Crashing ships as weapons

[Timerover51]

I don't believe Darrians or Sword Worlds to be willing to use them, even if only out of fear to retaliation (also, the political effects on their allied I and Zhodani would have to be considered).

In theory, the Darrians have the "Star Trigger" so they would not need C-ships, if such ships could be built, for which I have severe doubts, using them would advertise that they do not have a "Star Trigger". The Sword Worlders would have to take the possibility of the Darrians triggering a Nova on them if attacked.

As for the K'Kree, they are the ones more likely to use them against meat eaters, but as this would involve also to exterminate their fellow herbibores in the planet, not sure if they woun't try it, except as a last resort.

Any other herbivores on the planets would be viewed as undesired competition by the K'kree, to be removed along with the carnivores. Besides, if the herbivores have not gotten rid of the carnivores, then clearly they are of an inferior species, to be exterminated as well.

 

[Lycanorukke]

The Sword Worlders would have to take the possible of the Darrians triggering a Nova on them if attacked.

True. Though I wouldn't be surprised in the Swordies had some C-ships loaded with radioactive waste ready to be pointed at Darrian worlds - "You Star Trigger us we C-ship you".

Any other herbivores on the planets would be viewed as undesired competition by the K'kree, to be removed along with the carnivores. Besides, if the herbivores have not gotten rid of the carnivores, then clearly they are of an inferior species, to be exterminated as well.

The K'kree are mentioned several times as having exterminated carnivore species on other worlds, so they may have C-shipped worlds in the past if the world wasn't suitable for them or not worth the trouble of conventional attacks - it does suit their mindset.

However when the Hivers threatened to use 'mind control rays' to turn them into carnivores and vastly superior computer technology (hack the C-ship computer and give it a 'return to sender' command) the K'Kree would back off fast - at least until they had the upper hand.

 

[aramis]

I'm not yet convinced a ship could intercept and redirect an incoming threat.

Remember: if it's small enough, the fireball of your missile will detonate it, spreading it's bits out over a MUCH MUCH larger area. How large depends upon how fast the two are going, relative masses, and how long after impact and shatter before arrival.

On the other hand, a 1m/s deviation on an object with a week left before impact is 604 KM diverted.
If you can put 10 m/s deviation, it's moved 6040 km.

If you deflect it correctly, you can also result in a longer time to impact and thus cause it to go elsewhere.

Suicide a shuttle into it at 2G, and BANG.

Also, remember, any maneuvers it does to avoid impact cause it to miss.

And if you disable it's acceleration, it misses.

You don't need to deflect a "burn-in-to-impact course"... you just need to stop the burn, which makes the extant trajectory off by 0.5 A T_r ²

 

[kilemall]

5:1 time dilation, slowtime people have 5 seconds to fire/react, frac-c ship has 1 second.

 

[Kilgs]

Kinetic Energy:
E_k = m_oc² * {(1 / sqrt [1 - R²]) - 1}, where are R = v/c.

If I assume that 1.0 dton = ~ 10.0 metric tons:

A 100 dton ship masses ~ 1000 mt = 1 * 10⁶ kg​

Therefore:

E_k = (1*10⁶)*(3*10⁸)² * {(1 / sqrt [1 - (0.9)²) - 1} ==>

E_k = 1.165*10²³ J​

Since 1.0 ton of TNT = 4.184 GJ:

E_k = ~ 27.8 Teratons for a 1000 mt mass moving at 0.9c.​

IIRC, the Gigaton range is sufficient to crack a planetary crust and/or alter a planet's orbit in a noticeable way.

Sonuva... you just ruined my entire logic behind complex mathematics. I told my teacher that I would never find a use for it given my career aspirations and that I never wanted to find a use for it!

I stand corrected.

 

[Quint]

Remember: if it's small enough, the fireball of your missile will detonate it, spreading it's bits out over a MUCH MUCH larger area. How large depends upon how fast the two are going, relative masses, and how long after impact and shatter before arrival.

