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Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 01:59
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If a rifle barrel is held 100% level with the ground and 1yard above the ground, how far would a bullet, fired at 2850ft per sec, travel before hitting the ground?
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 06:53
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cant give you an exact answer unless you give us a BC of the bullet, but its going to be somewhere around 350yd to 425yds on just a rough guess (assuming my brain is working this morning).
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 07:00
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Approximately 417 yards. 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 07:03
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Rough guesses will be OK.
A brief explanation of your answer will count for extra points, a drawing or graph will be awarded extra extra points.
A you-tube video clip demonstrating your answer will give you a cum lauda pass.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 07:17
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im assuming kickboxer is giving answer for his load only and whatever he is shooting is one heck of a round also (or he confused the path of the bullet with a ##yrd zero for total bullet drop).
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 07:19
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X = 180 Y = 90 (X+Pyro)+(Y-Pyro) = ?

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Rough guess, 367 yards, assuming perfectly flat ground, a .4 b.c. bullet, barrel perfectly level, no line of sight over bore adjustments, yada, yada...........

Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 07:28
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find you a ballistic calculator and enter the data (g1 or g7 b/c there shouldnt be much difference with these short distances) and see what the total bullet drop shows you. There are many of them
JBM/Ballistics FTE, Shooter, Strelok, simple excel spread sheets, etc.
Also make sure you are looking at total bullet drop and not path of the bullet (based on how you worded the question) which will be affected by where you are zeroed at. If you want the path of the bullet then we will also need your zero distance and scope height.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 07:31
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Since initial velocity is zero (0), and we KNOW that the bullet begins to fall to the earth as soon as it leaves the barrel (regardless of angle), but angle is zero, therefore the equation becomes:
 
 t= sqrt (2h/g) where t = time of travel,  h = the height above ground, g = earthward velocity due to gravity.  Since g is more accurate in m/s^2, I always use metric terms and convert to "English".  Therefore, rounding error can result in some difference in end result.
Using the above equation, we get a result of approximately .43 seconds travel time at 2850fps.  Depending on the conversion/rounding travel is from 1222 to 1251 feet or 407-417 yards.  I choose the high end...
 
In this case, for me, a picture would add little to the understanding. 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:03
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Because the barrel is level, zeroed distance must be at the muzzle. For the sake of this calculation 1meter.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:05
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I get 419 yards. @ 30% humidity and 1204 altitude with a 30cal 168 berger
 
 
 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:17
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How long will it take a bullet to hit the ground if dropped from 1 yard? (180gr SGK)
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:28
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Originally posted by Kickboxer Kickboxer wrote:

 Using the above equation, we get a result of approximately .43 seconds travel time at 2850fps. 
This calculation is assuming a constant velocity (which is not the case), therefore your time of flight is off. 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:32
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The only thing that matters IMHO is speed of the bullet and that the angle is zero. G is constant so distance traveled is going to depend on velocity period. Bullet wieght and BC are almost nil in comparision to velocity and good ol' gravity.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:40
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Originally posted by Bigdaddy0381 Bigdaddy0381 wrote:

I get 419 yards. @ 30% humidity and 1204 altitude with a 30cal 168 berger
trying not to split hairs here, but there is no way youre only getting 36" of bullet drop with a 168 berger launched at 2,850fps at 419yards. 
+1 on what Trigger said, its gonna take you to about 375yards. 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:53
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Originally posted by lumberjack149 lumberjack149 wrote:

Originally posted by Bigdaddy0381 Bigdaddy0381 wrote:

I get 419 yards. @ 30% humidity and 1204 altitude with a 30cal 168 berger
trying not to split hairs here, but there is no way youre only getting 36" of bullet drop with a 168 berger launched at 2,850fps at 419yards. 
+1 on what Trigger said, its gonna take you to about 375yards. 
 @ 375 yards i'm dropping 25.5inches. with a flight time of  0.450
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:53
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Originally posted by 8shots 8shots wrote:

How long will it take a bullet to hit the ground if dropped from 1 yard? (180gr SGK)
0.495
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 08:56
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thebullet will take the same amt. of time as if youhad just dropped the bullet. given gravity at 32 ft/sec. calculate the time, bullet travels at given ft per sec, how many sec, (same as drop) then in that fraction contains how many feet.   ---  your answer
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 09:08
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Originally posted by Bigdaddy0381 Bigdaddy0381 wrote:

Originally posted by lumberjack149 lumberjack149 wrote:

Originally posted by Bigdaddy0381 Bigdaddy0381 wrote:

I get 419 yards. @ 30% humidity and 1204 altitude with a 30cal 168 berger
trying not to split hairs here, but there is no way youre only getting 36" of bullet drop with a 168 berger launched at 2,850fps at 419yards. 

