Physics, bullet trajectories, and a minivan.

[sturmruger]

My friend and I were talking about shooting and the like. Some how we go onto the topic of shooting out the open door of a minivan (don’t ask me how). The question we had is how would the forward movement affect the bullet trajectory.

For instance if I were to throw a baseball at something out the window at a 90 degree angle while driving at say 50 mph the baseball would want to move away from the vehicle, but the forward momentum would also force it to move in a forward direction the same as the vehicle.

The question is this. How much would the point of impact change if you were to shoot a .308 rifle out of a moving vehicle at a target 50 yards away, assuming you could shoot when the vehicle was at an exact 90 degree angle with the target, and you aimed at the center of the target????

My buddy thought that since the bullet can overcome the earth’s gravitational pull, it will also break the forward momentum generated by the speeding car, and impact on target at 50 yards. I didn’t pay any attention in physics class so I have no clue.

I know there are some people on this board that paid attention in class and will know the answer to this question. I hope I phrased the question so it is understandable. THANKS

 

[YammyMonkey]

Same principle as leading a moving target, difference it you're the one moving. If your friend were correct the gun would shoot to one side when standing still.

How much of an effect is a question for another brain.

 

[JohnKSa]

Let's simplify things a bit.

Let's assume that the shooter is shooting out the door at exactly right angles to the direction of motion.

Then the velocity of the bullet can be found by:

Square root ( Mv x Mv + Bv x Bv)

Where Mv is the velocity of the minivan and Bv is the muzzle velocity of the bullet.

Now it gets a tiny bit complicated.

The bullet won't be going at a right angle to the minivan, it's going to be going at an angle that is a combination of the two directions of travel.

If the minivan's speed and the bullet's muzzle velocity were exactly the same speed then the bullet would travel at a 45 degree angle (angled in the direction of the travel of the minivan).

But that's not likely.

The direction of bullet travel can be calculated by:

Arctangent of (Mv/Bv)

That will give the angle that the bullet will travel (angled from the original right angle path of the bullet toward the direction of travel of the minivan).

Examples:

Muzzle velocity -- 3,000fps
Minivan speed-- 60mph = 88fps
Bullet velocity--3,001.3fps (That minivan's not helping much)
Bullet angle -- 1.68 degrees toward the direction of travel of the minivan


Muzzle velocity --1,000fps
Minivan speed--150mph = 220fps
Bullet velocity -- 1023.9fps (minivan's helping a bit here)
Bullet angle -- 12.4 degrees angled toward the direction of travel of the minivan

I'll leave it as an exercise for the reader to determine what effect this would have on point of impact @ 50 yards from the minivan.

 

[tcdrennen]

Exactly the same as if the minivan were stationary and the target were traveling parallel but backwards at 50 mph; you'd lead the target by the same amount.

 

[voilsb]

Also take into consideration that at 3000fps or 2790 or whatever you've got from that .308, it's going to cover that 50 yds pretty quick.

Also take into consideration that wind resistance is likely to slow down the bullet's lateral drift due to the initial velocity imparted by the van.

The bullet will not be moving laterally at 73fps by the time it hits the target. It probably won't have slowed down much either, being only fifty yards out. At 2790fps it will have covered that 150ft in about 0.054s.

 

[Kevlarman]

I hated vector math in highschool.

 

[Mal H]

My buddy thought that since the bullet can overcome the earth’s gravitational pull ...

Mssrs Newton, Einstein, Thorn and Hawking would like to meet your buddy.

The bullet is under the full influence of the earth's gravitational pull at all times, whether standing still or moving at high velocity. In fact nothing can escape the effects of the earth's gravity. The moon, the sun, even our galaxy are all affected by it; infinitesimal as it may be for the extreme examples.

 

[AllLockedUp(tm)]

Mal, on the same note:

If you were to simuntaneously fire a bullet from a gun (parallel to the ground), and drop a bullet from the same height as the gun barrel, wouldn't both hit the ground at the same time?

I seem to rememeber this is true.

 

[sm]

Never took physics, math is not my strong point.

I just kinda have an idea from shooting rabbits from the bed of a pick-up. We also tried a dealie with an open jeep and windshield down shooting teddy's and steels. [Just cause we could and wanted to know]

Skeet shooting experience helped, did better with SG and 1911style than I did with a rifle..., spend more time with first two than the later though.

 

[Art Eatman]

3,000 ft/sec MV. 88 ft/sec lateral (vehicle's forward) component of motion. 50 yards to target.

The bullet will take (simplified) 1/20th of a second to go 50 yards (150/3,000). In 1/20th of a second, it will have a lateral movement of 4.4 feet. So, aim 4.4 feet to the vehicle's rear, of the target.

The gravity thing: The usual theoretical deal is that it's a vacuum and you're shooting across a flat plane. Both bullets hit the ground at the same time.