On the other hand, a 1m/s deviation on an object with a week left before impact is 604 KM diverted.
If you can put 10 m/s deviation, it's moved 6040 km.

If you deflect it correctly, you can also result in a longer time to impact and thus cause it to go elsewhere.

Suicide a shuttle into it at 2G, and BANG.

Also, remember, any maneuvers it does to avoid impact cause it to miss.

And if you disable it's acceleration, it misses.

You don't need to deflect a "burn-in-to-impact course"... you just need to stop the burn, which makes the extant trajectory off by 0.5 A T_r ²

My math/geometry isn't nearly good enough for this, but wouldn't there be a increased deviation from the targeted point of impact due to the relative movement of the ship and the planet - given the course of the planet remains the same.

I'm trying to picture this in my head, but I'm not braining well today.

D.

 

[kilemall]

In the course of dealing with the frac-c impactor issue, let's not forget normal battle speed ships ramming ships/stations.

Try this quick resolution I just made up on for size-

In CT/HG terms, I'd say target roll is 10+ with the following DMs-

+ rammer agility

- target agility

+/- target ship size DM

Phasing is during movement, attack can only occur at short range and the rammer having higher agility.

Target ship gets a free round of firing against the incoming rammer(s), and fires only against ships that succeed in the above to-hit roll. If the ramming ship is destroyed or loses agility to hit, the rammer 'misses'.

Damage is 1 hit per 100 tons of rammer ship tonnage at spinal/nuclear weapon rolls (no +6 DM), x (rammer agility-target agility if higher then one).

So a Type S at max agility 2 against an agility 0 target would generate 2 spinal hits.

If there is some question as to whether the rammer would survive such an impact, treat the rammer as having damage inflicted as though the target ship rammed them, with the same agility multiplier.

 

[Timerover51]

Ek = ~ 27.8 Teratons for a 1000 mt mass moving at 0.9c.

Could someone perhaps explain where all of this energy is going to come from for a ship of 1000 dTons, with a maximum fuel payload of 1000 Tons of Liquid Hydrogen?

By my calculations, based on 3 kilotons of energy from the complete fusion of 57 grams of Deuterium, 3 Megatons of energy will require 57 kilograms of Deuterium, so 3 Gigatons will require 57 metric tons of Deuterium. I am assuming that 1000 Gigatons equals a Teraton, so therefore, you need 57,000 metric tons of completely fused Deuterium to generate 3 Teratons of energy.

Then it would appear that you will need 9 times 57,000 metric tons to equal 27 teratons, or 513,000 metric tons of Deuterium, complete fused and used with 100% efficiency in propelling the ship. Now, if my assumption about Teratons is incorrect, it gets even worse.

I submit that accelerating a mass to 0.9 C is not possible.

 

[Carlobrand]

Could someone perhaps explain where all of this energy is going to come from for a ship of 1000 dTons, with a maximum fuel payload of 1000 Tons of Liquid Hydrogen?

By my calculations, based on 3 kilotons of energy from the complete fusion of 57 grams of Deuterium, 3 Megatons of energy will require 57 kilograms of Deuterium, so 3 Gigatons will require 57 metric tons of Deuterium. I am assuming that 1000 Gigatons equals a Teraton, so therefore, you need 57,000 metric tons of completely fused Deuterium to generate 3 Teratons of energy.

Then it would appear that you will need 9 times 57,000 metric tons to equal 27 teratons, or 513,000 metric tons of Deuterium, complete fused and used with 100% efficiency in propelling the ship. Now, if my assumption about Teratons is incorrect, it gets even worse.

I submit that accelerating a mass to 0.9 C is not possible.

It does indeed get even worse. A ton of liquid hydrogen is a dTon volume, so they don't even have so much as a hundred metric tons of fuel, given that they need room for a bridge and engines and stuff. 3 gigatons or so is probably as much as they can manage.

