An Odd Ballistics, Trajectory question…
- Thread starter IM_Lugger
- Start date
Superhouse 15
New member
Ballistics
Can't quite remember enough high school physics to figure it now, but try here:
http://www.eskimo.com/~jbm/calculations/traj/traj.html
Can't quite remember enough high school physics to figure it now, but try here:
http://www.eskimo.com/~jbm/calculations/traj/traj.html
But of course,
You can do it statically or dynamically. Statically by calculating the energy lost through friction during the ascent, then calculate the height at which the bullet's potential energy would equal the remaining kinetic energy. That's the point, in energy management terms, when the kinetic energy has all been converted into heat from friction and the potential energy of the mass of that bullet at that height.
Dynamically, you can write a differential equation that calculates the deceleration of the bullet based on the "drag" of gravity (32 ft /sec /sec) and the aerodynamic drag of the bullet. The latter will vary with air diensity, as air density decreases with height.
Julian Hatcher, in Hatcher's Notebook, goes into this subject with emprirical testing. The good news is that the terminal velocity of most bullets—particularly since they fall base-first—is quite low. If the bullet has indeed gone straight up & come straight back down, you generally have to be looking up and take it in the eye to be injured. However, must "Fourth of July" shooting-into-the-air injuries and fatalities are caused because it's impossible to shoot exactly straight up.
One example from Hatcher is the 150-grain .30-'06, which rises to about 9,000 feet in the atmosphere, reaching a terminal velocity of 300 fps on the way down.
Regards,
Walt
Hatcher's terminal velocity for the 150gr bullet was between 300 & 400fps. That is based on the bullet remaining spin-stabilized and falling base first which is what his experiments showed was the likely case for this particular bullet.
Hatcher's assessment of lethality was based on a threshold of 60ftlbs that the military used at the time. I'm not sure how the military arrived at that threshold, but I assure you that 150gr at 300-400fps is nothing to sneeze at even if it doesn't achieve 60ftlbs. Subsequent studies have shown that even at velocities below 150fps this type of bullet can easily penetrate the skin and lodge in the underlying tissue.
Here's an article talking about airguns which states that a pellet (typical pellets are under 20gr) at 150fps can break the skin and at 200fps can fracture bone (I'm assuming they mean the skull since head wounds seem to be the focus of the article). Taking a 150gr bullet in the skull at 300 to 400fps can definitely be lethal, the idea that you have to get hit in the eye to even be injured is not consistent with reality.
http://findarticles.com/p/articles/mi_m0BUM/is_3_83/ai_n6077599
I statred out with airguns and I can tell you that .177 8gr pellet at 400fps would REALLY hurt; definatelly penetrate the skin and probably embed itself in the flesh....
but I'm more interested in how high a bullet would fly up before it would start to drop. Obviously it would depend on velocity, bullet weight, shape etc.
but I'm more interested in how high a bullet would fly up before it would start to drop. Obviously it would depend on velocity, bullet weight, shape etc.
curmudgeon1
New member
snip> "not consistent with reality."<snip
That's a polite way of saying its used oats. Right?
That's a polite way of saying its used oats. Right?
OldShooter
New member
ballistics etc.
What makes you think the bullet will not tumble? There is air resistance to the spinning action. It is unrealistic to assume that the bullet will fall tail first.
What makes you think the bullet will not tumble? There is air resistance to the spinning action. It is unrealistic to assume that the bullet will fall tail first.
Unless it's on a ballistic trajectory it should fall base first since that's the heavy end (no HBWC, please). The math would be easy if the bullet was fires in a vacuum and only gravity came into play.
The Mythbusters did a falling bullet test but never found the bullets fired from the M1. They had time of flight calculations, estimated height of trajectory, and terminal velocity but I don't remember the details. I do recall them using a level (on a tapered barrel???) to set the rifle plumb and I wondered how that was gong to work....
