Proof of bullet drop in the game

[Thunderbolt]

Re: Proof of bullet drop in the game

I was under the impression that the horizontal and vertical components of the vectors could be treated as seperate, individual particles. ie. firing something from a height of 0.5m at a speed of 900m/s paralell to the ground will take the same amount of time to hit the ground as it would if it was fired at 1m/s.

Trying to shoot a bot 5 meters away with a bullet travelling at 1m/s will not take 5 seconds to hit him.

Say we are shooting from a height of 0.5m

s = ut + 0.5at²
*0.5 = 0t + 0.5 x 9.8t²
0.5 = 4.9t²
0.10 = t²
root0.10 = t
t = 0.31 seconds.

The bullet would hit the ground in 0.31 seconds.

Sorry, I just had to correct what I believe was a mistake in your assumptions.

I'm probably wrong, I mean I've only done senior high school physics, and you have a PhD.

Dear Jesus, my brain hurts. Can you explain how did that?

[DLWebmaestro96]

Re: Proof of bullet drop in the game

wasnt this topic like already proven some time ago?... how many debates are there going to be about it?

27 times

[evilzucchini]

Re: Proof of bullet drop in the game

The reason that this is being debated again is because new evidence arose. . . so it deserved a new trial.

I rule that Bullet Drop exists!

*slams gavel*

ALL BOW BEFORE THE HONORABLE JUDGE ZUCCHINI!!!!

BOW DAMN IT!

[Spodeboy]

Re: Proof of bullet drop in the game

Dear Jesus, my brain hurts. Can you explain how did that?

That's just the formula to calculate the amount of time it takes for an object to drop to the ground with earth's gravity of 9.8 meters per second squared.

[F-1]

Re: Proof of bullet drop in the game

I was under the impression that the horizontal and vertical components of the vectors could be treated as seperate, individual particles. ie. firing something from a height of 0.5m at a speed of 900m/s paralell to the ground will take the same amount of time to hit the ground as it would if it was fired at 1m/s.

Trying to shoot a bot 5 meters away with a bullet travelling at 1m/s will not take 5 seconds to hit him.

Say we are shooting from a height of 0.5m

s = ut + 0.5at²
*0.5 = 0t + 0.5 x 9.8t²
0.5 = 4.9t²
0.10 = t²
root0.10 = t
t = 0.31 seconds.

The bullet would hit the ground in 0.31 seconds.

Sorry, I just had to correct what I believe was a mistake in your assumptions.

I'm probably wrong, I mean I've only done senior high school physics, and you have a PhD.

But he is not trying to hit the ground! he is hitting a wall in front of him! so the horizinal distance is fixed, but still the vertical drop distance changes with the initial speed because it takes more time to hit the target in front of you, which means you have more time to travel verticaly. Right?

What you calculated is the time for the bullet to travel in the air to have a drop of 0.5m.

What you should do is the following:
-Calculate the travel time: x=v*t sp t=x/v where x is the distance to the target.
-calculate the drop distance y=v0*t+0.5*a*t^2
-v0=0 verticaly so
-so now y=0.5*9.8*t^2 where a =9.8 is the gravity constant
(or y=0.5*9.8*x^2/v^2, homework! verify this!)

now assuming v=1000m/s like the sniper's rifle in the game then
-if the target is 200m away then t=0.2sec and y= 0.196m (not that significant)
-if the target is 500m away then t=0.5sec and y= 1.225m (very significant)

[F-1]

Re: Proof of bullet drop in the game

and BTW, when I say "hit a wall in front of him" I mean hitting a target at a fixed horizontal distance.

What does not matter here guys is the mass of the bullet because the game assumes a perfect world with no friction or air-drag.

[F-1]

Re: Proof of bullet drop in the game

More calcs!

t is the time to hit the target
y is the bullet drop

-if the target is 100m away then t=0.10sec and y= 0.05m
-if the target is 150m away then t=0.15sec and y= 0.11m
-if the target is 200m away then t=0.20sec and y= 0.20m
-if the target is 250m away then t=0.25sec and y= 0.31m
-if the target is 300m away then t=0.30sec and y= 0.44m
-if the target is 350m away then t=0.35sec and y= 0.60m
-if the target is 400m away then t=0.40sec and y= 0.78m
-if the target is 500m away then t=0.50sec and y= 1.23m
-if the target is 550m away then t=0.55sec and y= 1.48m
-if the target is 600m away then t=0.60sec and y= 1.76m

and one more thing! I hate snipers! so if you are sniping one day and saw me Black_Duck_1 then please please do not snipe me! let me know that you saved me and I wont hit you with my AntiTank rocket! I promise!

