Jump to content
SASS Wire Forum

How High is UP


Aunt Jen

Recommended Posts

I'm guessing a .357 might make a half mile before the earth sucked it back down. Time going up in about 3 seconds. .38's and .45's not as far because of velocity to start is much slower. Just sayin'

 

Big Jake

Link to post
Share on other sites

How high will a bullet go if fired straight up? What variables would be involved? just digging into it more deeply out of curiosity. I am not a physicist, not an engineer. But I am a curious person.

 

1. Gravity: Gravity would pull against it at, what, 32 feet per second per second? So speed of the bullet would matter.

 

2. Drag:

 

a. Shape of bullet: Taking into account the shape of the bullet and its width (drag coefficient), which would be (I'm guessing ) parasite drag on the bullet? I believe drag increases as the square of the speed, so it'd decrease markedly as the bullet slowed...

 

b. Air Density:

Altitude fired from (1013 millibars at mean sea level, 942 at 2,000, about 500 at 18,000')

Temperature of the air

Presence of High or Low pressure systems

Moisture content in the air

 

I'm sure some of these are minor or negligible. But can someone run a few numbers? Or do we have any physicists or engineers here who have played with this idea?

Link to post
Share on other sites

How high will a bullet go if fired straight up? What variables would be involved? just digging into it more deeply out of curiosity. I am not a physicist, not an engineer. But I am a curious person.

 

1. Gravity: Gravity would pull against it at, what, 32 feet per second per second? So speed of the bullet would matter.

 

2. Drag:

 

a. Shape of bullet: Taking into account the shape of the bullet and its width (drag coefficient), which would be (I'm guessing ) parasite drag on the bullet? I believe drag increases as the square of the speed, so it'd decrease markedly as the bullet slowed...

 

b. Air Density:

Altitude fired from (1013 millibars at mean sea level, 942 at 2,000, about 500 at 18,000')

Temperature of the air

Presence of High or Low pressure systems

Moisture content in the air

 

I'm sure some of these are minor or negligible. But can someone run a few numbers? Or do we have any physicists or engineers here who have played with this idea?

 

Hi. It's not a recommendation. It's a curiosity, a question in problem-solving, a pondering, a what-if in physics. I think it'd make a good topic for a documentary. :)

Link to post
Share on other sites

Hi. It's not a recommendation. It's a curiosity, a question in problem-solving, a pondering, a what-if in physics. I think it'd make a good topic for a documentary. :)

 

....maybe the folks in Iraq, Afganistan, or middle east can answer it :lol: They love to shoot straight up :wacko:

 

GG ~ :FlagAm:

Link to post
Share on other sites

Technically, it would not start a re-entry unless it left the atmosphere. ;)

 

If you find out, tell these guys. They don't know either.

 

I',m sure it's a fairly simple math problem factoring in gravity, velocity, and bullet weight. I doubt air density would have much effect. Alas I'm to simple to do math. :lol:

Link to post
Share on other sites

....maybe the folks in Iraq, Afganistan, or middle east can answer it :lol: They love to shoot straight up :wacko:

 

GG ~ :FlagAm:

 

HAR , I have always wondered how many of those people get konked by falling bullets at one of thier big celebrations. Doesn't say too much about the thought process beforehand. :lol::lol:

Let 'er rip with that AK , Abdul , Rex :D

Link to post
Share on other sites

Ok, just playing with my mental blocks here:

 

suppose initial bullet velocity is 1024 ft/sec (why? math is easier). 1024 ft/sec = 698 mph

 

 

Accelleration due to gravity is 32ft/sec/sec (why? because it is at sea level at least)

 

Suppose there is no air resistance (Why? too many other factors to consider)

 

The bullet will rise for t=v/a = (1024 ft)/(32 ft/sec/sec) = 32 sec (we are overlooking the effects of air resistance)

 

using the formula t^2 = 2*distance/(32 ft/sec/sec) we get

 

distance = t^2*(32 ft/sec/sec)=1024*32 ft = 32768 ft = 6.2 miles

 

Why should you ignore this result?