The Mythbusters did a falling bullet test but never found the bullets fired from the M1. They had time of flight calculations, estimated height of trajectory, and terminal velocity but I don't remember the details. I do recall them using a level (on a tapered barrel???) to set the rifle plumb and I wondered how that was gong to work....
The rotational losses are very low, so the bullet tends to remain at least somewhat gyroscopically stabilized.
It will fall base first if not disturbed since the base is heavier and base first is more stable in flight.
It is the gyroscopic stabilization that overcomes the overturning moment in flight and keeps the bullet point on.
Since the velocity is going to range from muzzle velocity to zero even the typical ballistic coefficients are not going to work.
Any particular BC is only valid over a limited range of velocities.
Some bullet makers supply more than one BC to try and overcome this problem.
OldShooter
New member
ballistics
Okay, who has data on rotational losses of a bullet over time? I'm not ready to believe the bullet will necessarily return to earth base first.
Okay, who has data on rotational losses of a bullet over time? I'm not ready to believe the bullet will necessarily return to earth base first.
bullet Balistics
All about the caliber, grain, charge and gravity. If you were to shoot at level plane, the bullet is affected immediately by gravity when it leaves the barrel. If a BB was dropped at muzzel level at the same time the bullet leaves the muzzel, both the bullet and the BB would strike the ground at the same time, provided the ground was level. Shooting on a 90 degree line of fire, the bullet has to loose it's energy and momentum before it begins its decent with gravity.
All about the caliber, grain, charge and gravity. If you were to shoot at level plane, the bullet is affected immediately by gravity when it leaves the barrel. If a BB was dropped at muzzel level at the same time the bullet leaves the muzzel, both the bullet and the BB would strike the ground at the same time, provided the ground was level. Shooting on a 90 degree line of fire, the bullet has to loose it's energy and momentum before it begins its decent with gravity.
One need not assume anything. Hatcher's testing proved conclusively that some rifle bullets will remain spin-stabilized and land base first.
Hatcher's tests also showed that some rifle bullets will tumble and the only pistol testing I've seen indicates that they tend to tumble as well.
As for the original question of maximum height, you can pretty much plug numbers into 1st year physics equations to get an idea for how high it would go before falling. Take these equations:
Final Velocity = Initial Velocity + Acceleration * Time
-and-
Distance = (1/2 * Acceleration * Time Squared) + (Initial Velocity * Time)
Acceleration due to gravity is about 9.8 meters per second squared, or about 32.2 FPS per second. So let's take a pistol round fired at 1000 FPS (for now, I'll ignore mass - it's ok, Newton did too). It will take 31 seconds for the bullet to reach 0 FPS, at which point it will start falling. Plugging that into the second equation, you get distance = .5*-32.2*31^2 + 1000*31 = about 15,500 feet.
The most precise way to figure it is following the energy loss of the bullet (as previously mentioned). Trouble with that is the rate of energy lost to drag varies with velocity (ie the rate is constantly falling). Figuring that would require knowing the drag coefficient of the round, and I'm not sure if that's the same as the ballistic coefficient, nor do I remember the calculus formulas to plug it into. Putting drag into the calculations would lower the height, so my method over-estimates the height. Now if the drag slows the spin of the bullet enough and it starts to tumble, you're looking at a separate equation to model the changing drag coefficient, and you start getting to some genuinely interesting formulas; at least I would assume so, that's over my head.
Sorry I can't remember the energy method, but I'm two years of business major courses removed from my last physics class.
