:headscratch:
... but I can't figure out how far it travels in that time.
I imagine it'll travel all the way to the ground.
The bullet will hit the ground in 51.7 seconds ( if you meant 43k alt and ignoring the possibility of uneven ground or bullet rise or terminal velocity). Does that help any? :D
(39.4 seconds from 25k)
:) I watch that show also, anyway what your missing is deceleration (or more properly drag ) varies with the sqaure of vel.
The baisic equations forms are (combining constants all into BulletDragCO ).
Drag = Vel^2 * AirDensity * BulletDragCO.
and
Acceleration = Drag / BulletMass
<Snip>
HiTech
Without a lot of time to browse, I quickly found one site (wikepedia, I believe) that listed the muzzle velocity of a WW2 Browning .50 Cal MG to be 3,050 feet per second, or about 2,080 mph. :O
E25280: Intial deacceration on the 50cal would be in the 1400 FPSPS Range
HiTech
Are you accounting for drag Challenge?
I,ll run a sim on it tomorrow.
HiTech
Are you accounting for drag Challenge?
I,ll run a sim on it tomorrow.
HiTech
I disagree Challenge
The bullet is falling at its terminal velocity as soon as it leaves the gun. Falling for that long, the drag is going to eventually stop its forward motion, or nearly so. Leaving you with only its downward terminal velocity.
I think a .50 is going to lose half its muzzle velocity in the first second, or 800 yards, beyond that it will be slowing rapidly.
Falling for over 30 seconds gives the bullet lots of time to shed minor forward movement via drag.
Interesting exercise!
Is that going straight down at impact?
With the assumption that the bullet does not destabilize , I would think it would given the numbers below.Alt 25k 45k
Launch Speed 450mph 528mph
Secs to impact 65 97
Horizontal Distance 30858 55107
Plane Dist Traveled 42880 75082
Speed at impact. 444FPS 446FPS
HiTech
In the vertical direction it is accelerating downward as soon as it leaves the gun, but terminal velocity is not achieved instantaneously. Gravitational acceleration is 32 ft/s^2. Terminal velocity would be when it accelerates to a speed (vertical) where the vertical component of drag is equal to the gravitational force.
Exactly. I didnt believe it would hit the ground at terminal velocity (not having slowed to that point) but according to HiTechs results it does.
Yes and no , what happens is it never hits a terminal vel because that terminal vel is chaining as it is descending. It still had about 2fpsps deceleration at impact.
I think in real life the bullet would start tumbling fairly soon, I.E. around 20 secs. And then the terminal vel would be around 200 - 300 fps. And the drag co would increase immensely.
HiTech
This may be a little off topic, but were there ever any injuries reported on the ground from falling bullets or spent cases from planes?knowing full well it was a lie the russians accused Heartman of killing a lot of people with the bullits that "missed" as he shot down aircraft.. seing as it is clearly possable (most notably from "gun claping" accedents) but I have never read of a recoded incedent of some one being killed from a bullit intended for air to air shots. But you would think... all them rounds... someone had to have been hit.... and what about all the flack as it fell? I never hear about that eather..
knowing full well it was a lie the russians accused Heartman of killing a lot of people with the bullits that "missed" as he shot down aircraft.. seing as it is clearly possable (most notably from "gun claping" accedents) but I have never read of a recoded incedent of some one being killed from a bullit intended for air to air shots. But you would think... all them rounds... someone had to have been hit.... and what about all the flack as it fell? I never hear about that eather..
It had to have happened. We know it happened at Pearl Harbor.
This may be a little off topic, but were there ever any injuries reported on the ground from falling bullets or spent cases from planes?
I not questioning your statement but do you have sources? I'd very muchly like to read about it.
I guess a tater shell would still be dangerous after falling from 42k
The ZZ1589A/B fuzes used the initial spin from the MK108 barrel to unlock the fuse for contact detonation. After a certain amount of time passes the speed of revolution will slow down or stop for the Minengeschoss. At which point lacking sufficient spin to keep the firing spring locked up in the housing, it self detonates the Minengeschoss to keep it from potentialy destroying something or someone on the ground. The earlier AZ1587 unlocked in the same manner but had no self detonation function once armed. It had a few pre detonation issues.
Calculus will work as well, as I used to solve a similar problem for a squaddie awhile back. Here is the basic setup, just change the values. Same equation will answer your questions.
http://i1229.photobucket.com/albums/ee468/Derektutor/1Basicdiagram.jpg
http://i1229.photobucket.com/albums/ee468/Derektutor/3PartsandPieces.jpg
I am simply amazed at this thread... It makes me feel so dumb!
I am simply amazed at this thread... It makes me feel so dumb!
Some how Airborn brought back this proof in my head. Learned it in High School.
http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html
HiTech
So Hitech, what figures do you use for atmospheric drag in the game? How is it scaled through altitude (linear, logarithmic, etc.)? Is this information even something we are privy to? This question has come up before within our squad actually, when trying to answer some questions (as seen on the white board pic I posted up with the equation answering the OP's question). I'm not being a jerk here, just asking because I work with mathematics. Oh, and no, I don't plan to go through and check every aspect of the game, lol.
what about a HVAR rocket? how fast does it travel, when will it hit the ground at 45k and will it take a nose dive or go at a little bit of a slope?
Just a guess, but there are standard external ballistics calculators around (i.e. Sierra and some freeware) that can do this calc.
The only "tricks" are to input atmospheric density at the firing point, and to use the true airspeed as a full value 0 degree wind.
Hitech - how did you account for changing atmospheric density as the projectile drops?
Just curious
Just a guess, but there are standard external ballistics calculators around (i.e. Sierra and some freeware) that can do this calc.
And I had my answer weeks ago.Oh? :headscratch: I read through all the posts and I didn't see what I thought was a definitive answer. And please don't think that statement means that I have the answer.
With the assumption that the bullet does not destabilize , I would think it would given the numbers below.Alt 25k 45k
Launch Speed 450mph 528mph
Secs to impact 65 97
Horizontal Distance 30858 55107
Plane Dist Traveled 42880 75082
Speed at impact. 444FPS 446FPS
HiTech
interesting thread. maybe i missed it as i do from my bad eyesight, or not reading carefully enough, but in the game do the bullets or cannon shells accully hit the ground or does the computer determen to just poff the round? :salute
They have a max time of flight after that they disappear.:salute ty sirs have been wondering that for awhile
HiTech