Quote:
|
Originally Posted by tater
I don't think the game takes depth into account at all. In fact, since DCs can do hitpoint damage, which makes your crush depth more shallow, the deeper you are when hit, the more dangerous it is for you.
Cool pic I found:
A quote from a ww2 USN manual:
Quote:
|
The effective radius of the percussive wave depends upon the structural strength of the attacked vessel, and no definite values can be stated. Approximate information indicates that a 600-pound charge may cause moderate damage at 80 feet, but to be fatal it must explode within about 30 feet. The 300-pound charge may prove fatal within 20 feet. It is to be noted that doubling the weight of charge does not double the effective radius.
|
May prove fatal at ~6m for what would be the biggest IJN DC (162kg warhead). That maps to D I guess. 2 or 3 and D...
tater |
Ok, first you are right about depth being a factor. But pondering on this I started to wonder if it doesn't have as big an effect effect on cumulative damage you take as how easy it is to exceed a catastrphic stress. I.e. less than catastrophic and the hull rebounds, but because of the increased depth and stress on the hull, the added pressure and uneven application of pressure makes it easier to exceed a catastrophic point. So if it's not catastrophic, you don't take significantly more damage, only a smaller percentage more, but you exceed the catqstrophic failure point easier. Wish I could find a good paper on this.
The reason doublig the charge doesn't double the range is because of the way the shock wave dissipates.
Here's an example I was able to dig up talking about a shock wave in air:
http://www.makeitlouder.com/document...stimation.html
In water, there is more force dissipated faster because of the density of the fluid.
So for grins lets say a 300 lb and 600 lb DC have 100% effectiveness at 1m and we'll use a loss factor of 0.5 for every 2 meters past that. (These are merely for illustration, not actual numbers and rounded of to the nearest 0.X)
So here is the theoretical dropoff of 0.5 for our example:
distance . . . 300lb . . . 600lb
3m . . . . . . . 150 . . . . .300
5m . . . . . . . 75 . . . . . 150
7m . . . . . . . 37.5 . . . . . 75
9m . . . . . . . 18.8 . . . . . .37.5
11 . . . . . . . 9.4 . . . . . . 18.8
13 . . . . . . . 4.7 . . . . . . .9.4
15 . . . . . . . .2.4 . . . . . . .4.7
17 . . . . . . . 1.2 . . . . . . .2.4
19 . . . . . . . 0.6 . . . . . . .1.2
21 . . . . . . . 0.3 . . . . . . .0.6
So even though you start of with a charge twice as powerful it only takes a distance 10% greater, not twice as great for the force to drop below 1 lb
This is an extremely simplified example because I'm leaving out possible log or natural log functions which would make the drop off faster past a certian point, and I'll guarantee the factors are off. But this shows why doubling a charge does not make the DC capable of creating the same damage at twice the distance of a DC half it's size.