Attention all ballisticians

[Vibe]

Here we go again OOhhhhh I'm gonna; get allover in trouble A'gain......

The driving force is the bullet "backing up" from it's frame of reference.
al

I thought you'd got over that "stump broke" boolit idea.

Oh. OK

OK, I dunno if this will work but:

The energy to start the bullet laterally "comes from" ... the wind blowing on the nose of the bullet

al

 

[mks]

Everything grows at the same rate

My question
I’ve always wondered: WHY is it that crosswind deflection is precisely proportional to lag time? It would be intuitive if wind deflection were proportional to total time of flight, that would be like the ‘feather in the wind’ visualization. Even if wind deflection were not neatly proportional to any ‘easy’ quantity, I could accept that as a non-intuitive complexity of ballistics. BUT, since it’s mathematically provable that wind deflection is directly proportional to lag time, I’ve always wanted to know why, and what exactly that means.
I’m almost satisfied with my understanding of the concept when thinking about the ‘no deflection for thrust = drag; zero lag time’ case, and the ‘upwind deflection for thrust > drag; negative lag time’ case. But there’s still something mysterious about the fundamental nature of lag time that I can’t quite put my finger on.

So, why’s it so simple?

Brian,
That's a good one. Maybe this will help. A bullet in flight is not in static equilibrium, but drag moves it toward a state of equilibrium. For a muzzle velocity of Vx0, the bullet velocity in the downrange direction decays toward an equilibrium velocity of zero. In a cross wind, the bullet starts out with cross range velocity Vz=0 and approaches an equilibrium cross range velocity equal to the cross wind speed Wz. The fractional rate at which these two components of velocity progress from their initial to final states is exactly the same

Vz/Wz = delta Vx/Vx0

The fractional extra time that it takes the bullet to get to the target (lag time delta t) is also exactly correlated to the change in velocity

Vz/Wz = delta Vx/Vx0 = delta t/t

where t is time of flight to the target if no drag occurred.

Furthermore, cross range deflection Z grows from zero to a final (maximum) value equal to the cross wind velocity times time of flight Zmax=Wz t in the same proportion

Vz/Wz = delta Vx/Vx0 = delta t/t = Z/Zmax

By rearranging the last part of this equation, we get the familiar formula that deflection is equal to crosswind speed times lag time

Z=Wz delta t

but the rest of the equation tells us more. A bullet that has slowed down by 1% of its muzzle velocity takes 1% longer to get to the target and has acquired a cross range velocity equal to 1% of the cross wind velocity, and is deflected by 1% of its maximum value. Deflection is proportional to lag time because both grow toward their maximum values at the same rate.

Cheers,
Keith

PS. Deflection IS proportional to time of flight if you use the actual cross range velocity of the bullet

Z = Vz t

Since Vz is the same fraction of Wz as delta t is of t, these proportions are interchangeable, so

Z = Vz t = Wz delta t

 

[Boyd Allen]

As this thread has proven beyond a shadow of a doubt, such clarity is rare. Thank you.

 

[bsl135]

Keith,
That's a lot to think about.
Here's what I did.
I ran a trajectory from 0 to 1000 yards in 1 yard increments (G7 BC = 0.233, Vxo=3000 fps, ICAO sea level standard, 10 mph crosswind), pasted the output into MS Excel, and performed your equations to see if the loss fractions were actually equal for the 4 quantities you identified:

Vz/Wz = delta Vx/Vx0 = delta t/t = Z/Zmax

The picture that shows how well the relations hold (it's about time we have a picture) is shown here:

As you can see, the Vz/Wz = dVx/Vxo relation is solid. The 'noise' in the Vz/Wz (blue) line is due to round off error in the lateral velocity calculation; I only have two decimals for wind deflection in my output...
Anyway, you can see that the forward velocity loss fraction is equal to the lateral velocity fraction, which is growing to approach Wz.

Now the problems start.

You stated that the Tlag/Tvac ratio would be equal to the above fractions. Sounds good, but it's plotted in green, and clearly doesn't track. Furthermore, the Z/Zmax fraction is plotted in purple, and doesn't track with anything either.

