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A good friend of mine (very educated and old enough to know better), shot a squirrel off the top of his fence with a .22 rifle in a subdivision. He said he "laid down on the grass and shot up" so the bullet wouldn't hit his back neighbor's house. When I reminded him that the bullet would come down, he was embarrassed and said he would never do it again. Stuff like this does happen in populated areas. He is obviously not a physics major.

ScottDLS wrote:

glock27 wrote: ...
i watch mythbusters they did a terminal velocity test on dropping a penny from "empire state building" all in scale of course. but a pennies maximum speed is something like 62 miles per hour. a .22 would weigh less than a penny and have less terminal velocity
....

Oh...please tell me you are joking. I believe it was Galileo c.1634 that proved that all objects regardless of weight, fall at the same rate of acceleration (absent wind resistance). Newton then described the physics in much more detail somewhat later. So, drop an anvil and a coyote from a plane and both will fall at the same speed unless coyote spreads out to have a bigger surface area than the anvil.

A .22 would have less wind resistance than a penny and so have a terminal velocity somewhat higher, but I would volunteer to catch it in my hand at 1 mile.

What you're both missing is the difference between drop speed and forward speed, and the role of mass. All are in play. The forward speed is imparted by the expanding gas produced by the burning powder charge, and the drop speed is that imparted to the bullet by gravity. The forward speed is what makes the .22 dangerous instead of merely an annoyance.

That's also why you might not mind catching a penny dropped from the empire state building in a sturdy hat, but you wouldn't want to catch that same coin from a supersonic fighter in flight because the penny would also carry the fighter's forward velocity.

The anvil's mass is what makes that much more dangerous than a coyote when dropped straight down. The formula for terminal energy is mass x velocity squared. If you add a lot of mass, it bumps this number up substantially.

Correspondingly, if you take two items of the same mass and impart lots more speed to one than the other, the faster one will deliver significantly more energy to the target, all other things being equal.

Excaliber wrote:

ScottDLS wrote:

glock27 wrote: ...
i watch mythbusters they did a terminal velocity test on dropping a penny from "empire state building" all in scale of course. but a pennies maximum speed is something like 62 miles per hour. a .22 would weigh less than a penny and have less terminal velocity
....

Oh...please tell me you are joking. I believe it was Galileo c.1634 that proved that all objects regardless of weight, fall at the same rate of acceleration (absent wind resistance). Newton then described the physics in much more detail somewhat later. So, drop an anvil and a coyote from a plane and both will fall at the same speed unless coyote spreads out to have a bigger surface area than the anvil.

A .22 would have less wind resistance than a penny and so have a terminal velocity somewhat higher, but I would volunteer to catch it in my hand at 1 mile.

What you're both missing is the difference between drop speed and forward speed, and the role of mass. All are in play. The forward speed is imparted by the expanding gas produced by the burning powder charge, and the drop speed is that imparted to the bullet by gravity. The forward speed is what makes the .22 dangerous instead of merely an annoyance.

That's also why you might not mind catching a penny dropped from the empire state building in a sturdy hat, but you wouldn't want to catch that same coin from a supersonic fighter in flight because the penny would also carry the fighter's forward velocity.

The anvil's mass is what makes that much more dangerous than a coyote when dropped straight down. The formula for terminal energy is mass x velocity squared. If you add a lot of mass, it bumps this number up substantially.

Correspondingly, if you take two items of the same mass and impart lots more speed to one than the other, the faster one will deliver significantly more energy to the target, all other things being equal.

Excaliber is correct in many ways, however, I am a mechanical engineering student with significant experience in Newtonian mechanics and agree that this problem is about bullet energy. There are two basic types of energy. Kinetic energy, which is the energy of objects in motion, and potential energy, which can be in various subtypes (chemical, elastic, gravitational). When the round is sitting in the chamber, it has an abundance of chemical potential energy, which is rapidly converted into kinetic energy (KE=0.5mv^2) as the bullet accelerates down the barrel. Now that the bullet has left the barrel, it is subject only to gravity and resistance from the fluid it is travel traveling through.

