- Thread starter BrokenBiker
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BrokenBiker
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Have you allowed for the curvature of the earth? What about air pressure? How tall is the shooter?
i dont care...lol
BrokenBiker
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i only ever use chairgun or strelok but they are tailored to my rounds and chair gun is only good for air power or rimfires really
HMB
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It's industrial software so is the bees knees. But it's been spot on to 2500yds
BrokenBiker
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i only really used it for the air rifle to work out drop in inches...set a target up with inch squares and then you know how many squares to hold over at certain ranges, so you can draw a little table suited directly to your scope markings etc...i went off it when i found strelok but only really use that for pushing the ranges a bit when i know i have time to put info in and so on...it allowed me to take a 310 yard shot that i wouldnt have otherwise taken and i hit bang on the mark after adjusting with the strelok info.
norlander
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I was digging in me memory banks as at one point I was lumped with Medium Machine Gun kit. Found this lesson plan for MG instructors. If you speed thru it, you see trajectory shown-coz bullets don't fly flat, and 7.62 MGs will drop rounds out to 2000metres.
Anyway, you might pull a nugget out to argue the case from the lesson plan, but you really need the elevation bit of the elevation and deflection tables for MGs on fixed lines to get the strike-you'd get an average though coz MGs produce a beaten zone.
http://m.semo.edu/pdf/showmegold_Machine_Gun_Theory.pdf
Anyway, you might pull a nugget out to argue the case from the lesson plan, but you really need the elevation bit of the elevation and deflection tables for MGs on fixed lines to get the strike-you'd get an average though coz MGs produce a beaten zone.
http://m.semo.edu/pdf/showmegold_Machine_Gun_Theory.pdf
Rudi McAnichal
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You didn’t do anything wrong. The bullet would indeed travel 63 miles, if only you could shoot in a vacuum.
Muzzle velocity 3285 ft/s, 45 degree elevation, so vertical and horizontal velocities are both 2323 ft/s.
Gravity decelerates the bullet at 32.2 ft/s2, therefore it travels upwards for 72s (=2323/32.2), and then comes back down again which takes another 72s.
Total flight time 144s. Horizontal distance travelled = 144s * 2323 ft/s = 334512 ft = 63.35 miles.
The one big factor omitted from those calculations is air resistance, which we cannot easily avoid in practice. It would be difficult to build a vacuum chamber large enough to test even a toy catapult, and I seem to remember school teachers telling us that naturists abhor a vacuum cleaner.
If you do ever get the chance to try shooting in a vacuum, remember the bullet will come back to earth with the same velocity that it left the muzzle. Ouch.
On the moon, where gravity is only about one sixth of what we have on earth and there is virtually no atmosphere, calculation suggests your bullet would travel about 380 miles. Do astronauts carry any firearms, I wonder?
Almost 100 years ago, during the WW1 bombardment of Paris, Germany was using the “75 mile gun” which a muzzle velocity of about 5400 ft/s. That is something to ponder on if anybody is thinning out the branchers in May with an air rifle.
ronan
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i seem to remember that the germans were missing their target, consistently striking an area they were not aiming at. The reason was the rotation of the earth while the shell was flying through the air. The coriolis effect as far as i remember.
Rudi McAnichal
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You look quite young in that photo, but I reckon it must have been taken some while ago if you remember WW1.
Maker
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American ones do
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malcolm
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I did a physics project on this at school. The aim was to figure out the drag coefficient of an air gun pellet. This was before the invention of health and safety and I got to stop on after hours to fire an air gun down a corridor into a ballistic pendulum with strict instructions not to hit the cleaners who kept popping out into the corridor without warning. I think the drag coefficient of an air gun pellet turned out to be about 0.3 but don't remember any helpful equations.
Rudi McAnichal
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School was never as much fun as that, but I did have to work with the equations many years ago when designing grain cleaning equipment (winnowing to separate seeds from chaff).
It is fairly straightforward to estimate trajectory, if you assume drag coefficient to be constant.
Step 1 - Calculate drag force on the bullet as it leaves the muzzle:
Drag force = 0.5 * (air density) * (velocity)^2 * (drag coeff) * (x-sect area)
Step 2 - Calculate the deceleration due to air resistance:
Deceleration = (force) / (bullet mass)
Step 3 - Choose a short time interval, work out what the speed and position of the bullet will be at the end of that time interval, and keep repeating the exercise over and over again. All the sums have to be done for both vertical and horizontal directions. Must have been a tedious job in the pre-computer age, but nowadays it is very simple to replicate the equations on a spreadsheet.
Similar procedures can be used when designing settling chambers to remove particles from air, water, oil, etc.
I seem to remember that the drag coefficient is not really a constant (it varies somewhat according to the velocity) and there are probably other factors taken into account in the ammunition manufacturers’ software. More than my brain can cope with nowadays.
Maker
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Pfft, is that it? We get to drop steel balls through lightgates and hang weights off brass wire!
If you assume a drag coefficient of 0.1 you get a deceleration of 170ms^-2, it's got to be less than that surely?*
Rather than repeat a simple calculation over and over, can't you just use differentiation/integration?
*EDIT: According to wiki, the drag equation you mention only works for blunt objects with turbulence behind it.
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