The Effect a Raindrop Has On a Bullet

[Win_94]

I had asked a question on the effect Recoil has on a bullet's trajectory and got wonderful help from this site. Thank you! Now I have another question...

I used to think rain had no effect on a bullet until I realized I wasn't actually hitting any raindrops, when I started hitting them, I noticed a big difference in the bullets point of impact.
Unfortunately, this has caused an argument on specific firearm forum and I came here to ask if someone can do the math on such an instance.

First a video of me shooting in the rain, hitting a raindrop, whereas my point of impact has changed being about 12 inches lower than normal at 350 yards.

https://www.youtube.com/watch?v=dFyQLS2KXBc

The picture displayed above shows the cloud made by hitting the raindrop. The strike was maybe 50 yards downrange of the muzzle, whereas it needed to then travel 300 yards further.
Here is the variables of that particular shot.

Bullet weight: 150grains
the bullet design is like the one on the left.

Muzzle velocity; 2360fps, every 50 yards it drops in velocity about 100fps.
Normally, it drops height-wise 33 inches by the time it reaches 350 yards.

I don't know what else you would need, just ask and I may be able to help.

I know that a twig can make a huge difference in trajectory with the same setup at 575 yards.

https://www.youtube.com/watch?v=wMvXm_GWUoc

That looks to be about 35 inches low; which would be somewhat more than the raindrop.

Thanks in advance!

Edit by Borek: fixed the youtube tags.

 

[Drakkith]

What's your question? I didn't see one.

 

[Win_94]

What's your question? I didn't see one.

Is there an effect? If so, too what degree?

They are saying there is no effect.

 

[Win_94]

Others have suggested that the bullet's shockwave doesn't let water touch the bullet.
You can see that in this video.
"www.youtube.com/watch?v=Num9TR7wlrw"

Which I think is preposterous. If the air is hitting the bullet then reflecting off forming the shockwave, then something denser than air should be able to also hit it.
Sound reasonable?

 

[Drakkith]

If the pullet passes through a water drop, it must move the water out of the way to do so. This requires that it expend kinetic energy. Whether the water actually touches the bullet or not is irrelevant. Even if it doesn't, the water is still touching the air and the air is still touching the bullet. If the water is moved it MUST come from the bullet, as there is nothing else there to move the water.

Now, as to how much effect it has, I cannot answer that.

 

[Win_94]

If the pullet passes through a water drop, it must move the water out of the way to do so. This requires that it expend kinetic energy. Whether the water actually touches the bullet or not is irrelevant. Even if it doesn't, the water is still touching the air and the air is still touching the bullet. If the water is moved it MUST come from the bullet, as there is nothing else there to move the water.

Now, as to how much effect it has, I cannot answer that.

Thank you!

 

[Andrew Mason]

I know that a twig can make a huge difference in trajectory with the same setup at 575 yards.

https://www.youtube.com/watch?v=
Link:

That looks to be about 35 inches low; which would be somewhat more than the raindrop.

Twigs do not deflect bullets in the sense of changing the bullet's momentum enough to change the bullet's direction significantly. Twigs can destabilize bullets slightly (slight yaw). This causes the bullet to prescribe a spiral path over a long distance (the 575 yards in that video is a pretty long distance for a shot) which can cause it to miss the target. This has been tested: see L. F. Moore, "Bullets in the Brush - Where do they go?", American Rifleman, Vol. 116 No. 9, Sept. 1968 p. 61.

As far as a raindrop is concerned, I don't see how a single raindrop would cause the bullet to deflect or destablize. The momentum of a single raindrop is insignificant compared to the bullet momentum.

AM

 

[olivermsun]

As far as a raindrop is concerned, I don't see how a single raindrop would cause the bullet to deflect or destablize. The momentum of a single raindrop is insignificant compared to the bullet momentum.

I don't follow the "momentum" argument. If a pool ball were to strike a glancing blow on a stationary ball with the mass of a large marble (little mass, no momentum at all), would it be so surprising if the trajectory of the ball were very slightly, but measurably, changed?

 

[mikeph]

If you're talking about a cue ball striking another at 45 degrees and suffering a noticeable traverse deflection I'd agree- the mass ratio of a raindrop to a bullet could easily be big enough to give a noticeable (~cm) deflection over a path of 350 m. However, the bullet is not a sphere and the raindrop is not rigid, so I don't think you'll get anywhere near that deflection. Most likely the bullet will glance the droplet, and only a small fraction of water will deflect sideways, with most of it being pushed forwards.

