For you ballisticians.

[Pete Wass]

What about momentum

Ohhh, and BTW, in the example of a large slow bullet VS a small fast bullet it can be noted that the large slow bullet presents a larger sail....... more surface for the wind to blow against.

but it drifts less...

al

it would seem to me that momentum might play into the whole bullet drift thing. I have found it interesting that 115g bullets shot at say 2980 seem to be more wind resistent than say a 50g .22 caliber bullet shot at 3600. Momentum?

 

[Gene Beggs]

Good question Tony,,

What Al said got me to thinking. So a bullet, plane and boat all leave the 0 yard line to take a 1000 yard trip. Between 500 and 600 yard they have a 3 oclock 10mph wind...all other ranges there is 0 wind. So we have discussed how all three will act in that wind but what happens when they get to 601 yards? We know the plane and boat under their own power can then make corrections (take a new bearing) and the bullet will be affected by MOA rules but since it is already in motion (from the side push) would it stay in motion or would the spin forces overcome that motion?

I don't know about you guys but the plane, boat, sail, & Al on a bullet just makes this harder for me. Hard enough to get my pea brain wrapped around one thing let alone three or four. Now I have all my books out like I am writing a term paper just trying to keep up!

,, let's forget about the boat and airplane for now and concentrate on the bullet.


Remember; an elongated bullet streamlines itself to the relative wind created by its movement through the airmass. If the air mass is stationary (no movement relative to the ground) flight path and ground track are the same.

Tony, in your 1000 yard example, there would be no drift in the first 500 yards, but as the bullet enters the air mass at 500 yards which is moving from right to left at 10 mph, it experiences drag on the right side of the bullet and must yaw slightly to the right (into the crosswind) in order to streamline itself with the relative wind. According to Vaughn and Jackson, this takes place almost instantly; in maybe one or two nutations.

As the bullet progresses through the airmass between 500 and 600 yards it is flying clean (streamlined) relative to the airmass and if not for the fact that it is decellerating constantly, there would be no drift over the ground. But the bullet IS decellerating due to atmospheric drag; it has no onboard means of propulsion and cannot maintain its airspeed, so it drifts to the left a little (relative to the ground) between 500 and 600 yards. How much? That depends on rate of decelleration. A very efficient, low-drag projectile will shed velocity slower and drift less.

So,, emerging from the moving airmass at 600 yards and entering the stationary airmass between there and the target, the bullet must once again realign itself with the relative wind. In this case, the bullet experiences drag on its left side and must yaw slightly to the left. Since the airmass between 600 and 1000 yards is not moving relative to the ground, there is no drift. The bullet drifted to the left, relative to the ground, only between 500 and 600 yards. This displacement remains and will show up on the target, but from 601 yards to the target there is no additional drift.

Whew,, that got my head to spinning. Let's pause here for a moment and see if everyone is still with us.

Questions? Comments?

Gene Beggs

 

[mks]

Tangential viscous stresses turn a round ball

Without a moment arm there is nothing to produce a torque, I'm not sure that the round ball does adjust it's spin axis. Irrelevant to the question but interesting....

al

Al,
I am afraid I oversimplified my explanation. A round ball that is not spinning has normal pressure and tangential viscous surface stresses that are symmetrical around the axis of the resultant velocity vector. But a ball spinning on a axis not aligned with the velocity vector has asymmetrical pressure and viscous stresses. Pressure, by definition, always acts through the center of a sphere, and thus has no moment arm. Tangential stresses, by definition, each have a moment arm equal to the radius of the sphere. The resultant of the tangential stresses is offset from the center by an amount dependent on the magnitude of the crosswind, analogous to the center of pressure, but not exactly the same. Pressure and tangential stresses both contribute to the turning of a pointed bullet, but only tangential stresses can turn a round ball. Good question, Boyd!

Cheers,
Keith

 

[Gene Beggs]

Thanks Keith, this helps.

Al,
I am afraid I oversimplified my explanation. A round ball that is not spinning has normal pressure and tangential viscous surface stresses that are symmetrical around the axis of the resultant velocity vector. But a ball spinning on a axis not aligned with the velocity vector has asymmetrical pressure and viscous stresses. Pressure, by definition, always acts through the center of a sphere, and thus has no moment arm. Tangential stresses, by definition, each have a moment arm equal to the radius of the sphere. The resultant of the tangential stresses is offset from the center by an amount dependent on the magnitude of the crosswind, analogous to the center of pressure, but not exactly the same. Pressure and tangential stresses both contribute to the turning of a pointed bullet, but only tangential stresses can turn a round ball. Good question, Boyd!