I'm also confused by your PS.

You said:

PS. Deflection IS proportional to time of flight if you use the actual cross range velocity of the bullet

Z = Vz t

But Vz isn't constant, it changes continuously.

You also said:

Since Vz is the same fraction of Wz as delta t is of t, these proportions are interchangeable, so

Z = Vz t = Wz delta t

But earlier you defined delta t as the time of flight in the absence of drag (vacuum tof). We know that Z = Wz*Tlag.

I get the sense that your idea is consistent, but I'm not able to understand fully from your explanation. Thanks for taking the time. So far, your relation:

Vz/Wz = delta Vx/Vx0

is new to me, and is a big step toward unraveling the mystery and significance of lag time.

Thank you,
-Bryan

 

[Gene Beggs]

Yeah, you guys are gettin' close!

Yeah, what Bryan and Keith said. I knew it all along; wondered how long it would take you guys to see it.

And then someone said, "Beggs, get outta' here; you are so full of it."

Seriously guys, I'm impressed! When formulas, equations and higher math are introduced old Beggs is in over his head.

Sure would like to know more about Keith a.k.a. MKS

Gene Beggs

 

[mks]

Corrected time and average velocities

Now the problems start.

You stated that the Tlag/Tvac ratio would be equal to the above fractions. Sounds good, but it's plotted in green, and clearly doesn't track. Furthermore, the Z/Zmax fraction is plotted in purple, and doesn't track with anything either.

I'm also confused by your PS.

You said:


But Vz isn't constant, it changes continuously.

You also said:


But earlier you defined delta t as the time of flight in the absence of drag (vacuum tof). We know that Z = Wz*Tlag.

I get the sense that your idea is consistent, but I'm not able to understand fully from your explanation. Thanks for taking the time. So far, your relation:
is new to me, and is a big step toward unraveling the mystery and significance of lag time.

Thank you,
-Bryan

Bryan,
I made two simplifications, one of which I now see wasn't needed.

1) I assumed delta t << t where t is the no drag time, just to make the equations simpler, but this restricts the solution to short range (note that the green and purple curves start with the same slope) and doesn't affect the simplicity much. If this assumption is relaxed, t is the total time of flight including drag. Take the deflection equation Z = Wz delta t and divide through by Zmax = Wz t and you get

Z/Zmax = delta t/t

Using t as total time of flight should make your green and purple curves overlap for all ranges.

2) The other assumption, to avoid calculus, was to use average velocities. McCoy equation 7.26 is Vz/Wz = 1-Vx/Vx0 or Vz/Wz = delta Vx/Vx0, which confirms that your red and green curves should overlap. This equation works whether Vz and Vx are their values at the range or their average values over that range, so long as they are both treated the same. Now Vx0=X/t0, where t0 is the no drag time of flight, and using average velocity Vx=X/t, so the equation becomes

Vz/Wz = 1-t0/t = delta t/t

which is the equation I originally showed, but with corrected t.

Now for velocities based on their values at range, for linear increases in Vz and delta Vx (not exactly correct, even for constant BC), their values at the range are double the average. So using values at range, the equation becomes approximately

Vz/Wz = 2 delta t/t

and the red and blue curves should have about double the slope of the green and purple. It looks like they are close to that, and with the correction for t, should be closer.

I used the average Vz (lets call it Vzbar) in the PS also. As you recognize, deflection is actually the integral of Vz(t) with time, but by defining Vzbar=integral[Vz dt]/t, the equation in the PS works.

Sorry for the confusion. I was trying to make the explanation as simple as possible, but you quickly found the oversimplifications.

Cheers,
Keith

 

[alinwa]

Man!!! This is cooler than wrasslin'!!!!!!!!

Brian and Keith...... keep it up men!

You're making the math sensible.

rooahhhhh



al

 

[D R Greysun]

Oh for pity sakes, are you guys from outer space? It starts here X it goes there Z, Y I don't need to know. Z is all that matters!

D R

 

[Tony Shankle]

drag model

Sorry I haven't been on for a few nights, had 100 necks to turn and I was taking my time.

