# The Effect a Raindrop Has On a Bullet

1. Mar 18, 2013

### 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.

Last edited by a moderator: Mar 23, 2013
2. Mar 18, 2013

### Staff: Mentor

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

3. Mar 18, 2013

### Win_94

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

They are saying there is no effect.

4. Mar 18, 2013

### 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" [Broken]

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?

Last edited by a moderator: May 6, 2017
5. Mar 18, 2013

### Staff: Mentor

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.

6. Mar 18, 2013

### Win_94

Thank you!

7. Mar 21, 2013

### Andrew Mason

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

Last edited by a moderator: Sep 25, 2014
8. Mar 21, 2013

### olivermsun

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?

9. Mar 21, 2013

### 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.

10. Mar 21, 2013

### olivermsun

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. Mar 21, 2013

### Staff: Mentor

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. Mar 21, 2013

### 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.

Last edited: Mar 21, 2013
13. Mar 21, 2013

### 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?

14. Mar 21, 2013

### Win_94

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.

Last edited: Mar 21, 2013
15. Mar 21, 2013

### OmCheeto

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. Mar 21, 2013

### Win_94

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.

Last edited: Mar 22, 2013
17. Mar 21, 2013

### Staff: Mentor

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. Mar 21, 2013

### Andrew Mason

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

Last edited: Mar 21, 2013
19. Mar 21, 2013

### Staff: Mentor

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. Mar 22, 2013

### Andrew Mason

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. Mar 22, 2013

### A.T.

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.
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. Mar 22, 2013

### Staff: Mentor

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.

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. Mar 22, 2013

### Andrew Mason

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. Mar 22, 2013

### Staff: Mentor

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. Mar 22, 2013

### Andrew Mason

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

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