COH: Europe In Ruins

Something I've been looking to figure out, so far no luck.

If a weapon does 10 damage per second and another weapon does 20 damage per second then the second weapon is twice as good.

If the first weapon, that deals 10 damage per second, also has splash damage, lets say it has a radius of 5 splash equivalent to COH units, than what is the relative value of the first weapon to the second?

If you can come up with a method to solve this problem I will reward you in some way.

The answer is still the same, the dps would only increase if you specified the amount of targets that the splash was actually hitting.

Now assuming that the splash is the same as the full damage then one could presume that the dps would equal (10*x) dps where x is the number of units that the splash touches.

Like he said, as far as my math goes...

(x = targets splashed)

Value of weapon A: 10*x
Value of weapon B: 20

You can't really put a direct mathematical value to A as it's situational depending on targets.

Are we talking about the Ostwind or Scoutcar upgrade in CoH2? :D

Right, I'm no math guru but the way I'd think about it is, if both weapons are of the same stats aside from splash then I'd do this. Again, I ain't no math guru but what you seem to be looking isn't something set in stone either.

Weapon A, 20D = 100% Efficiency
Weapon b, 10D*s = xx% Efficiency

If s = 1 then b is 50%
If s = 2 then b is 100%
If s = 3 then b is 150%
If s = 4 then b is 200%
If s = 5 then b is 250%
If s = 6 then b is 300%



So, if it is a Russian weapon it will be fighting primarily 4 man squads. You'll want to take the chance of it splashing the entire squad into account but let's assume it catches 2-3/4 on most shots. This is how you'll have to judge what weapon is stronger. If against 6 man squads that the russians roll with, the splash weapon will be stronger than when going against the smaller german teams due to the greater number of targets.

I'd base the likelihood of how many members splashed by formation distance, out of curiosity how far is the spacing between each man in a standard formation?

Another thing to keep in mind is damage focusing, if splashing 2 men then it will kill them at the same rate the 20 damage weapon does. However 1 man will go down in half the time and this cuts the opposing squads (if 4 men) firepower sooner, this would also have a value in the number you're trying to get to.


Still, at the end of the day all you can do is assign your own "Value" to splash based on these parameters, as far as I know you can't really get a hard mathematical value for this given it depends entirely on how it behaves within the game system. All I can do anyway is make a best guess with the info above. Then again, I ain't no maths wizz.

How i would do it to be more precise would be.

Dps = (b+(x*(b/d)

b would be base damage.

x would be number of units caught in blast

d would be distance from target

So lets say we add a stuh into the mix and its firing upon 5 rifle man. Say the stuhs base damage is 25. and the rifle men are standing about 5 meters away. We will set x to be 4 because 1 rifle is hit head on

(25+ (4*(25/4) = 50

However that would only apply to an optimal scenario were you are splashing an entire squad dead on and counting on this variable to be a constant will lead to an unrealistic performance value. One would have to take other factors into account as well when comparing it to a weapon with no splash but twice the damage.

Well Spartan and Niko are right, but the problem is unfortunately not so simple IMO. You can't really determine a weapon's value after its DPS, even if you consider all other circumstances ( accuracy, etc) the same.
Just a brief example so you know what I mean: let's say two particular weapons have the same DPS (100 for instance), and they are supposed to fight infantry (which have relatively low hp value). One of the weapons fires only once per minute, and it deals 6000 dmg, the other one fires every 6 seconds dealing 600 dmg per shot. Now since they are supposed to fight infantry which have relatively low hp, the 2nd weapon's value is bigger in this case, since a 600 dmg-shot will kill an infantry just as much as an 6000 dmg-shot will, but the second one takes 1 minute to reload. Probably the easiest way to determine their value is what Niko said, but it's not exactly correct.

By default you can't compare them since both wapeons have in physical term to different dimension. The dimension of the first value would be [dmg*s^-1*m^-2] (spoken "damage per area and time") and the dimension of the second wapeon would be [dmg*s^-1] (just "damage per second)

It might be useful to introduce a new dimension like averaged damage per area since even bullets have a certain area of effect. But for that wapeons new introduced value would be really huge for projectils, since the damage is concentrated in a small area. Additionally you need to understand the difference between the integrated damage over time, weighted average/root mean square/arithmetic average -value to be able to compare different wapeons.

