So, will critical damage mod sets now be useless?

The additive nature of critical damage explains why with high critical damage, the gains from the critical damage set may not be worth the gains from an offense set. But I already detailed that scenario above.

Using your own numbers,

(4000+263)*1.9*1.5*1.92 = 23,327.136.
4000*1.9*1.5*(1.92+0.3) = 25,308.

So the critical damage set provides 1.0849 times as much damage as the offense set in your example with in game offense bonuses.

Dropping the offense multipliers out, you have,

(4,000+263)*1.92 = 8,184.96, and

4,000*(1.92+0.3) = 8,880.

It should be no surprise that the critical damage set in this hypothetical example provides 1.0849 times as much damage as the offense set.

The in game offense multipliers here are a constant, therefore all they serve to do is change the scale. Arguing that they cannot be dropped from the equation is like arguing that 30 centimeters is always better than 12 inches.

To explain my inequality better: Taking the two separate lines on a graph and setting them equal to each other yields the point where the lines cross (linear algebra). I have represented this as Offense - CD =0. Since CD is subtracted from Offense, if the value is positive, offense is greater (that section on the graph where the offense line is higher), and if the value is negative, it represents where the offense line is below CD. Using two lines as an example is of course an oversimplification, as it is a multivariate problem and we are therefore discussing two surfaces, or at best, two lines with respect to a single value of a third variable. But the same principal holds.

"The idea is to determine different values and then compare them to each other. If the crit damage is greater, we determine what % greater it is and that gives us the amount of crit we need to have before average damage would be higher with the crit damage set."

This ratio does not seem at all related to critical chance. The average damage would be CD*(CC)+1*(1-CC). The offense set equation would then have a modifier. You would then compare the two lines, as you say.

To simplify things, let's use current set bonus numbers and ignore offense bonuses from other mods. 1.1*1.5 = 1.65, so for the offense set equation, we have 1.65 (c)+1.1 (1-c). The CD set is 1.8 (c) + (1-c). You might brute force these two equations to find that a value of c= 0.4 yields a result of 1.32 for each equation.

Or you could subtract one from the other to obtain -0.15c + 0.1 (1-c)=0, or -0.15c +0.1-0.1c=0, or -0.25c + 0.1 =0. This is what I have done in my calculations above. Solving for c, you would find c=0.4 is the cutoff.

If I understand your method correctly, you would take 1.8/1.65 = 1.09. Since you cannot have greater than 100% cc, I'm guessing you would either go with 9% cc, or perhaps 91% if you reverse the equation, and these seems more consistent with your previous assertions. Neither, of course, would be correct.

In short: The additive nature of critical damage bonuses absolutely affects the determination of which set to use, because of the relative increase vs. the multiplicative increase of the other set. However, it is additive to a MULTIPLIER, and this addition takes place before the in-game offense bonuses are multiplied by that multiplier. Those in game bonuses become a constant, and can be dropped out of the equation since all they do is change the scale, and do not create an interaction effect between the sets (as in game CD additives would).