So we know gravity and DRAG are the major/most powerful forces acting on our bullet and we see pictures of flowfields but the bullet is always in a "no wind" condition in these pics and charts. So what do you think that flowfield looks like when the bullet is nose into the wind and "skidding" sideways downrange? What happens to that BC model in the "skidding" position?

Here is a good visual for shapes...

Now here is a series of pics of a supersonic bullet coming close to a solid object and the flowfield disruption...

These pictures are as close as I can find to something disrupting a flowfield. What does it look like in a stiff wind? So you think when the bullet angles left or right maybe that drag model starts looking more like the round ball in the top diagram? Since there are so many different bullet shapes, and so many angles of wind how would you apply mathmatics to a wind compentent due to the bullet shape changing entirely?

In our other thread the first post mentioned a side BC to which I was quick to say was a "flawed" question...now that side BC in this case is starting to look like a good thing to me. Along with the angle of the wind wouldn't that increase the accuracy of my ballistic programs?

I just couldn't let it go Al, Bryan didn't call me back last week so I thought I would start some trouble! Get that chopper ready, we may take another ride soon!

 

[bsl135]

Keith,
I haven't had a chance to really look into your response. I'm sure I can get the green and purple to match if I use actual tof in the denominator, because then both ratios are just scaled by the wind speed and the fractions are the same (but different from the red and blue).

Tony,
Sorry I didn't return your call.
Wind doesn't blow on the side of a bullet. The idea of a 'side BC' is popular in discussions like this, but incorrect. The shock structure around a bullet flying in a crosswind is purely conical, and axisymmetric with respect to the bullets axis, and the oncoming airflow, albeit at an angle with the ground. You only need to know the bullet's axial drag coefficient (regular BC) to know it's crosswind deflection.
The bullet has an initial angle of attack when it's fired into a crosswind, but aligns itself within a few precession cycles (few yards). Then at long range, the bullet develops a yaw of repose and creates 'spin drift' as gravity bends the trajectory, but that's very minor, only steering the bullet off course ~1MOA at 1k yards.
There are reasons for normal stable bullets to fly with small angle of attack, but wind isn't one of them.

Al,
Your round balls blowing in the wind only complicate the visualization of drag and wind deflection, but it's no different. The drag on the round ball is still angled to the LOS when it's flying in a crosswind, regardless of whether it's spinning, or what axis it's spinning on. Drag and deflection still happen the same way.
Now; a round ball that's spinning on an axis not aligned with the oncoming air flow will be generating some magnus force, which could be more dominant than wind deflection (depending on spin rate and how misaligned the spin axis is) and veer off in a direction dictated by magnus (like a curve ball). This doesn't mean the round ball is immune from wind deflection, it just means that magnus can be the dominant force.
I believe I read somewhere (probably on here in TMOAWDT) that the reality of a 'round lead ball' is that when fired, it obturates (smushes) to fill the bore somewhat, and emerges from the muzzle with a hemispherical front and rear, and with a cylindrical midsection. From a rifled bore, the 'ball' is spun, and actually has a 'long axis' that is parallel to it's trajectory just like a normal bullet, so spin on random axis would be unlikely even for a 'round' ball.

-Bryan

 

[alinwa]

From a rifled bore, the 'ball' is spun, and actually has a 'long axis' that is parallel to it's trajectory just like a normal bullet, so spin on random axis would be unlikely even for a 'round' ball.

-Bryan

But.....

A decelerating gyroscope (roundball) precesses all on its own. I doesn't act like a spin stabilized bullet which is constantly reacting to and overriding this tendency. (irrelevant to this discussion, but I'm just sayin'.... ) and a round ball takes 10 seconds to go 1000yds!!

Not too awful hard to visualize "curve ball" there......

al

 

[Tony Shankle]

Tony,

Wind doesn't blow on the side of a bullet. The idea of a 'side BC' is popular in discussions like this, but incorrect. The shock structure around a bullet flying in a crosswind is purely conical, and axisymmetric with respect to the bullets axis, and the oncoming airflow, albeit at an angle with the ground. You only need to know the bullet's axial drag coefficient (regular BC) to know it's crosswind deflection.


