If there are 600 mines at Truk on turn 1 how long will they last and when will they need to be...

[ShakyJake]

Okay, I see what you're saying. Instead of a damage percentage applied to a facility that it must work off, it's just a generic production points system? That's actually a bit easier to calculate (while making a lot more sense, too). In fact, we wouldn't even really need to do any fancy math for this, but just use a geometric approach.

If you were looking for the daily production points output of a facility for any day "n", your formula would be:

X = (starting production) + (n - 1)

And the daily number of damaged production points would be:

Y = D - n + 1

Where "D" is the starting number of points which are damaged.

Pretty simple. Plugging n=1 for day one into both these equations, you'll find that the daily production output on day 1 is simply the original starting production, while the number of damaged prodcution points is "D", being whatever the beginning number of damaged points was.

For the sums of these equations you are looking for, for the total production points produced over a period and also the total damaged points over a period, we get into that geometric approach I mentioned. If you were to chart the output and the damaged points for each day, you would notice that they are both really just triangles, and the area of those are easy enough to calculate.

The formula for the sum of all production points over a period until day "n" becomes:

XX = n*[ (starting production) + 0.5*(n-1) ]

Using your example from another thread, of a facility that starts with 100 undamaged points + 100 additional points which are damaged, you can plug in n=1 into the equation to find that the total output for day 1 is the 100 points from the original undamaged points on the facility alone. If you want to know what the total output of the facility is on day 30, then you'll find XX = 30*[(100)+0.5*(30-1)] = 3,435 production points have been produced.

For the total number of damaged points in the facility over a period until day "n", your formula would be:

YY = 0.5*n*(2*D-n+1)

Again, D is the beginning number of damaged points, like 100 from the above example. If you plug n=1 into the equation, you'll get YY = 0.5*1*(2*100-1+1) = 100, so that on day 1 there are 100 total points damaged. Plug in n=2 for day 2, and your total will come out to be 199. Plug in day 30, and you'll see the facility has totalled YY = 0.5*30*(2*100-30+1) = 2,565 damaged points on day 30.

*long breath, and a beer*

So, does this help you out at all? Again, I'm not sure if this is exactly what you're looking for, but these formulas will give you the sums of all production and damaged points over a period. If each production point and repair point entail a specific supply cost, then just call the total cost:

Total Cost = XX*(supply point cost) + YY*(repair point cost),

and you're all set. Good luck, hope this helps.

-Jake

[Mr.Frag]

Thanks, thats exactly whats needed!

[Bradley7735]

Well, I haven't done any real testing, but...

I'm playing allies vs the AI in the long campaign (dec7th to 1946)

I haven't added any mines to the ones that start on the map. When I get to about mid to late February (approx 70 - 80 days), the mines have decayed to single digits.

Now, I'm not a big mine fan but I think that if mines degrade to approx zero in 70 days in a friendly port, it's too fast. I think the calculation is more like 2%, not the 1% the manual states.

Of course, I don't really know how long mines lasted historically. I know the sea can be tough on anything in it. Maybe these are realistic decay rates?

Just thought I'd post what I saw.

bc

[Damien Thorn]

I loved mines in UV so I think you can guess my views on having mine decay. [:@]


1 What is the rate of decay in a non-friendly hex? Shallow water and deep water?

2 What ever happened to the air-dropped mines we heard about when WitP was in development? There is no air mission to drop mines. Did that get cut from development?