 

[olivermsun]

IMost likely the bullet will glance the droplet, and only a small fraction of water will deflect sideways, with most of it being pushed forwards.

The OP does have an "observation" of the bullet trajectory apparently being pushed down 12" compared to a dry day. So if it's not the raindrops, then why?

I suppose one might ask how much net momentum change the bullet sees during its flight from the sum of all raindrops "encountered" plus any downward airflow associated with the rain.

If you being to consider airflow, one make also like to ask whether the OP has any information as to whether the wind is exactly the same as usual, e.g., is there a headwind associated with the rainy weather?

 

[mfb]

If the collision between the bullet and the raindrop is asymmetric, it can deflect the bullet - water is pushed away at one side, but not (or less) at the other side. The motion of the water is negligible.

I don't have values for those bullets, but here is a quick check: Approximating the rain drop as cube, 10mm^2 contact area, 3mm thickness (in flight direction), and 20 degree contact angle (only determined by the shape of the bullet). Let's assume that just one side of the bullet hits the water. 10g mass of the bullet, which has a velocity of v.
The displaced water has a mass of 30mg, and gets a velocity of sin(20°)v=0.34v. The bullet gets a velocity component of 30mg*0.34v / (10g)=0.0010v orthogonal to its direction of motion. After 110m (~350 yards), this gives a deflection of 11cm (~4 inch).
Not so far away from the experimental value, so it might be possible. You can replace my guessed values with more realistic values if you like.

The reduced horizontal component leads to a slightly longer flight time, this can give an additional deflection (always downwards).

 

[Win_94]

Would this be consistent with the consensus?

224 second video; Water Droplet POI Shift Test

I stupidly took the first shot rushed and didn't hold for wind; That determined where my POA would be for the entire test.

This video is to provide data as to, "Can a raindrop affect POI? ...if so, "By how much, and in what way?"

I think the video lends credence to my assertion that there is an affect on point of impact[POI] in the rain. The affect can be by 4MOA in amplitude. The placement of the impacts are random.
I see no pattern in height wise deviation; I no longer think there is a velocity issue. The random pattern dictates that the bullet is being deflected dependent on which side of the bullet's axis hits water.

Something to note: of all the shots that hit water, only one was inside the kill-zone of a deer; as per the first 3 shots being the center point.

The only shot that I called "pulled" was shot 4; I have total confidence that all other shots had the same POA.
I'm a competent enough marksman. The first shot was after estimating change of POI from a Hornady 208gr A-Max, to the 150gr FMJ. I made a 1/2 inch error in height...

The furthest windflags are set at 20 yards from the target, then 100 yards, then 220 yards. We can't see the 150 yard flag from here.

This test specifically eliminates "shooter error" from the equation of shooting in the rain. The age old rule is, "Rain affects the shooter far more than the gear."
The temperature variation was 6ºF.
The wind was from close to 12 o'clock most the time. The strongest crosswind was during the best group.
No visual impairments whatsoever.
I do agree now that the effect is random.
This statement, "deflection is not significant enough" is incorrect. One needs to know one's limitations; knowing this "affect" is real, I can then take that information into consideration if the need arises.

I'll upload the close-ups of the water hits soon.

 

[Win_94]

Also, I have an experiment on if a shockwave has enough energy to vaporize a raindrop before the bullet hits.

Shockwave Energy Test

If it has the energy to vaporize water, what would it do to a wet napkin?

 

[Win_94]

is there a headwind associated with the rainy weather?

Headwinds and tailwinds have negligible affect on bullets, especially that slight wind in the original video. Wind is more a lateral than vertical, unless the wind is blowing fast or steep, uphill or downhill. When marksmen talk about "windage" they are talking about horizontal drift.

 

[OmCheeto]

If the collision between the bullet and the raindrop is asymmetric, it can deflect the bullet - water is pushed away at one side, but not (or less) at the other side. The motion of the water is negligible.

I don't have values for those bullets, but here is a quick check: Approximating the rain drop as cube, 10mm^2 contact area, 3mm thickness (in flight direction), and 20 degree contact angle (only determined by the shape of the bullet). Let's assume that just one side of the bullet hits the water. 10g mass of the bullet, which has a velocity of v.
The displaced water has a mass of 30mg, and gets a velocity of sin(20°)v=0.34v. The bullet gets a velocity component of 30mg*0.34v / (10g)=0.0010v orthogonal to its direction of motion. After 110m (~350 yards), this gives a deflection of 11cm (~4 inch).
Not so far away from the experimental value, so it might be possible. You can replace my guessed values with more realistic values if you like.

