Boyd Allen
Active member
Obviously they are talking about rising or falling relative to line of sight (LOS)...not so confusing to me.
Butch Lambert
Active member
Thanks guys,
Another forum the so called expert said not to worry about 50fps spread as it had very little effect.
Butch
Another forum the so called expert said not to worry about 50fps spread as it had very little effect.
Butch
Boyd, this is true BUT.....LOS is an artificial imposition, there's nothing different about the "rising" VS the "falling" sides....... the bullet is just falling.
Let's say you're standing on the edge of a precipice..... firing "down"......
you see what I mean??
al
Boyd Allen
Active member
Look, a picture.
This is why I think that all of this discussion about whether a bullet turns into or away from the wind by less than a half degree is a bit silly.
http://www.nennstiel-ruprecht.de/bullfly/fig21.htm
This is why I think that all of this discussion about whether a bullet turns into or away from the wind by less than a half degree is a bit silly.
http://www.nennstiel-ruprecht.de/bullfly/fig21.htm
T
Tony Shankle
Guest
They are correct. The bullet's nose will never move as high above the trajectory curve as it was on the previous turn because of axis shifts and changes in pressure. The nose drops as does the trajectory curve. The spinning of the bullet is acutally in a small swinging arc. This cycloidal movement is more down and toward the direction of rotation as it takes it trip down range.
Rinker is my hero!
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Gene Beggs
Active member
Thanks Boyd
Boyd, I knew you were very busy today, hunting down the details of the error in the Sierra manual.
Well done my friend, well done.
Gene Beggs
Boyd, I knew you were very busy today, hunting down the details of the error in the Sierra manual.
Well done my friend, well done.
Gene Beggs
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T
Tony Shankle
Guest
Ever noticed that the flowfield is somewhat smaller (more gentle) on the bottom of the bullet in shadowgraphs like this one???
They are correct. The bullet's nose will never move as high above the trajectory curve as it was on the previous turn because of axis shifts and changes in pressure. The nose drops as does the trajectory curve. The spinning of the bullet is acutally in a small swinging arc. This cycloidal movement is more down and toward the direction of rotation as it takes it trip down range.
Rinker is my hero!
Well, no they're not "correct" ......... correct is that the nose drops as does the trajectory curve, as you said above. What Sierra is saying is that a bullet behaves differently when fired "upward" than if say fired flat......
Here's the end of the paragraph ref'd:
"As the bullet flies the trajectory curves downward, and φ decreases, going through zero at the summit of the trajectory,
then going negative and increasing in magnitude as the trajectory steepens. The bullet noses downward, and this......" etc etc.
The summit of trajectory???
Are you going to agree that the bullet acts differently when fired say 10degrees "uphill" than fired flat?
What about a bullet fired 10degrees below horizontal, DOWN hill, this bullet never "rises above" anything..... does it act differently than one "launched at an upward or 'positive' angle"????
ALL that's really happening is that the bullet reacts to the changes in airflow by searching for center of flow. "Gravity" has got nuttin' to do with nosing over except to generate a wind..... and as the bullet reacts to find the wind it kicks out at right angles, it develops a nervous tic...... it's like steering a cat with a branding iron..
irrelevant to the winddrift thread BUT..... relevant to understanding the flight of a flung rock.
I still can't believe you got that picture Tony!!!
LOL
al
Gene Beggs
Active member
Al, read again, my post #150
Al, in your first paragraph above, you stated,
"You'se guys are getting close but still figgering planes and not decellerating projectiles."
(GB) No, that's not true. If you will read again, my post #150, I made it clear, I was referring to a decellerating bullet, not an airplane.
In paragraph three above you stated,
"Now, when it pops out into clear air on the other side it does NOT "resume course"....... it renegotiates a new velocity which retains some of it's divergence.... not ALL, but some of it."
(GB) What do you mean by, "it renegotiates a new 'velocity? Do you mean a new heading? Ground track? What?
Al, you said you would try to draw a picture to explain what you're talking about. That would help.
Gene Beggs
You'se guys are getting close but still figgering planes and not decellerating projectiles
Let's fuh'GEDDABOUT the part of the trajectory before the stream of crosswind at 500-600..... we just pick up the bullet crabbing across the stream, it's DEFINITELY on a downwind course and fairly crabbed over..... and sl o o w i n g as it goes....
Now, when it pops out into clear air on the other side it does NOT "resume course"....... it renegotiates a new velocity which retains some of it's divergence.... not ALL, but some of it.
I'll try to draw a picture since this is the ONE salient point which puts about three years argument into perspective!
This goes back to months ago with "you and me on two sides of the ribber" Gene
Let's Get 'Er Done Eh!!!!
LOL al
Al, in your first paragraph above, you stated,
"You'se guys are getting close but still figgering planes and not decellerating projectiles."
(GB) No, that's not true. If you will read again, my post #150, I made it clear, I was referring to a decellerating bullet, not an airplane.