 

If there were no really air resistance, the same formulas would be in effect when the bullet started falling and it would reach a speed of 1024ft/sec (=698 mph) when it hits the ground 64 seconds after the shot is fired. In reality the falling bullet reaches terminal velocity of around 200 mph. You have to figure that if air resistance slows the falling bullet, it also slows the rising bullet so that the bullet stops rising in less than 32 seconds and never reaches an altitude of anywhere near 6.2 miles.

 

SO, to answer your question of "How high does the bullet rise?"

 

Hell, I don't know, but I do know that it is less than 6.2 miles.

Link to post
Share on other sites

Half way :P

 

Nope! Goes ALL THE WAY up and ALL THE WAY back down. The top is only half the total distance. Let's be scientific. Let's say you drive from your place to the local toy store and back. When you get there you have driven all the way or half the total distance. When you get home again you are all the way home and have completed the total distance for the trip.

 

I have simplified this by removing a few variables such as trips to the gas station, the market, the hardware store, to visits friends, etc, but the basic principle is the same.

 

I have also not calculated the cost of any guns, ammo, leather goods, liquor, clothing, and other baubles and play pretties you accumulate on the trip.

 

Prof. F. Rod, DF (That's Doctor of Foolishness, dammit. Get you minds out of the gutter.)

Link to post
Share on other sites

Here is another bit of physics to blow your mind. If you fire a bullet, any bullet from any rifle, 22LR to 50 BMG, parallel to the ground, and drop a bullet from the height of the barrel at the same time, the bullet you fired will hit the ground at the same time as the bullet you dropped.

 

And if that doesn't really blow your mind, try and wrap your thoughts around this one. Picture a record player (dangit, most of us are old enough to know what a record player is). OK, so compare the innermost ring and the outside edge. They both make one complete turn in the same amount of time, but a point on the outside ring has to make a lot bigger circle than the inside ring in the same amount so it has to move a lot faster, so 2 points on the same record are moving at two different speeds. There's something to contemplate when you are half the way to emptying a whiskey bottle!

Link to post
Share on other sites

Here is another bit of physics to blow your mind. If you fire a bullet, any bullet from any rifle, 22LR to 50 BMG, parallel to the ground, and drop a bullet from the height of the barrel at the same time, the bullet you fired will hit the ground at the same time as the bullet you dropped.

 

 

If the ground in front of you is flat for the entire distance of travel of the bullet, where flat is defined as being an arc a uniform distance from the center of the earth.

Link to post
Share on other sites

on one of the episodes of myth busters they went out in the middle of the salt salt flats to be away from every one, i dont remember what they where trying to prove but they fired several rounds straight up in the air then ran for cover i know they where timing it and trying to get it to come back in close proximity to where it was fired just dont remember why was a year or 2 ago when i saw it ...... just thought i would bring this up!

Link to post
Share on other sites

Actually, the ground between you and your impact point can be as uneven as you can imagine as long as it does not intersect the trajectory of the projectile until it hits a point at the same altitude as the point you are standing upon, but these are all the standard footnoted assumptions listed in fine print at the bottom of the page or in the back of your physics book. :rolleyes: I did enjoy physics, even in high school. My high school instructor enjoyed using bullets at the basis of examples, and that caught my attention from day one.

Link to post
Share on other sites

If you are seriously interested in this subject check out Hatchers Notebook.

He was some form of armorer or arms expert for the US Army and explored this and many other interesting subjects.

Link to post
Share on other sites

The Mythbusters were testing the myth that falling bullets can kill, if fired straight up. They set up a 9mm pistol, fired remotely, while they were under ballistic glass. If I remember correctly, it took about 30 seconds before they heard bullets hitting around them. Then they measured the depth of penetration in the dirt when they found a bullet hole. While all of the bullets found would put a knot on your head, none would have been fatal.

Then they repeated the experiment using either a .308 or .30-06, cain't remember which. Results were the same, except it took the rifle bullet longer to hit the ground. So they busted the myth that bullets fired STRAIGHT UP are fatal when they come back down.