 

[aramis]

My math/geometry isn't nearly good enough for this, but wouldn't there be a increased deviation from the targeted point of impact due to the relative movement of the ship and the planet - given the course of the planet remains the same.

I'm trying to picture this in my head, but I'm not braining well today.

D.

The white line is the orbit.
The red line is the course, assuming continued acceleration.
The blue dot is the point/time of impact with the world on the red course.
The blue X is the intercept.
The green is no change in vector, just kill the acceleration (it doesn't go quite as far)
The fine blue is a sample visualization of the edge of the debris cloud.

If you hit it at X, based upon a projected impact if it continues acceleration, then stopping its engines at X means it's off course - at the time the planet is at the blue dot, it's at the near end of the green portion of the line, rather than being at the impact point.

If you shove it off course, it has to correct for that AND the impact time/place change, as the planet isn't sitting still.

 

[esampson]

. . .Also, every brachistochrone trajectory looks like a normal approach until the halfway point, if the ship fails to flip and burn. By that time, you could have a ship accelerating at 1G for 20 AU (about 2 billion miles or 3 billion km) without anyone really caring. . .

Actually, that is not really true. The issue is this, if you are flying a brachiostone trajectory then it will take a certain amount of time for you to reach the planet (which is in motion). Therefore the 'correct' trajectory will aim you to where that planet will be (or reasonable close) when you finish your deceleration maneuver.

On the other hand if you are flying a suicide dive trajectory you will reach the planet significantly sooner.

Using your 20 AU at 1G as an example a ship on a suicide dive trajectory would reach the planet 89 hours, 54 minutes, and 33 seconds earlier than the ship flying a 'proper' brachiostone trajectory. In that time Earth would have moved nearly 9.7 million kilometers.

Now, you could fly the proper trajectory for the first half so it looks like you are trying to intercept the planet properly and then when you should turn over fly a new vector that will take you into collision with the planet but you are going to have to waste some of your initial energy in order to correct to the new vector which will lessen your impact.

A bigger issue than that, however, is that from as soon as you hit that halfway mark your intent is now clear. I don't have the time to do the exact calculations you would need to fly since you have to make a course correction but if you were to fly an absolutely straight intercept path for the suicide dive the time would be about 217 hours. For a normal brachiostone intercept your turn around point is at around 153 hours. This means that you will be over 64 hours away when it becomes obvious that you are not coming in to land, more than enough time for defenses to destroy and divert the incoming ship.

 

[aramis]

A couple minor corrections to my above posts:

• "Any maneuver to avoid results in a miss"
  should be
  "Almost any maneuver to avoid results in a miss" - there is always a narrow range of deflection that merely delays the impact by requiring a correction in place of a some of the terminal velocity...
• "At 0.1C, Time Dilation is ignorable"
  Well, that's actually T_ref=T_moving/√(1-(V/C)), which at 0.1C, gives us
  T_ref=T_moving/√(1-0.1)
  T_ref=T_moving/√(0.9)
  T_ref=T_moving/0.9486832980505138
  T_moving=T_ref*0.9486832980505138
  So it's losing 6% of its speed by the end. Averaging out a square function is beyond my math (except iteratively), but I suspect it works out to be about a 2% loss of vector...
  It's enough to make serious hit or miss.

And in Re Adam's Assertion that "every brachistochrone trajectory looks like a normal approach until the halfway point,"...

Wrong. Because every brachistrochrone trajectory is aiming for a future position. And because time crossing is a square function.

