The Effect a Raindrop Has On a Bullet

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Rain can significantly affect a bullet's point of impact, as evidenced by observations of a 12-inch drop at 350 yards when hitting raindrops. The discussion revolves around whether the kinetic energy required to displace water affects bullet trajectory, with some arguing that the momentum of a raindrop is negligible compared to that of a bullet. Calculations suggest that a bullet could experience measurable deflection due to water droplets, potentially leading to a downward shift in impact. The randomness of the observed deviations indicates that the effect of rain on trajectory is real and should be considered by marksmen. Overall, the conversation highlights the complexities of bullet dynamics in wet conditions.
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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

ShootRaindrop_zps7404e873.jpg


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.
309n310n311.jpg

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.
 
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What's your question? I didn't see one.
 
Drakkith said:
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.
 
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?
 
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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.
 
Drakkith said:
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!
 
Win_94 said:
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
 
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Andrew Mason said:
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?
 
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.
 
  • #10
MikeyW said:
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?
 
  • #11
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).
 
  • #12
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.
 
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  • #13
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?
 
  • #14
olivermsun said:
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.
 
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  • #15
mfb said:
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!
 
  • #16
mfb said:
and 20 degree contact angle (only determined by the shape of the bullet).

This is the bullet.

150gr308bullet_zps418b6dfe.jpg


.308 inches in diameter, 1.125 inches long, 150 grains, (there are 7000 grains in a pound.)
It has a ballistic coefficient of .398.
 
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  • #17
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.
 
  • #18
olivermsun said:
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
 
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  • #19
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?
 
  • #20
jim mcnamara said:
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
 
  • #21
Andrew Mason said:
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.
Andrew Mason said:
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.
Andrew Mason said:
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.
 
  • #22
Andrew Mason said:
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.

Win_94 said:
(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.
 
  • #23
A.T. said:
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
 
  • #24
Andrew Mason said:
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.
 
  • #25
mfb said:
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
 
  • #26
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.
 
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  • #27
Andrew Mason said:
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.
 
  • #28
A.T. said:
What matters is the mass ratio, not the momentum ratio.

To elaborate further on this:

attachment.php?attachmentid=56990&stc=1&d=1363995389.png


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.
 

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  • #29
Andrew Mason said:
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.
 
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  • #30
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.
 
  • #31
MikeyW said:
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.
MikeyW said:
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.
MikeyW said:
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.
 
  • #32
mfb said:
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.
 
  • #33
MikeyW said:
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.

mfb said:
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
 
  • #34
Andrew Mason said:
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.
 
  • #35
Andrew Mason said:
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.
 
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  • #36
Win_94 said:
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.
Perhaps you could describe how you are conducting your experiments and how you are determining the deviation.

One way to do this would be to see how much a small vertical stream of water deflects a bullet. In other words, do the same experiments as the American Rifleman did but substituting a water stream for the dowel.

AM
 
  • #37
Andrew Mason said:
Perhaps you could describe how you are conducting your experiments and how you are determining the deviation.

One way to do this would be to see how much a small vertical stream of water deflects a bullet. In other words, do the same experiments as the American Rifleman did but substituting a water stream for the dowel.

AM

The experiment is outlined in Post 12.

Here is a video of the test. Water Droplet POI Shift Test
Here is a follow-up video to see the droplet hits better. Zoomed In, Slow Motion, Water Drop Hits

I created a dropper devise as to shoot through a stream of drops. I wanted to preform the test in a round robin method but my devise was freezing-up on the first 4 of 10 shots. So the last 6 shots I set out to hit water.

I don't have a gun vise so I preformed the test as I would usually shoot. the first 4 shots at 300 yards made a 3.25 inch group, (the 4th shot was a called flier, the first 3 shots were what I would expect, a 1.25 inch group at 300 yards.) I was confident on on the technical aspects of all the shots except shot 4. I was shooting a 30-06 using ball ammo, (fill metal jacket bullets,) per my own hand-loaded cartridges. The velocity was 2600fps at the muzzle, with an estimated 2500fps at the dropper devise.

The dropper devise was set 40 yards forward of the muzzle to insure muzzle blast could not skew the test. So the bullets would travel 260 yards before hitting the target.

The temperature was 15ºF at dawn, when the test was over, (approximately 2 hours later,) the temperature was 21ºF. Ammo can be adversely affected by large temperature swings; a 6ºF difference would not skew the test.