But there much more interesting values which are more useful to compare wapeons. In mathematical terms, the first time derivative of damage per second could is much more interesting - it could be interpreted as something like "damage different per time intervall". If you analyze a HMG with this, the deveriate value would be close to zero, since the damage output is really constant (which mean there is only a low difference over time) while a tank maingun whil have a big value for the first time diverate because it shots really slow and therefore there are big timely differences.


Maybe you don't even understood what I wrote but overall you can't compare this two values - and when you do it it's not meaningful and has no technical content.

Hang on, give me a minute

For example: It's the same when you would compare pressure with force. Both have something to do with force (transversal accelerated mass := force, angular accelerated moment of inertia := torque) but the second (pressure) is defined with force per area.

Now you can adapt this example of the opening question.

(http://www10.pic-upload.de/16.04.13/nblhqumst6ee.png)

(http://www7.pic-upload.de/16.04.13/u2f3lzrgxtwy.png)

More information for you to work with:

Both weapons are identical in how they function. Same ROF, accuracy, etc. The only difference is the damage and splash. For arguments sake, lets assume there is no overkill on the damage and each shot divides evenly into the units health.

The splash for the first weapon function as follows.

Distance = 4; this is the max radius of the splash area.

The AOE applies differing damage values depending on distance. At 0-1 the splash deals full damage; at 1-2 the damage scales from 100% to 25%; from 2-4 the damage deals 25% of the base amount. Accuracy is 100% at all distances.

I realize there is other contextual data to take into consideration such as the proximity of other units to the target; lets assume there are 2 units in proximity at all times - you can use any scenario duration you wish.

Given those variables, what formula would you use?

Well you are basically claiming the same as I am, no need to derivate tbh, just use common sense. But the question is still open: how to overcome this problem and give the best estimate for weapon's value?

That special szenario can be easily solved if.. the two targets are significant smaller in space then the inner splash radius.
Since you have 100% accuracy you will always hit the target directly and with the second condition you can always say that the second target will always stay in the inner splash radius. Which overall means that you will always deal the same amount of damage.

The first gun can only hit (lets say member of a of squad) at the time and dealing 20 damage, while the second gun deals 10 damage per shot, but will still be able to hit two member which sums up to 20 damage per shot. But this is just a ideal scenario - If you start to throw area accuracy, scatter, the different AOE damage modifier, squad movements... etc. into this mix it's really hard to build a formula for that. You need a complicated distinction of many different cases which will make you got nuts. And even if you have that formula, it will only works if you do a infinite time of analysis of this "experiment" because otherwise the statistic elements behind this will act as the variance component in this calculation.

To be honest: What is the real deal behind your question?

Figure out how to value splash into a weapons value.

According to these assumptions I calculated the following:

First weapon deals 20 DPS, no problem with that.

The second weapon will do the following DPS, assuming that there are 2 units in proximity:
1, both units are in close range : damage dealt per second: 2x10=20
2,both units are in medium splash range: dmg dealt per second: 2x10x0.625= 12,5   ( 0.625 comes from the average of 100%-25% dmg output, (100-25)/2 + 25%, assuming that its totally random)
3, both units are in long range radius: 5  (im just gonna write the number of DPS from now on)
4, first unit close range, second medium: 16,25
5, first close, second long: 12,5
6, first medium, second long: 8,75

If you add these number up you get 75, divide it with 6 ( number of different possible outcomes) you get 12,5 avg DPS.

Now you compare it with first weapon's 20 DPS you can see, that CP (cateris paribus- all other circumstances left untouched) the first weapon is more effective if they are facing 2 man squads.

I calculated the avg DPS for the 2nd weapon with the same method if there are 3 units in proximity, and I got 18,75 DPS. This clearly shows what common sense already knows: by increasing the number of units affacted by splash radius, the better avg DPS the second weapon will get, and therefore the 2nd weapon is getting more and more advantage over the first one.

Conclusion: it's important to determine the average squad size of infantry the weapon is facing + the space between the individual units. ( since if the squad size and the space between units is too large the splash wont affect all of them) Can you give these information aswell?

From the get-go - there is no absolute way to derive the real constant value of a weapon in this kind of scenario. Weapons that deal splash damage will forever be subject to a floating value that depends on which enemy unit is being hit (that unit's position relative to other units around it), how many units are within range of splash and other factors. However, we can simplify some of these values to get an estimated value.

