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Bullet drop formula?

Frisco

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#1
I've been searching for like an hour online and I can't find a formula to calculate bullet drop.

I don't need tables, I'm doing it for an assignment for my trig class so I have to do the math myself.
Any ideas? I think I've calculated all the actual data I need, I just don't know exactly how to put it together.. Thanks.
 

x SF med

the Troll
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#4
It's a slope formula with the added distractors of velocity, gravity, time - it could be even more complicated with spin and wind over distance. It's a 4 dimensional slope equation. Remember before there is drop there is rise (ballistic jump) so you have a parabolic/ballistic slope.
 

Marauder06

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#5
It's a slope formula with the added distractors of velocity, gravity, time - it could be even more complicated with spin and wind over distance. It's a 4 dimensional slope equation. Remember before there is drop there is rise (ballistic jump) so you have a parabolic/ballistic slope.
:eek:
 

Frisco

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#6
I figured out the drop with a perfectly flat trajectory, the ballistic jump is where I'm having trouble I don't know how to factor it in. I've got the effects of wind, spin drift, and the coriolis effect calculated, so I will factor those in once I find the drop...
 

Frisco

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#8
If you calculated it for a prefectly flat trajectory out of the muzzle it would work if your barrell was exactly level with your scope.. IE your reticle is perfectly parallel to your barrell to begin with.. The parabolic slope only comes into effect if your already sighted in at a given distance.. right?
 

x SF med

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#9
You need to calculate the 'vacuum parabola' from muzzle to the crossing point of the 'flat trajectory' if you figure poa/poi is essentilly flat at 100m, you can interpolate the rise by using 150m drop as being the height of rise at 50m (high point for ease of calculation (friction/velocity change/spin change/gravity actually pull the high point back toward the muzzle - but using center of flat is close enough) . This changes slightly with barrel angle for drop at greater ranges. Remember as the projectile loses energy/velocity/spin, the effect of gravity increases.

Think a long tailed student's t distribution or a long tailed z distribution as the basic shape, the steep slope of the curve being the terminal side of the ballistic projection.
 

Mac_NZ

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#10
You need to calculate the 'vacuum parabola' from muzzle to the crossing point of the 'flat trajectory' if you figure poa/poi is essentilly flat at 100m, you can interpolate the rise by using 150m drop as being the height of rise at 50m (high point for ease of calculation (friction/velocity change/spin change/gravity actually pull the high point back toward the muzzle - but using center of flat is close enough) . This changes slightly with barrel angle for drop at greater ranges. Remember as the projectile loses energy/velocity/spin, the effect of gravity increases.

Think a long tailed student's t distribution or a long tailed z distribution as the basic shape, the steep slope of the curve being the terminal side of the ballistic projection.
Who are you and what have you done with the Troll?
 

x SF med

the Troll
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#12
sometimes you actually have to do the math because you don't have the charts or a fancy computer.

I didn't get into charge versus projectile weight calculations and how that changes the actual initial parabola from muzzle, did I? You guys always forget I was an 11B/18B before I got better and became a medic. I still like making boomsticks do their magic.
 

Crusader74

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#13
You need to calculate the 'vacuum parabola' from muzzle to the crossing point of the 'flat trajectory' if you figure poa/poi is essentilly flat at 100m, you can interpolate the rise by using 150m drop as being the height of rise at 50m (high point for ease of calculation (friction/velocity change/spin change/gravity actually pull the high point back toward the muzzle - but using center of flat is close enough) . This changes slightly with barrel angle for drop at greater ranges. Remember as the projectile loses energy/velocity/spin, the effect of gravity increases.

Think a long tailed student's t distribution or a long tailed z distribution as the basic shape, the steep slope of the curve being the terminal side of the ballistic projection.

Does that change during extreme weather? Ie Rain or Hot Weather? I remember a rd rises in rain and drops in hot weather..