Von Mudra;4804907@ Fuchs: Find me a place to get some 7x57mm, and then we're in business :P There's no surplus off the stuff, so I end up having to buy from remington or such, which is horribly expensive.... And my M24 is, other then my K31, my best shooter. Its so beautiful>__<

Alright, get a shovel and a plane ticket. They once blew up an old ammo dump here, it's all in the ground. 20% of the rounds is still intact.

To clarify, in game, we have two things you must lead for. Ping delay and projectile speed. BF2's projectile physics engine I don't know to much about, but you must lead a bit. Then comes the lead you must do with ping aswell, which is much larger usually. So both arguements might be valid, if you're playing with a high ping, you might have to lead more then realistically. If you play on LAN or against bots, the lead isn't much. So for the sake of arguement, go play some LAN on any 64 player map and try sniping bots. For minimizing the ping delay, try setting up the toolbox with the ping correction utility.

Kev4000;4805986To clarify, in game, we have two things you must lead for. Ping delay and projectile speed. BF2's projectile physics engine I don't know to much about, but you must lead a bit. [/quote] I have some info on this. These are Scouty's calculations on how bullet drop works in BF2, which are implemented in the WoOkie sniper mod. If FH2 could use these for more realistic ballistics it would be amazing :naughty: Warning: Nerdy/56k overload.EDIT: I haveSpoiler:Show[quote=Scouty]I have finally obtained a definitive result, no guesses, everything was tested and/or derived. Much WOoKie man love goes to Jonny, a PR Contributor who is doing the rifle zeroing for their mod. He supplied me with much advice during my quest. Enjoy the long read! ---------- Starting with real life. We have two different formulas for drag, applicable for different cases.1) LINEAR DRAG, LOW SPEEDSStokes' drag, also known as linear drag. The drag force is: Fd = – Kl · V (where Kl is a constant, and V is the velocity of the moving object) (also notice that it opposes the velocity) And with Newton, acceleration in the y-direction is: M · A = – M · G – Kl · V (where A is the acceleration, M is the mass of the object, and G is the gravitational constant)2) QUADRATIC DRAG, HIGH SPEEDSRayleigh's drag, also known as quadratic drag. The drag force is: Fd = – Kq · V² (where Kq is a constant, and V is again the velocity of the moving object) (also notice that it too opposes the velocity) And with Newton, acceleration in the y-direction is: M · A = – M · G – Kq · V² For reference:wiki. ---------- Then going to BF2. Assuming Dice has coded a reasonably realistic model, my main task was to figure out which of the two formulas is applicable. So I did some free fall experiments, and made an interesting discovery. As it turns out there's a relation between the terminal velocity, Object.Mass, Object.Drag and Physics.Gravity. Vt = – C · M · G / D (where Vt is the terminal velocity, C is a constant, M is Object.Mass, G is Physics.Gravity, and D is Object.Drag) (incidentally, I measured C to be about 31.83) ---------- Now, when the terminal velocity is reached, the acceleration is zero, so that gives:1) LINEAR DRAG, TERMINAL VELOCITY– M · G – Kl · Vt = 0, and thus: Vt = – M · G / Kl But the experimental relation also holds: Vt = – C · M · G / D So that gives a relation between Kl and C: Kl = D / C (C is a true constant, Kl changes with variable D)2) QUADRATIC DRAG, TERMINAL VELOCITY– M · G – Kq · Vt² = 0. and thus: Vt² = – M · G / Kq But the experimental relation also holds: Vt = – C · M · G / D So that gives a relation between Kq and C: Kq = – D² / ( C² · M · G ) (C is a true constant, Kq changes with variable D, M and G) ---------- Now we have expressed Kl and Kq in the measured constant C, we can solve the two differential equations: (A = d²y/dt² and V = dy/dt):1) LINEAR DRAG, DIFFERENTIAL EQUATIONM · d²y/dt² = – M · G – Kl · dy/dt So y as a function of time is:2) QUADRATIC DRAG, DIFFERENTIAL EQUATIONM · d²y/dt² = – M · G – Kq · (dy/dt)² So y as a function of time is: ---------- So that leaves us with two nice and complicated formulas for bullet drop. But the question remains: which one is correct? To determine that we must compare them to the experimental data. I've plotted both in the following figure: In these graphs, both linear drag and quadratic drag seem to give accurate results (quadratic looks to be a bit better even!). It's hard to see since the experimental data is a bit erratic (I was recording at 30 fps, and this is a graph of the first second only). So I tried plotting over a greater interval: Now you can start to see the difference between the two kinds of drag. The quadratic drag starts to diverge from the experimental data. But it becomes painfully clear in the long run: Quadratic drag gives a lot more drop over time, and that simply isn't the case in BF2. And the error of the linear drag solution is generally well under 5%, so it must be correct. Note that the error does blow up near the start, but that has to do with the form of the formula. The measured constant C weighs very heavy at the start, so any error in measuring that will give an exponential error during the first second or so. ----------Conclusion: BF2 uses linear drag!Result: We now know the formula to predict bullet drop under any circumstance!Notice that this is a figure of height over time, not height over distance, it was only a free fall experiment (zero forward velocity). But we did show that BF2 uses linear drag, and that applies to velocity in any direction. So it's an easy matter to derive a formula for bullets fired at a certain angle and muzzle velocity.anotherhawt Jagdpanther in my avatar in case you didn't notice.