The fundamental problem is figuring out how many units will be splashed upon with each shot. Although there is no absolute way of knowing this (one could simply try and observe a hundred shots and see what happens each time to get a resonable estimate - but that will still be a chance-based occasion) let's try and decide upon a basic formula for this. I would propose the following one:

Effective splashed units = number of units splashable * (1+Splash Radius - Formation spread) / Formation spread.

0<=(1+Splash Radius - Formation spread)/Formation spread<=1

Why this formula? Well, it is a fairly simple that can give a reasonable insight onto what is going to happen. Units do not maintain constant formation spread - they are subject to a bound that they try to maintain, but they pretty much never do it perfectly. Hence certainty over splashing over them (or not splashing over them at all) can not be claimed. However, if the squad tends to have a spread that is more or less equal to the splash - some of the units will probably be hit, while most won't. That's why the number 1 is entered into the equation. The reason why the simple formula is bound between 0 and 1 is because you can never splash less than 0 units, and never more than there are which are splashable in the first place.
Let's give an example:

There are 5 units in a squad. We have a splash radius of 5 and a formation spread of 3.

Only 4 units are splashable (one is targeted directly), so substituting in the values we get :

ESU= 4*((1+5-3)/3)=4*1=4

Effectively, you can expect that all 4 units out of the 4 will be splashed.

Assuming that there is no splash drop-off in damage, the unit is enjoying the value of 50 DPS with a base damage of just 10 DPS.

If you know that the splash has periodic drop-off, all you need to do is simply calculate the ESU's for each bound, and adjust each ESU beyond the short-range one by subtracting the previous one to obtain the realistic ESU at each range.

E.g.:

Base ESUs:

Short - 1
Medium - 2
Long - 4

Real ESUs:

Short = 1
Medium = 2-1=1
Long = 4-2=2

(1 person will be expected to be hit in the short-range splash, 1 in the medium-range blast, and 2 in the long-range blast).

And then you simply multiply each Real ESU by the appropriate damage modifier.

If you are going to implement a continuously scaling function (I.E. Damage continuously drops from 100% to 25% as range goes from 1 to 4) then you can still use this sort of function, but you will need to define it as an integral - and frankly put I'm not really that interested in doing that, as it's kind of tedious without really giving much benefit. It simply makes more sense to define it as a single bracket with a damage modifier of 62.5% - the difference will be absolutely negligible except in very, very precisely defined cases.

That equation equals 4 ? 4*1

Taking the second scenario that you gave with different info - no proximity, but guaranteed number of units always being SOMEWHERE in proximity.

This is much simpler to calculate.

10 DPS base (the target itself) - ignore for now, it's simplistic.

We know that 2 units will be in proximity of the target at all times. They are perfectly randomly distributed in the splash radius and have an equal possibility of landing at any continuous point (0;4)

Hence it all becomes about probability. 0-1 is 25% of the distance, so there's a 25% chance to land there and be dealt full damage.

2*0.25*10=5

1-2 is another 25% of the distance, with an average damage within this distance of 62.5% Units can be dealt 100%, or 25%, or 33.4559% - it doesn't matter. On average, due to the perfectly random nature of the distribution it will always be 62.5%. So:

2*0.25*6.25=3.125

2-4 is 50% of the distance, 25% damage taken in this bound. Ergo:

2*0.5*2.5=2.5

Total splash damage effective = 5+3.125+2.5=10.625

Total DPS = 20.625

You could also draft up a damage-probability table that would simply write out the joint probability of each event happening to each soldier and multiplied by total damage taken by both soldiers, giving Expected damage for each possibility. Writing out just the possible outcomes is simple:

(Legend - 1,2 denotes soldier, L,M,S denotes which range he was hit at).

1. 1S 2S - 20DMG 0.25^2 probability = 1.25
2. 1S 2M - 16.25 DMG 0.25^2 probability = 1.015625
3. 1S 2L - 12.5 DMG 0.25*0.5 probability = 1.5625
4. 1M 2S - 16.25 DMG 0.25^2 probability = 1.015625
5. 1M 2M - 12.5 DMG 0.25^2 probability = 0.78125
6. 1M 2L - 8.75 DMG 0.25*0.5 probability = 1.09375
7. 1L 2S - 12.5 DMG 0.25*0.5 probability = 1.5625
8. 1L 2M - 8.75 DMG 0.25*0.5 probability = 1.09375
9. 1L 2L - 5 DMG, 0.5^2 probability = 1.25

Added up = 10.625 EDPS from splash

So a total of 20.625 DPS for the weapon.