Art

 

[MrFreeze]

Ahhh....Physics questions on THR. Ya gotta love 'em! They probably end up with the 2nd most posts, right behind SHTF!

In regards to the question of bullet dropping vs shooting, there was quite a thread about that not too long ago on here.

From physics, one thing I do remember is that perpendicular vectors do not affect the velocity of each other.

Travis

 

[sm]

Dang Art !

Heck I just led the darn thing > this far< well heck that didn't work lets try >> this far<< . Akin to "Kentucky Windage" you understand.

[thanks for showing the math, please tell me no exam on Fri tho']

 

[Art Eatman]

I never shot at road signs, as a kid, but riding in the back of a pickup and throwing rocks at them taught me a lot about "vector analysis".

Art

 

[MaterDei]

Seems to me that Art's math takes into account a round projectile, equally influenced by cross and head wind (friction). Most bullets, however, are not round and 'want' to fly straight. I believe this would negate some of the lateral movement.

Of course, I'm no expert and furthermore I didn't stay at a Holiday Inn Express last night so I could be totally wrong.

 

[Art Eatman]

MaterDei, since the rifle is travelling the same speed as the vehicle, the bullet will initially have that same forward (lateral) component of motion. This will change with wind resistance, of course, just as does the forward motion. For simplicity's sake, I've assumed that 1/20th of a second is not a significant factor affecting either vector.

, Art

 

[JohnBT]

I can't get past the thought that there will be a layer of dense air flowing over the surface of the vehicle that will affect a shot from within the van versus a shot with the muzzle stuck out the window or door. But then if the muzzle was outside wouldn't it whistle going down the road like the sound you get blowing across a pop bottle? That sort of turbulence has got to foul up something? Right.

But mostly I want to know where I can find some of those bullets that "want' to fly straight"? I don't think they sell them around here, or maybe they're under the counter and I don't know the secret handshake.

John

 

[MrAcheson]

Art's analysis is as correct as we can really make it without getting the math really complicated. In the real world there is going to be a lot of aerodynamic phenomenon going on with the bullet generating lift and drag which will alter its flight path considerably. Plus as has been said, air turbulence around the minivan is non-trivial.

 

[Zak Smith]

Here is a practical method to get an answer, using only basic geometry and an accurate external ballistics calculator (which are abundant).

I think it is most straightforward to approach the problem with the minivan as the frame of reference. Then bullet movement in the direction of the van's movement will be translated to a crosswind of 50mph, and the target will be translated as a wall sliding to the side at 50 mph.

If the target is 100 yards away from the muzzle, parallel to the car's path, then simply plug the bullet's BC and MV into a ballistic calculator with a 50 mph crosswind. From this calculation, you will get the flight time and the location of the bullet impact on the target - if it were standing still.

During the flight time, the target plane was translated 50mph * the flight time, for the translation distance. This will be only a left/right distance.

The actual bullet impact position can be calculated using this translation applied to the position from the ballistic calculator.

-z

PS. This method does not take into account the change in air density around the vehicle to do its travel through the air - the airflow around the car. It should be accurate if shooting from "clean air", though.

 

[Mal H]

Zak, you have left out the forward movement of the bullet imparted to it by the forward movement of the van. You are correct that the 50 MPH wind will be a large influence on the trajectory of the bullet, but it can't be substituted for the forward momentum.

To illustrate the fallacy of your model, imagine the same van/rifle/target situation in a vacuum. By your thinking, with no air to worry about at all, the bullet will hit the target if you are aiming directly at it from the moving van. Not so. As Art said you will miss by around 4 feet due to the forward movement of the bullet as it travels toward the target.

 

[Zak Smith]

Mal,

I am taking that into account my translating the target plane. In my model, the van is stationary and the air and target plane are moving. You missed this part:

[...] and the target will be translated as a wall sliding to the side at 50 mph.
[...]
During the flight time, the target plane was translated 50mph * the flight time, for the translation distance. This will be only a left/right distance.

The actual bullet impact position can be calculated using this translation applied to the position from the ballistic calculator.

regards
Zak

 

[Mal H]

Ah! So you are. That should give a good result. I guess I was fixed on the ballistic calculator and wind statement. My apologies.

