Hi, micro, thanks for the update! I can say from my experience with the carrier-ILS-capable HUD, that the ball is set at exactly 4.00 degrees, for both the default and the Nimitz carrier. I don't know if that helps. It's pretty high...
I had issues with this all the way back in FS9. I was having troubles landing, needing 1000-1100 fpm descents to stay on slope sometimes. In my investigation, I found the ILS was set to 4.00 degress, not the 3.xx that a land based is. At first I thought this was really stupid. However, I was neglecting the 35 kts of wind over the deck (whether it's from the carrier moving or a static carrier with wind down the deck).
If you think about what 4.00 degrees means in terms of tangent that angle represents some vertical height in proportion to some horizontal distance. The problem is the frensel system is static...it assumes the carrier is stationary.
But it's not. Let's assume we start at 600' MSL and trace to touchdown. Then a 4 degree glide slope means we are 8,580' from touchdown (tan(4degrees)=600/base).
The problem is that at a speed of approximately 130 kts (making an averageish calculation here). This comes out to somewhere on the order of 35 seconds to touchdown (time=distance/velocity calculation).
Well, in that 35 seconds, a carrier moving at 30 kts (again, rounding) will have moved moved about 0.3 NM or another 1,800 ft (distance=velocity*time).
However, our starting height is still the same. So now the two legs of are triangle have changed from 600' MSL and 8580' to touchdown to 600' MSL and 8580+1800' to touchdown (10380'). So if we recalculate the new angle, 3.25 degrees (arctan(600/10380)--in orther words the dynamic glideslope is your basic 3 degree glideslope. The 4 degree static glideslope is necessary to accound for this.
Note, I'm doing this math using the google calculator from a hotel room, you'll likely get different...and more accurate...results if you repeat them with a real scienctific calculator.