The wind did change during the test. At first there was a 6mph wind from 10 0'clock, resulting in a 4 inch deviation to the right. As the day went on the winds got a bit stronger but mainly from 12 o'clock. Wind has a horizontal element, a light to moderate wind from 12 o'clock will affect the bullets trajectory minimally. You can see three different wind flag positions as to estimate if wind had an affect on the point of impact.

I tried to get a trusted marksman to preform the experiment but they are not going to go against the grain of common belief. Firearm enthusiast are a strange crowd, if they were wrong about an issue they have been vocal about for years, it would call their expertise into question.
Case in point, I suspect Adam Ant is from a firearms site which I raised the issue at first. I assume they Googled the quote I posted from Drakkith, and came here to disrupt the conversation.
This is why I don't link to here, because it will turn it into a melee.
[edit]On viewing of Adam's profile... He joined before the my post was created, so he isn't from that firearms site; but he is showing the bias of firearms enthusiasts. It is common belief that the laws of physics apply to all firearms and associated cartridges, except for the lever action 30-30.[/edit]

I also have a video displaying the energy of a shockwave. Shockwave Energy Test
I figured if a shockwave could vaporize a raindrop before the bullet could hit it, it should shred a wet napkin, (which is the most delicate thing I could imagine.)
This page display a good picture of a bullet bow shockwave.
Here is a good video of one.


I have contacted a YouTube channel with a high speed camera to record a bullet hitting a raindrop. Unfortunately the video will not be recorded until the cold weather breaks in New England.

If there are any more questions, I would be glad to answer them.

Thank you for all the efforts, all of you have contributed!
 
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  • #38
Win_94 said:
I created a dropper devise as to shoot through a stream of drops. I wanted to preform the test in a round robin method but my devise was freezing-up on the first 4 of 10 shots. So the last 6 shots I set out to hit water.

I don't have a gun vise so I preformed the test as I would usually shoot. the first 4 shots at 300 yards made a 3.25 inch group, (the 4th shot was a called flier, the first 3 shots were what I would expect, a 1.25 inch group at 300 yards.) I was confident on on the technical aspects of all the shots except shot 4. I was shooting a 30-06 using ball ammo, (fill metal jacket bullets,) per my own hand-loaded cartridges. The velocity was 2600fps at the muzzle, with an estimated 2500fps at the dropper devise.
Unless you can compare the trajectory prior to striking the water to the trajectory after, it is impossible to tell whether a shot was deflected. The distribution of shots at the target with the water off appears to be comparable to the spread of shots through the water. Even if there was some significant difference to the distribution, that difference could be the result of other factors since you are relying on just your aim.

You might try using the paper screen technique used by Mr. Moore in the American Rifleman article.
AM
 
  • #39
Andrew Mason said:
Unless you can compare the trajectory prior to striking the water to the trajectory after, it is impossible to tell whether a shot was deflected. The distribution of shots at the target with the water off appears to be comparable to the spread of shots through the water. Even if there was some significant difference to the distribution, that difference could be the result of other factors since you are relying on just your aim.

You might try using the paper screen technique used by Mr. Moore in the American Rifleman article.
AM

There is no way I could see the target with paper in the way.

The 4 shots I took without hitting water, including the flier is more accurate than the 5 shot group in the test you've provided. If their group was shot at 300 yards it would have been 5.1 inches as opposed to my 3.25 inches. My rifle is more accurate even when I screw-up.

That is the problem with m doing these tests, no one will ever believe the results. No matter what I do.
I wasted 10 shots on that test and am not able to buy more ammo. There is no ammo to be had. That is the only test on the planet on this subject; you're saying it is completely irrelevant?
Even when I shoot cartridges with varying charge weights, the dispersion isn't 12 inches.
I don't have a budget on this project to buy equipment as you you deem necessary. I shot my last 10 rounds of the same load components. I can't preform another test, and no one else on the planet will touch it.
 
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  • #40
Andrew Mason said:
The distribution of shots at the target with the water off appears to be comparable to the spread of shots through the water.

No. The 4 shot group that didn't touch water are 1/4 the size of those that hit water.
 
  • #41
Andrew Mason said:
Unless you can compare the trajectory prior to striking the water to the trajectory after, it is impossible to tell whether a shot was deflected.

That is also wrong. You are assuming the bullets striking the water will tumble and corkscrew to the target, I don't think it will happen, and it isn't happening. If it were happening there would be sideways strikes on the target; there are none.
I think the strike on a liquid, can't be half as violent as a hit on a solid. You can't assume the same corkscrew results.
 