1. Mudra you might be better off taking up reloading lol...

2. Sniping isn't magic. If you have the luxury of setting your elevation and windage before you take the shot, you can aim right on target. If not you still lead and raise the "Shot". Section V (3.19) will show you what I mean... FM 23-10 Chptr 3 Marksmanship

3 All bullets follow projectile motion. This means that its vertical position depends on Time and not Weight of bullet. The longer it takes for a bullet to get from point A to point B the more time gravity (32feet/s^2)(9.81m/s^2), has to affect its final position. The faster the initial velocity of a projectile the flatter the projectile curve. delta x= Vot+1/2 at^2 delta x=change in position Vo=initial velocity a=acceleration t=time in seconds.

Now say you have a round traveling 800m/s and you fire it completely horizontally at something 800m away. t=1s Vo=0 (Because we are considering vertical velocity) a=9.81m/s^2 downwards = gravitational acceleration therefore the bullet drop would equal to 4.91m. That is the best you will get for a 800m/s horizontal shot at a target 800m away. Inorder to compensate you have to angle your barrel up a certain number of degrees so that you have a vertical velocity component which would equal to Vo X sin(angle). Unfortunately as you increase your angle your velocity in the horizontal plane also decreases following Vo X cos(angle). With a 10 degree angle you would end up with a horizontal velocity of about 788m/s instead of 800. You would have to now take into account the extra time.

Problem is that in RL you have other factors that affect your bullet too. Mainly drag, which itself involves the coefficient of drag of the medium and the surface area of the projectile that goes against the "resistance".

All in all it causes your projectile to slow down which increases the time for it to get where its going even more which causes even more Drop due to gravity...

So To conclude this overly written post.

Scopes/sights arn't magic, If you have ever actually fired a real rifle/pistol at a good distance you would know this. If your Idea of shooting with a scope is letting someone zero it for you and then let you shoot bulls-eyes all day at 100 yards then you shouldn't be talking about knowing about shooting.

Reloading ammo is always the best way to save money if you do alot of shooting.

And finally, although you thought you would never use math and physics when you were in highschool, It does come in handy, even if only to help explain video game principles... Imagine how many famous mathematicians/scientists must be rolling in their graves hearing us talking about Video games in relation to their subject of interest.

Yeah, I know. I've thought a lot about taking up reloading, but I don't have the time or the space honestly, plus its still an odd caliber round.

Persornaly i like the balistics as they are. Here is a comparison to realworld balistics in my own mind. When you take in to account the fact that when you line up you sight with your eye it might be off, ever so slightly. Action all around you, bullets buzzing past you, adrenaline pumping through your veins, breathing heavily. Nervous that every step you take might be your last, making you jumpy. The wind blows changing the bullets trajectory in mid flight. You sprint, your tired and unable to hold your rifle as true as the last time. As you scan intensly for your next target you accidentaly hit the elevation knob. The wind blows again, this time bringing with it a grain of sand that lands in your barrel. Word comes that the enemy is flanking, you fall back. Cautiously you work your way to the fall back point when suddenly a strange man in a german uniform jumps out in front of you, scared shitless you jump at the sight of him and franticly fire your rifle, a miss. Your hand races to the bolt ripping it back an slamming it foreward. A sharp pain in your abdomen, another in your chest, you buckle from the pain. A bright light, large pearly gates, a deep voice, "welcome". well thats a good story actually. These are all things that could happen in a real life scenario that would affect you aim, so when you miss a shot in fh, "pretend" its one of those factors kicking in. once you do that the balistics are pretty realistic.

Von Mudra;4806233Yeah, I know. I've thought a lot about taking up reloading, but I don't have the time or the space honestly, plus its still an odd caliber round.

British name for it is the .275 Rigby right?

Anyways cabelas has it at 20$ for 20.

And cheaper than dirt a bit better Cheaper Than Dirt - America's Ultimate Shooting Sports Discounter

Yep, stiill very expensive for me, considering I get 8mm and 7.62 at a rate of 10 bucks for 100 rounds....

Also, to respond to Pointblank, you are very correct. Its like when I did my write up on how hard it is to reload a clip in real life...there's a lot of factors that aren't in a game.