Oh yeah, rolsc, thanks, did it too fast and didn't check my work.

Fixed now.

I like Myst's approach, but there is a problem. Let's assume the standard formation for infantry is more or less a line. ( not a straight one but still) Now your equation on ESU only applies if you hit a unit which is on the end of the formation. But if you hit a unit which is in between others? In that case more units are going to be splashed, since the splash radius applies for both directions.

Keep Myst's basic concept but add probabilities to it. I mean, if there are 5 guys in a squad, the chances of hitting each is 20% =0,2.  So there is 20% chance that guy number 1 is going to get hit( see illustartion below), splashing guy number 2. There is 20% chance that guy number 2 is gonne get hit, splashing guy 1 and 3. etc    Calculate dmg for all possibilities, multiply them with the chance, add them up and u get the DPS you wanted. And with it you have considered the formation spread and chances of hitting each guy factors.

I think you should expand your theory with this idea.

 1                      3                   5
            2                     4

I get where you're coming from - my assumption there is indeed that the shot lands in the middle (tbh not that terrible an assumption for many units - particularly something like a StuH which you want to attack ground with anyway).

I'd say a simpler way to account for it is by simply implementing a discounting term.

 1                      3                   5
            2                     4

Assume hitting 3 allows splash on 1,2,4,5. (100% efficiency)
Hitting 2 allows splash on 1,3,4. Hitting 4 allows splash on 2,3,5. (75% efficiency)
Hitting 1 Allows splash on 2,3. Hitting 5 allows splash on 3,4. (50% efficiency)

If we assume there's an equal chance to hit any of those men then the discounting term is simply:

1*0.2+ 0.4*0.75+0.5*0.4=0.7

So simply multiply your obtained ESUs by 0.7 to get an even more realistic answer.

Great work guys, that answers my first question perfectly! My second problem is to calculate the impact of scatter on a weapons value. I understand you can't get an absolute value, what I'm trying to do is get an idea of how things will be impacted with some accuracy but not total accuracy.

The next problem is determining the impact of doubling a weapons scatter; what value is lost or gained on the weapon. So a weapon with 2.5 AOE and 5 scatter has a value of 10, if I double the scatter does the value become 20? I've attached an image illustrating how scatter is calculated, the engine randomly selects an area within the yellow section as the center of the explosion.

The actual area is calculated by taking the weapon range and adding/subtracting the scatter distance value to get 2 numbers which represent the inner and outer radius. The scatter angle value determines what section is cut out to become the scatter area.

I'm trying to determine how much of a damage drop off I'm losing when I double scatter distance, the value that determines the radius of the circle. Let me know if you need more information.

My head hurts.....

Scatter will always, unequivocally, make the weapon worse off against single targets where the target is small, stationary and the weapon is aimed precisely at it, although at a benefit against targets which are not stationary and thus the shots are not aimed precisely at it - and that is frankly put almost impossible to estimate without a large bank of empirical data (but frankly put I'd say the losses outweigh the gains almost all the time, for reasons you'll see in a second).

How much worse off this weapon will be against units that are small, stationary and the weapon can be easily seen by this set of formulae:

The weapon has a splash area of pi*x^2, where x is the AoE radius.
The weapon has an effective potential splash area of pi*(y+x)^2 where y is the scatter radius of the shell.

Assuming you are firing accurately (I.e. casting the barrage on a position the enemy is in) - each shell effectively has the following chance to hit the target:

(pi*x^2)/(pi*(x+y)^2)

This simplifies, simply, to

p=x^2/(x+y)^2

x then always increases the chance, y always decreases the chance. If y=0, then probability is=1, if x=0, then probability is=1 (for obvious reasons).

Implementing scatter angles changes the formula as such (note - simplified to a large extent as there is nothing I hate more than trigonometry which is required to give a more precise set of equations. Specifically the problem is that the AoE radii will almost always go outside the confines of the cone of potential hits within the scatter cone. Unless y is very large relative to x, at which point the scatter angle becomes inconsequential anyway).

