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jonoMT View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote jonoMT Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 09:30
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.
 
Reaction time is a factor...
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Post Options Post Options   Thanks (0) Thanks(0)   Quote jonoMT Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 09:36
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.
Reaction time is a factor...
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Dale Clifford Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 09:47
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.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kickboxer Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 10:52
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. 
 
 
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Alan Robertson Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 10:55
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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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kickboxer Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 10:59
Originally posted by Dale Clifford Dale Clifford wrote:

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
 
This is basiclly true... however the most "commonly stated" terms 9.81m/s^2 vs. the most common English of 32 ft/s^2, the metric is most accurately expressed and what I most often use.  I believe I could come close at 32.2 ft/s^2, but just never have used it. 
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Dale Clifford Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 11:46
ballistic curves are basically log equations and as such have closed forms. When doing numerical analysis using diffference equations or differential equations, the analyst will try to fit the computed data with a closed form to arrive at an error. This is the basis of Patagonia's software and there complaint against other forms of software on the market. Patagonia computes using curve fitting techniques derived from real time flight velocities vs. location.
 
 
Now for a really, really hard math question   ---
 
Prove that -1 x -1  =  +1    (seriously this is hard) the answer is not the obvious cartisian graph solution
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kickboxer Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 13:45
Originally posted by Dale Clifford Dale Clifford wrote:

ballistic curves are basically log equations and as such have closed forms. When doing numerical analysis using diffference equations or differential equations, the analyst will try to fit the computed data with a closed form to arrive at an error. This is the basis of Patagonia's software and there complaint against other forms of software on the market. Patagonia computes using curve fitting techniques derived from real time flight velocities vs. location.
 
 
Now for a really, really hard math question   ---
 
Prove that -1 x -1  =  +1    (seriously this is hard) the answer is not the obvious cartisian graph solution
 
the proof is not difficult, just tedious... time consuming... after you prove -1 * x = -x, it is just details...
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Dale Clifford Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 15:15
won"t work , it assumes that -1 x -1 is correct--it will lead to a logical nor statement stating
if you believe neither this or that is incorrect then so and so must be
 
make a long story short and save the suspense of everyone hanging on their sets, (opps seats)
 
the statement isonly true in convergent algebras and must be proven using algebraic ring theory.
 
thus the saying "algebraists love closed sets"    (have to go to set theory with the proof)
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Alan Robertson Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 16:40
Hope you guys aren't closet topologists trying to tell the difference between your coffee cup and your donut hole.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Dale Clifford Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 16:52
they are the same ---   2 - manifolds
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Post Options Post Options   Thanks (0) Thanks(0)   Quote dsr Quote  Post ReplyReply Direct Link To This Post Posted: February/06/2012 at 22:19
410 yrds assuming constant velocity from muzzle. ( as first order approximation) with g=32.17 ft/s^2. The solution is a vector quantity and can be solved independently for vertical and horizontal components. The vertical solution controls the answer by being the first at a time ~ 0.431 seconds which sets the time of flight before the one second mark. Providing a distance approximately 410 yards down range. Further information only refines the the accuracy of the answer.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote 8shots Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 02:30
t= sqrt(0.9144x2/9.81)
=.432sec flight time
at 950yds per sec
=410 yds
.9144m being 1yard
9.81 being speed of gravity in m/s^2.
Essentially I was looking for equating flight time of the bullet with the flight time of a dropped bullet.
Ok , back to coffee and doughnuts Howdy


Edited by 8shots - February/07/2012 at 04:05
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Kickboxer View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kickboxer Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 06:48
Originally posted by Dale Clifford Dale Clifford wrote:

won"t work , it assumes that -1 x -1 is correct--it will lead to a logical nor statement stating
if you believe neither this or that is incorrect then so and so must be
 
make a long story short and save the suspense of everyone hanging on their sets, (opps seats)
 
the statement isonly true in convergent algebras and must be proven using algebraic ring theory.
 