[Thunderbolt]

Re: Proof of bullet drop in the game

-Calculate the travel time: x=v*t sp t=x/v where x is the distance to the target.
-calculate the drop distance y=v0*t+0.5*a*t^2
-v0=0 verticaly so
-so now y=0.5*9.8*t^2 where a =9.8 is the gravity constant
(or y=0.5*9.8*x^2/v^2, homework! verify this!)

now assuming v=1000m/s like the sniper's rifle in the game then
-if the target is 200m away then t=0.2sec and y= 0.196m (not that significant)
-if the target is 500m away then t=0.5sec and y= 1.225m (very significant)

I want to understand you, but I just can't... Please teach me, oh great PhD holder!

[F-1]

Re: Proof of bullet drop in the game

I want to understand you, but I just can't... Please teach me, oh great PhD holder!

Are you being sarcastic or what?

We need a chalk and a board! stop by my office one day and I will teach you.

[SupUnd]

Re: Proof of bullet drop in the game

Think of it this way. According to Newton, everything falls at the same constantly growing speed in a perfect vacuum. Now, the BF2 world is probably a prefect vacuum to simplify things. Assuming they are using these equations to simulate reality, everything will fall at the same ever increasing speed regardless of its weight or vertical velocity. Also, the rate of increase in speed, aka the acceleration, will also be constant.

So, if you jump off a building you'll start to fall slowly and then pick up speed until you hit the ground (or go splat). You throw out a grenade as you fall and it will stay in front of you as you fall. It will slowly move farther away horrizontally but it will fall at the same growing speed. It will grow in speed as quicly as you grow in speed verticaly so it will always be in the same position in front of you. Same thing if you shoot a bullet. It may be traveling vertically near or above supersonic speeds but it will fall the exact same speed as you and will hit the ground at the same time as you (assuming the ground is flat for miles around and its path to the ground isn't obstructed).

Again, this is assuming they are using these equations in the game. They very well may not be. The best way to find out is to go into a plane, go WAY WAY up into the sky, jump out of the plane (without using your parachute), and then lob a grenade. Try to throw the grenade straight ahead or ever so slightly up, never down. If you throw it straight ahead it will stay at the same height as where you threw it relative to your falling body. I don't throw grenades often so I'm not sure how easy this experiment is.

[Thunderbolt]

Re: Proof of bullet drop in the game

It's all this math... Ugh

[Lizzle*47]

Re: Proof of bullet drop in the game

I was just calculating the time it would take for a bullet to fall from a height of 0.5m, with a horizontal speed of 0m/s. It takes 0.31 seconds. Because you said trying to shoot a person 5m away would take 5 seconds if the bullet had a horizontal speed of 1m/s, this was not true, so I was simply showing you how it could not be so. So whatever distance the bullet has to travel (once again assuming that in the world of bf2 Newton's laws still apply, which I don't think they explicitly do), it has to do it in less than 0.31 seconds. There is a formula for this:

d = v/t

(For all the non physics chaps: d is distance, v is velocity, and t is time)
= 1000/0.31

3.225km is the maximum distance it could travel, if shot paralell to the earth from a height of 0.5m.

[Lizzle*47]

Re: Proof of bullet drop in the game

In his defense, he did say "if you ever hit him."

However, if you ignore the fact that it won't reach the bot because it would hit the ground first or you take away the vertical acceleration, it would take 5 seconds to reach the bot on the horizontal vector (1 m/s * 5 s = 5m).

The question is, if the bullet were travelling 1 m/s, how high above the head of the bot would the gun need to be to get a head shot from 5 ms (horizonatlly) away?

Extra credit: At what angle do you need to fire the gun from level with the bot's head to get a head shot when you are 5m away?

Well, this is taking me back a couple of years to the very first stuff I did in high school physics...

The bullets total travel time will last no longer than 5 seconds. It will have to take 2.5 seconds to rise to its greatest height, and another 2.5 to fall, in order to make up the 5 seconds and not fall short or overshoot.

At maximum height the vertical velocity = 0m/s.

v² = u² + 2as

where v is the final vertical velocity
u is the initial vertical velocity
a is the acceleration (-9.8m/s²)
s is the vertical displacement, which is what we are trying to find.

s = v² - u²/2a
= 0² - 1²/2 x -9.8
= 0.05m

The bullet could only reach a height 0.05m above its starting point, so 0.55m, if fired at 1m/s directly upwards. Haha, I'm beginning to think this poor bullet will never reach it's intended target.

It's like shooting a very low powered hose at an object very far away. No matter what angle you use it will never reach the object because it just doesn't have enough power.

[imported_Fig]

Re: Proof of bullet drop in the game

SupUnd, your forgetting about terminal velocity.

Things will accelerate at the same speed (two balls galileo dropped of leaning tower of pisa), but given enough time, the lighter object will eventually stop accelerating its falling speed, and level off, while the heaver object will keep accelerating longer untill it reaches its terminal velocity.

[UnrealAlex]

Re: Proof of bullet drop in the game

No offense but this thread is the definition of dorkiness lol.
Dork myself, don't hate lol.

COuldn't understand anything though really o.o

I think the guy proved it.