360/scatter angle * p. (The larger the scatter angle - the closer the probability is to being equal to the standard circle solution. The smaller the scatter angle - the more effective accuracy can be found in the shell).

For especially small scatter angles (where SA/360*2yr<1/4x is what I would use as a good benchmark, though the actual solution will always be between the two formulae regardless of the SA, y, r unless SA is 360) the formula can simply be simplified to a simple linear probability. Assume that the shell falls somewhere along this line with a distance y:


|-------------------------------------|

Then there are three solutions for this:

Solution 1 where the shell was targetted to fall on either edge of the line:

P=x/y

Solution 2 where the shell was targetted to fall in the centre of this line:

P=2x/y

Solution 3 where the shell was targetted to fall not on the edge of this line, but close enough so as to have the explosion be slightly "wasted" by going partly outside of the y confine

P=(2x-z)/y (where z<x)

Assuming the three targeting decisions are more or less equal and randomly distributed the formula becomes a rather simplistic:

P=1.5x/y

As always the limit of the answer is min(1;P)

Which formula you should chose will depend pretty much entirely on the size of the scatter angle. For the sake of relative accuracy, barring extreme cases of the scatter angle being close to 0 or close to 360 I would suggest using:

P(true)=[360/scatter angle * x^2/(x+y)^2 + 1.5x/y]/2

That would be my quick general case solution for the relationship between scatter angle, radius and AoE without taking into account differing AoE boundaries that produces the probability of an AoE hit. I think by this point you may have realised that solving for how this would change under different AoE boundaries would be simply by solving the equation for each individual boundary and then subtracting the previous sized-one from the larger sized one. Then simply multiply each probability you obtain by the relevant splash damage level, add them together and you got yourself a decent estimate of expected splash damage.


Obviously, yet again, this is for shooting at immobile targets. It is very easy to predict one for shooting at non-stationary targets, but to a very large degree that is simply pointless. Assuming the target is acting rationally and given a speed that allows them to get out - they will always be able to move out of the way of the explosion(hence p=0). Assuming they are not rational and act randomly - they simply get the modifier of scatter angle/360 to their overall probabilities to indicate they chose the direction they run in randomly. Assuming a speed that can let them partially get out of dodge and rationality - simply replace y with O=tv+y, where t = shell travel time, v = speed of target.

None of this is perfectly accurate, but will give you good estimates of what's going on.

Oh, and to see what would happen to expected probability (and thus, damage) - simply differentiate those formula with respect to y. If you're differentiating by Y while using O, then simply use the chain rule when differentiating.

If you asssume semi-rationality, by the way - the target is equally likely to get it right as he is to get wrong, then the t*v modifier in O should be divided by 2.

Oh and actually - assuming no scatter angle the benefit of having more scatter can actually be partially quantified (though I can't be arsed deriving a proper formula):

When y+x>tv (I.e the unit does not have a chance of leaving the area before the barage ends) dP/dy<0 (there is no benefit to increasing scatter for the weapon).

When y+x<tv (I.e. The unit CAN leave the circle in time) dP/dy>0 (there IS a benefit to increasing scatter in the weapon).

Now I want you all to make schematics for a rocket capable of traveling to mars and dispersing coh2 copies on the surface without damaging them.

I will need this on my desk before Friday.

Modify a real rocket schemat and add bomb doors for the copies :)

When I sat in leason I thought about the fact that I could do a 3D-simulation with Mathematica. It would be possible to create a simple model for the splash damage (three cylinder with different radius and heights) and a target (lets say a another cylinder or two cylinder next to each other).


With a manipulate command you could move the target at any point in the layer (x,y coordinates). The dealt damage to the target would be the convolution integral from the overlapping solids. It would be possible to do it in realtime since mathematica can solve this integrals easily in a numeric process and modern computers have easily enough process capability to calculate this. (Overall its still a easy task for a computer)


With that method you could desing any splash damage and its possible to design even a complex target-solid with this method. But I'm not sure if it's worth it to programm such thing. - I could get some help and It might be done in a hour.

Yeah, modelling it in some good Maths software would work out quite well and give a more precise solution to the equation.

Is this for COH 2?

Ofcourse it is, probably Ostwind or Scout Car Upgrade which i have mentioned and were ignored :D

The the problem with the SC 20mm gun is that his has a very low damage splash of 0.5, and an effective chance to hit inf sitting at 2.5-5%, go figure why it sucks, lol

With a splash of 0.5 your chance of actually splashing is effectively 0. We're talking roughly StuG/Panther level splash lol.