Some of that potential energy was also converted to rotational energy, which makes the bullet fly straight, but also reduced the shape factor that produces "air resistance." As long as we can reasonably make the assumption that the bullet is still spinning along an axis parallel to the direction of travel we can also assume a high terminal velocity. When that bullet leaves the gun and travels at an angle from the ground, an amount of energy that was formerly kinetic, builds up as gravitational potential (GPE=mgh) . As the bullet descends, that energy is once again transferred back into kinetic energy, less the amount lost to air resistance. Like I said before, if that bullet maintains its rotation the losses will be small. As that bullet is coming down, it accelerates at a constant 9.8 m/s^2 downward, which means the farther past the apex of the trajectory it goes the greater the angle at which it will be coming down. So for long shots that strike a target after the apex of the shot they will have to be at steeper angles.

When that shot hits, its energy is NOT either going to be the dropping energy or the forward energy that hits you, but a vector sum of the two combined. Since those two energies are merely converted, the magnitude of the energy imparted on the target will be the same.

I hope this helps!
SIDENOTE: if you are interrested, spheres actually make for terrible ballistic objects because of the low pressure wake that trails them in flight.

Lonest4r wrote:

Excaliber wrote:

ScottDLS wrote:

glock27 wrote: ...
i watch mythbusters they did a terminal velocity test on dropping a penny from "empire state building" all in scale of course. but a pennies maximum speed is something like 62 miles per hour. a .22 would weigh less than a penny and have less terminal velocity
....

Oh...please tell me you are joking. I believe it was Galileo c.1634 that proved that all objects regardless of weight, fall at the same rate of acceleration (absent wind resistance). Newton then described the physics in much more detail somewhat later. So, drop an anvil and a coyote from a plane and both will fall at the same speed unless coyote spreads out to have a bigger surface area than the anvil.

A .22 would have less wind resistance than a penny and so have a terminal velocity somewhat higher, but I would volunteer to catch it in my hand at 1 mile.

What you're both missing is the difference between drop speed and forward speed, and the role of mass. All are in play. The forward speed is imparted by the expanding gas produced by the burning powder charge, and the drop speed is that imparted to the bullet by gravity. The forward speed is what makes the .22 dangerous instead of merely an annoyance.

That's also why you might not mind catching a penny dropped from the empire state building in a sturdy hat, but you wouldn't want to catch that same coin from a supersonic fighter in flight because the penny would also carry the fighter's forward velocity.

The anvil's mass is what makes that much more dangerous than a coyote when dropped straight down. The formula for terminal energy is mass x velocity squared. If you add a lot of mass, it bumps this number up substantially.

Correspondingly, if you take two items of the same mass and impart lots more speed to one than the other, the faster one will deliver significantly more energy to the target, all other things being equal.

Excaliber is correct in many ways, however, I am a mechanical engineering student with significant experience in Newtonian mechanics and agree that this problem is about bullet energy. There are two basic types of energy. Kinetic energy, which is the energy of objects in motion, and potential energy, which can be in various subtypes (chemical, elastic, gravitational). When the round is sitting in the chamber, it has an abundance of chemical potential energy, which is rapidly converted into kinetic energy (KE=0.5mv^2) as the bullet accelerates down the barrel. Now that the bullet has left the barrel, it is subject only to gravity and resistance from the fluid it is travel traveling through.

Some of that potential energy was also converted to rotational energy, which makes the bullet fly straight, but also reduced the shape factor that produces "air resistance." As long as we can reasonably make the assumption that the bullet is still spinning along an axis parallel to the direction of travel we can also assume a high terminal velocity. When that bullet leaves the gun and travels at an angle from the ground, an amount of energy that was formerly kinetic, builds up as gravitational potential (GPE=mgh) . As the bullet descends, that energy is once again transferred back into kinetic energy, less the amount lost to air resistance. Like I said before, if that bullet maintains its rotation the losses will be small. As that bullet is coming down, it accelerates at a constant 9.8 m/s^2 downward, which means the farther past the apex of the trajectory it goes the greater the angle at which it will be coming down. So for long shots that strike a target after the apex of the shot they will have to be at steeper angles.

When that shot hits, its energy is NOT either going to be the dropping energy or the forward energy that hits you, but a vector sum of the two combined. Since those two energies are merely converted, the magnitude of the energy imparted on the target will be the same.

