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Anyone have any good math for getting some ball parks on how many average hits you can expect from large blast weapons?

Maybe bs3 and 3 vs 10 MEQ and vs 5 TEQ vs 5 bikes etc.

Or just people's general feeling for how many hits can be expected etc

I expect to get 3 hits when I roll a hit on the scatter die. After that, it's just guesswork.

Personally, I take the number of hits and then add 1 BS to the firer to account for the chance you might wing one or two with a near miss.

Oh, unless you're talking about something on a bigger base than the small infantry base. In that case, I assume it gets one hit.

I completely agree with Ailaros. Three hits for standard infantry bases, and a single hit for anything larger. I sometimes use tank shocks to help pull units together and increase my numbers, but it usually means I'll lose the vehicle in the process.

Just look at what u're gona shoot down. Place a blast over them and u'll see. No point in counting averages, cause if the blast doesn't scatter - you hit all, if it does, you hit less. This average numbers won't give you anything.

I tend to assume 4 or 5 for a large blast on a hit averaging with 2 on a scatter for normal infantry bases. Even give that, I do tend to assume 4. The partial coverage rule really improves it, and I rarely see fully expanded units. You just don't have the board space to keep multiple units of Boyz or other swarm units fully spread out, and if they are spread out, shoot the blast at something else. That said, I find alerios and chaos ' s points completely valid as well, it's a matter of preference.

The problem with the horde thesis of extra hits comes when you start looking at points cost.

Yes, it's going to be more likely to hit more models in a horde because they have less space to put their miniatures in, thus the greater density (though I've found I can still put models at better than 1.5" coherence without much problem even at larger points values, myself). However, the more density, then, the less per-model-hit you're actually killing.

To put it another way, if your opponent has to be twice as packed because they're bringing twice as many models, but the models are half as expensive, then you're not actually killing more, even if you're hitting more models.

As such, the "but they're bringing a horde, so you're more effective" isn't strictly true, even before talking about canny horde players properly displacing.

I also agree with Ailros. In my experience a large blast will average 3 hits. This ignores complete misses (which are rare with spread out models) but the times when you get 4 models should balance them out.

It is never a good idea to assume poor spacing even though it may happen. Even in a relatively closely spaced horde (1") you will still only get (3x1" base + 2" spacing = 5" in a line) 3 models if they are lined up and 5 models if stacked. When you consider you could scatter badly and clip less than optimum and these assumptions are the maximum posible then 3 models is still not a bad prediction.

You asked for math so I'll give a shot at a somewhat rigorous approach. The complication here is that it depends on several factors. BS and base size are less factors than the most obvious orientation of the target unit. However, just because there's some variables doesn't mean we can't get something out of doing some example formations as long as you keep in mind its an approximation at best.

With those caveats its not all that hard of a task. You could do the straight up calculation but given the nature of the scatter (especially the direction) its more effort than I care to put in. The simple route is to just put the rules into a simulation, run it a million or so times and count out the number of average hits. Assuming I did everything correctly (fairly confident) the results aren't really that surprising. I won't list out a huge number of things but I will give some rough qualitative observations and one specific example.

From a few different test unit formations the distribution of number hits all take on the same-ish general shape: A large peak at the maximum number (representing hits and scatters of same # of hits) and then a (very) roughly even distribution back to zero. In other words, besides the maximum amount any less number of hits is about equally likely (not true for an individual formation but as a general trend across multiple formations). I would've expected 0 hits to be the second most likely but it isn't more likely than any other number between 0 and max. I guess its not really that surprising given you have to roll a 7 or so to even scatter off your original center let alone off nearby models.

For a unit of 10 infantry bases organized in a 2x5 rectangle thats columns and rows are seperated base to base by 1.9" with a large blast placed over either of the two most central models (4 hits if no scatter) at BS3, I get an overall average of 2.5 hits. The distribution is:
0 hits: 17%
1 hit: 10%
2 hits: 21%
3 hits: 10%
4 hits: 42%

Note that with a rectangular orientation odd numbers of hits require pretty specific scatters hence being somewhat suppressed. These roughly 10% fluctuations depend extremely heavily on the unit orientation. For instance if you have a more staggered formation instead of regular the odd numbers would be more likely than the even. So not considering exact formations these features would mostly average out into the roughly flat portion described above.

