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[Mysthalin]

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

Fixed now.

[rolcsika0128]

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

[Mysthalin]

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.

[pqumsieh]

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.

[tank130]

My head hurts.....

[Mysthalin]

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.

[Mysthalin]

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.

[Mysthalin]

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).

[Sachaztan]

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.

[nikomas]

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

[Ahnungsloser]

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.

[Mysthalin]

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

[Unkn0wn]

Is this for COH 2?

[Baine]

Is this for COH 2?

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

[nikomas]

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

[Mysthalin]

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

[nikomas]

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

It's like a 20mm marder, lul

[pqumsieh]

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

[Ahnungsloser]

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.



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.

[pqumsieh]

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.