.22LR extreme range trajectory

[Tony Williams]

I've been asked to help with a problem which I can't solve with the info I have, so I'm hoping that someone on this grate sight will know!

A man was struck by a .22LR bullet, and calculations showed an angle of descent of the bullet of 40 degrees. What range would it have been fired at? It isn't known whether a rifle or pistol was used, or whether standard or high-velocity ammo was fired, so the answer can't be precise. I'm really looking for a maximum and minimum range bracket which would cover these variables.

Can anyone help?

 

[Fast Frank]

More info is needed.

Not all .22 ammo is the same, and a stinger starts life going quite a bit faster than a target round.

Coming down at 40 degrees seems to indicate a fairly long range, but not maximum.

I would guess 200 to 300 yards, just based on my experience shooting .22s.

You might find somebody willing to go through all kinds of math gyrations to get an answer, but in the end it's all guess work until bullet weights, launch angles, and velocities are known.

 

[ceetee]

I can't answer the question, but I do know that a lot will depend on:

--The bullet weight, and
--Approximate velocity at impact...

 

[MrBorland]

I think the shooter must've been relatively close:

If a shooter holds the gun level and fires into the air, the bullet will go, say 1000 meters before hitting the ground. As a first approximation, the bullet travels in a parabolic arc, traveling 1000 horizontal meters. Since it only drops 2 meters (height when fired), it'll be going through a pretty shallow angle at all times.

OTOH, just going by the mathematical formula for a parabola (i.e. no ballistics data), the shooter must've shot 40 degrees into the air for the victim to be be hit by the bullet at 40 degrees. Some on-line ballistics calculator may be able to tell you how far a .22LR will travel if shot 40 degrees into the air, but my bet is that it's closer than several hundred yards.

 

[Vern Humphrey]

Using the Oheler Ballistic Explorer, a .22 hypervelocity round (1550 fps) zeroed at 10 yards will hit the ground at around 250 yards (that approxmiates holding the rifle "level.")

If zeroed at 250 yards (that is, pointed higher) it will hit the ground at around 325 yards.

The angle of fall is dependent on the angle of launch, not the range -- after all, if you pointed the rifle straight up, the bullet would come straight down. The "range" would be zero and the angle of fall would be ninety degrees.

 

[Neo-Luddite]

OTOH, just going by the mathematical formula for a parabola (i.e. no ballistics data), the shooter must've shot 40 degrees into the air for the victim to be be hit by the bullet at 40 degrees.

I second this--I don't have the right vocabulary to express myself in scientific terms--and this would only apply if you disregard air resistance of course--it is an idealized situation. It should hold true regardless of speed at the muzzle or size of the projectile; a 5" naval gun should yield that same shape of trajectory.

Without having some more data, you really can't know the range over ground (horizontal). If you knew the speed at the muzzle, you could compute the predicted maximum arc becasue you can assume that the angle it was fired at was also 40 degrees. Now, I don't remember the equation you 'plug' these nemuber into. The thing to remember to simplify the idea of the bullet traveling in a parabolic path is that it is actually following a path along an 'x' and 'y' axis simultaneously and gravity is always acting upon it to accelerate its motion toward the ground--that is what causes the arc.

Bottom line--you CAN'T know anything more WITHIOUT the speed at the muzzle.

 

[rangerruck]

i think a standard 40 grainer, at about 1250 fps, would be between 500 and 600 yds away, depending on wind, maybe out to 700 yds away.

 

[MrBorland]

I think the shooter must've been relatively close:

OTOH, just going by the mathematical formula for a parabola (i.e. no ballistics data), the shooter must've shot 40 degrees into the air for the victim to be be hit by the bullet at 40 degrees.

...but my bet is that it's closer than several hundred yards.

OK, now that I'm titrated with my daily caffeine fix, I have to back-track a wee bit: Still think the shooter must've been shooting into the air at around 40 degrees (or from a rooftop), but I agree this would extend the range, not shorten it. My foggy head saw 40 degrees as closer to Vern's 90 degrees.

 

[Marbles]

On the box of the Remington 500 and 525rnd bricks of .22LR they say: "Range: 1-1/2 miles. Be careful."

Maybe that means something?

(36gr HP at around 1150fps)

 

[Bendutro]

You have to make too many assumptions about the initial conditions to get any real accuracy...

BUT... 22LR ballistic calcs place a 45° positive angled shot (max range artillery style) around 1.5 miles. Slightly shallower 40° should be slightly less. Is this one of those celebratory gun firing cases?

This all assumes that the shooter and target were on the same horizontal plane, all bets are off if shooter was elevated.

 

[Vern Humphrey]

One point -- the elevation for maximum range for the .22 is about 30 degrees. This is because of the low ballistic coefficient -- higher angles of launch produce much longer flight lines, and drag extracts a toll.

So once you go beyond about 30 degrees elevation, the range begins to shorten.