  • #42
A few things I'm not clear on, how can you tell the bullet is hitting water? I don't see any splatter on the videos, and I'd assume the water gets vaporised anyway.
 
  • #43
MikeyW said:
A few things I'm not clear on, how can you tell the bullet is hitting water? I don't see any splatter on the videos, and I'd assume the water gets vaporised anyway.

Some turned into vapor, some were splatter. I assume vapor is more of a center hit than the splatter. I'm working on another post as to attempt another test. The new test will be using another camera as to capture the water hit more clearly. Please be patient.
 
  • #44
Win_94 said:
There is no way I could see the target with paper in the way.

The 4 shots I took without hitting water, including the flier is more accurate than the 5 shot group in the test you've provided. If their group was shot at 300 yards it would have been 5.1 inches as opposed to my 3.25 inches. My rifle is more accurate even when I screw-up.

That is the problem with m doing these tests, no one will ever believe the results. No matter what I do.
I wasted 10 shots on that test and am not able to buy more ammo. There is no ammo to be had. That is the only test on the planet on this subject; you're saying it is completely irrelevant?
Even when I shoot cartridges with varying charge weights, the dispersion isn't 12 inches.



I don't have a budget on this project to buy equipment as you you deem necessary. I shot the last 10 rounds that I had, which are of the same load components. I can't preform another test, and no one else on the planet will touch it.


I would do the test, if I had a rifle, and the conditions you have. Unfortunately it is not likely to be below freezing in my area for another 8 months.

I don't have much of a budget either, but I never let that stop me from doing science experiments.

First off, the raindrops velocity, compared to that of the bullets, can be effectively modeled as zero. Secondly, the raindrop probably appears to be solid from the bullets point of view, as pointed out by jim mcnamara in post #19: 740 m/s = 1655 mph.

If you can find, or make more bullets, I would perform the following experiment:

Supplies needed
1. powdered graphite
2. a bunch of paper
3. a roll of duct tape
4. an old ice cube tray
5. bullets
6. toothpicks
7. an eye dropper
8. pencil
9. clipboard
10. tape measure

Setup:
1. put a bit of powdered graphite into each of the old ice cube trays cavities
2. in the first cavity, place one drop of water, in the second cavity, place two drops of water, continue as in this manner with all cavities.
3. stir graphite with toothpick
4. leave a toothpick in each cavity
5. stick ice cube tray in freezer, making sure toothpicks are in the center of each cavity, and will make a wonderful graphite popsicle stick thing for tomorrow morning.
6. set up a linear paper wall, one foot away and parallel to the bullets paths, 11" high x 56" wide, with an old piece of plywood and duct tape 5 feet in front of where you are firing from
7. set up a paper wall with an old piece of plywood and duct tape 250 yards away. Same dimensions as above.
8. set up a stand between your gun and the paper wall nearest you that will hold steady your toothpick-graphite-sicles, that is in line with your target paper wall.

Experiment:
1. go out when it is still below freezing the next morning, as otherwise your graphitesicles will melt.
2. get another piece of paper on a clipboard, create a matrix of drops per graphitesicle,

Drops_____LOGD____DOBFB
1__________8"_____3"
2__________9"_____4"
3_________10"_____5"
etc
*LOGD = Length Of Graphitesicle Dispersion Pattern
*DOBDFB = Displacement Of Bullet From Bullseye

3. align the graphitesicles, such that they will be glanced by the bullets you are firing at the target
4. after each shot, circle the graphite marks on the near paper wall. Measure the length of the track with your tape measure. Write down the length in column 2.
5. after each shot, circle the bullet hole at the target with your pencil. Measure the distance from the bullseye. Write down that distance in column 3.

Modifications to experiment:
1. after each firing, change the paper on the near paper wall. It may be difficult to distinguish subsequent splatter patterns.


Assumptions on my part:
1. I theorize, that the graphitesicles will turn into water vapor once struck by the bullet, due to the tremendous pressure.
2. The graphite will adhere to the near paper wall. (Perhaps use paper towels, rather than notebook paper)
3. You'll be able to come up with more bullets.



When complete, show us your results.

ps. This very much reminds me of how I once metaphorically described high energy particle physics to someone.

Om said:
Take a Colt 45, shoot a watermelon, and have three blind men interpret what the watermelon consisted of by having the blind men feel around for seeds, goopy stuff, and outer shell bits.
 