The reduced horizontal component leads to a slightly longer flight time, this can give an additional deflection (always downwards).

I like your answer the best.

Does this problem remind anyone of the "deflecting an asteroid" problem?

If you can deflect an asteroid by a tiny fraction from its original trajectory, then over the course of millions of miles of the remainder of its flight, its ultimate relevant point in space(URPIS), compared to its un-disturbed URPIS, will tend not to eliminate all life as we know it, on our humble little planet.​

----------------------
In all honesty, I have never worked on the "deflecting an asteroid" problem.
And my new acronym indicates that I've been hanging out with young kids a bit too much.
un-DURPIS?
What's with this durp speak nowadays?
Derp!

 

[Win_94]

and 20 degree contact angle (only determined by the shape of the bullet).

This is the bullet.

.308 inches in diameter, 1.125 inches long, 150 grains, (there are 7000 grains in a pound.)
It has a ballistic coefficient of .398.

 

[jim mcnamara]

Using mfb's very good estimate with a likely candidate projectile:
150 grains == 9.7198365g
308 Hornady 150gr BTSP lists 2560fps muzzle velocity ~= 780m/s. So there should be a lot of drop: 32.47 inches at 350 yards ignoring wind effects and ignoring varying transit times due to drag: exterior ballistic coeffcient.

mfb's "guess" deflection of ~11cm @110m is pretty darned good. It assumes the bullet encounters the rain drop on exiting the barrel. I get a deflection of ~11.6cm @110m or ~4.5 inch. This model also ignores any disturbance to the stability of the rotating projectile.

So 4+ inches, which, relative to drop, is not all that great but is definitely there. Let's not consider a rainstorm, please.

 

[Andrew Mason]

I don't follow the "momentum" argument. If a pool ball were to strike a glancing blow on a stationary ball with the mass of a large marble (little mass, no momentum at all), would it be so surprising if the trajectory of the ball were very slightly, but measurably, changed?

Let's start with a 6.5 mm bullet that is 3 cm long weighing 10 g. with a muzzle velocity of 2000 fps = 610 m/sec. The momentum of the bullet is, therefore, 6 kgm/sec.

Now take a raindrop. mass = 100 mg. (10 raindrops = 1 gram) which has a diameter of about 6 mm. I think that is a fairly big raindrop. see: http://hypertextbook.com/facts/1999/MichaelKodransky.shtml And let's say it is falling at 10 m/sec which appears to be the terminal velocity of large raindrops: http://hypertextbook.com/facts/2007/EvanKaplan.shtml

The mass of the raindrop is .1 g = .0001 kg. So its momentum is .001 kg m/sec. The momentum of the bullet is, therefore, 6000 times greater than the momentum of the raindrop. So if you add the momentum vector of the raindrop to the momentum vector of the bullet you get a vector that differs from the bullet momentum vector by an angle of 1/6000 radians = .01 degrees (1 rad. = 57.3° ≈ 60°). So, over 100 metres, the bullet will deflect 100tan(.01°) = .015 m = 15 mm. So over 500 m it will deflect about 75 mm.

And that is only if the entire raindrop imparts its momentum to the bullet. That can only happen if the raindrop has enough time in contact with the bullet to impart its entire momentum to the raindrop. But since the bullet is traveling at 610 m/sec or 33x610 bullet lengths per second (ie. about 20,000 bullet lengths per second), the raindrop cannot be in contact with the bullet for more than 1/20000th of a second. In that time, the raindrop falls 10000/20000 = .5 mm or about 1/12th of its diameter.

The bottom line is that a single raindrop has too little momentum and too little time to impart that momentum to the bullet to cause any material deflection of the bullet.

AM

 

[jim mcnamara]

Andrew -

Consider the momentum transfer from the "projectile system" to the raindrop. What happens if the raindrop is broken into large numbers of very high velocity small droplets?

The projectile has to have lost momentum, the raindrop fragments have gained momentum. The momentum change is proportional to the very large change in velocities of thousands of small water droplets. And the mass. Which component of momentum, mass or velocity, is represented as the the squared value?