 

[Zak Smith]

Let's do an example of 50 yards, a muzzle velocity of 2700, BC of 0.3, and a van velocity of 50mph. Rifle zero'd at 50 yards. Assume shooting out of right side of van.
50mph is 73fps or 876 inches/sec. First, the ballistic solution:

# Maximum range is 50.999999 
# Zero range is 50.000001 
# Max time of flight is 1.000000 s
# Ballistic coefficient is 0.300000
# V0 is 2700.000072 
# Elevation is 0.001171 radians (0.067088 degrees) 
# Height of sight is 1.500000 
# Crosswind is 73.000002 
# Drag model is G2 (3)
#
#      t          V         Vx          X         Vy          Y         Vz          Z    
    0.0012  2698.3833  2698.3815     1.0936     3.1204    -1.4542     0.0437     0.0004
[...]
    0.0567  2625.9939  2625.9929    50.3062     1.2742     0.0054     2.0009     0.6848

The deflection due to the wind is 0.68".
The flight time is 0.0567 seconds.
The target moved 0.0567 * 876 = 49.7" in that time (the translation).
Add the deflection to the translation and you get the bullet impact 49.7 - 0.68 = 49" inches or 4'1" to the left of the point of aim when the bullet was fired.

If we repeat the same exercise for 200 yards (and a 200y zero):

# Maximum range is 201.000003 
# Zero range is 200.000005 
# Max time of flight is 1.000000 s
# Ballistic coefficient is 0.300000
# V0 is 2700.000072 
# Elevation is 0.001638 radians (0.093862 degrees) 
# Height of sight is 1.500000 
# Crosswind is 73.000002 
# Drag model is G2 (3)
#
#      t          V         Vx          X         Vy          Y         Vz          Z    
    0.0012  2698.3833  2698.3797     1.0936     4.3814    -1.4358     0.0437     0.0004
[...]
    0.2353  2410.1363  2410.1215   200.1312    -3.2202    -0.0063     7.8373    11.3529

Flight time is 0.2353 seconds, wind deflection is 11.35.
The translation is 876 "/s * 0.2353 = 206.1".
Subtract the deflection of 11.35, for impact 195" to the left of the point of aim.

-z

 

[voilsb]

Over the course of only 50 yards, yes, you can probably simplify the situation and assume a stationary van and a moving target because of the short distance and therefore time involved.

In reality though, it is much more complicated than that, because you're not dealing with inertial frames of reference. You're dealing with accelerating frames.

As soon as the bullet leaves the van (or the muzzle, if the muzzle is outside the van), it's 50mph (73.3fps) lateral speed will start to decrease due to wind resistance. It's forward speed also decreases, slightly increasing the time-to-target, and therefore increasing the amount of time it has for lateral deceleration.

Also, as pointed out earlier, there will be turbulant factors if the muzzle is outside the van in the wind. And there will be pressure factors (due to the aerodynamics of the van) if it is inside the van, which will add some deflection to the bullet. And all these fluid dynamics are further complicated because it's a supersonic projectile, which fiddles with how fluids (like the air) behave, so you don't get the luxury of using a normal wake model.

So if you *really* want to assume a stationary van, the van is in a wind tunnel where the windspeed decreases as it gets farther from the van, and the target moves at a speed equal to the windspeed where the bullet is located.

 

[Zak Smith]

voilsb,

What do you mean by we have accelerating frames of reference? The car is moving at constant speed relative to the target and the atmosphere through the duration of the shot.

The lateral deceleration of the bullet is taken into account by the stationary-van-with-crosswind model.

I think the only thing not taken into account my model is the pressure bubble around the car. My back of the envelope guess is that the effect is not significant at car speeds: if you approximate the pressure layer around the car as a heavy wind "gust" of 2x or 3x the car's speed (just a guess on my part), it's only going to act on the bullet for a few feet of distance as the bullet passes through it. At 2700fps, that's only 1.85 milliseconds for a layer thickness of 5 feet. The additional work done on the bullet by this force is equivalent to only an additional 10-15' of normal 50mph wind, if there were no pressure layer.

So if you *really* want to assume a stationary van, the van is in a wind tunnel where the windspeed decreases as it gets farther from the van, and the target moves at a speed equal to the windspeed where the bullet is located.

No. Since we are dealing with non-relativistic speeds, we can look at the problem from any frame of reference. I believe using the van as the rest frame is the easiest given the tools we have readily available.


-z

 

[voilsb]

The lateral deceleration of the bullet is taken into account by the stationary-van-with-crosswind model.

No it doesn't. It gives the bullet an initial lateral speed of zero and a slowly increasing lateral speed up to a constant 50mph. Over 50 yards, with a 50mph cross-wind, the bullet will likely never reach 50mph laterally.

Over 50 yards, it's probably a good estimate. Over 200 yards, it probably is not, as the bullet might have even stopped moving laterally before then. Your model will have it going 50mph laterally still. How much distance does it take for the bullet to lose 73fps? After it loses that lateral speed, it will no longer drift farther off-target. Don't forget that the side of the bullet is larger than the cross section, therefore it will be more greatly affected by wind resistance.


And yes, car speeds the pressure differences are pointless. And the reason it's not an inertial frame is because as soon as it exits the van the bullet begins slowing down laterally. So as it leaves the van it's 50mph laterally, at 10yards it might only be going 43mph laterally, and at 50 it might only be going 15mph laterally.