Final Velocity = Initial Velocity + Acceleration * Time
-and-
Distance = (1/2 * Acceleration * Time Squared) + (Initial Velocity * Time)
Acceleration due to gravity is about 9.8 meters per second squared, or about 32.2 FPS per second. So let's take a pistol round fired at 1000 FPS (for now, I'll ignore mass - it's ok, Newton did too
). It will take 31 seconds for the bullet to reach 0 FPS, at which point it will start falling. Plugging that into the second equation, you get distance = .5*-32.2*31^2 + 1000*31 = about 15,500 feet.The most precise way to figure it is following the energy loss of the bullet (as previously mentioned). Trouble with that is the rate of energy lost to drag varies with velocity (ie the rate is constantly falling). Figuring that would require knowing the drag coefficient of the round, and I'm not sure if that's the same as the ballistic coefficient, nor do I remember the calculus formulas to plug it into. Putting drag into the calculations would lower the height, so my method over-estimates the height. Now if the drag slows the spin of the bullet enough and it starts to tumble, you're looking at a separate equation to model the changing drag coefficient, and you start getting to some genuinely interesting formulas; at least I would assume so, that's over my head.
Sorry I can't remember the energy method, but I'm two years of business major courses removed from my last physics class.
Last edited:
chris in va
New member
I'd be very surprised if your average bullet would go three miles straight up.
But I've been schooled before.
But I've been schooled before.
He says that he's over-estimating the height since he's neglecting drag and he's correct.
He's also right when he says that drag varies with velocity, but it also varies with air density and air density varies by about 25% over the likely range of altitudes we're dealing with so a precise calculation would probably have to take that variation into account as well.
The short answer to the original question is that it can be calculated but it's not a simple problem.
He's also right when he says that drag varies with velocity, but it also varies with air density and air density varies by about 25% over the likely range of altitudes we're dealing with so a precise calculation would probably have to take that variation into account as well.
The short answer to the original question is that it can be calculated but it's not a simple problem.
You guys are either a helluva lot smarter than me or you're really good at BS. Anyway, I read an account (can't remember where) of an experiment in which a bullet was fired straight up from a platform in a small lake. When the bullets came back down they could be more easily detected by the 'splash' they made, thus they could be timed and a variety of conclusions drawn from the calculations. One bullet (maybe more) came back down and hit the seat of the rowboat used to reach the site. The imprint in the wood showed that the bullet did indeed fall base-first and it was still spinning.
Freedom528
New member
Falling projectiles
Oldphart,
From a website: www.loadammo.com, I had found this account of the experiment you had referenced.
In 1920 the U.S. Army Ordnance conducted a series of experiments to try and determine the velocity of falling bullets. The tests were performed from a platform in the middle of a lake near Miami, Florida. The platform was ten feet square and a thin sheet of armor plate was placed over the men firing the gun. The gun was held in a fixture that would allow the gun to be adjusted to bring the shots close to the platform. It was surmised that the sound of the falling bullets could be heard when they hit the water or the platform.
They fired .30 caliber, 150 gr., Spitzer point bullets, at a velocity of 2,700 f.p.s. Using the bullet ballistic coefficient and elapsed time from firing until the bullet struck the water, they calculated that the bullet traveled 9,000 feet in 18 seconds and fell to earth in 31 seconds for a total time of 49 seconds.
Based on the results of these tests it was concluded that the bullet return velocity was about 300 f.p.s. For the 150 gr. bullet this corresponds to an energy of 30 foot pounds. Earlier the Army had determined that, on the average, it required 60 foot pounds of energy to produce a disabling wound. Based on this information, a falling 150 gr. service bullet would not be lethal, although it could produce a serious wound.
Oldphart,
From a website: www.loadammo.com, I had found this account of the experiment you had referenced.
In 1920 the U.S. Army Ordnance conducted a series of experiments to try and determine the velocity of falling bullets. The tests were performed from a platform in the middle of a lake near Miami, Florida. The platform was ten feet square and a thin sheet of armor plate was placed over the men firing the gun. The gun was held in a fixture that would allow the gun to be adjusted to bring the shots close to the platform. It was surmised that the sound of the falling bullets could be heard when they hit the water or the platform.
They fired .30 caliber, 150 gr., Spitzer point bullets, at a velocity of 2,700 f.p.s. Using the bullet ballistic coefficient and elapsed time from firing until the bullet struck the water, they calculated that the bullet traveled 9,000 feet in 18 seconds and fell to earth in 31 seconds for a total time of 49 seconds.