(Using subscripts to indicate temporal positions in the following)

A course from point A to B is actually a course from A₁ to B₂ not from A₁ to B₁.
Let's say the course is 1,000,000,000 m and acceleration is a mere 10 m/s²
Standard is find the halfway point, flip, and do twice. D=0.5AT² for each leg...
leg is 500,000,000, so 10,000 seconds, for a course length of 20,000 seconds.
For the straight burn course, however, 1000000=0.5AT² the total course is only 14142 seconds. A difference of 5858 seconds.
So given start at A₁, if it's aimed for B₂ it's an impact, but if it's aimed at B₃ it's brachistrochrone. If it's at Earth-orbital speed, 30 km/s, that's 175736 km of difference from B₂ to B₃... If I worked the calculator right, a hair shy of 1.4° difference in course aim.
​

There will be some variance due to other bodies gravities, as well... which makes an impact-possible possible-thrust cone different from a brachistrochrone course's, and once past the midpoint, that very quickly collapses.

Also, if the course and current burn look even close to an impact and current burn is insufficient for a turn-around to stop, you've got grounds to whack them. The 3I will commend you for it, even.

 

[Lycanorukke]

At what point would the Port or local IN ship make the call to 'nuke it'?

Would they attempt to establish contact? Give a wave off? Or do they simply say "Only warning. Change your course to this, stop your engines, and await boarding. You have 10 seconds to comply".

Establishing contact/ascertaining the problem is nice, but given every second is important how long would they delay before firing? Is there a procedure or would it fall under 'commanders discretion'.

 

[Carlobrand]

At what point would the Port or local IN ship make the call to 'nuke it'?

Would they attempt to establish contact? Give a wave off? Or do they simply say "Only warning. Change your course to this, stop your engines, and await boarding. You have 10 seconds to comply".

Establishing contact/ascertaining the problem is nice, but given every second is important how long would they delay before firing? Is there a procedure or would it fall under 'commanders discretion'.

Are we assuming the "starts in the target system" scenario?

 

[esampson]

. . .Well, that's actually T_ref=T_moving/√(1-(V/C)), which at 0.1C, gives us
T_ref=T_moving/√(1-0.1)
T_ref=T_moving/√(0.9)
T_ref=T_moving/0.9486832980505138
T_moving=T_ref*0.9486832980505138
So it's losing 6% of its speed by the end. Averaging out a square function is beyond my math (except iteratively), but I suspect it works out to be about a 2% loss of vector...
It's enough to make serious hit or miss.
[/list] . . .

Actually, the formula is T_ref=T_moving/√(1-(V²/C²)) which is a time dilation of only about .501%. Doing a quick spreadsheet and assuming 6G's of acceleration I broke it down into 1000 increments and averaged the time dilation to .167%.

However, to accelerate to .1C even at 6G's takes such a long time (nearly 6 days) that the time dilation ends up causing a difference of about 14 minutes. Earth would move about 25,000 km in that amount of time which is around 2 times the width of the Earth, so you would likely still need to take relativity into account (though coming in tangent to the planet's orbit could be how you take it into account)

 

[Adam Dray]

Actually, that is not really true. The issue is this, if you are flying a brachiostone trajectory then it will take a certain amount of time for you to reach the planet (which is in motion). Therefore the 'correct' trajectory will aim you to where that planet will be (or reasonable close) when you finish your deceleration maneuver.

Oh, good call. I hadn't put all that together.

Yeah, you're right that a planet could detect a suicide trajectory almost instantly because of that consideration.

 

[Adam Dray]

And in Re Adam's Assertion that "every brachistochrone trajectory looks like a normal approach until the halfway point,"...

Wrong. Because every brachistrochrone trajectory is aiming for a future position. And because time crossing is a square function.

Yeah, you're right. I hadn't put that together in my head.

 

[aramis]

Are we assuming the "starts in the target system" scenario?

If you're over 7 days vector to orbital when you arrive from outsystem, expect to be plasma as fast as missiles can get to you.

The most accessible versions are doing it with either a scoutship or a type A...