  @ 375 yards i'm dropping 25.5inches. with a flight time of  0.450
i agree with your time of flight, just not the drop. Are you sure youre talking true bullet drop and not your calculated flight path with a 100yd zero?
 

Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 09:12
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That is true tested bullet drop. The drop time is by calculation as I have no way to test that with my set up's. Besides It's what my rifle will do. Yours might be different. I test a few rifles a year when i do loads for them. This was one of my personal rifles.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 09:13
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kickboxer answer is correct however the degree of precision depends on the significant figures used for the inputs, in other words gravity in metic sys. in not necessarily more accurate than gravity in british. the only real variable is where you are measuring gravity as it changes at different places on the plantet. Galileo is hanging his head
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 09:30
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As has been mentioned, it's going to depend on the BC and the atmospheric conditions. But, generally, it's going to be somewhere in the vicinity of of 375 yards for a really good .308 bullet @ 2850. I entered a sight height of 0.0 and a zero of 1 yard in JBM (the minimum allowed) and 0 yards in Ballistica and came up with the same results: At about 363 yards, my Berger 175 gr. OTMs (G7 BC .259) @ 2690 and a station pressure of 24.00 Hg, temp 59F is going to hit the dirt.
 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 09:36
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It's a good question, BTW, because it reinforces the concept that we are never shooting a bullet in a straight line to a target out at distance...something you see anytime you watch a quarterback lob a hail mary pass...although not so much from the Pats yesterday. Spraying water through a nozzle on a garden hose is a good way to demonstrate both gravity's effect on a trajectory and the effects of compensating by aiming higher.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 09:47
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this question is asked in most beginning physics classes, the assumptions made about the bc in "flight" are the same as the bc when its drops,  (cancel out) the entire arguement is cateris paribus(all things being equal) because the orginal quesion does not contain information outside the given parameters. One of the reasons the profs ask it -- is to see how much shick gets drug into the problem, not related to the basic question' If outsideinformatin is allowed in, when do you stop? It couldbe argued that neither exist because quantum reality is a plasma, or that the numerical methods in the coupled differential equations and the integration steps used in the computer program aren't accurate enough to reflect the difference.
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 10:52
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Originally posted by lumberjack149 lumberjack149 wrote:

Originally posted by Kickboxer Kickboxer wrote:

 Using the above equation, we get a result of approximately .43 seconds travel time at 2850fps. 
This calculation is assuming a constant velocity (which is not the case), therefore your time of flight is off. 
 
Being a tester of munitions and their performance by profession, I am not allowed to "create" data.  I answered the question as asked... discounting resistance/drag, wind.  There was no mention of it and no facts were included in the problem statement. 
 
 
Post Options Post Options   Thanks (0) Thanks(0)     Back to Top Direct Link To This Post Posted: February/06/2012 at 10:55
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Originally posted by Dale Clifford Dale Clifford wrote:

this question is asked in most beginning physics classes, the assumptions made about the bc in "flight" are the same as the bc when its drops,  (cancel out) the entire arguement is cateris paribus(all things being equal) because the orginal quesion does not contain information outside the given parameters. One of the reasons the profs ask it -- is to see how much shick gets drug into the problem, not related to the basic question' If outsideinformatin is allowed in, when do you stop? It couldbe argued that neither exist because quantum reality is a plasma, or that the numerical methods in the coupled differential equations and the integration steps used in the computer program aren't accurate enough to reflect the difference.
Since you brought up quantum reality, it could be that the bullet fired from the gun gets there before you pulled the trigger and when you last walked down range, thus you end up shooting yourself in the foot, or that the bullet you drop by hand doesn't show up until your great- great grandson happens by and it hits him in the foot... all happening simultaneously, of course.
Actually, time isn't construed as changing in different quantum realities/dimensions/levels, so much as distance/size, with distance being the only illusory aspect and time as a constant.
Or something.

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