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I posted the below in the WoT Forums and I wanted to offer it up to the experts here to make sure my understanding and math were correct.  It is a basic watered down version of the complexities of the aiming mechanics with some examples to help illustrate the wording.  Thank you.

 


 

The aiming algorithm is different than the pen/damage algorithm as it utilizes the gun value known as distribution...or effective accuracy.  It is a value given of where your shell will land at 100m from the target.  The higher the value, the wider the shot.  The further the object, the chance for a miss increases on a linear scale.

 

If I understand on a basic level this means that a gun with a .40 dispersion will land, around 95% of the time, within a .40m circle at a distance of 100 meters.  Think about that.  Two men of average height standing at opposite ends of a football field looks roughly to be about the size of half a dime to each other.  This gives us a rough ratio of 0.375" : 72" or a scale factor of .00521.

 

https://www.youtube.com/embed/HpiZEo_QYlc

 

Now, move your target to 200 meters and if I understand this correctly, roughly 95% of your shells will land in a .80 meter diameter circle.  At 300 meters, 95% of your shells should land in a 1.20 meter diameter circle.  Remember 1 meter = 3.28084 feet.

 

If I did my math correctly:

 

The size of a TOG II facing you at 0 meters is 10.13m wide X 3.05m tall.  At 100 meters it would appear to be approximately 2" wide x 5/8" tall.  This gives you a rectangular area of 1.2504".  So, a .40m distribution is a circle 1.31234 feet (or 15.74808 inches) in diameter.  

 

TOG II Maths:

Width = 10.13m = 33.2' = 398.4" * .00521 = 2.076"

Height = 3.05m = 10' = 120" * .00521 = 0.6252"

 

I am surprised we hit anything at all!  Thank goodness that sniper mode gives us a 8x or 16x zoom for magnification (couldn't find stated value).

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Next time, for the sake of all that is healthy, use either metric or SAE, not both..... 

Even if you do convert them, it's still not preferable.

 

ok...I will make a note of that.  Thank you.

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If I understand on a basic level this means that a gun with a .40 dispersion will land, around 95% of the time, within a .40m circle at a distance of 100 meters.

 

Someone please correct me if I'm wrong but AFAIK the dispersion is clamped at the extremes, so that would be 100%. Same going for damage/penetration rolls, they're clamped at +/- 25%.

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Someone please correct me if I'm wrong but AFAIK the dispersion is clamped at the extremes, so that would be 100%. Same going for damage/penetration rolls, they're clamped at +/- 25%.

I think you're right, I remember reading that something outside the 2 standard deviations would be recalculated.

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I based my 95% from the Wiki:

 

 

 

Gun Accuracy

The accuracy value for a gun is given in meters at a range of 100m. The lower the value the more accurate your gun is. The value describes 2 standard deviations σ from the center of your aim. In other words, for a gun with 0.32m effective accuracy at 100m, 95.45% of all shots will land within 0.32m of the center of your aim at that distance. Dispersion amount increases linearly with distance, i.e. 0.32m effective accuracy at 100m translates to 0.64m at 200m and 1.28m at 400m.

 

If I messed up on my understanding, I would love for someone to point it out so I can make sure I modify my other post and get it right.

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