Based on the results of these tests it was concluded that the bullet return velocity was about 300 f.p.s. For the 150 gr. bullet this corresponds to an energy of 30 foot pounds. Earlier the Army had determined that, on the average, it required 60 foot pounds of energy to produce a disabling wound. Based on this information, a falling 150 gr. service bullet would not be lethal, although it could produce a serious wound.
If you ignore aerodynamic drag (a big assumption), the max height reached by a bullet fired straight up at a muzzle velocity of v is approximately v*v/64, where v is in feet per second and the max height is in feet. So for 1000 fps, the bullet gets to 1000 * 1000 / 64 = 15625 feet, which is consistent with RCJ's result.
O.k., so I am a nerd engineer. I calculated the maximum height reached by a bullet by directly solving the differential equation of motion. The assumption I made was that the drag force was proportional to the square of the instantaneous velocity of the bullet. This should be a good model for subsonic trajectory, according to my old fluid mechanics textbook.
I then took a 180 grain .40 S&W round with an initial velocity of 900 ft/s, and used the density of air at sea level to make a plot of the maximum height the bullet can rise as a function of the ballistic coefficient. Here is the plot (I included the formula if anyone wants to use it, be careful of units, the software I use automatically converts all units to metric, then I convert it to miles at the end.)
http://i217.photobucket.com/albums/cc229/spamanon/ballist_3.jpg
I had a hard time finding values for the average ballistic coefficient, but I GUESS for a FMJ, we would be around .4, which would put the maximum rise at about a half a mile. Note the case where the ballistic coefficient is zero (no drag) gives a maximum rise of almost 2.5 miles. If anyone wants to see the derivation of the formula, I will send it to them. PM me, but be warned it is math-heavy.
Also, a model is only as good as the assumptions. I believe the approximation that the drag force is proportional to the square of the velocity is about as good as you are going to get without having to solve things numerically.
I then took a 180 grain .40 S&W round with an initial velocity of 900 ft/s, and used the density of air at sea level to make a plot of the maximum height the bullet can rise as a function of the ballistic coefficient. Here is the plot (I included the formula if anyone wants to use it, be careful of units, the software I use automatically converts all units to metric, then I convert it to miles at the end.)
http://i217.photobucket.com/albums/cc229/spamanon/ballist_3.jpg
I had a hard time finding values for the average ballistic coefficient, but I GUESS for a FMJ, we would be around .4, which would put the maximum rise at about a half a mile. Note the case where the ballistic coefficient is zero (no drag) gives a maximum rise of almost 2.5 miles. If anyone wants to see the derivation of the formula, I will send it to them. PM me, but be warned it is math-heavy.
Also, a model is only as good as the assumptions. I believe the approximation that the drag force is proportional to the square of the velocity is about as good as you are going to get without having to solve things numerically.
Last edited:
Drag increases as the square of velocity until about Mach 0.95, then levels off until about Mach 1.05 and then starts to slowly decrease (the Mach angle is getting smaller).
This behavior is caused by the change from uncompressed flow to compressed flow.
It is normally only solved using numerical techniques since the drag function is very difficult to reduce to a continuous function.
Three zones, <0.95 Mach, >1.05 Mach, and between these the third zone with nearly constant drag vs. speed.
Pejsa developed one of the better models, an has a book available for $25 and software for $40.
I have not used his SW so I do not know how well it would handle the variation in density with vertical travel.
This behavior is caused by the change from uncompressed flow to compressed flow.
It is normally only solved using numerical techniques since the drag function is very difficult to reduce to a continuous function.
Three zones, <0.95 Mach, >1.05 Mach, and between these the third zone with nearly constant drag vs. speed.
Pejsa developed one of the better models, an has a book available for $25 and software for $40.
I have not used his SW so I do not know how well it would handle the variation in density with vertical travel.