Yup, with a much lower chance of hitting than either of them, and waaaay less damage.

It's like a 20mm marder, lul

Its so we better understand the impact of modifying scatter distance, scatter angle, and AOE distance.

Im getting more used to this stuff but this is just a basic concept. In that actual type you can change the AOE ranges and the damages and the positions of the two targets. That could be a scenario where one soldier is sitting in the low range splash while his buddy is staying at the mid range damage.

(http://www10.pic-upload.de/17.04.13/mkiu7qltkvnq.png)

Question: How does CoH works? Does the Soldier get the damage from the lowrange splash and the midrange splash or just from the higher one?


At the next few days I'm really busy with University and stuff (beside from that I'm staff [voluntary] of a special agency for technical reliefs and have a big practice scenario at the weekend) but I will TRY to get some real results out of that thing. Maybe its possible to put some statistics in it(scatter, accuracy - whatever) to go completly nuts with this calculation. I'm still getting used to that programm since I'm just started to learn it right now... ..and it's possible that I'm overstrained with this and have scoop. I will find out.

Scatter Area =(((((PI()*(Y+X)^2)-(PI()*(Y-X)^2) ) )*Z/360))
Y = weapon range, X = scatter distance, Z = scatter angle

That is the formula for calculating the scatter area.

A common set of numbers we use is as follows which gives us:

Y = 30
X = 5
Z = 7.5

Scatter area = 39.2

If a weapon has 2.5 AOE Distance, then by determining the area of a circle that value gives us 19.63.

If we assume I am aiming directly at an object, then I would imagine my probability of hitting that object is 19.63/39.2 = 0.5.

So you guys just wrote up a completely new formula in this thread?
http://www.youtube.com/watch?v=FfPFtbJAEm8

Why would you do any of that? You don't need a 3-dimensional simulation (2 would suffice) and I don't even know how a convolution would come to this. You could model it stochastically fairly simply, but I don't think you'd be able to accurately model the position of squad members, especially when they dart around under fire. Maybe that could be done stochastically, but I don't know how you'd do it accurately, and now the model's getting quite complex (and not much different than just testing it in-game.

It can't be done mathematically there are just to many variables.

The most imposing being that inf combat ai seeking cover, moving under fire.

The 3D stuff ist just for the visualization purpose. I would implement the scatter in this way:
Not sure, but I think mathematica has a random number generator. So I will use one which generates a random number from <0-1> and will
multiply that with the scatter value, followed by a shift from (just by adding the scatter values) on the original coordinates. Now the gun will "never"
shot at <x=0,y=0> and will have always a small difference to it.


@Masacree:
Yeah, wasn't fell thought from me. The calculation itself could be easily done with a simple if command. I just need to transform the karthesic coordinates into polar coordinates and check in which radius I'm at. Dealt damage so the target is summed up to a group is the dealt damage to each induvidual of the group.

except that the damage scale is gradual and not set in stone, lol. If he is just inside the .5 value he likely got .52 damage.

I'm am pretty damn sure the damage is applied from the range of the closest hitbox point, so if a square it would be the outer edges distance to the explosion, not the center and as per above, it's a linear scale, not hard numbers.

If 10m is 1.0d and 20m is 0.75d, then 15m is 0.825d

So basically what you're saying is that the damage is not constant within a given radius since it has a starting value and the beginning of the radius and starts
to fall down to the value of outer radius. Which means that each AEO has a damage difference per length intervall and isn't flat out like in the picture I've posted before.

Are 100% aware of that or you just damn sure about it?

A developer on the CoH2 sega forums stated that all range based values like AoE and Accuarcy are linear.

While the AoE modifier is not called out specifically, I'd bet on it behaving the same.

Well no, the "point blank" or "short" range is still full damgae, it all depends on what you set it to.

For example, on a 5m splash it's 100% out to 1.5, then starts dropping.

At least I've learned something new today.

Actually, I'm not even sure why I said "well no" as I don't think I even read your post when I made that comment, I think this might have been out of reflex  :P.

Edit, actually...

Likely that, it's either way to oversimplified or just not understandable, lol.

Someone PM I2ay he pulled some crazu shit in ventrilo when we played war thunder.