 

[kBob]

First I believe Vern is correct about max ordinate angle for some .22LR cartridge/ gun combinations. Certainly when launching long range artillery the angle is seldom 45 degrees even with the best shaped, weighted and launched rounds. There was always a space in the arc through which no appearent change in range in either direction was noticable. One I recall provided the same point of impact for everything from about 42 degrees to almost 48 degrees out at extreem range and the shells were falling at a much steeper angle than they left the gun at.

Not enough data and to many variables.

This is liike asking if a GMC product hits someone, how much energy is transfered.

If we do not know the model or speed this is not answerable in any meaning ful way. We do not know the size weight or vector and velocity of the person hit.

No correct answer but, "it can not be reliably determined from the available data"

.22 Bullets come in a variety of shapes, and weights. The cartridges are loaded to produce a variety of pressures and the weapon they are fired from greatly changes starting velocities even if all else is the same.

As stated by others you do not meantion if the shooter was firing at an angle other than zero degrees to the horizon or plumb. You also do not menation if there is any vertical difference between the shooter and victim.

Before anyone starts figuring parabollas again remember that when fired in an atmosphere bullets do not have a mirror image path before and after apogee. For instance while shooting the M-109 155 mm howitzer at extreme range the older models of HE fell at a near vertical angle.

I hope this is not a Criminology/ Law Enforcement Education class for there is truely not enought data to act on.

On the other hand I once had a basic Physical Science Review class at a community college where it took me over a week to convice the professor that a ".22LR Bullet" fired straight up would not return to earth with a velocity equal to the muzzle velocity.

Then he got weirded out when he found out just how low the velocity of returning bullets was thanks to terminal or balance velocity in the atmosphere.

Made me wonder why the state had me taking the class. (returning to school on GI bill they gave me some required courses, most of which were a pathetic wasted of time, fortunately spelling was not one of them)

-Bob Hollingsworth

 

[benEzra]

Lightweight bullets do not follow parabolic trajectories in this atmosphere. They follow impetus trajectories (drag dominated), like a golf ball or ping-pong ball, i.e. a 10 or 15 degree elevation can result in a 45 or 50 degree descent as the bullet's forward motion is slowed by air resistance. For most of a bullet's flight regime, drag is the overwhelmingly dominant force altering the bullet's motion.

Maximum range for bullets from most small arms is obtained with an elevation around 30 degrees, though something shooting very heavy bullets at low velocities might benefit from a higher elevation. For a .22, maximum range at 30 degrees elevation will be around a mile (1760 yards), and at the end of that it will be coming down quite steeply.

Try going here:

http://www.eskimo.com/~jbm/calculations/traj/traj.html

and plug in 40 grains, 1100 ft/sec, and a ballistic coefficient of 0.18 (that's a guess, but a 55gr .223 is around .25, so it's probably close). Then set the zero range to 1000 yards, 1200 yards, 1500 yards, and 1760 yards and see what the trajectories look like. Or go to the main menu on that site and look at the maximum range calculator.

 

[627PCFan]

Dont forget to add another 100 yards for the obligatory bouncing after its first grounding-

 

[slogfilet]

+1 Kbob and BenEzra. You cannot make any assumptions on the initial angle of the projectile based on the data given.

 

[CYANIDEGENOCIDE]

+1 ezra
projectiles only travel in a parabola when neglecting air resistance.

 

[Zak Smith]

You cannot neglect air resistance.

In an extreme drop regime, even the most common "flat fire" trajectory methods fall apart. However, if we were to use the Sciacci method to compute the trajectory and then look at drop rate (inches per yard):

Drop angle of 40* down from horizontal is a drop rate of 0.84 down for every 1 across. In terms of inches per yard, this would be 30" of drop per yard.

Using a BC of 0.10 and a MV of 1400 fps, the bullet has this drop rate at approx 1300 yards at 1000' altitude.

With a BC of 0.20, the distance is now approx 2100 yards.

Of course, the accuracy of the computed angle (40*) is probably not great.

_Bullet_           _BC_ _MV_         0     200     400     600     800    1000    1200    1400    1600 | YARDS
22LR              0.100 1400 >    0.09   -0.56   -1.76   -3.71   -6.88  -12.19  -21.42  -38.90  -79.81 | drop per yard (inches)

 

[Jubjub]

I happen to have a Lyman manual handy, and Lyman lists a bullet mold with a very similar profile to a .22 rimfire bullet, #225438. It shows a weight of 41 grains and a ballistic coefficient of .094. Plugging that into a ballistic calculator along with a muzzle velocity of 1250 fps yields an error message for anything past 1300 yards. I assume that means maximum range is around there. It shows a velocity in the low 200 fps range, which I hope means the wound in question wasn't too serious.

From what I've read regarding extreme range shooting, a wide variation in firing elevation will end up at around the same max range, varying in angle of fall pretty widely. My official guesstimate is 1100-1300 yards.

 

[sarduy]

A man was struck by a .22LR bullet, and calculations showed an angle of descent of the bullet of 40 degrees. What range would it have been fired at?