This is analogous to 150 pounds of wet sand in a plastic bucket at the end of a guardrail colliding with a 2500lb vehicle going 50mph. The vehicle loses lots of momentum very, very quickly. This works because the sand is at rest and then accelerated in all directions, even though the mass differences are large.

This is not to say the project/raindrop effect is massive, but I believe you are ignoring impulse = J. And the magnitude of the subsequent momentum change.

Or. Have you ever ridden on any kind of open vehicle at highway speed in rain? Hurts doesn't it. If the raindrops are so "puny" why does it hurt? What would the same experience do to your skin at 740m/s?

 

[Andrew Mason]

Or. Have you ever ridden on any kind of open vehicle at highway speed in rain? Hurts doesn't it. If the raindrops are so "puny" why does it hurt? What would the same experience do to your skin at 740m/s?

Perhaps you could show us what you mean using some numbers. We are talking about one raindrop colliding with a bullet. The raindrop simply cannot apply enough force to the bullet for a long enough time to impart any significant momentum to the bullet. The bullet simply passes through the edge of the raindrop and pushes a few of its molecules out of its way.

AM

 

[A.T.]

The bottom line is that a single raindrop has too little momentum...

The initial momentum of the drop is not relevant. What matters is how much momentum is transferred to it. An obstacle with the initial momentum of zero, could still have a significant effect.

The momentum of the bullet is, therefore, 6000 times greater than the momentum of the raindrop

What matters is the mass ratio, not the momentum ratio.

So if you add the momentum vector of the raindrop to the momentum vector of the bullet

This assumes that there is only vertical momentum transfer, and that the drop has zero momentum after collision. If the drop hits the upper side of the cone-like tip the water will be up flying up again, so you have more vertical momentum transfer than if it just stopped falling. And you have a horizontal component.

I'm not saying it has a significant effect, but your approach to show this is not convincing. But I admit that it is a complex situation. Here an idea:

Instead of the drop, consider an elastic collision with a small falling ball of the same mass as the drop. Assume the force acts at 45° to the vertical at the upper side of the tip-cone. Solve for momentum conservation in 2D.

The effect of increased precession of the spinning bullet, affecting it's aerodynamics during the rest of the flight, might be the main issue here, just like with the twigs.

 

[mfb]

The bullet simply passes through the edge of the raindrop and pushes a few of its molecules out of its way.

And this gives those "few molecules" (can be the full rain drop) a high velocity (much more than 10m/s, as it is proportional to the bullet velocity) orthogonal to the bullet velocity.

(there are 7000 grains in a pound.)

I'm converting (nearly) everything to SI units anyway.


Using my value of 30mm³ for a rain drop, assuming 10m/s downwards velocity and 2cm/hour of rain, we have about 20 drops per cubic meter. The cross-section for a collision is approximately 1cm^2, therefore the probability of a hit is about 0.002/m. Within 100m* (or ~110 yards*), this gives a total probability of ~20% to hit a rain drop. Hits close to the target will lead to smaller deflections, of course.

*hmm, I misread the original distance of 350 yards as 350 feet (where is the point in having 4 different length units with weird conversion factors anyway?), but my calculations are still consistent.

 

[Andrew Mason]

The initial momentum of the drop is not relevant. What matters is how much momentum is transferred to it. An obstacle with the initial momentum of zero, could still have a significant effect.

What matters is the mass ratio, not the momentum ratio.

What really matters is the bullet speed.

To deflect the bullet you have to cause it to move a significant distance in a direction perpendicular to its initial path before it reaches the target distance. At 610 m/sec the bullet takes .164 seconds to travel 100 m. And at that speed, the time of contact between the bullet and water drop is 1/20000th of a second (.03m/610msec^-1)

So to deflect the bullet 1 cm from the initial path at the target distance the drop has accelerate the bullet's lateral speed of from 0 to 1cm/.164 = .061 m/sec. in the 1/20000th of a second in which the water drop and bullet are in contact. So it has to impart an acceleration of a = Δv/Δt = .061/(1/20000) = 1220 m/sec or a force of 12 Newtons. I would be interested in knowing how the raindrop can exert that amount of lateral force on the bullet.

AM

 

[mfb]

What really matters is the bullet speed.

As shown before, the bullet speed cancels in the equations if it has to push water away with its conical section.

The bullet has to push the rain drop away with that force (12N? Does not matter!) in order to proceed in its approximate path.

