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You are watching: How far will a bullet travel at 45 degrees


I"m wondering how far a 230gr cartridge fired indigenous 5" barrel will travel. For instance (If I"ve done my physics right): If i fire a bullet at a height of 6 feet over level ground in ~ 930 fps, ns calculate the bullet will travel around 1 second before hitting the ground, or approximately 930 direct feet. However, walk the .45 ACP cartridge have enough "pop" to actually shoot the bullet the far, or will it operation out of heavy steam long before?Just curious.Thanks
Assume I"m shooting parallel come the ground, no wind, and, come quote my physics instructors in college, "ignore the effects of waiting resistance".I"m just looking for a ballpark figure.
As long as we"re splitting hairs, go ahead and factor in relative humidity, temp and elevation also. Oh, and also there"s yes, really no such thing as parallel to the earth.take a 230 gr bullet and also drop the from 6"...time it. Gravity is a constant, so it will fall at the exact same rate as soon as fired "level" to the ground. Multiply her velocity by the moment you obtained for her test drop, and that should tell girlfriend how numerous feet it will travel. THis is, of course, grossly over-simplified. I"m no physics prof, and it"s to be 7 yrs since I had actually it in college. I didn"t stay at a holiday inn refer lastnight either. As a issue of fact, i didn"t sleep at all...been clearing roads from this damned hurricane. Many thanks Katrina!!
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I"m sure the military has operation tests that this nature before (they"ll carry out just about anything once they"re bored). You may want to submit this question to the NRA"s American Rifleman newspaper if you"re really that interested to know. I"ll bet they have the answer filed far somewhere.
Avoid the temptation to replace whatever on your brand-new 1911 just to do it "better". Understand what you"re an altering out and also why. You might spend a the majority of money addressing things the weren"t damaged to begin with. Shoot at least 500 rounds with it first, then decide what friend don"t like and also want to improve.
For valuable purposes, of course, the pistol deserve to be organized parallel to the ground. I don"t know the exactly answer, yet I do know that multiply the f.p.s. In ~ the muzzle by estimated time of travel won"t get you there. Also know the the number posted as that released by the military is incorrect for these purposes. There room plenty of ballistics program around. Simply plugging the info in would acquire you yes, really close.
take a 230 gr bullet and drop it from 6"...time it. Gravity is a constant, so that will loss at the very same rate once fired "level" come the ground. Multiply her velocity by the moment you got for her test drop, and that must tell friend how countless feet it will travel. THis is, the course, grossly over-simplified.
D = 1/2at^2 for zero early stage velocitysolve because that t:t = sqrt( 2D/a) for D = 6ft and a = 32fps/st = .612sec if wind resistance is negated, and gun is horizontaldt = .612 x 960 = 587ft down range
vapors750
You just need to be a physics teacher. however Forstr has actually it right and your formulae too. Good old Newton strikes again v his apple.
Bob
http://www.brucemontague.ca/Be thankful we"re not obtaining all the government we"re paying for.They that would provided up crucial liberty for short-term security, deserve no liberty or security.(Benjamin Franklin)
I was reasoning along the lines the original write-up was questioning how much can one obtain a 230 grain bullet to go. What is the the furthest you can obtain a bullet to go the end of a 5" pistol.I read someplace the ideal angle to obtain the farthest distance is 45 levels up.So if you pointer up the barrel X degrees (whatever is optimum for this purpose) what"s the farthest you deserve to expect to acquire a cartridge to travel? perhaps that"s the 1,955 yards?I always read ~ above the ago of a crate of .22 bullets that they can travel over 1 mile (1,780 yards.) If a .22 have the right to do the it"s no unreasonable the a .45 can do it. This is a an ext interesting inquiry with more practical applications (at the very least for the Army.) I have some really old rifles that have actually sights adjustable out to 2000 yards and also beyond.
You just have to be a physics teacher.