Maybe not the answer you (or anyone) wants but I find this kind of stuff mildly interesting

How did you get that distribution? If you put the large blast template on the left-middle model, and scatter almost anywhere in the left arc with a roll of higher than 6, you're going to be catching little more than empty air. Likewise, for a rightwardsish scatter with a die roll over 9.

I put in a set of coordinates for the models in the unit then I'm simply throwing random direction and scatter rolls. Then I loop through the every model in the unit and see if it is within the blast radius. Count up the hits, and repeat 1 mil. times to get a distribution.

its true that if it scatters perpendicular to the long direction of the unit you will miss with a roll greater than six with BS3. I used 1.9" seperation so the width of the base means a roll of 6, minus 3 for the BS still hits the original model. If you call these directions left and right (I think this is what you're saying above) there are still significant arcs in the other directions (up and down?, north and south?) where you still hit at least some models.

In these north and south directions you have to scatter really far to not hit anything; roughly 8 inches meaning a roll of 11 with BS3. Then ofcourse there's the inbetween directions (north west etc.) that fall in between the two extremes. Apparently according to the simulation these roughly cancel out.

It would be interesting to do the number of hits vs the angle of scatter with respect to long direction of the unit. I suspect it would be something close to a sine function. If the bug strikes I'll do that some other time

An impressive feat.

I am a bit confused about this, though:

JubbJubbz wrote: Then of course there's the in-between directions (north west etc.) that fall in between the two extremes. Apparently according to the simulation these roughly cancel out.

Why would they cancel out? It's not like you're practically always getting a hit with a narrow scatter from the file, and basically never hitting in a scatter along the rank, and then it's a linear progression in-between.

That would be true, perhaps, if the shape of the infantry was a square, but it's not in the case of infantry in a rectangle. Just think, for a moment, if those 10 infantry were in 1 rank of 10 files. A scatter of sufficient distance along practically any scatter angle except for those nearly along the rank would be a complete and total miss.

I'd think, at least.

Also, I'd be curious about my old anti-blast rings I used to make. I used to put my 20-man blob squads in O-shapes, the idea being that a scatter of any direction (except for a scatter along the curve) would lead to a complete miss, just as if I'd have put the guardsmen in a line (except for scatters along the line). Except the O-shape let me use the old 5th ed's cover rules better.

But my intuition has lead me astray on more than one occasion. I would be very interested if I were wrong on this one (and, of course, what the "correct" shape of displacement actually is).

All I meant by 'cancel out' is that the extreme of scattering along the long direction of the unit is the opposite of extreme of scattering perpendicular to this direction and directions in between will have a number of hits between those two directions. Thus the average is somewhere between the two

As you suspect its definitely will not be linear. You would need a very weird formation to give a linear change with respect to the angle of scatter. something like an arc thats center is where the blast starts. With a single file line as you mention you'd get a narrower angle that will result in non zero hits, but even that isn't a tiny angle. Even with a single file line there's still about a 6" wide strip you can scatter and still hit the line. This is why I expect (though haven't investigated) a sine function as you vary the angle of scatter (for a given radius). Its hard to talk about such things in exact terms because chunky discrete nature but the data don't lie (unless there's a bug in my simulator )

For visualization I think its useful not to consider the size of the unit but rather the size of the 'box' that you can scatter in and still hit something. Just roughly using BS3, 1" base, and 2" separation, that box for a unit of 2 x 5 is roughly 9"x18". Thats a big ol' box considering your most likely scatter distance is 4" and caps out at 9".

You bring up an interesting point approaching it from the defensive POV; looking for a formation that minimizes the expected hits. Without looking into it, I suspect it being a single file line. The circle is an interesting choice and could be better. Given that the blast has to start on a model and thus on the circle itself, I don't know how this would work. A circle of 20 guard I would imagine is quite large.