 

[Andrew Mason]

And this gives those "few molecules" (can be the full rain drop) a high velocity (much more than 10m/s, as it is proportional to the bullet velocity) orthogonal to the bullet velocity.

This would be true if the raindrops were steel balls. What holds the raindrop together while part of it withstands the force of the bullet impact?

AM

 

[Atom Ant]

I'll take a shot at it

Wait! I've had time to think about this. First of all, there's no way your hitting a 6" target at 350yds with a lever action 30/30 and open sights. You mean 350ft, right? Next!

I'm going to say that there's no way that a rain drop can hit a speeding bullet, since the force out in front of the bullet will be much more than the weight of the raindrop.
I'm thinking.

 

[A.T.]

What holds the raindrop together while part of it withstands the force of the bullet impact?

The raindrop doesn't stay together, but the water still needs to get out of the way.

If the drop hits one half of the tip-cone, it will be sprayed in a 180° arc laterally. Compared to single object (of the same mass as the drop) deflected laterally this means less net lateral momentum transfer (because some of it cancels), but it is still the same order of magnitude.

 

[A.T.]

What matters is the mass ratio, not the momentum ratio.

To elaborate further on this:

In the rest frame of the bullet the obstacle has the momentum $p$. $\Delta p$ is the minimal change in momentum to move it out of the way. $\Delta p$ just removes the obstacle's momentum perpendicular to the cone surface. The obstacle doesn't bounce off, just slides along the cone outwards after impact. $\Delta p_{\perp}$ is the lateral component of $\Delta p$.

$\Delta p_{\perp}=cos \alpha \cdot sin \alpha \cdot p$

$\Delta v_{\perp bullet}m_{bullet}=cos \alpha \cdot sin \alpha \cdot v_{bullet}m_{obstacle}$

$atan \beta = \frac{\Delta v_{\perp bullet}}{ v_{bullet}} = cos \alpha \cdot sin \alpha \cdot \frac{m_{obstacle}}{m_{bullet}}$

Where $\beta$ is the deflection angle, which depends only on the mass ratio and shape. For $\alpha = 45°$ and the mass ratio 1/100 you get $\beta=0.3°$ or 0.5m offset after 100m.

Obviously a water drop will be sprayed in different lateral directions, reducing the lateral deflection. On the other hand we assume a perfectly inelastic collision, where the water just flows along the cone. If the water bounces off $\Delta p$ and thus the deflection will increase.

 

[A.T.]

force of 12 Newtons. I would be interested in knowing how the raindrop can exert that amount of lateral force on the bullet.

Talking about forces here doesn't offer much intuition, because for such short impacts they are always large. Do your calculation for a 0.1g drop, that hits a wall vertically at 610m/s. The deceleration time is the diameter divided by the speed. What is the average force there? The lateral force on the bullet is some fraction of this, depending on the shape and the impacted area.

 

[mikeph]

AT, those calculations make sense, but I think you overestimate the effect by only considering momentum in the plane of the two centres of mass- for a mass ratio of 1/100 you need a pretty huge raindrop, and you're assuming that the entire mass of this raindrop will be diverted with upwards momentum along the side of the bullet.

This is only strictly true for the water which lies in the plane- a lot will be deflected outwards of the paper/screen, and the same amount will be deflected inwards towards the screen. Although these two z-direction impulses cancel, you are counting them in your calculation.

Furthermore, the bullet angle you use is only 45 degrees at one radial point, and that point might be a tiny fraction- if you average the angle with respect to the bullet's radius (ie. proportionate to where the raindrop might impact) I think the effective angle will be much smaller, as they tend to narrow to a single point at the top rather than a perfect hemisphere. Some of the water will be deflected more, but I think most of it will be deflected far less than along a 45 degree slope.

If your mass ratio is 1/100 then assuming the bullet weight is about 5 g, the raindrop radius must be about 2.5 mm, whereas the bullet radius is about 3.8 mm. They're pretty similar in size and that makes me think this overestimate is quite big.

 

[A.T.]

This is only strictly true for the water which lies in the plane- a lot will be deflected outwards of the paper/screen, and the same amount will be deflected inwards towards the screen. Although these two z-direction impulses cancel, you are counting them in your calculation.

Yes I mentioned the spraying.

I think the effective angle will be much smaller,

45° was just an example. Also note that we assume perfectly inelastic collision. If the water bounces off then the effective angle is different from the shape of the bullet.