I hope this helps!
SIDENOTE: if you are interrested, spheres actually make for terrible ballistic objects because of the low pressure wake that trails them in flight.

That's what I had in mind - but explained MUCH better. Thanks!

Exactly, the penny in the Mythbusters "experiment" was starting from a velocity of zero and accelerating by the force of gravity opposed by air resistance, the bullets in Mythbusters were fired (supposedly) straight up (or close enough anyway) and also were starting from essentially zero velocity, also opposed by air resistance.

The exemplar .22lr is launched at (1300fps for the sake of arguement) and starts to: decelerate immediately on leaving the muzzle, due to air resistance; and drop to the ground due to gravitational acceleration. The bullet's velocity at arrival is determined primarily by how rapidly it loses its velocity due to air resistance because gravitational acceleration will not accelerate ANY object beyond terminal velocity due to air resistance.

I seriously doubt that the .22 would do the kind of damage reported at a range of a mile, not enough kinetic energy retained.

The .50bmg round at TMS a while back is another matter, while the holes it created seem to indicate that it was at least yawing significantly if not tumbling, indicating it was close to terminal velocity, still had lots of kinetic energy left due to its mass.

I wish there was something in the report more than ".22 caliber bullet". If it were definitely a .22LR round, then I'd have serious doubts about the possibility of its origin from the range. If it were .223 Remington/5.56 NATO, it's significantly more plausible, as those are generally somewhat heavier (55 grain up to 75 grain or so), and travel at much faster velocities of around 3000fps.

If it was a jacketed bullet, rather than lead/plated, then we could more certainly assume a higher velocity and it would mean the origin could more plausibly be from the range with sufficient velocity to cause the injuries sustained. Then again, they don't seem to be able to determine any kind of trajectory, and it really could have come from anywhere within a calculated radius. If trajectory can be narrowed down at all, I'm sure that information will come out in the suit.

The guy certainly does deserve some kind of satisfaction if the responsible party can be found, but suing the range because "maybe" it came from there isn't right. The person responsible for firing the round is the ultimately the responsible party, and I certainly do hope the person is found. I think that's unlikely to happen, given the large number of unknowns here.

I know I'm late for the party, but I couldn't resist this challenge:

CrimsonSoul wrote:... the shooter would have had to be aiming extremely high in order to make it that far ...

According to my HP 50g Graphing Calculator and the included formulas for projectiles, I compute the following:
The initial angle would only be +3 degrees and range would be 5124 feet when the bullet once again arrived at 5ft above ground 3.9 seconds later.
assumption: bullet leave barrel at 1300 fps from 5ft above ground (shoulder height)

here is some more interesting data.

At ~150 yds and .35 secs the bullet would be 25 ft above ground level
At ~300 yds and .7 seconds the bullet would be 40 ft above above ground level

I can recompute the data if someone can provide the actual berm distance from shooting position and height of the berms.

Also, I can provide the actual formulas that I used, but I won't bore you with that.

Lonest4r, perhaps you could double check my figures.

Has anyone heard any more on this?

Paul

psehorne wrote:I know I'm late for the party, but I couldn't resist this challenge:

CrimsonSoul wrote:... the shooter would have had to be aiming extremely high in order to make it that far ...

According to my HP 50g Graphing Calculator and the included formulas for projectiles, I compute the following:
The initial angle would only be +3 degrees and range would be 5124 feet when the bullet once again arrived at 5ft above ground 3.9 seconds later.
assumption: bullet leave barrel at 1300 fps from 5ft above ground (shoulder height)

What BC did you use? What's your calculated velocity at 100 yards and 1000 yards?

cbr600 wrote:What BC did you use?

I assume you mean bullet caliiber. For this example, lacking the facts, I used 1300 fps initial velocity.

What's your calculated velocity at 100 yards and 1000 yards?

I did not calculate velocity.

HP calculator formulae:
range = initial velocity sqared / gravity acceleration x sin(2 x angle)
distance =initial velocity x cos(angle) x seconds in flight
height = initial height + initial velocity x sin(angle) x seconds in flight - 1//2 x gravity acceleration x seconds in flight squared

Using the range formula, I set the range to 5280 ft and solved for the angle. The result was less than 3 degrees. I then used 3 degrees in the other formulae.

