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How big a bomb

shield

SOC-14 1K
If you planted a large bomb (assume a nuke or similar) against the hull of a decent sized ship (20-40knts) how big would it have to be to a) knock out the ship, b)convert the ship into an expanding cloud of debris?

Ye, some PCs will probably be trying something similiar soon :(

Cheers
Richard
 
I worked out the damage a contact detonating nuke missile would have on a ship in TNE using the demolition rules on page 303.

One stick of dynamite = 0.25kg = 1 DP
therefore 1kg = 6 DP

The smallest nuke in SS3/FF&S has a 0.1kt warhead, which is 100x6 = 600DP.

This equates to C=86.

A 100kt nuke would have C=2739.
 
Sigg

TL11-14, bonded superdense or superdense.

As for armour thickness, I imagine three armoutr levels: unarmed merchant hull, cruiser thickness armour (old high guard armour factor 5ish) anf heavily armoured (armour factor 10+)

cheers
Richard
 
The smallest nuclear pumped x-ray detination laser warhead in TNE is the 10kilotonne.
Assuming this equates to 10,000kilos of TNT this equates to 60,000DP, which results in a concussion/penetration of 866. Doubled if "Tamped" and halved if not. Reading TNE's description of "Tamping" I'd say it would be unlikely to get a tamped charge against a starship hull and I wouldn't consider a hull hit from a Nuclear Det laser warhead missile to be considered tamped either as they're not designed to actually "hit" hulls.
Thus we get a penetration of 433. Divide that figure by the toughness of your targets hull material (in TNE superdense is 14/cm) and you get 30.9. Subtract the thickness in cm and you arrive at the diameter of the hole the explosion made. So if that 10k-20k cruiser had 10cm thick hull armour (which would be AV:140 and perhaps a tad light) you've put a 20.9cm diameter hole in the vessel. Interpret that how you will.

However, if you took your penetration 433 result straight to the damage cards for Brilliant Lances you would have to go to the very end of the critical hits table (the 201+ column) and see that in addition to scoring the maximum 5 major damage results in the one location, 101-120 points, plus 15 major damage results in adjacent locations(433points), you've also scored atleast 4 critical hits on a large (10,000kt+) target.
You've also got a chance to have simply vapourized the target as inflicting any critical hits gives you this chance. Roll number of critical hits or less on a d20. 201+ column says 4 for a large target but remember we've done 433 so you'd be within rights to say 8 critical hits if you chose. 8 or less on a d20 is destinctly possible hmmmmn? And that is the smallest nuke det warhead! No wonder they shyed away from having missile hull hits in TNE.
 
Actually as Andrew has pointed out my Kilotonnes to tonnes to kilos is wrong. The damage would be much, much, well a heck of a lot more
 
Boy, do I now feel stupid
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Andrew, you are quite correct, and the error lies with me copying something across without thinking about it properly.

0.1kt = 100t = 100,000kg.

So the 0.1kt nuke should have C=2739.
 
The 100kt nuke would be 600,000,000DP, which is C=86,602 :eek:

This penetrates 3092cm of superdense armour, assuming an untamped explosion.

The 0.1kt nuke would put a hole in up to 97cm of superdense armour, assuming an untamped explosion.
 
gents

thanks for the replys / thoughts. I think I will use a 100ktn nuke (effectively un-tamped) which should punch through 30m of superdense. Which should go throught the hull, throught the interior (and allowing for the blast / heat to expand through the hull volume) and probably out the otherside.

Considering the fireball will cover a fairly large area of the hull, I think it should burn through the hull over a large area rather than just burning a 1/2 meter hole, as each square metre of hull will be exposed to the same plasma fireball.

Cheers
Richard
 
I knew I was going to do something else...

Battle Rider has counters for heavily armoured ships, and conversion rules for FF&S/Brilliant Lances.
So it should be possible to work out how many cm thickness the various armour factors you've given would be, and then calculate how wide a hole the nuke puts in the hull.

Be back in a little while...
 
Ok then...

Battle Rider gives the following ships that may be relevent:
</font>
  • TL15 CL AV10 = 26cm BSD</font>
  • TL15 ED AV10 = 26cm BSD</font>
  • TL15 BR AV12 = 44cm BSD</font>
  • TL15 BB AV16 = 89cm BSD</font>
  • TL14 BR AV12 = 44cm BSD</font>
  • TL13 BB AV12 = 89cm SD</font>
 
That's about a 30m diameter hole in a TL13 battleship...

and only a 15m diameter hole in a TL14+ armoured ship :eek: :eek:

Now, as to internal system damage...
 
Just how far will the warhead have moved at interception velocities? The delay from a contact fuze to warhead detonation can cause the warhead to flatten against the hull before it can detonate.
Modern bombs have fuze extensions or radar range finders to allow proper standoff distance for detonation. Those do factor in a close approximation of closing velocity, something very difficult to do for an intercept missile. I am not saying it can't be done, just that it will be difficult and prone to errors.
But on the bright side, a hunk of depleted uranium moving at .0001C would still make a mess of anything it touched going the other way.
 
Ship to ship missiles might be the one place where collapsing rounds could work. Due to high velocities, trying to get a fuze to detonate and trigger a nuke at hull contact would still be very difficult with electronic fuzing and impossible with mechanical fuzing. A collapsing round warhead should be able to get into hull contact before going supercritical.
If a hardened penetrator is attached to the missile nose, the collapsing round warhead may be able to start it's fission reaction after passing the outside surface of the hull armor, very effectively tamping the explosive reaction inside the doomed ship. "expand to view of shiny bits of metal in a fast expanding cloud" :eek:
 
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