In paragraph three above you stated,
"Now, when it pops out into clear air on the other side it does NOT "resume course"....... it renegotiates a new velocity which retains some of it's divergence.... not ALL, but some of it."
(GB) What do you mean by, "it renegotiates a new 'velocity? Do you mean a new heading? Ground track? What?
Al, you said you would try to draw a picture to explain what you're talking about. That would help.
Gene Beggs
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Boyd boyd BOYD..... the degrees don't MATTER! The question at hand is, WHAT DRIVES THE BULLET OVER....... and it ain't the wind blowing on it.
It IS the non-linear suctive force, the negative pressure present in the lee of the bullet as it tears a hole in the ambient
('AT oughtta' get 'er goin' ag'in!!)
LOLOL
al
Gene Beggs
Active member
Al, in your first paragraph above, you stated,
"You'se guys are getting close but still figgering planes and not decellerating projectiles."
(GB) No, that's not true. If you will read again, my post #150, I made it clear, I was referring to a decellerating bullet, not an airplane.
In paragraph three above you stated,
"Now, when it pops out into clear air on the other side it does NOT "resume course"....... it renegotiates a new velocity which retains some of it's divergence.... not ALL, but some of it."
(GB) What do you mean by, "it renegotiates a new 'velocity? Do you mean a new heading? Ground track? What?
Al, you said you would try to draw a picture to explain what you're talking about. That would help.
Gene Beggs
A new heading would work, a new ground track, pilotspeak. Velocity is defined as speed coupled with direction and I used it as such. I'm kind of thrashing about for words here because we all "hear" differently.
In some conversations 'velocity' just means speed, but in physics it contains direction.I'll draw the pic and see if my kids can help me upload it.
But meantime here's something to consider....
The bullet is coasting along and enters the 500-600yd crosswind. It quickly realigns it's nose and for 100yds it negotiates this cross-current. After 100yds it's direction has changed........ it's headed off in a new direction...... the only way to bring it back into parallel is to let it coast for ANOTHER (prox) 100yds through an opposite crosswind. It took 100yds to get it's direction, you can't fix it by just popping it back into still air. It will enter the still air with its crabbed attitude, it WILL be buffeted back into an alignment, but not back to its original path. The true correction would take somewhere near a hundred yards again.
I'll go off and draw my picture
al
Ever noticed that the flowfield is somewhat smaller (more gentle) on the bottom of the bullet in shadowgraphs like this one???
No I hadn't.... are you proposing that this is indicative of an attitude? Or just a random cyclical yaw. Are there other pictures?
al
Well, I've got the picture.... my problem is #$%^ Vista, my wife's printer is hooked to week-old comp which came w/Vista.
She's fighting it. The whole last week was changing over her business files and editing programs, now she's patching in the scanner..
Give us a little time.
al
She's fighting it. The whole last week was changing over her business files and editing programs, now she's patching in the scanner..
Give us a little time.
al
Boyd Allen
Active member
Al,
Did you look at the picture? In this thread much was made of whether the bullet tilts very slightly toward or away from the wind. I posted the drawing as a reminder that the model that has been discussed is evidently too simple, which has seems to have caused mountains to be made of molehills. I repeat, round balls are the worst case for wind drift, therefore it must not depend on slight angular variations in elongated bullets, especially since these slight variations would seem to disappear into the noise of the gyrations that the picture shows.
Did you look at the picture? In this thread much was made of whether the bullet tilts very slightly toward or away from the wind. I posted the drawing as a reminder that the model that has been discussed is evidently too simple, which has seems to have caused mountains to be made of molehills. I repeat, round balls are the worst case for wind drift, therefore it must not depend on slight angular variations in elongated bullets, especially since these slight variations would seem to disappear into the noise of the gyrations that the picture shows.
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Yup, I looked at the picture. It's not to scale
and it doesn't illustrate all of the actions, only one.But anyway.... YES a roundball is worstest of all.
And YES the "nose to the wind" is only an artifact.
And YES the fact that drag is "along the axis of the spin stabilized bullet" is ALSO just an artifact.....
But what is important is that bullets DO DRIFT and this drift is much greater than can be explained by wind blowing. And in several years we've yet to come up with verbal imagery which satisfies..... and math won't do it because it's simply not descriptive enough (witness Toby's vector analysis above with its ".48degrees in and out" .... doesn't MEAN anything, it's just the calculated attitude of the bullet at two points in it's trajectory....)
The attitude of the bullet doesn't dictate drift. UNlike an aeroplane. The attitude of a bullet only indicates its reaction to external forces, wind drift is CAUSED by deceleration. My "backing up" story is dead-on accurate.
mebbeso pix soon....
al
Boyd Allen
Active member
The sketch that Al asked me to post.
Gene Beggs
Active member
Alinwa's vocabulary!