I can recalculate this once we know the initial velocity of the round that was actually used and the berm distance and berm height and the actual to the victim’s location.

psehorne wrote:

cbr600 wrote:What BC did you use?

I assume you mean bullet caliiber. For this example, lacking the facts, I used 1300 fps initial velocity.

What's your calculated velocity at 100 yards and 1000 yards?

I did not calculate velocity.

HP calculator formulae:
range = initial velocity sqared / gravity acceleration x sin(2 x angle)
distance =initial velocity x cos(angle) x seconds in flight
height = initial height + initial velocity x sin(angle) x seconds in flight - 1//2 x gravity acceleration x seconds in flight squared

Using the range formula, I set the range to 5280 ft and solved for the angle. The result was less than 3 degrees. I then used 3 degrees in the other formulae.

I can recalculate this once we know the initial velocity of the round that was actually used and the berm distance and berm height and the actual to the victim’s location.

It looks like you are missing a factor or two. Remember that air resistance is significant (BC is Ballistic coefficient BTW) it appears that your formulae would only apply in a vacuum.

The path of a projectile fired in our atmosphere does not describe the perfect arc your program predicts, because the projectile has significantly less velocity at the terminal end.

Here's a quick test: Calculate the maximum range and see what the incident angle would be. If it's anywhere near 45 degrees then it's wrong for a projectile in air.

Julian Hatcher's experiments are dated, but may be enlightening nonetheless. In chapter 17, table 15, we see "1906 Ball Service" has a muzzle velocity of 2700 fps but that drops to 958 fps at 1000 yards.

jimlongley wrote: It looks like you are missing a factor or two. Remember that air resistance is significant (BC is Ballistic coefficient BTW) it appears that your formulae would only apply in a vacuum.

The path of a projectile fired in our atmosphere does not describe the perfect arc your program predicts, because the projectile has significantly less velocity at the terminal end.

Here's a quick test: Calculate the maximum range and see what the incident angle would be. If it's anywhere near 45 degrees then it's wrong for a projectile in air.

You are correct. Do you (or anyone following this) know how to incorporate air resisrance into the formulea? In the meantime I will do some research.

Once we have complete formulae, it will be a simple matter to plug in various data to run different scenarios, and eventually plug in the actual initial velocity and BC of the round used. For now it will suffice to use characteristics of a representatice round (22lr or 223 for example).

It is important to know berm height and distance from the shooting station in order to determine the feasibility of a round leaving the shooting range being the culprit.

cbr600 wrote: What BC did you use? What's your calculated velocity at 100 yards and 1000 yards?

Looks like the BC of a typical 22 long rifle would likely be between .150 and .200, but I don't yet know how to incorporate BC into the formulae. Still researching...

Did you try something like this? http://www.winchester.com/learning-cent ... lator.aspx" onclick="window.open(this.href);return false;

psehorne wrote:

jimlongley wrote: It looks like you are missing a factor or two. Remember that air resistance is significant (BC is Ballistic coefficient BTW) it appears that your formulae would only apply in a vacuum.

The path of a projectile fired in our atmosphere does not describe the perfect arc your program predicts, because the projectile has significantly less velocity at the terminal end.

Here's a quick test: Calculate the maximum range and see what the incident angle would be. If it's anywhere near 45 degrees then it's wrong for a projectile in air.

You are correct. Do you (or anyone following this) know how to incorporate air resisrance into the formulea? In the meantime I will do some research.

Once we have complete formulae, it will be a simple matter to plug in various data to run different scenarios, and eventually plug in the actual initial velocity and BC of the round used. For now it will suffice to use characteristics of a representatice round (22lr or 223 for example).

It is important to know berm height and distance from the shooting station in order to determine the feasibility of a round leaving the shooting range being the culprit.

The actual solution incorporating air resistance is a differential equation of the third order. I would have to dig out my textbook for this particular type of question, but I am really busy with other classes at this point. It is best to do such a problem in Mathematica or Matlab because it has the tools you would need built into the software. If I get any time to work the problem out I will let you know, but I have two tests coming up this week and will therefore probably be very busy.