A method I may try to tackle is this is to look at how frequently models are hit vs how close they are to the starting blast position. Due to the nature of polar coordinates, I suspect the models closes to the starting position are hit more followed by models that lie near the most likely radius of scatter (7" - BS). If this suspiscion is true a formation that maximizes the average distance between a given model and every other model in the unit would be good. One that does that while at the same time avoids distances of most probable radius would be even better.

Out of curiosity I looked at the distribution for a straight line. The overall average becomes 1.7 with a similar distribution but with the max at 3 instead of 4. 1 is still supressed not suprisingly since in order to hit only one model you have to pretty narrowly graze the line.
0 hits: 27%
1 hits: 11%
2 hits: 24%
3 hits: 37%

Sorry for the novel but others' interest spurs my curiosity onward.

JubbJubbz wrote:All I meant by 'cancel out' is that the extreme of scattering along the long direction of the unit is the opposite of extreme of scattering perpendicular to this direction and directions in between will have a number of hits between those two directions. Thus the average is somewhere between the two

Yeah, the average is between the two, but the boundaries don't necessarily cancel each other out.

If you look back at that 2x5 rectangle again. If you scatter perpendicular, you only need to roll a 6 to hit no models, while scattering parallel, you need to roll a 12 to completely miss everyone.

As those numbers aren't equal, it doesn't seem right to discard them as controlled.

JubbJubbz wrote:You would need a very weird formation to give a linear change with respect to the angle of scatter

I imagine a filled-in circle would accomplish this, right? Put your blast marker on the model in the middle and angle of scatter would (nearly) not matter. Right?

I think perhaps this was the idea behind my O-shape, as it's sort of the opposite of a filled-in circle for our purposes. Put in more blunt language, it's to create the possibility of the template scattering into the middle of the circle and not hitting anybody.

JubbJubbz wrote:A circle of 20 guard I would imagine is quite large.

Well, it's not TOO bad:



That's 4 units of 20 guardsmen in a chain. Of course, 6th ed rendered this formation useless, but that's for reasons unrelated to blast weapons.

I guess that would be a further question - the best way to avoid blast weapons while also taking into account take-from-the-front wound allocation, but that would be rather out of scope of this topic.

JubbJubbz wrote:numbers

Out of curiosity, what does that look like in standard deviation notation?

'Cancel out' is bad language on my part. Perhaps work against each other is more accurate but I think we're in accord on that part. Its just my bad attempt to describe in words what the data is showing.

If you place the blast directly in the center of a model, and the blast has a radius of 2.5" it needs to scatter 2.5" plus the radius of the base in order to miss. I figure you're correct here because the bases are just shy of an inch giving you a needed scatter of 2.9" or so instead of precisely three. I do indeed use a base radius of less than 0.5" so I think it is correctly modeled in the simulation that a roll of 6 would indeed miss the original center.

Any circle, filled in or not, would give you a linear distribution if the blast can be placed in the center, this is true, but more precisely than linear it would be pretty well flat. I was thinking linear in the more general sense of hits = constant * angle. The weird thing about a hollow circle is that because of the blast rules you can't place the blast in the center so you must place it on the edge of the circle. This does give some angular dependence. I will test this in the nearish future because now im interested

Standard deviation can be misleading here as this is a very non-normal, asymmetric distribution. Notice the peak is at one extreme. For these types of cases the std dev does not really represent 'error bars' as they are commonly interpreted to be. You don't have the +/- 2*sigma = 95% type conclusions. Those only hold for normal distributions.

That being said the std dev is perfectly well defined and not so hard of a calculation given the above percentages. If memory serves it would the square root of the sum of the squares of the differences of the values weighted by that values probability. Or:

sqrt [ 0.27*(0-1.7)^2+0.11*(1-1.7)^2+0.24*(2-1.7)^2+0.37*(3-1.7)^2 ] = 1.2

Variances are weird, I usually find the probabilities more intuitive.