-Bryan

Bryan,

My comment about the side BC was not the point and was tounge in cheek if you will. My point is that paper BC (and even effective BC) changes when that bullet points into that wind. While that COG is continuing on its flight path the COP is turned into the wind which presents a different form factor as it travels forward. Mathematically, BC is the ratio between sectional density and coefficient of form. Even if it is just 1 degree how could it NOT effect the coefficient of form thus affecting the BC? Also, how could the compression area and shock waves remain exactly the same when the forward moving form factor has changed? If what you are saying is true then we could fire them sideways and backwards and still have the same BC.

 

[alinwa]

Bryan,

My comment about the side BC was not the point and was tounge in cheek if you will. My point is that paper BC (and even effective BC) changes when that bullet points into that wind. While that COG is continuing on its flight path the COP is turned into the wind which presents a different form factor as it travels forward. Mathematically, BC is the ratio between sectional density and coefficient of form. Even if it is just 1 degree how could it NOT effect the coefficient of form thus affecting the BC? Also, how could the compression area and shock waves remain exactly the same when the forward moving form factor has changed? If what you are saying is true then we could fire them sideways and backwards and still have the same BC.

Tony, tony, TONY........


IMO this whole presumption is flawed!

The coefficient of form is always centered up on the flowfield if the bullet's stable. Left wind, right wind, NO wind it's all the same to the BC........

From the bullet's point of view, the FLOWFIELD'S point of view, the bullet IS NOT "presenting sideways".............. Absolutely the ONLY "sideways" is yaw of repose or equilibrium yaw.

From a ground view the bullet LOOKS to be sideslipping but it ain't! This ain't no aeroplane!

COG is NOT "continuing on it's flight path".....it's trying to but it can't. COG is constantly right behind COP in the flow.....(except for the lag effect of equilibrium yaw) ..... so the compression area and shockwaves DO stay the same except for fishtail fluctuations or flutter as the nose adjusts to changeups.


No sideslip... NO sideslip... nonoNO!!

NO shockwaves on the side of the bullet.......


And please pleasePLEASE don't make me ride the lightnin' again!!!!! I still feel like Satchmo from you tucking in.....

Nor take the chopper out neiyther. I haven't even put the seat back in yet....... (((I hadda' wear it for several days..... in fact, I was down at Talladega and dudes were like "HEY! That's a great idea! Gotta' GET me one 'a those! The Ol' Lady gets tired of draggin' our chairs around and this way I wouldn't have to put my beer down! Whar'dja pick that up!??"

I told 'em "that's a long story harold".......)))






al

 

[mks]

hula hoop dance

Bryan,

My comment about the side BC was not the point and was tounge in cheek if you will. My point is that paper BC (and even effective BC) changes when that bullet points into that wind. While that COG is continuing on its flight path the COP is turned into the wind which presents a different form factor as it travels forward. Mathematically, BC is the ratio between sectional density and coefficient of form. Even if it is just 1 degree how could it NOT effect the coefficient of form thus affecting the BC? Also, how could the compression area and shock waves remain exactly the same when the forward moving form factor has changed? If what you are saying is true then we could fire them sideways and backwards and still have the same BC.

You may be talking about secondary effects. While the first order approximation is that there is no wind-induced yaw after the first few cycles of precession, the nose of the precessing bullet is looping around the direction of the resultant wind that it sees. And the shock wave cones are doing a hula hoop dance around the bullet, but the BC doesn't change much, at least for small angles. For instance, one degree of yaw on a 50 cal bullet increases its drag coefficient by a little less than 1% [McCoy page 82]. The increase is proportional to the square of the angle, however, so it does grow quickly as the bullet gets more sideways.

Cheers,
Keith

 

[alinwa]

You may be talking about secondary effects. While the first order approximation is that there is no wind-induced yaw after the first few cycles of precession, the nose of the precessing bullet is looping around the direction of the resultant wind that it sees. And the shock wave cones are doing a hula hoop dance around the bullet, but the BC doesn't change much, at least for small angles. For instance, one degree of yaw on a 50 cal bullet increases its drag coefficient by a little less than 1% [McCoy page 82]. The increase is proportional to the square of the angle, however, so it does grow quickly as the bullet gets more sideways.

