Crashing ships as weapons

[esampson]

I did. 1/20 the cited energy for Chicxulub, 5 x 10²³ Joules, gave the numbers I provided, not the numbers you provided. 1/20 the lower value gave 1 x 10²². 800 metric tons at 0.5c yielded 9 x 10²¹ joules. But, given the magnitudes involved, a quibble.

Yes, but again, the numbers I had for Cicxulub were 1e+23. Anyway, you're right, it's a quibble.

Yup, the baseball could fracture your skull, stop your heart - but I'm reasonably sure you understood the analogy and are not trying to say the planet behaves like a human body. So, again, quibble.

Correct. Bullets actually exploit the mechanics of the body more than the baseball (bleeding, organ damage, shock). A planet doesn't have internal organs.

Also, the damage cause by delivered energy is considerably more complicated than energy over the cross section. The bullet from a .45 pistol has a higher energy over its cross section than a deer slug fired from a shotgun. Which one inflicts more damage?

Interesting question. How much damage will a grain of sand do if it were fired from a particle accelerator that could shoot grains of sand? Because, really, that's what we're looking at here.

No, it's not. That would be a grain of sand with considerably more energy than the baseball. This is a case where the grain of sand has considerably less energy. Yes, it is travelling faster but because of the mass it still has only 5% of the energy of the baseball. And let's not forget that the grain of sand is not being carried on high speed winds or anything like that. This grain of sand exits whatever is accelerating it, has to cross intervening air, and then impact its target.

I'm not suggesting a massive crater either. Again, grain of sand fired from a particle accelerator. In - through - maybe out, but my knowledge fails once it starts interacting with the crust and lithosphere. I'd suggested a bullet through an apple once, but there's a scale problem that gives a more violent picture than I'm trying to portray. This is more like that 0.5c grain of sand fired through an apple. That scout has a 90 square meter cross sectional area that it's driving the energy through. Certainly as it encounters matter and reacts to impact with air, that cross section is going to change, but at 0.5 c it goes from Karman Line to impact in 6.67 x 10^-4 of a second. My thought is it spends the bulk of its energy within the planet.

Actually, if the ship were to pass through the planet that would mean the planet takes less damage because the ship, which already had far less energy, didn't transfer all of its energy. As you correctly pointed out, a planet isn't a human body. It doesn't go into shock, bleed, or suffer organ failure. Energy transferred is your primary issue. I will agree it isn't the only issue, but it's the primary one.

What happens with the atmosphere, that's another story. If a lot of energy gets transferred to that column of atmosphere, it's going to be rather like a nuke went off.

Absolutely. As I said, just because it's an airburst doesn't mean there's no effect on the planet. What it does mean is that you don't form the massive crater with all the ejecta that causes a global winter. You still have massive damage and loss of life and again, if it appeared before that I was suggesting that the airburst would mean that the suicide attack was nothing more than an annoyance I didn't mean it to. It would still be a very serious event, just not quite as world ending as people were talking about earlier.

I'm getting something like 900 metric tons mass of air in a column 90 square meters in cross section, ground to space. 3/4 of that is in the bottom 11 kilometers. Physics-wise, what happens when an 800 ton solid mass at 0.5c interacts with a 333 ton 89 kilometer tall column of "trace" to "very thin" air? What happens when an 800 ton solid mass at 0.5c interacts with a 667 ton 11 kilometer tall column of air? Does the scout's configuration make the slightest difference? If we call the hull iron, then at 0.5c the atoms in the air are encountering impact energies at 6.5 GeV, if I have the math right.

What happens when the scout ship meets that upper column of air is that it reacts as if it was slammed into by 333 tons travelling at .5c (as per Newton's law of equal and opposite reactions). That means the ship loses a lot of forward momentum (about 40%) and in that fraction of a second it decelerates from .5c to .38c. If the scout's configuration mattered at any point during it's flight it no longer matters because its current configuration is as a formerly scout shaped object.

One big thing to note is that you said that this was an "800 ton solid mass". Actually, it isn't. A scout ship is filled with lots of empty space. 800 tons divided by 1350 cubic meters gives it a density of .59. Asteroids such as the Cicxulub impactor are a great deal more dense.

 

[Carlobrand]

... Asteroids such as the Cicxulub impactor are a great deal more dense.