 

If a bullet is fired at an angle where it retains its spin and stability it will retain enough velocity to be fatal.

 

They also tested the dropped bullet and fired bullet hitting at the same time. They didn't quite, but it was only like 2 milliseconds difference. They used a .45ACP for the test.

 

BTW I'm no physicist by any means, but some folks will argue that the bullet dropped/fired will never hit at the same time if fired from a rifled barrel. They say the rotation of the bullet affects the speed of falling. I don't understand how, since to me the force of rotation is equal in all directions, but the smart folks can put a whole bunch of funny letters and symbols on a chalk board and come up with an answer.

Link to post
Share on other sites

If you are seriously interested in this subject check out Hatchers Notebook.

He was some form of armorer or arms expert for the US Army and explored this and many other interesting subjects.

 

Hatcher expressly looked into the lethality of bullets falling back to earth after having been fired straight up (like you would do if you were trying to shoot down an airplane).

 

"And if that doesn't really blow your mind, try and wrap your thoughts around this one. Picture a record player (dangit, most of us are old enough to know what a record player is). OK, so compare the innermost ring and the outside edge. They both make one complete turn in the same amount of time, but a point on the outside ring has to make a lot bigger circle than the inside ring in the same amount so it has to move a lot faster, so 2 points on the same record are moving at two different speeds. There's something to contemplate when you are half the way to emptying a whiskey bottle!"

 

And by the same token the wing tips of an airplane will be moving at a different speed when it is making a turn.

 

Duffield

Link to post
Share on other sites

On dropped. Vs fired bullet hitting ground at same time...

 

I'd guess same time.

 

BUT. What if there were some circumstance that could give the bullet a little lift? Aerodynamically.

 

Example:

 

Bullet fired. Straight. Spinning. All forces same top n bottom

 

Then as bullet falls in flight, yet still pointed straight ahead, it develops POSITIVE ANGLE of ATTACK as still pointed straight ahead but relative wind has a downward motion.

 

So........bullet could develop some lift during flight. The more it slows in flight. The more lift is generated due to increasing angle of attack.

 

iPhone

Link to post
Share on other sites

I'm gonna try and take a stab at this. If'n my college fysiks is right....

 

To calculate distance use the following formula : V(f)^2 = V(i)^2 + (2GD)

 

Where

V(f) = final velocity in feet per second

V(i) = initial velocity in feet per second

G = gravity = 32.174 ft/sec^2

D = Total distance traveled in feet

 

In this case since the bullet is being shot upwards the final velocity will be equal to 0 since it will be stopped on the ground.

 

The initial velocity is the velocity out of the barrel, depending upon bullet mass, bullet shape, barrel twist and length, powder load, etc. For example purposes let's say we are using .38s with cowboy loads and the velocity is 800 feet per second.

 

So the equation becomes

0 = (800 ^2) + (2 x 32.174 x D)

 

Solving for D

D = (800^2) / (2 x 32.174)

D = 9946 feet or 1.88 miles

 

To calculate the time in seconds to reach that height use the following formula : V(f) = V(i) + (G x T)

 

Where

T = Total time in seconds the bullet is in the air

 

Solving the equation for T

T = [V(f) – V(i)] / G

T = 800 / 32.174

T = 24.9 seconds

 

Time to reach maximum elevation of 9946 feet would be half of total time or (24.9/2) = 12.45 seconds

 

If'n there are any professors or ballisticians out there than can shed any corrections to the above have at it. I'm neither a professor or ballistician...just an engineer who blew the dust offn' my text books the schoolmarm let me use in my persuit of hier edukamation!

Link to post
Share on other sites

Hatcher expressly looked into the lethality of bullets falling back to earth after having been fired straight up (like you would do if you were trying to shoot down an airplane).

 

"And if that doesn't really blow your mind, try and wrap your thoughts around this one. Picture a record player (dangit, most of us are old enough to know what a record player is). OK, so compare the innermost ring and the outside edge. They both make one complete turn in the same amount of time, but a point on the outside ring has to make a lot bigger circle than the inside ring in the same amount so it has to move a lot faster, so 2 points on the same record are moving at two different speeds. There's something to contemplate when you are half the way to emptying a whiskey bottle!"

