero? If so, it fits both round balls and elongated bullets with lots of shades of grey in between.

Boyd Allen said:
Center of pressure does not mean that the actual pressure is somehow magically applied to one point.
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Actually it does, sort of. The center of pressure is defined such that if you replaced all the distributed pressure forces with one point force of appropriate magnitude that caused the bullet to respond exactly the same, it would be located at the center of pressure. Magic?? Eh, just math.

Cheers,
Keith
 
Keith,
I think that we are playing with words. For purposes of calculation, the bullet reacts as if the pressure is applied at that point, but the pressure is applied over the profile of the bullet, just like having a center of mass does not mean that the mass only exists at that point. As far as a theory for simple drift goes, I believe that the one in the paper that I referred to covers that. Did you read it? To answer your question, of course I would have to accept a model that explains the behavior of all bullet shapes as long as it did not involve the action of mythical creatures, or a mechanism that did not make sense for any.
Boyd
 
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An intermission......

I wish I had a video to post here:
It would be a neighbor that lived down the dirt road I used to live on. He asked to go with me to "check out his Marlin thutty five" when I needed to fire form a few test rounds in a 250 Savage Ackley Improved I had just chambered. He threw his 35 Rem Marlin up to his shoulder (Refusing my suggestion to at least use a sandbag on the hood of the truck), and popped off ONE round at a Budweiser can about 20 yards away - (it jumped about 5 feet, possibly from a bullet in the dirt near it, or maybe even a direct hit,....he didn't walk over to inspect it)..
He said "she's still 'right on" and put the rifle back in the truck. I paused and asked if he might want to send a few more down range to which he replied "why?,... She's never failed me".

OK, just thought I'd give everyone's grey matter a 'reset'.... LOL-!
 
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Boyd Allen said:
Keith,
I think that we are playing with words. For purposes of calculation, the bullet reacts as if the pressure is applied at that point, but the pressure is applied over the profile of the bullet, just like having a center of mass does not mean that the mass only exists at that point. As far as a theory for simple drift goes, I believe that the one in the paper that I referred to covers that. Did you read it? To answer your question, of course I would have to accept a model that explains the behavior of all bullet shapes as long as it did not involve the action of mythical creatures, or a mechanism that did not make sense for any.
Boyd
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Boyd,
There are two significant errors in Leupold's paper. First, he says that the projectile path in a cross wind is described by the same triangle as formed by the muzzle velocity and the cross wind velocity (Fig. 2). This is like saying that the bullet is blown sideways at 10 mph in a 10 mph cross wind, which is obviously not correct. He ends up with the correct equation 4, but the explanation is bogus. Second, in Fig. 4 he explains that pointed bullets have greater air resistance in a cross wind because the bullet flies askew to the wind. Again, this is incorrect. In the last paragraph of this section he mentions precession, but having spent the entire rest of the section throwing the reader off, the damage is done.

Models such as presented by Leupold are called 3 degrees of freedom (3DOF), because they only describe bullet motion in the forward, sideways and upward directions. But note that Leupold doesn't include the upward deflection, for a good reason. One can get by with a 3DOF model to calculate horizontal drift, but it's hard to explain vertical deflection (the original question of this thread) without modeling precession. And modeling precession requires 6 degrees of freedom (6DOF), where the extra three correspond to rotation of the bullet around each of the forward, sideways and upward directions.

Keith
 
r44astro said:
Some of you may find this helpful
http://helicopterflight.net/gyroscopic P.htm
Bill
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Bill,

I've got a little bit of helicopter flight time myself.... RC type. No way in heck I'd get in a real one after all the stuff I broke.:eek:

Earlier I think you were saying because of the gyroscopic forces, the bullet resists being moved off course and therefore does not move into the wind. Flying an RC copter outside was about impossible for me if there was any significant wind. Things happened so quick I couldn't keep up with what all was going on but I didn't get the impression it was resisting moving off course. Of what I remember, a wind would deflect the body and it would start twisting and turning in ways that didn't make sense to me so I didn't seem to be able to apply the proper control corrections. I think what I was seeing is what I now understand to be gyroscopic precession.

This was with an aerobatic copter with control being all with the operator. I had another copter with a "fly bar" on top of the rotor that when pitched by the wind it made rotor corrections to the swash plate making it MUCH easier to fly. Also, I believe in both case they had an electronic gyroscope that aided in the controls. Do the real helicopters have anything like this and do you experience the same precession in a side wind?
 
Joe Woosman said:
Bill,

I've got a little bit of helicopter flight time myself.... RC type. No way in heck I'd get in a real one after all the stuff I broke.:eek:

Earlier I think you were saying because of the gyroscopic forces, the bullet resists being moved off course and therefore does not move into the wind. Flying an RC copter outside was about impossible for me if there was any significant wind. Things happened so quick I couldn't keep up with what all was going on but I didn't get the impression it was resisting moving off course. Of what I remember, a wind would deflect the body and it would start twisting and turning in ways that didn't make sense to me so I didn't seem to be able to apply the proper control corrections. I think what I was seeing is what I now understand to be gyroscopic precession.

This was with an aerobatic copter with control being all with the operator. I had another copter with a "fly bar" on top of the rotor that when pitched by the wind it made rotor corrections to the swash plate making it MUCH easier to fly. Also, I believe in both case they had an electronic gyroscope that aided in the controls. Do the real helicopters have anything like this and do you experience the same precession in a side wind?
Click to expand...

I flew RC Helicopter before I flew full scale and full scale is so much easier. Some of the more advanced do have auto hover, auto pilot and so on, I am sure Billy and Bart could elaborate. The ones I flew did not. You just look outside and input control that is needed. I was just trying to point out the reason bullets have this incredible accuracy is because of the spin and therefore gyroscope. There are few on here that can elaborate to the mathematics I can not. As far as precession in side wind or force the rotor is operating in a different plane than a bullet. In a cross wind I would have to crab just like an air plane. When you do the mathematics a bullet spinning out of a 13 twist barrel, using 3000 feet a second would be rotating approximately 165,000 Revolutions a minute, at least when leaves the barrel. I think it's mostly about gyroscopic action of the bullet.
But then again DMA
Bill
 
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