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  • #45
That is an interesting approach.

Here is a simpler version of the experiment:
Create a thin stream of water, which fractures into individual drops quickly. Shoot through the point where you have a high drop density. Check that you can produce this in a reliable way (camera?) and switch it on/off.

Shoot n times without water, ~2n times with water, repeat.
For each set of shots, record the 2-dimensional positions of the impacts. Choose n to have very similar conditions for both shot types, but probably large enough such that you don't have to run to the target after every shot.

Post the raw data here.
The shots with water should have two components: One with the same distribution as the shots without water (no drop hit) and one component which is broader than the other (by some unknown amount) as it has the additional water deflection. I don't know how expensive those bullets are, so it might be impossible to see this splitting, but at least you can check the average deviation and something like the amount of outliers.
 
  • #46
I looked over my stocks and I can disassemble 14 30-30 cartridges, to make 10 30-06 cartridges of the same load as the previous test.

Before I preform another test, I want the parameters laid-out.

First; I'll hook up the dropper device to a hose at the house; so the range will be 200 yards this time.
I need to do this because when I pulled shot 4, it was due to rushing to get the shot off before the water ran out.
At that time, I decided to change my procedure as to not need to rush.

Second; I'll position the camera in such a way to capture the bullet hits on the water more clearly. I'll have a dark background with lighting on the water drops.
(This is the reason I am preforming another test, because the video quality was not good enough to be clear enough after the 2 edits needed to make the video, then upload it to YouTube.)
I only have one camera, the video of the water hits will be the only video.

Third; I'll use a large piece of cardboard as the target, so all bullet hits can be examined to see if there is a pronounced yawing affect. This will also aid in exact measurements since there will be no complete misses, (as with shot number 5.)

Fourth; on the issue of having paper at specific intervals.
The only way I could make that work, is to make a free recoil devise to cradle my rifle. See this video...

...and then I would need to make a concrete bench to rig the devise to.
That is the only way to insure the rifle is still on a relatively close aiming point without manually aiming each time.
I don't have either of those items.

Another issue with the paper is I have rolling hills on my range, I would need at least 32 12 foot poles and as many shorter ones to position paper at 32 specific areas to capture the bullet trajectories on the range I have chosen to shoot at. I could reduce the length to 100 yards, but then I would need to use the bottle to feed the dropper devise since it is not close enough to the house. If I use the bottle, I am sure it will cause me to rush again, resulting in another flier, and again, a completely worthless test. That is not happen this time.

The paper at intervals is not needed; the point of impacts of the shots not hitting water should suffice as a the norm. We need not know if yaw or deflection was the result of the deviation, we simply need to know if there is deviation.

Fifth; I am going to take the shots off of a bench in such a way that I get exceptional accuracy; since there is a hose fed dropper devise, I will not need to rush, therefore all shots will go as planned.
I'll shoot through water first; this way if I miss water, I can make up for it later down the road. When I know I've hit water 5 times, I'll then shoot the rest of the 10 shots without water.

Sixth; The measurement of all shots not hitting water determine the norm. The measurement of all shots hitting water will be weighed relative to the norm. Anything else is ridiculous! My shooting ability with the shots that didn't hit water in the first test, was more precise than American Rifleman's mechanical devise.

Seventh; if there is any other elements of the test that absolutely needs to be included, or there is anything in my proposed test that is unacceptable, speak-up now.
 
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  • #47
OmCheeto said:
I would do the test, if I had a rifle, and the conditions you have. Unfortunately it is not likely to be below freezing in my area for another 8 months.

I don't have much of a budget either, but I never let that stop me from doing science experiments.

First off, the raindrops velocity, compared to that of the bullets, can be effectively modeled as zero. Secondly, the raindrop probably appears to be solid from the bullets point of view, as pointed out by jim mcnamara in post #19: 740 m/s = 1655 mph.

If you can find, or make more bullets, I would perform the following experiment:

Supplies needed
1. powdered graphite
2. a bunch of paper
3. a roll of duct tape
4. an old ice cube tray
5. bullets
6. toothpicks
7. an eye dropper
8. pencil
9. clipboard
10. tape measure

Setup:
1. put a bit of powdered graphite into each of the old ice cube trays cavities
2. in the first cavity, place one drop of water, in the second cavity, place two drops of water, continue as in this manner with all cavities.
3. stir graphite with toothpick
4. leave a toothpick in each cavity
5. stick ice cube tray in freezer, making sure toothpicks are in the center of each cavity, and will make a wonderful graphite popsicle stick thing for tomorrow morning.
6. set up a linear paper wall, one foot away and parallel to the bullets paths, 11" high x 56" wide, with an old piece of plywood and duct tape 5 feet in front of where you are firing from
7. set up a paper wall with an old piece of plywood and duct tape 250 yards away. Same dimensions as above.
8. set up a stand between your gun and the paper wall nearest you that will hold steady your toothpick-graphite-sicles, that is in line with your target paper wall.