Cheers,
Keith

WHOAAHhhhhh back here......

And where in McCoy's text does he mention wind? The section mentioned (page 82) is about wind tunnel testing and as such refers to CYCLICAL yaw, not wind induced yaw..... Cyclical pitching DOES present sideways but wind-induced does not. In any event, cyclical pitching moments are transient whether "wind induced" or otherwise. ONLY equilibrium yaw is relevant to BC unless you're saying that firing through a switchy crosswind will cause a decrease in BC?

To put it another way, YES pitching yaw presents more frontal area but it does this wind or no wind..... it's not a wind effect.

And BTW, to say that 1% is negligible is somewhat misleading. 1% of WHAT? Like drop a .465 BC to .461BC???? And isn't this figured into the "coefficient" since coefficients of drag are generally measured empirically? 1% of a coefficient is somewhat hard to quantify but if one were to consider a 1% change in velocity it becomes apparent that it's huge, not insignificant. A 1% velocity change at 1000yds WILL knock you out of contention entirely! For a typical 1K load the result would be 6" of vertical. I think that a 1% difference (fluctuation) in BC if real is significant.

al

 

[Vibe]

From a ground view the bullet LOOKS to be sideslipping but it ain't! This ain't no aeroplane!


al

Heck Al, from the ground an aeorplane LOOKS to be sideslipping - but it is no more traveling sideways to the wind than our bullet is. And is, in reality, less able to.

 

[mks]

WHOAAHhhhhh back here......

And where in McCoy's text does he mention wind? The section mentioned (page 82) is about wind tunnel testing and as such refers to CYCLICAL yaw, not wind induced yaw..... Cyclical pitching DOES present sideways but wind-induced does not. In any event, cyclical pitching moments are transient whether "wind induced" or otherwise. ONLY equilibrium yaw is relevant to BC unless you're saying that firing through a switchy crosswind will cause a decrease in BC?

To put it another way, YES pitching yaw presents more frontal area but it does this wind or no wind..... it's not a wind effect.

And BTW, to say that 1% is negligible is somewhat misleading. 1% of WHAT? Like drop a .465 BC to .461BC???? And isn't this figured into the "coefficient" since coefficients of drag are generally measured empirically? 1% of a coefficient is somewhat hard to quantify but if one were to consider a 1% change in velocity it becomes apparent that it's huge, not insignificant. A 1% velocity change at 1000yds WILL knock you out of contention entirely! For a typical 1K load the result would be 6" of vertical. I think that a 1% difference (fluctuation) in BC if real is significant.

al

Al,
Just saying that since we all agree bullets reorient themselves to point, on average, into the wind, there are other things going on. Precession does cause the bullet to not be pointed exactly into the wind, and BC decreases because of it. Yes, as in 0.8% decrease in BC for the 50 cal bullet. And you are correct, that is a big deal at 1000 yds, but it occurs even without wind.

Here is a question to muddy the waters further. I haven't studied this enough to know the answer. How will the nose of a bullet respond to a vertical wind? It flies nose high and drifts right due to gyroscopic effect in response to upward pressure at the CP and downward weight at the CG when it gains downward velocity due to gravity. Wouldn't it do the same thing if the wind were blowing upward on it? It may be the lack of a force on the CG, like gravity, in the horizontal direction that allows it to turn into a horizontal wind.

Cheers,
Keith

 

[Lynn]

If 1% variation in a bullets BC number is significant let me know whose bullets you are using that are half that.
Waterboy

 

[alinwa]

Heck Al, from the ground an aeorplane LOOKS to be sideslipping - but it is no more traveling sideways to the wind than our bullet is. And is, in reality, less able to.

Yeahhh, I gotta' agree.

Poorly illustrated.

nice catch

al

 

[alinwa]

How will the nose of a bullet respond to a vertical wind?


Cheers,
Keith

Exactly like it does to a "regular" wind.

I guess I've thought I had this covered many times in prior posts??? It goes to show, communication, CLEAR communication is hard!

al