....actually computer system engineer...but ~ above the course to gaining my degree....I had actually to take it 3 soldier of physics.To deal with the "how far" at 45 degrees...you must solve vector equations (taking right into account the horizontal and also vertical velocites)....which ns don"t precisely remember right now! I"d simply recommend making use of a ballistics table and also take the easy way out.
....actually computer system engineer...but top top the route to gaining my degree....I had to take it 3 quarters of physics.To resolve the "how far" at 45 degrees...you must solve vector equations (taking into account the horizontal and also vertical velocites)....which i don"t exactly remember appropriate now! I"d just recommend making use of a ballistics table and take the easy way out.
well, if the is a 45 degree angle, the horizontal and also vertical velocities will certainly be the same. 45 levels will constantly yield the the furthest ballistic reach other than when considering missiles which are under power throughout the whole or most of the flight.
D = 1/2at^2 for zero initial velocitysolve because that t:t = sqrt( 2D/a) because that D = 6ft and a = 32fps/st = .612sec if wind resistance is negated, and gun is horizontaldt = .612 x 960 = 587ft under range
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Gen. William Thornson, U.S. Army​
Hatcher claims 1640 yards or .93 miles. The is not at 45 degrees, due to the fact that maximum range is completed at less than the in an atmosphere.He also gives the maximum variety of a .22 LR (40 serial bullet, muzzle velocity 1145 fps) as 1500 yards or 4500 feet. This is .85 miles, so the "Danger to one mile" warning is not lot of an exaggeration, considering that terrain may go into into the picture.Jim
if ns figgured that right
if i figgured it appropriate in my little balistic program I downladed previously after reading this post: v the barrel at 45 levels it have to be about 1600 yards = 0.909 miles = 4800 feet. The barrel in ~ 0 levels is approximately 1300 yards = 0.7386 mile = 3900 feet. Granted mine program only goes out to 1159 yards for this reason those numbers are aproximate, don"t organize me or the forum come them.all of this to be figgured utilizing the numbers, ie: balistic coeficiant (0.160), velocity (830 fps) and such the end of mine speer manual, because that a 230 FMJ .452 diameter. That and also the regimen was composed in the mid to late "80s. Alex
It"s a good thing as soon as your hobbies support each other, right? ie... Reload
to Shoot, Shoot come Reload.

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As every LymanReloading Handbook, 46th ed.,1984(yeah, I recognize it"s old, but a cartridge of a offered weight/shape/velocity/angle won"t go any type of farther today than it would certainly 20 yrs ago).45ACP/230FMJ(Ball)/30 deg angle:740fps=1533yds850fps=1700yds945fps=1833ydsTo quote among the other posters - "Your mileage may vary"
So if Newton is ideal the bullet will certainly hit the soil in around 587 " if the gun is level v the soil and around 1600 ft if canted 30 strange degrees. Nice come know......, now if I could hold the pistol level and also with no movement for 2 secs or less I might have the ability to hit the bullseye in ~ 25 yards. I won"t ask how much the bullet rises or falls in 25 yds.
Bob
http://www.brucemontague.ca/Be thankful we"re not getting all the federal government we"re paying for.They that would offered up essential liberty for short-lived security, deserve neither liberty or security.(Benjamin Franklin)
I won"t questioning how much the cartridge rises or drops in 25 yds.
Heck, that simply depends on if its looking for someone! :dope:Sorry....Anyway I thought the 1.5 miles to be pretty lot the max of any "standard" gun till I looked in ~ the earlier of a .17HMR and also it said 2 miles. (T or F?)Now, because that the physics knowledgeable in here...If girlfriend take away air resistance doesn’t the resemble the properties of a vacuum? Such that the bullet weight no much longer matters?So utilizing what I have learned in this thread, uneven my logic is flawed i beg your pardon it frequently is...Take any bullet of any type of weight and also without calculating for air resistance it will fall to the ground native 32" in 1 second, however due to the acceleration you can not say 8" in 1/4 second...So what is the duration the takes because that the "any" cartridge to hit the ground indigenous a conventional firearm at a level elevation of 54" (4.5", the approximate height of a firearm in a traditional isosceles view of one average elevation man)?