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Posted: 6/30/2002 11:44:23 AM EDT
My basic math not counting friction losses, and just using basic conversions tell me if I shoot at 45 degrees, something moving at 1400 fps will travel over 64000 feet or about 12 miles. Does this sound right? Or is it further than that?
Link Posted: 6/30/2002 12:06:28 PM EDT
Gravity man. I belive a high powered rifle bullet, lets say .308, drops i think a foot after half a mile? Damn, i've really forgotten these things. But yea, lets say your standing, and shoot a bullet straight ahead, the bullet will hit the ground way before 4 miles.
Link Posted: 6/30/2002 11:46:05 PM EDT
If I recall correctly, a .22LR has the most travel @ two miles. Be kinda interesting to have someone test this out on a lake somewhere. American Shooter or some such doing it so it's done safely.
Link Posted: 6/30/2002 11:51:37 PM EDT
Originally Posted By Andrewh: My basic math not counting friction losses, and just using basic conversions tell me if I shoot at 45 degrees, something moving at 1400 fps will travel over 64000 feet or about 12 miles. Does this sound right? Or is it further than that?
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yeah if there were no wind resistance or gravity. I think two miles is about it with a .22 because it is so light and small. My girlfirend is a physics major, so we argue all of the time (of course she is always right, I hate it)
Link Posted: 6/30/2002 11:55:52 PM EDT
Originally Posted By MAHABALI: My girlfirend is a physics major, so we argue all of the time (of course she is always right, I hate it)
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woah, never quoted myself before, kinda weird.. anyway, One of our biggest arguements is: If you shoot a bullet perfectly level, straight, and at the exact same time you drop a bullet from the same height, (no matter if it is 3' or 3", or whatever) that they will land at the same time. At least that is what she says, how can this be right??? we have argued about this countless times.
Link Posted: 7/1/2002 12:02:42 AM EDT
Link Posted: 7/1/2002 12:03:43 AM EDT
Thats what she says, still doesnt make sense to me.
Link Posted: 7/1/2002 1:28:48 AM EDT
Without repeating the many pages on these questions that have been posted here, and on other forums: 1 Your girlfriend is absolutely right. 2 Actually the optimum angle for maximum distance is around 27 degrees, not 45 3 A bullet from, say, a .30-'06 will touch down after 4-6 miles, generally speaking. Yes, there are exceptions, but that would cover 90% of the time.
Link Posted: 7/1/2002 2:36:00 AM EDT
Originally Posted By Andrewh: My basic math not counting friction losses, and just using basic conversions tell me if I shoot at 45 degrees, something moving at 1400 fps will travel over 64000 feet or about 12 miles. Does this sound right? Or is it further than that?
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That's about what I get: 61,000 feet
Originally Posted By MAHABALI: One of our biggest arguements is: If you shoot a bullet perfectly level, straight, and at the exact same time you drop a bullet from the same height, (no matter if it is 3' or 3", or whatever) that they will land at the same time.
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Force = Mass * Acceleration. The gravitational force causes an acceleration (change in velocity per unit of time). Forces (and accelerations) are vectors, meaning they act in a certain direction. That means that there is no way for the acceleration due to gravity (acting downward) can affect the horizontal velocity of an object. The converse must also be true; the horizontal velocity of an object has no impact on the time it takes an object to fall under the acceleration of gravity. Now the fall time [i]will[/i] be affected by vertical velocity, because that is the direction in which the acceleration due to gravity acts. So an object with more upward velocity will take longer to fall because gravity first has to reduce the upward velocity before the object starts moving down. An object which starts off with some downward velocity will hit the ground faster because it already has a head start. The bullet fired straight out and the bullet which is dropped each have no vertical velocity and thus will hit the ground at the same time (in an ideal situation).
Link Posted: 7/1/2002 3:05:38 AM EDT
How far will it go? There are two components of the bullet's path: vertical and horizontal. Lets look at the vertical first. Y = T X V X (sine of angle) - 16 X T X T find the sine of the angle of inclination of your weapon, and plug it in to this formula.The value of V is the muzzle velocity of the round. Set Y = 0, and solve for T. Use the larger of the two T values, and put it into this formula: H = T X V X (cosine of angle) Put into it the larger T value, the muzzle velocity, and the cosine of the inclination angle. This will tell HOW FAR DOWN RANGE she will travel. Finding out how HIGH she go is a little more difficult. If you want to find out, let me know and I'll tell you how. This approach is simplistic in that we neglect such variables as cross section area of round, temperature, wind resistance, and so forth. Hope this helps.
Link Posted: 7/1/2002 3:11:29 AM EDT
Originally Posted By Celt: 2 Actually the optimum angle for maximum distance is around 27 degrees, not 45
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I thought the optimum angle for distance was around 53 degrees? 27 seems awful low.
Link Posted: 7/1/2002 3:34:56 AM EDT
according to the book "understand ballistics" by robert rinker. the angle for maximum range varies between 27 and 35 degrees. the angle seems right iirc a max range for a base ball is in a similar range angle of departure range. he does list max range of some bullets that hve been tested... .22LR 40 grain 1145 fps = range 1500 yards .30 ball m2 152 grain 2880 fps = 3500 yards .30 boattail m1 172 grain 2600 fps = range 5500 yards there are other the table is on page 233. he says most of the data has been taken from military tests.
Link Posted: 7/1/2002 3:54:00 AM EDT
the rocket scientists here have ignored a very important variable in the real world that is often ignored in the basic physics experiments: the effect of wind resistence. Sure gravitiy is a constant in the vertical plane, but the effects of wind resistence, the bullet design (ballistic coeff.), muzzle velocity all factor in the deccelleration in the horizontal plane. At short ranges, ie under 300 yds, this is not a big deal to high power rifle, but it is at long range ie over 800 yd. the bullet listed above may theorectically go several miles, but in the real world, probably not. Std issue ball ammo (be it 55 or 62 gn) 5.56mm or even .308 ball ammo will not be supersonic at 1000 yds. Even 80 gn match bullets (maximal ballistic coeffecient) for 5.56 mm will not reach 1000yd with standard pressure loads out of a 20" AR15. When bullets drop below mach 1, the turbulence destroys any reasonable chance for accuracy. Of course, this is data gleamed from actual long range target shooting up to 1000 yds where we want to hit what we aim at, ie direct fire. If you just want to point it in the sky like a howitzer for indirect fire, that's a seperate issue.
Link Posted: 7/1/2002 4:17:36 AM EDT
[Last Edit: 7/1/2002 4:30:37 AM EDT by maxdram]
I DID mention at the end of my 2-cents worth that my equations do ignore wind resistance. FYI, a LONG RANGE projectile (not the puny bullets we fire from our rifles) shot at an angle of 47 degrees will have a greater horizontal range than if fired at 45 degrees, for that very same reason. Its pathway takes it through more less dense air tnan the 45 angle, whose pathway is longer through the more dense air. Comprenda?
Link Posted: 7/1/2002 10:44:34 PM EDT
Originally Posted By MAHABALI: One of our biggest arguements is: If you shoot a bullet perfectly level, straight, and at the exact same time you drop a bullet from the same height, (no matter if it is 3' or 3", or whatever) that they will land at the same time. At least that is what she says, how can this be right??? we have argued about this countless times.
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Because a bullet has no "lift" as it travels. Gravity pulls down on it right out of the barrel just as one drops it from their hand. I've never tested this but the brainiac folks like your GF swear it's true.
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