Cheers,
Keith

Thanks to Vaughn, Jackson, McCoy, Litz and others, I now have a good understanding of how wind affects elongated projectiles, but I have struggled with how wind effects round balls. Your explanation illuminated the puzzle considerably. I'm not quite there yet but I'm getting closer.

Thanks again,

Gene Beggs

 

[Pete Wass]

No

,, let's forget about the boat and airplane for now and concentrate on the bullet.


Remember; an elongated bullet streamlines itself to the relative wind created by its movement through the airmass. If the air mass is stationary (no movement relative to the ground) flight path and ground track are the same.

Tony, in your 1000 yard example, there would be no drift in the first 500 yards, but as the bullet enters the air mass at 500 yards which is moving from right to left at 10 mph, it experiences drag on the right side of the bullet and must yaw slightly to the right (into the crosswind) in order to streamline itself with the relative wind. According to Vaughn and Jackson, this takes place almost instantly; in maybe one or two nutations.

As the bullet progresses through the airmass between 500 and 600 yards it is flying clean (streamlined) relative to the airmass and if not for the fact that it is decellerating constantly, there would be no drift over the ground. But the bullet IS decellerating due to atmospheric drag; it has no onboard means of propulsion and cannot maintain its airspeed, so it drifts to the left a little (relative to the ground) between 500 and 600 yards. How much? That depends on rate of decelleration. A very efficient, low-drag projectile will shed velocity slower and drift less.

So,, emerging from the moving airmass at 600 yards and entering the stationary airmass between there and the target, the bullet must once again realign itself with the relative wind. In this case, the bullet experiences drag on its left side and must yaw slightly to the left. Since the airmass between 600 and 1000 yards is not moving relative to the ground, there is no drift. The bullet drifted to the left, relative to the ground, only between 500 and 600 yards. This displacement remains and will show up on the target, but from 601 yards to the target there is no additional drift.

Whew,, that got my head to spinning. Let's pause here for a moment and see if everyone is still with us.

Questions? Comments?

Gene Beggs

Makes perfect sense.

 

[Tony Shankle]

,,So,, emerging from the moving airmass at 600 yards and entering the stationary airmass between there and the target, the bullet must once again realign itself with the relative wind. In this case, the bullet experiences drag on its left side and must yaw slightly to the left. Since the airmass between 600 and 1000 yards is not moving relative to the ground, there is no drift. The bullet drifted to the left, relative to the ground, only between 500 and 600 yards. This displacement remains and will show up on the target, but from 601 yards to the target there is no additional drift.

Questions? Comments?

Gene Beggs

Thanks Gene, I still have a question. I was aware of the aligning of the bullet but still have the question of what force stops the side motion or drift after it has left the wind? Which effect or force overcomes the sideways momentum? Does an object in motion not stay in motion?

 

[tobybradshaw]

What Al said got me to thinking. So a bullet, plane and boat all leave the 0 yard line to take a 1000 yard trip. Between 500 and 600 yard they have a 3 oclock 10mph wind...all other ranges there is 0 wind. So we have discussed how all three will act in that wind but what happens when they get to 601 yards? We know the plane and boat under their own power can then make corrections (take a new bearing) and the bullet will be affected by MOA rules but since it is already in motion (from the side push) would it stay in motion or would the spin forces overcome that motion?

[See Gene's post for the bullet part of the answer.]

As long as the boat and the airplane maintain a constant speed and are fin-stabilized (so that they turn into the water/air flow) they will arrive at 1000 yards at exactly the same spot they were pointed at when they started. Their trajectory, seen from above, is just a straight line with a slowdown between 500-600 yards proportional to cos(angle into the flow). This assumes that the speed of the boat and plane are greater than the crossflow speed.

Toby Bradshaw
baywingdb@comcast.net

 

[Gene Beggs]

Thanks Gene, I still have a question. I was aware of the aligning of the bullet but still have the question of what force stops the side motion or drift after it has left the wind? Which effect or force overcomes the sideways momentum? Does an object in motion not stay in motion?

Tony, I'll try to explain. I may need some help so I hope Keith, Toby and other experts will stand by. Here goes.

You asked, "Does an object in motion not stay in motion? What stops the side motion or drift after the bullet has left the wind?"

Yes,, an object in motion stays in motion until acted on by an external force which in this case is the atmosphere or airmass. If the airmass thru which the bullet is flying is moving relative to the ground, the bullet is carried along with it; if the bullet encounters an airmass that is stationary relative to the ground, drift stops.