If you load nought but fuel in the type A, using Bk2 designs, 80 tons of fuel gets you your J1 and 8 months power. Your error will be ±3000 km and ±16.8 hours. You MUST then, essentially aim for a tangent which is 90° from radius. On earth, that's a chunk of arc about 3.7 degrees, and if I followed the formula right, only a 268 km if you aim for the point exactly 90° from your course given the sime, and ±3000 km - you can hit earth that way, if you get all the dilation math right. You can even be fast enough with the freighter that you come out of jump on a course that's going to impact within minutes.

There are probably regulations prohibiting building demountable tangage that big...

 

[esampson]

If you're over 7 days vector to orbital when you arrive from outsystem, expect to be plasma as fast as missiles can get to you. . .

No. Not at all. I would expect that if you assume a fairly good amount of traffic you probably have misjumps occur a couple of times a year where someone lands that far out, either because a navigator fouled up on their math, a far solar body that isn't tracked causes an issue, or a badly calibrated drive causes the ship to land in the wrong place (remember, just because something can't occur during the rolls players ordinarily make when they travel doesn't mean it never happens, unless you assume nobody in the entire universe ever trips walking down the stairs).

And someone being that far out isn't even really a problem. If they were to try a suicide run you would have over a day from the point where it becomes completely obvious that they don't plan to land to deal with them.

(Also, at 7 days out while a ship would still have a horrible amount of energy unless it is a 6G ship it would only have a small fraction of the speed of light. For a 2G ship a '7 day flight' would be about 1.8 billion kilometers. To cover that distance in a 2G suicide dive would take a little less than 5 days and the ship would have only reached .028 C. It's still enough velocity to be catastrophic but a far cry from the .1 C people are imagining reaching)

 

[aramis]

No. Not at all. I would expect that if you assume a fairly good amount of traffic you probably have misjumps occur a couple of times a year where someone lands that far out, either because a navigator fouled up on their math, a far solar body that isn't tracked causes an issue, or a badly calibrated drive causes the ship to land in the wrong place (remember, just because something can't occur during the rolls players ordinarily make when they travel doesn't mean it never happens, unless you assume nobody in the entire universe ever trips walking down the stairs).

And someone being that far out isn't even really a problem. If they were to try a suicide run you would have over a day from the point where it becomes completely obvious that they don't plan to land to deal with them.

(Also, at 7 days out while a ship would still have a horrible amount of energy unless it is a 6G ship it would only have a small fraction of the speed of light. For a 2G ship a '7 day flight' would be about 1.8 billion kilometers. To cover that distance in a 2G suicide dive would take a little less than 5 days and the ship would have only reached .028 C. It's still enough velocity to be catastrophic but a far cry from the .1 C people are imagining reaching)

You misread my meaning.
More clearly:
If you jump in, and your engines indicate you've mroe than 7 days burn needed to come to rest relative, andyour course indicates you're going to impact the planet, they are not going to care a whit why. THey will vaporize your, declare you a terrorist, and give your passengers posthumous purple hearts...

Out system means out of system, not outer edges.

And if you're spending more than 14 days total trip in n-space, they can tell if you're on a flip & stop or not, absed upon where you're heading.

Most systems, by the way, are only a few G-hours difference in local vector. The worst are a couple G-Days. A misjump isn't likely to give you 7+ G-days difference. And when it does... tough.

The canonical 3I scours worlds because of localized infections and/or threat of use of WMDs. A ship is a WMD, and gets treated as such. Come in on a collision course with a world, and they blow your hide out of their ssky... and do their best to ensure that if any hits, it's spread out as far as possible. Everyone getting a sky-flash tan is far preferable to killing the whole continent.

There's no room in those calcs for nice. Nor for Fogivness. Fail to do the right thing just once, and people die, in droves, so better to err on the side of a few noble sacrifices than allowing even one terror plot to ram a world.

It's not nearly as bad if using the T4 limits on T-plates (ISTR them being in T5, too, but I'm not checking)... but ti's still a ship vs a continent. Ship dies if it even looks slightly threatening.