And a great deal larger. You're suggesting that 4.5 x 10²⁰ joules over 90 square meters - 5 x 10¹⁸ joules per square meters is going to have roughly the same reaction as 9 x 10²¹ joules over 176.7 million square meters (or maybe less, I can't recall what value you were using for Chicxulub's diameter) - 5 x 10¹³ joules per square meter.

I can't help thinking that 1/20 the energy applied to 1/2 millionth the area isn't going behave the same or even similar to an asteroid impact, that it isn't going to make matter traveling at 0.5 c - matter with enough energy to break up the atoms in the air, those air molecules in turn breaking up the atoms of the ship's hull - decide to take a right angle turn and explode outward less than 6.67 x 10-4 of a second after impacting atmosphere, and I haven't actually heard a good explanation for why it would.

 

[JimMarn]

There is a planetary scientist who works for NASA who has an accelerator that fires pellets of different material into different materials.

He has fired stone spheres at high speed into sand, rock, clay, etc. I saw him on a show about how the universe was made or the Earth was made. One of those. Probably on Discovery Science.

He records them at very high speed and then plays them back.

 

[aramis]

There is a planetary scientist who works for NASA who has an accelerator that fires pellets of different material into different materials.

He has fired stone spheres at high speed into sand, rock, clay, etc. I saw him on a show about how the universe was made or the Earth was made. One of those. Probably on Discovery Science.

He records them at very high speed and then plays them back.

his speeds are still non relatavistic...

 

[whartung]

!

his speeds are still non relatavistic...

Sure they are! They're relatively slow!

 

[Carlobrand]

Sure they are! They're relatively slow!

So you'd give them a C-?

 

[esampson]

Everything is C-. It's just a question of 'minus how much?'

 

[Hal]

With a 1000 diameter limitation, you can still game it to accelerate as many times as you have jump fuel for 2000 diameters each pass.

Simply you start accelerating at the 1000 diameter (which is a radius, by the way) limit and you have 2000 diameters of distance to accelerate. By 1000 diameter limit, you clearly use the diameter of a star.

Once you've passed the limit, you jump.

Jump retains the vector, so you just jump back to beginning of the 2000 diameter run limit.

You do this several times. The final jump puts the ship on a course for the target, and it accelerates for the entire trip.

This is more exciting if you have a particularly large star within jump range of your target.

Sure, this'll take a few weeks to pull off, but, hey, you're either committed or you're not.

No time to do the math on how fast a 6G drive can get you going. Note you get less acceleration with each pass, since your speed is up and you're in the window of applicability for less time.

Just getting back into traveller after an absence, but the one thing that came up in your comment is that the jump vector is retained.

Problem is? Retained according to what? What frame of reference are you using?

A while back, I was working out a way to programmatically (using VB.NET) work a way to handle vector (on a 2d surface, not three!!!) description per ONE point of origin, or one reference point.

Think about it, if you want to pretend that CT was a simulation for 3d real space, what has to be true:

A) there has to be a central point in which the big bang originates, and is technically the center of all existence.

B) there has to be relative motion of the galaxies around that big bang point. So the Galaxies are in motion.

C) then there is the primary galaxy in which the Terran/Imperial/Zhodani star systems float about it. It too will have a center of mass for the entire "system" of stars as a galaxy.

D) then there is the central star that is the point of origin for the ship about to enter into jump space. It too has planets in motion, all around a center of mass for the star system in question.

So, what is the vector of the ship relative to? The sun it is in orbit about, and is directly affected by? What is its vector relative to another sun, whose orbit it taking its own sun in a given direction and velocity, along with its own planetary system's vectors - etc.

The ONLY way for your idea to work, is that the ENTIRE jump the ship is making regardless of how many jumps it is making, is on the correct vector 100% of the time for the target world relative to the original big bang moment's location (or center of the universe).

My next post is going to then ask "what happens when..." regards to what really happens to a ship in jump transit and the rest of the universe...

 

[Hal]

Then there is the issue of "time".

Let's take a simple example... A micro jump between Earth and a station out in orbit near Uranus.

The ship leaves Earth's orbit, and transits to the jump point 100.1 diameters away (for safety considerations). It enters jump space at precisely midnight on January First (pick a year, any year for your own mind's sake, but we're only interested in the time it takes to actually jump).

So, what happens when the ship enters jump space? Do the clocks on the ship agree with the clocks on Earth when it enters jump space? Let's say they do. What happens then, when the ship exits jump space near its target point. Do the clocks still agree with each other? If you answer yes, then we're going to hit you with the next problem...