Hope this helps. Maybe someone else can offer a better explanation.

Gene Beggs

 

[alinwa]

This displacement remains and will show up on the target, but from 601 yards to the target there is no additional drift.



Gene Beggs

So Gene,

Are you saying the bullet loses its sideward acceleration entirely and continues to the target parallel to its original course?

al

 

[Tony Shankle]

Tony, I'll try to explain. I may need some help so I hope Keith, Toby and other experts will stand by. Here goes.

You asked, "Does an object in motion not stay in motion? What stops the side motion or drift after the bullet has left the wind?"

Yes,, an object in motion stays in motion until acted on by an external force which in this case is the atmosphere or airmass. If the airmass thru which the bullet is flying is moving relative to the ground, the bullet is carried along with it; if the bullet encounters an airmass that is stationary relative to the ground, drift stops.

Hope this helps. Maybe someone else can offer a better explanation.

Gene Beggs

Atmosphere and airmass as force...are you sure about that? And even if airmass and atmosphere are forces wouldn't there have to be a reaction time? We know the nose repoonds quickly to wind or even back to no wind but do we know that drift does not continue for some time?

 

[Gene Beggs]

So Gene,

Are you saying the bullet loses its sideward acceleration entirely and continues to the target parallel to its original course?

al

Yep, that's exactly what I'm saying. The bullet is not free to just drift along sideways on its own; it is encapsulated if you will, by the airmass through which it is flying, which in this case is not moving relative to the ground, so there can be no drift.

Gene Beggs

 

[Tony Shankle]

don't think so Gene

Sorry to pose those questions (a bit of baiting for you and Toby) but that is not correct. You see a plane or boat (which I wish we would stop talking about) are under there own power and while they crab angle in a wind they do resume "the straight line" once they come out of the wind because their own power overcomes the momentum.

A bullet however does not. Newton holds true here and that bullet will continue its deflection because it is already in motion.

 

[Gene Beggs]

Atmosphere and airmass as force...are you sure about that? And even if airmass and atmosphere are forces wouldn't there have to be a reaction time? We know the nose repoonds quickly to wind or even back to no wind but do we know that drift does not continue for some time?

Tony you asked,

"Atmosphere (airmass) as force; are you sure about that?"

Yep, darn sure! And if you are flying thru that airmass at 3000 fps it exerts one helluva' force.

If the bullet is flying through an airmass that is moving laterally over the ground (a crosswind) and suddenly enters an airmass that is not moving, drift stops; immediately! Drift could not continue for any period of time or the bullet would be skidding sideways through the air mass, and I assure it cannot do that.

Maybe someone else can offer a better explanation.

Gene Beggs

 

[Tony Shankle]

Tony you asked,

"Atmosphere (airmass) as force; are you sure about that?"

Yep, darn sure! And if you are flying thru that airmass at 3000 fps it exerts one helluva' force.

If the bullet is flying through an airmass that is moving laterally over the ground (a crosswind) and suddenly enters an airmass that is not moving, drift stops; immediately! Drift could not continue for any period of time or the bullet would be skidding sideways through the air mass, and I assure it cannot do that.

Maybe someone else can offer a better explanation.

Gene Beggs

Acceleration of the deflection stops but the momentum deflection does not stop. Someone prove me wrong and I will inform Newton and Mr. Rinker.

By airmass are you referring to the flowfield?

 

[tobybradshaw]

Thanks Gene, I still have a question. I was aware of the aligning of the bullet but still have the question of what force stops the side motion or drift after it has left the wind? Which effect or force overcomes the sideways momentum? Does an object in motion not stay in motion?

Tony, here's an experiment and some arithmetic.

Experiment: Toss a Nerf football out of the side window of a parked car. Note how far from the car the ball lands. Now get the car going 100mph and toss the ball out. Will it go as far (laterally) as the ball tossed from the parked car?

And, no, I won't pay for your speeding ticket.

Arithmetic: Shoot a BIB 95gr FB (BC=0.435) at 3000fps from a 6BR, with a 10mph crosswind from 3 o'clock. In the 0.1sec it takes the bullet to go 100yd, it will slow down to 2775fps and move 0.67in downwind.

To move 0.67in in 0.1sec there must be a lateral component of acceleration = (2*0.67in)/(0.1sec^s) = 134in/sec^2 = 11ft/sec^2 = roughly a third of a g.

The deceleration due to drag is (3000-2775fps)/(0.1sec) = 2250ft/sec^2 = 70g. I.e., the force on the bullet due to drag in the "nose-first" direction is 200 times greater than the "side drag" the bullet is producing by its lateral motion, so it doesn't take long for the aerodynamic drag on the base of the bullet that is "sticking out into the wind" (the new relative wind, in this case) to zero out the lateral momentum.