If your mass ratio is 1/100 then assuming the bullet weight is about 5 g, the raindrop radius must be about 2.5 mm, whereas the bullet radius is about 3.8 mm. They're pretty similar in size and that makes me think this overestimate is quite big.

That is a good point. I think most of the effect could come from destablisation/precession, rather than from direct momentum transfer. But that is even more difficult to estimate.

 

[A.T.]

As shown before, the bullet speed cancels in the equations if it has to push water away with its conical section.

An intuitive reason that the transferred momentum cannot depend on the momentum ratio or the speed ratio is that the those ratios are frame dependent, while the transferred momentum is absolute. The transferred momentum can only depend on other frame invariant variables like the mass ratio.

 

[Andrew Mason]

If you're talking about a cue ball striking another at 45 degrees and suffering a noticeable traverse deflection I'd agree- the mass ratio of a raindrop to a bullet could easily be big enough to give a noticeable (~cm) deflection over a path of 350 m. However, the bullet is not a sphere and the raindrop is not rigid, so I don't think you'll get anywhere near that deflection. Most likely the bullet will glance the droplet, and only a small fraction of water will deflect sideways, with most of it being pushed forwards.

I tend to agree. But the "sideways" is not all the same direction. It would be all around the bullet so it would not prefer one "sideways" direction over the other.

As shown before, the bullet speed cancels in the equations if it has to push water away with its conical section.

The bullet has to push the rain drop away with that force (12N? Does not matter!) in order to proceed in its approximate path.

I agree that the bullet has to push away whatever is in its path. Whether it pushes the entire raindrop depends on whether it hits the entire raindrop. If the 6.5 mm bullet hits the entire 6mm diameter raindrop it will have to explode practically through the middle of it. In that case it is only the momentum of the raindrop before the collision that would affect the lateral momentum of the bullet.

So what you are suggesting is that the bullet just hits half the raindrop with half a bullet.

Analysing that is a bit more complicated. You are suggesting that it imparts momentum to the raindrop to one side of the bullet's path and the bullet experiences a net sideways momentum that is equal to the mass of the raindrop x a very high speed, which is sufficient to change the path of the bullet significantly.

I don't see that. I would expect that what happens is that it imparts momentum to the molecules comprising the part of the drop that it passes through over a range of angles from 0-180 degrees around one side of the bullet. It is the sum of all those momenta that imparts a net momentum to the bullet. And it will not all be sideways - much of it will be forward momentum. So it is only a small portion of the part of the raindrop that the bullet passes through that imparts much net lateral momentum to the bullet.

I expect the physics would be complicated to analyse accurately. The American Rifleman article that I referred to earlier shows the results of an experiment firing rifle bullets at wood dowels at 25 yards and observing the bullet orientation and path over the ensuing 75 yards using 32 paper screens. The paths of the bullets was spiral, a consequence of the loss of stability rather than lateral momentum imparted by the contact with the dowel. Some of the tests involved striking just the edge of the dowel (see Fig. 9 from the article). This caused a deflection of about 9 inches over 75 yards.

I just can't see a single raindrop having nearly as much effect as a hardwood dowel.

AM

 

[mfb]

So what you are suggesting is that the bullet just hits half the raindrop with half a bullet.

My calculations assumed a raindrop diameter of ~3mm, hitting one side of the bullet. I expect that sidewards deflection reduces the maximal net momentum by about 20-30% (if the drop covers exactly "one side" of the bullet in a uniform way, it would be a factor of 2/pi or a reduction by 36%).

much of it will be forward momentum

That does not matter, see A.T.'s sketch.

I expect the physics would be complicated to analyse accurately.

That's why I calculated a rough estimate, not a precise number.

I don't see how figure 9 corresponds to an asymmetric hit. At that speed, wood and water are not so different, if their area density is similar.

 

[Win_94]

The paths of the bullets was spiral, a consequence of the loss of stability rather than lateral momentum imparted by the contact with the dowel. Some of the tests involved striking just the edge of the dowel (see Fig. 9 from the article). This caused a deflection of about 9 inches over 75 yards.

I just can't see a single raindrop having nearly as much effect as a hardwood dowel.

AM

I'm seeing at the most, 8 inches of deviation at 260 yards; 1/4 of the dowels deviation at most, and 1/12 of the dowels deviation on average.

Oh, from viewing the bullet trace, I don't notice any spiral affect.