So you're saying that a Megathron cannot track another Battleship MWDing at 15km? Wow, that's a pretty impressive amount of misinformation coming out of a single sentence....Yes, mea culpa. I did completely ignore that MWD increases the size of the target's signature and this not only makes it easier to target quickly, it also helps negate the advantages that moving fast transversally has against tracking.
My error lies in that I think of the two separately but the formula to determine hits has them combined in the equation. From Evelopedia:
But its not a two part process, its a one step calculation and turning on a MWD may increase transversal speed but it also increases the radius. So the question is, how much effect does the two changes have?
(Side Note: I use "transversal" velocity but the correct value to use in the calculations is "Angular" velocity which measures apparent movement of something in radians per second.)
So let's go back to our Megathron versus Maelstrom example, using current blaster values. Let's also assume that we are using Caldari Navy ammo in Neutron Blaster Cannon IIs, and that the Maelstrom is at optimal so the second part of the equation is zero. The stock Maelstrom (skills all level V) moves at a top speed of 118 m/s and a sig radius of 460 meters; with the MWD on it moves at 835 m/s and a radius of 2760 meters. The blaster on the Megathron has a tracking of 0.07442 radians/sec and a signature resolution of 400 meters.
So what's the angular velocity of the Maelstrom? Well, assuming the ship remains at optimal range, the a single radian is 4500 meters (i.e. optimal range of the blaster) and thus the angular velocity is 0.02622 rad/second at normal top speed and
So chance to hit Maelstrom without MWD on = 0.5 ^ (( (0.02622/0.07442) x (400/460) )^2 )
= 0.5 ^ ( (0.3523 x 0.8696) ^ 2)
= 0.5 ^ (0.3064 ^ 2)
= 0.5 ^ 0.0939
As Luccul pointed out below, I got the angular velocity wrong with MWD on. Looks like I used the sig radius as the velocity, sigh. The actual angular velocity is 0.1856 to give us:
And chance to hit once that Maelstrom punches the MWD button?
0.5 ^ ( ((0.1856/0.07442) x (400/2760)) ^ 2)
= 0.5 ^ ( (2.4940x 0.1449) ^2)
= 0.5 ^ (0.3614^2)
= 0.5 ^0.1306
So, as everyone keeps pointing out to me, the 500% increase in speed is cancelled out by the 500% increase in signature. Gah.
At 15000 meters the angular velocity of the Maelstrom is 0.00787 r/s normally and 0.05567 r/s with MWD. We need to take the second part of the equation into play now so fall off range is 13000 meters.
No MWD at 15km:
0.5 ^ ((( (0.00787 / 0.07442) x (400/460) ) ^2 + ((15000 - 4500) / 13000 )^2)
= 0.5 ^ ( (0.1058 x 0.8696 )^2 + ( 0.8077)^2 )
= 0.5 ^ ( 0.00846 + 0.6524 )
And MWD at 15 km:
0.5 ^ ((( ( 0.05567 / 0.07442) x (400/2760 ) ) ^2 + ((15000 - 4500) / 13000 )^2)
= 0.5 ^ ( (0.7481 x 0.1449 )^2 + ( 0.8077)^2 )
= 0.5 ^ ( 0.01175 + 0.6524 )
So assuming my math is correct (and I'm running out of time to check my calculations so someone please comment if I'm wrong), the extra velocity at 15 km is negated by the ballooning size of the signature. In close ranges (i.e. optimal), the MWD speed makes a significant difference.
Now the question is, how much does that 20% extra tracking actually help blasters?
A big thanks to Suleiman Shouaa for pointing out my glaring mistake.