 

And by the same token the wing tips of an airplane will be moving at a different speed when it is making a turn.

 

Duffield

 

I've been told that that is why helicopters can't (yet) fly faster than the speed of sound. At the moment that the SOS is reached the blade moving forward will be going a tad faster than the blade going backward and the result will destroy the rotors rather instantly at the moment one breaks the barrier and the other doesn't.

Link to post
Share on other sites

That is a easy answer. Till it stops LOL

The Lord Jesus Christ is the only way!

Link to post
Share on other sites

I think I have a headache!! Not from shootin up but from readin all these figgers!!:wacko:

Link to post
Share on other sites

Wow! I'm hearing estimates that are way more than I guessed. Who would have thought?

 

Then, the next question is:

 

how much powder would you need to put into a .357 so the bullet could reach the moon?

Link to post
Share on other sites

At least a double charge! Actually it takes 17500mph to achieve a stable orbit, then bump that up to 22500mph to fully escape earth's gravity. If you only get 22499mph you'll just fall back into orbit again.

Link to post
Share on other sites

At least a double charge! Actually it takes 17500mph to achieve a stable orbit, then bump that up to 22500mph to fully escape earth's gravity. If you only get 22499mph you'll just fall back into orbit again.

Bummer. So if I get 22500 I can head toward the moon?

Hmmm.. So if a .357 goes 1450 fps, or 988 mph, then I'd have to put 22.8 .357 charges in it? Or is not that linear? Probably not. There's probably some equation about it, like how power required equals the cube of the speed, as it does in airplanes.

 

If so, then the cube of 22.8 is about 11,811 .357s worth of charges in the thing to make it go?

 

I am not a techie, but I've eaten lunch at Caltech.

 

Did I get the right figure within a half million?

Link to post
Share on other sites

Jen, I'm not the one that can put all them funny symbols and letters on a chalkboard, then tell you the answer. I do think that if a bullet was fired from earth at 22,500 it still wouldn't make it out of orbit, as friction would slow it down. If I remember right from 7th grade science (a whole lot of long time agos :lol: ) you already have to be in orbit when you accelerate to the required 22,500.

 

Another problem, the expansion rate of smokeless gunpowder is limited to somewhere 'twixt 5,000 and 6,000 fps, so a bullet can never leave the barrel at a higher speed that those numbers.

 

So I reckon you'll have to come up with a new propellant! :lol:

Link to post
Share on other sites

If you took out your revolver with regular loads and shot straight up, how high would the bullet go, before beginning its re-entry?

 

For a

 

.45

.357

.38

 

?

 

Aunt Jen . . . do you live next door to "Hootin' Hiedi" ? . . .

 

The threads that you start have me suspicious that you are getting into her corn squeezin's ... :blink:

Link to post
Share on other sites

I'm going to come along and toss (I think) a monkey wrench into the discussion. I read a lot of military history and the threshold for an aircraft to be above small arms fire (rifle caliber or lesser, presuming 30-06 or it's rough equivelent) is 3000 feet. That doesn't seem to match up with some of the calculations already posted.

 

:blink::wacko:

Link to post
Share on other sites

I'm going to come along and toss (I think) a monkey wrench into the discussion. I read a lot of military history and the threshold for an aircraft to be above small arms fire (rifle caliber or lesser, presuming 30-06 or it's rough equivelent) is 3000 feet. That doesn't seem to match up with some of the calculations already posted.

 

:blink::wacko:

 

I can assure you, from personal experience, that 2500 feet is above the effective range of an AK-47. However, ground attack aircraft, both helicopters and fixed wing, operate low enough that small arms fire is effective against them. Look at what the Brits did to the Argentine A-4s in the Falklands campaign!

Link to post
Share on other sites

We must not forget to factor in the height of the person

pulling the trigger! :rolleyes::unsure::wacko::blink:

Link to post
Share on other sites

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

By using this site, you agree to our Terms of Use.