A. The bullet was fired no more than 1.2 miles away if fired from a rifle. (max range for a .22)

A. if the man was able to hear the shot that hit him, there is a good chance that the bullet was fired from a 40 degrees point higher, meaning from a building or a plataform.

but a lot of data is missing, so i think that will cover it all.

 

[rogn]

Was the shooter standing overhead or up in a tree? Then the range could be rather short. I think we can nmake a reasonable assumption that the range was close to terminal if the rifle(?) was fired for maximum distance(question seems directed that way.) The 40* downward path would suggest that the projectile was not going to have a lot of further forward travel. 40* suggests max range as bullets reaching that range tend to ricochet off the target into the ground; Sandy Hook, NJ abt 1890 tests with Creedmore type target rifles. So my vote goes for close to but not max range, if fired for max distance--maybe 12-1400yd

 

[kcmarine]

OP


"Hey... I'm gonna make a topic filled with a bunch of fancy words..."

 

[30Cal]

He's looking for a range bracket. Not THE EXACT range.

Pick some nominal .22LR subsonic and go with a velocity nominal for a short pistol (call it 40grs RN, 800fps). What's the range that will give you a 40º angle? This is the minimum of our bracket.

Pick some nominal .22LR high velocity ammo (say CCI stingers--1400fps) and go with a velocity out of a fairly long rifle barrel. What's the range that will give you a 40º angle? This is the max range of the bracket

.22LR BCs vary from about .105 to .150. Apply the low number to the pistol case (min range) and the high number to the rifle case (max range).

A 40º down angle means 300" of drop over a 10yd interval. I used JBM online ballistics calculator set with a 10yd increment and looked for a 300" delta in drop. Thus max range of roughly 1700yds.

Range Drop Drop Windage Windage Velocity Mach Energy Time Lead Lead
(yds) (in) (moa) (in) (moa) (ft/s) (none) (ft•lbs) (s) (in) (moa)
1690 -14169.5 -800.6 0.0 0.0 303.9 0.272 8.2 10.627 0.0 0.0
1700 -14483.5 -813.6 0.0 0.0 302.9 0.271 8.1 10.758 0.0 0.0

Using the min case parameters, I get about 1050yds. Call it 1000yds min.

So flat ground, between 1000-1700yds.

 

[elmerfudd]

Assuming the bullet was not fired at a downward angle, the forward velocity of the bullet should be 1.19 times the terminal falling velocity of the bullet. This is arrived at by taking the cotangent of 40 degrees, (1.19), and assuming that an object as light as a rimfire bullet would quickly reach it's terminal falling velocity. Once you've got the velocity, you should be able to use the bullets BC to get a VERY rough idea of how far away it was fired from. But, you're dealing with hugely different potential initial velocities. You still don't know if this was a short fired from a 2" pistol barrel or a hypervelocity bullet fired from a 20" rifle barrel. The difference in initial velocity would be enormous.

 

[Tony Williams]

Thank you for your comments, gentlemen. This isn't an exercise, it is an enquiry from a police officer about an actual incident. Since he identified the bullet as a .22LR I have assumed that it is the standard lead round-nosed type, but I will check that with him.

I do have some information about ballistics, including a couple of maximum-range tables. One of these is for a 7.5mm Swiss military rifle round (streamlined bullet, 2,600 fps), which reveals that the maximum range is 4,457 yards achieved at an elevation of 34 degrees 42 minutes, with a terminal descent angle of 71 degrees 21 minutes. The other I have is for a 20mm Madsen AP round, which is surprisingly similar: max range 4,700m achieved at an elevation of 36 degrees 40 min, with a terminal descent angle of 80 degrees 19 min. The range at which the descent angle is 40 degrees for the Madsen is 3,800m, which is just over 80% of the maximum range. By interpolation, the equivalent distance for the 7.5mm Swiss seems to be in the region of 3,800 yards, which is around 85% of maximum range.

I know that the .22LR 40 grain solid when fired at 1,255 fps has a maximum range of 1,623 yards (source: NRA Factbook) which, if the 80-85% ratio applies, would suggest that the bullet could have been fired at up to 1,300-1,380 yards. This would presumably be for a high-velocity loading fired from a rifle. For a standard-velocity from a pistol, I assume that the likely range is about 1,000 yards (that's just a guesstimate). However, I do not know how accurate these figures might be. The ballistic calculator I normally use (JBM) gives lots of valuable data but not the descent angle. I was hoping that someone might have some ballistic tables for .22LR, or alternatively, a ballistic programme which does give the descent angle. However, .30Cal seems to have produced similar results to my rough estimates, so 1,000-1,500 yards is probably as close as we can get.

 

[futureranger]

if the bullet hit with at a 40 degree downward angle, the shooter was aiming less than 40 degrees into the air with air resistance, if the shooter was on a level plain with the victim. But if the shooter fired the round from a elevated position, on a roof perhaps, the "launching angle = landing angle" rule goes out he window. Was the incident in a urban environment?