Experiment:
1. go out when it is still below freezing the next morning, as otherwise your graphitesicles will melt.
2. get another piece of paper on a clipboard, create a matrix of drops per graphitesicle,

Drops_____LOGD____DOBFB
1__________8"_____3"
2__________9"_____4"
3_________10"_____5"
etc
*LOGD = Length Of Graphitesicle Dispersion Pattern
*DOBDFB = Displacement Of Bullet From Bullseye

3. align the graphitesicles, such that they will be glanced by the bullets you are firing at the target
4. after each shot, circle the graphite marks on the near paper wall. Measure the length of the track with your tape measure. Write down the length in column 2.
5. after each shot, circle the bullet hole at the target with your pencil. Measure the distance from the bullseye. Write down that distance in column 3.

Modifications to experiment:
1. after each firing, change the paper on the near paper wall. It may be difficult to distinguish subsequent splatter patterns.


Assumptions on my part:
1. I theorize, that the graphitesicles will turn into water vapor once struck by the bullet, due to the tremendous pressure.
2. The graphite will adhere to the near paper wall. (Perhaps use paper towels, rather than notebook paper)
3. You'll be able to come up with more bullets.



When complete, show us your results.

ps. This very much reminds me of how I once metaphorically described high energy particle physics to someone.


Ice would skew the test. No one would doubt that ice causes a deviation.
 
  • #48
mfb said:
That is an interesting approach.

Here is a simpler version of the experiment:
Create a thin stream of water, which fractures into individual drops quickly. Shoot through the point where you have a high drop density. Check that you can produce this in a reliable way (camera?) and switch it on/off.

The reason I have a lot of drops is because the wind blows the drops quite a lot. ven with the 10 holes I had to shift the rifle left and right to chase the drops insuring to get a hit.

mfb said:
Shoot n times without water, ~2n times with water, repeat.
For each set of shots, record the 2-dimensional positions of the impacts. Choose n to have very similar conditions for both shot types, but probably large enough such that you don't have to run to the target after every shot.
I was going to do it that way, but was missing the drops due to the stream being too sparse.
But, The only conditions that would make a difference, would be wind, and that would only deviate side to side. Any differences in height, will be due to drop deflection.
mfb said:
The shots with water should have two components: One with the same distribution as the shots without water (no drop hit) and one component which is broader than the other (by some unknown amount) as it has the additional water deflection. I don't know how expensive those bullets are, so it might be impossible to see this splitting, but at least you can check the average deviation and something like the amount of outliers.
I'm not quite sure what you are saying here. I'll take a stab at addressing it.

Hopefully, I have figured out how fast the flow should be as to be able to hit a drop each time I want to. If I don't have that understood, I might only get 4 or even 3 hits on water; resulting in a skewed test. But this time I'll be close enough to the house to view the video on the computer, I'll probably not be able to see on the camera viewer, I couldn't see them before.
Conditions are not going to be exact, no matter what I do. That is why it needs to be taken into account the condition that will make the largest difference, wind; and that will be a lateral deviation.

It isn't that bullets are expensive, it is that Obama's war on firearms has caused a run on everything firearms. I can't find components anywhere, and when an online store gets them in stock, within minutes they are gone again. I don't expect to be able to buy bullets for months, quite possibly longer.
 
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  • #49
Win_94 said:
We need not know if yaw or deflection was the result of the deviation, we simply need to know if there is deviation.

That I've thought about it more... I think yaw must be present. The only way it couldn't be present is if the drop deflected the bullet at a precise balance point. Unless it is a head-on collision, that isn't possible.
I suspect the yaw was slight and was dampened well before it hit the target.
 
  • #50
Andrew Mason said:
One way to do this would be to see how much a small vertical stream of water deflects a bullet. In other words, do the same experiments as the American Rifleman did but substituting a water stream for the dowel.
This would be an awesome Mythbuster's episode. Especially with the high speed camera.
 
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