Al, you're a real piece of work, and I mean that in a complimentary, brotherly love sort of way. Conversations with you are always lively, entertaining and in some cases, even educational. I especially appreciate your comment,
"The question at hand is, WHAT DRIVES THE BULLET OVER?....... and it ain't the wind blowing on it, it IS the non-linear suctive force (italics are mine GB) the negative pressure present in the lee of the bullet as it tears a hole in the ambient."
Al,, that one is a classic! "Non-linear suctive force."
Thanks for brightening my day. I haven't had such a good laugh in a long time.
Have a good one today my friend.
Later,
Gene Beggs
Boyd boyd BOYD..... the degrees don't MATTER! The question at hand is, WHAT DRIVES THE BULLET OVER....... and it ain't the wind blowing on it.
It IS the non-linear suctive force, the negative pressure present in the lee of the bullet as it tears a hole in the ambient
('AT oughtta' get 'er goin' ag'in!!)
LOLOL
al
Al, you're a real piece of work, and I mean that in a complimentary, brotherly love sort of way.
Conversations with you are always lively, entertaining and in some cases, even educational. I especially appreciate your comment,"The question at hand is, WHAT DRIVES THE BULLET OVER?....... and it ain't the wind blowing on it, it IS the non-linear suctive force (italics are mine GB) the negative pressure present in the lee of the bullet as it tears a hole in the ambient."
Al,, that one is a classic!
"Non-linear suctive force." Thanks for brightening my day. I haven't had such a good laugh in a long time.
Have a good one today my friend.
Later,
Gene Beggs
M
mks
Guest
More crow eating
OK, so I went to McCoy and checked the point mass trajectory equation:
dV/dt=-BC |V|(V-W)+g
where BC is the ballistic coefficient, V is the bullet velocity vector, |V| is the scalar magnitude of the bullet velocity, W is the wind velocity vector, g is the gravity vector and t is time. The the z (crosswind) component of this equation is
dVz/dt=-BC |V|(Vz-Wz)
where Vz is the velocity of the bullet in the crosswind direction and Wz is the crosswind component of the wind. Since the decelerating bullet acquires a Vz velocity between 500 and 600 where Wz=15fps, it emerges with this same velocity. Without doing the calculus, Vz grows to about the same proportion of the crosswind as the deceleration is to the original velocity, so about 9% of 15fps or 1.38fps. After 600 where Wz=0, Vz decays at the same rate (percentage-wise) as the forward velocity. That is, it gets smaller, but doesn't disappear until both velocities go to zero, and results in an estimated 11.2" deflection at the target (close to Toby's estimate).
Al, you are right and I was wrong! (Glad I didn't have a box of bullets riding on this)
Cheers,
Keith
This would be so much easier with a blackboard.
The scenario:
BIB 95gr FB BC=0.435
muzzle velocity=3000ft/sec
crosswind @ 3'oclock @ 10mi/hr (15ft/sec) from 500-600yd only
At 500yd the bullet is going 1964ft/sec. When it hits the 15ft/sec crosswind the bullet turns into the wind [the point turns to the right] by arctan(15/1964)=0.44degrees.
The nose angle continues to increase until the bullet has reached 600yd, at which time it is going 1783ft/sec and has a nose angle of 0.48degrees. It has been deflected downwind (left) about 1.2 inches between 500-600yd.
Upon entering the "calm" air past 600yd, the bullet's nose turns leftward into the relative wind by arctan(15/1783), which is 0.48degrees. So, the bullet is now pointing 0.04degrees to the left of the line upon which it entered the crosswind, and thus continues to diverge downwind from the original line of departure from the bore, which amounts to about 2.5 inches leftward displacement for every 100yd thereafter (3600inches/100yd*tan(0.04degrees)) for a grand total of about 11.2 inches at 1000yd (of which only a little over an inch happened during the "push").
Resolved: I will shoot more and do less typin' and figgerin'.
Toby Bradshaw
baywingdb@comcast.net
OK, so I went to McCoy and checked the point mass trajectory equation:
dV/dt=-BC |V|(V-W)+g
where BC is the ballistic coefficient, V is the bullet velocity vector, |V| is the scalar magnitude of the bullet velocity, W is the wind velocity vector, g is the gravity vector and t is time. The the z (crosswind) component of this equation is
dVz/dt=-BC |V|(Vz-Wz)
where Vz is the velocity of the bullet in the crosswind direction and Wz is the crosswind component of the wind. Since the decelerating bullet acquires a Vz velocity between 500 and 600 where Wz=15fps, it emerges with this same velocity. Without doing the calculus, Vz grows to about the same proportion of the crosswind as the deceleration is to the original velocity, so about 9% of 15fps or 1.38fps. After 600 where Wz=0, Vz decays at the same rate (percentage-wise) as the forward velocity. That is, it gets smaller, but doesn't disappear until both velocities go to zero, and results in an estimated 11.2" deflection at the target (close to Toby's estimate).
Al, you are right and I was wrong! (Glad I didn't have a box of bullets riding on this
)Cheers,
Keith