thus the saying "algebraists love closed sets"    (have to go to set theory with the proof)
 
totally incorrect
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kickboxer Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 06:52
Originally posted by 8shots 8shots wrote:

t= sqrt(0.9144x2/9.81)
=.432sec flight time
at 950yds per sec
=410 yds
.9144m being 1yard
9.81 being speed of gravity in m/s^2.
Essentially I was looking for equating flight time of the bullet with the flight time of a dropped bullet.
Ok , back to coffee and doughnuts Howdy
 
As I said earlier, depends upon rounding assumptions.  410 yds is as good as any... and very acceptable..
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trigger29 View Drop Down
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X = 180 Y = 90 (X+Pyro)+(Y-Pyro) = ?

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Post Options Post Options   Thanks (0) Thanks(0)   Quote trigger29 Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 08:32
Originally posted by Kickboxer Kickboxer wrote:

Originally posted by 8shots 8shots wrote:

t= sqrt(0.9144x2/9.81)
=.432sec flight time
at 950yds per sec
=410 yds
.9144m being 1yard
9.81 being speed of gravity in m/s^2.
Essentially I was looking for equating flight time of the bullet with the flight time of a dropped bullet.
Ok , back to coffee and doughnuts Howdy
 
As I said earlier, depends upon rounding assumptions.  410 yds is as good as any... and very acceptable..
In a vacuume.......
I understand the bullet will hit the dirt at exactly the same time as a bullet dropped straight down. This is not a mystery to any of us. In the real world though, at least where I live, there is air that will slow BOTH bullets....... The fired one will not go as far, as drag will slow it down, and the one dropped will also encounter wind resistance before hitting the dirt.

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Post Options Post Options   Thanks (0) Thanks(0)   Quote Alan Robertson Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 13:33
Originally posted by trigger29 trigger29 wrote:

Originally posted by Kickboxer Kickboxer wrote:

Originally posted by 8shots 8shots wrote:

t= sqrt(0.9144x2/9.81)
=.432sec flight time
at 950yds per sec
=410 yds
.9144m being 1yard
9.81 being speed of gravity in m/s^2.
Essentially I was looking for equating flight time of the bullet with the flight time of a dropped bullet.
Ok , back to coffee and doughnuts Howdy
 
As I said earlier, depends upon rounding assumptions.  410 yds is as good as any... and very acceptable..
In a vacuume.......
I understand the bullet will hit the dirt at exactly the same time as a bullet dropped straight down. This is not a mystery to any of us. In the real world though, at least where I live, there is air that will slow BOTH bullets....... The fired one will not go as far, as drag will slow it down, and the one dropped will also encounter wind resistance before hitting the dirt.
Not only that, but in the real world, the bullet tip is corkscrewing around the rotating path of the bullet, itself another corkscrew. A plot of the bullet tip in flight looks like a very long, small- diameter coil spring formed into a larger diameter coil spring, which represents the actual flight path, all as a reaction of the imperfect bullet accelerated through the rifled bore and then acted upon by atmosphere. With a very imperfect bullet, or a bullet loaded/fired very much out of axial alignment, the flight time of the bullet could be radically enhanced or attenuated. A plot of the actual rotation of the bullet would be yet a third corkscrew- like plot underlying the first two mentioned. A 3-in-1 screw. How do we ever hit anything?

What the hell am I doin' in this thread?
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Dale Clifford Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 15:29
Jeez- a 3-in-1 screw. Thats a lot to think about.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote SD Dog Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 16:51
Where's Ed?  Kind of his speed.
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X = 180 Y = 90 (X+Pyro)+(Y-Pyro) = ?

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Post Options Post Options   Thanks (0) Thanks(0)   Quote trigger29 Quote  Post ReplyReply Direct Link To This Post Posted: February/07/2012 at 17:28
A 3-in-1 screw maybe.......... The math, not so much.

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