NB: The above calculations are linear approximations, so I can do them!

Gene has described perfectly what the bullet will do if it encounters a crosswind 500-600yd into its flight.

Toby Bradshaw
baywingdb@comcast.net

 

[Tony Shankle]

Toby,

I am not arguing that the drag will correct the nose, I am arguing that it will still continue laterally even though the drag corrects it axis vector.

Let take your BIB with its .67" defection at 100 yards, now it enters a tunnel with no wind. You and Gene are saying it straightens out and will be .67" to the left of the X at 1000 yards. I don't agree. (correction to my poor writing skills...I do agree it straightens just not that the lateral movement stops)

That .67" will enter the tunnel at 100.1 yards with a sideways movement of 1.34" per 100 yards. It will then continue its lateral movement for 12.06 inches at the 1000 yard mark when it strikes the "target in the tunnel". This is of course ignoring Spin Drift, Coriolis, and all the other effects we have yet to touch on in this thread.

That 12.06" is of course much less that if it stayed in the wind the entire 1000 yards but it is a result of lateral momentum.

So you and Gene disagree with Rinker? (for those of you who have Understanding Ballistics turn to page 248)

 

[Gene Beggs]

Tony, you're right,,

Sorry to pose those questions (a bit of baiting for you and Toby) but that is not correct. You see a plane or boat (which I wish we would stop talking about) are under there own power and while they crab angle in a wind they do resume "the straight line" once they come out of the wind because their own power overcomes the momentum. A bullet however does not. Newton holds true here and that bullet will continue its deflection because it is already in motion.

,,,,let's quit talking about planes and boats.


Tony you stated, "Newton holds true here and that bullet will continue its deflection because it is already in motion."

WRONG!

Yes, Newton holds true and a bullet, once set in motion, drifting sideways, would continue moving in that direction forever if, and it's a big IF; you were shooting on the moon!

But we are not shooting on the moon where there is no atmosphere to encapsulate our bullets. Here on planet earth, God wisely and graciously provided for us a marvelous and perfect atmosphere in which to live, but that same atmosphere presents some challenges when it comes to shooting.

Early man probably never gave a thought to the airmass that surrounds him and his planet. Even if he did think about it, he probably regarded air as a weightless, formless nothing, but it's far from "Nothing!" At mach 2.7 it's one helluva' force!

Maybe someone else can do better.

Gene Beggs

 

[tobybradshaw]

So you and Gene disagree with Rinker?

Tony, I've never read Rinker, but I do agree with Gene.

Throw a Nerf football past a fan and see for yourself. Or get Gene to take you for a plane ride.

A spin-stabilized bullet (like an airplane) can't be drifting sideways in the air mass and still have its nose pointed straight ahead, because the nose MUST point into the relative wind (the vector sum of the downrange and crossrange velocities).

The lateral momentum of the bullet is negated (quickly) by collisions with air molecules on the side of the bullet when the direction of the relative wind changes.

Toby Bradshaw
baywingdb@comcast.net

 

[Gene Beggs]

Great discussion; huh?

Where else could we enjoy such an opportunity to share knowledge and comaraderie with so many shooters from all over the world? Shooters who, just like the rest of us, want to know exactly how things work.

Thanks Wilbur, for hosting the forums. This is great!

Later,

Gene Beggs

 

[Boyd Allen]

Earlier this evening, I posted this on the wrong thread

Let us introduce momentum into the discussion. It is a vector that decreases in length as a bullet decelerates. The vector that represents the force of the cross wind on the bullet remains constant (assuming a constant crosswind). increasing in influence as the down range vector decreases. The instantaneous direction of the bullet can be calculated as the sum of these vectors. Even though two bullets leave and arrive at the same time, the differences in their momentum as they proceed down range will cause a difference in their drift plots. This is exaggerated as drift is added to drift. A dropped bullet has an entirely different momentum profile, since it starts with zero downward momentum and is accelerated by gravity till it hits the ground or reaches a maximum value. There is a limit to the velocity that can be achieved by a falling object, and from that point on the downward momentum vector stays the same length which is entirely different from a fired bullet.

Another point worth noting is that as a tank of given proportions is increased in size, it has more volume per surface area. If we apply this to a solid shape, and hold density constant, larger solids of a given shape and proportion have less sail area per weight, giving the wind less influence per unit of weight, which is why it is easier to blow dust than boulders.