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Why that title?

Concerning the size of the ships and now the water depth.  1. I've had a number of situations where my "fleet" got tied up in knocks and I could not move them until I untangled them. "Smaller" ships may resolve this type of concern. 2. I've also built a dock where the water depth was too shallow for the larger boats, so I ended up with ships that could not go anywhere and had to be destroyed.  

Further thinking about my 1D simplication case, I find that we can obtain the actual probability quite easily.  Since x and y are independent, we can just take the square of the probability, which means probability of x being within +-1.5m of target times the probability of y being within +-1.5m of target.  Next, since infantry is considered as a circle, we further multiply it with the area of circle divided by area of square. So, the exact probability of archer hitting target at 60m is 73.35% x 73.35% x pi x 1.5 x 1.5 / (3 x 3) = 42.26%.  A spread upgrade would make it 54.77%.  Pretty low!  Would anyone perform experiment to prove it correct or prove it wrong?

Thanks for trying to understand my imprecise description.  Random normal distribution is a bell shape of standard deviation of 1.  If we multiply it by a spread of 2, it becomes a fatter bell shape, with standard deviation of 2. Spread is calculated like 2.25 x 60 / 100.  It doesn't change like the way you relate with area.  Consider it like a radius of a circle, centered on the landing point of projectile   At 100m the circle is large, so shooting is inaccurate, at 50m the cirlce is smaller, more accurate.  You are right that the area changes by 4 times, but spread is more like a radius, not area.  This is a 1D value, not 2D.  The random distribution is 2D.  Multiply the 2D distribution with the 1D spread factor to make the distribution fatter, larger, more spread out. x y are not position of the landing point of projectile, but the random values to be added to the position of the landing point to make it spread out randomly.  With spread, the landing position is no longer a point, but a 2D normal distribution centered on that landing point. You are right to point out that they "shouldn't" be independent.  But as fact, the source code calculated them independently, probably to speed up computation.  So the condition x^2 + y^2 = R^2 did not hold.  x is just a random value with normal distribution, and y is just another random value with normal distribution.  Together, x and y has no relationship.  That's why the CDF is not circular.  It is weird, but it is the current fact. Also seems you tried to understand "independent" as separate, and counted 3 spreads.  When I mentioned independent, it is in the sense of random variables.  Since it is 2D, there are only two independent random variables.  Spread is a constant factor, like 2.25 for archer.  At distance of 60m it is adjusted to 2.25 x 60 / 100.  Multiplying the spread would make the random normal distribution larger, fatter, more spread out. And I guess your final intuitive interpretation is correct.

In 1D, but in 2D, with 2 spreads, at twice the distance inaccuracy should increase by 4 times (given how areas change with distance).   What is super difficult is trying to understand what is that you are trying to say. A pair of independent spreads, then you multiply them by the spread? Do you have 3 spreads then? Not even counting target movement? What are x and y, exactly? Please, define things. In any case, assuming x and y are positions, of something in reference to a center, they shouldn't be independent, you should have the condition x2+y2=R2, with R being the displacement from that center, and this changes your integration limits and makes the CDF circular. If I understand correctly, that something is the arrow, so that circular CDF gives you landing point probabilities (this is a simplification, because the linear horizontal velocity, both longitudinal and transversal, and the quadratic vertical velocity of the parabolic movement will result in an elongated shape on the floor, which is also a simplification when not accounting for air drag, wind, etc). I guess then with some random generator you decide where the arrow fell, and evaluate proximity to nearby units.