If I jump to where the space station will be in 1 week's journey time, and I spend an extra 10% of the time in jump space, will I still exit at the originally aimed point - or will I miss the planet by the distance it travels in the extra time I spent in jump space? If I exit jump space earlier by up to 10%, will I have arrived further away from the planet because it hasn't moved the distance it was supposed to in 1 week, but in .9 x 1 week?

NONE of this stuff was discussed with CT or even later versions of Traveller that I'm aware of. If, from the point of view of the Universe, all ships spend EXACTLY 1 week in jump space transit, but the INTERNAL clocks can show a variance of 1 week +/- 10% - then the navigation issues go away in a large part because the ship will always exit at the plotted point relative to the external universe's point of time, not the internal universe of Jump Space.

However, if the ship clocks are always in synchronization with the external universe even while in jump space - then the time it takes for a planet to move along its orbit in 10% of a week's time period is not a MINOR distance to travel. It is also not a minor distance for a STAR to travel. It is also not a minor distance for a GALAXY to travel. In theory, as best as I can "imagine", the worst distance any ship can miss its planned exit point by, is equal to the FASTEST velocity and/or vector aspect of the point of view the ship is operating within.

So, if a galaxy is moving at immense speeds, jump variances (time wise) of 168 hours is not a trivial thing.

 

[mike wightman]

The big bang theory does not imply the universe has a centre.

The theory is that spacetime and energy came into existence from a singularity, the space time began expanding and some of the energy "condensed" into matter - imagine blowing up a balloon where the skin of the balloon is spacetime, now draw swirling patterns on the balloon to represent the galactic clusters then continue blowing up the balloon. \notice how the galaxies arr all moving away from each other as spacetime expands.
Within local clusters of galaxies gravity can make galaxies move towards each other.

So relative motion is relative to how your galaxy is moving relative to the galaxies that surround it, because in the grand scheme of things spacetime is continuing to expand, with most evidence pointing to the expansion getting faster.

 

[Hal]

The big bang theory does not imply the universe has a centre.

The theory is that spacetime and energy came into existence from a singularity, the space time began expanding and some of the energy "condensed" into matter - imagine blowing up a balloon where the skin of the balloon is spacetime, now draw swirling patterns on the balloon to represent the galactic clusters then continue blowing up the balloon. \notice how the galaxies arr all moving away from each other as spacetime expands.
Within local clusters of galaxies gravity can make galaxies move towards each other.

So relative motion is relative to how your galaxy is moving relative to the galaxies that surround it, because in the grand scheme of things spacetime is continuing to expand, with most evidence pointing to the expansion getting faster.

However, from WHAT point is the expansion occurring?

Put another way? The smallest unit in a star system is a "body". This body orbits the sun. But the sun is moving and is in orbit around a center of mass for the Galaxy. This center of mass for the Galaxy is in orbit around some gravitational point along with other galaxies. If there is no central orbital point for the galaxies themselves, then they are moving apart from each other - relative to what frame of reference?

No matter how you slice it - there will always be the "last external frame of reference" that is discovered, and thought to be the "center" for the final point of reference (even if it is wrong). Just going up THREE levels (star system, Galaxy, Multiple Galaxies), the issue of refining your vector relative to proper motion and when you exit jump space means that the distance missing your exit point due to time, can provide for a HUGE distance indeed.

Case in point:

What is the vector of a planet relative to the sun it orbits?
What is the vector of the sun relative to the galaxy it orbits in?
What is the vector of the "universe" where multiple galaxies move (assuming it isn't the center of existence)?

Even TWO of the variables above can have some rather extreme velocities involved. Quick search on the web:

"The orbital speed of the Solar System about the center of the Galaxy is approximately 220 km/s. At this speed, it takes around 1,400 years for the Solar System to travel a distance of 1 light-year, or 8 days to travel 1 astronomical unit "

So, let's say you're off by 1% of the 168 hours travel. At 220 km per second, our jump error due to TIME only, is 1,330,000 km. That doesn't even take into account how well you aimed your ship initially.

 

[mike wightman]

There is no singularity point anymore - that's the whole point - there is no centre of the universe because the spacetime that makes up the universe is expanding in every direction, hence the balloon skin model; the centre of expansion of the balloon skin is effectively whichever point you decide to measure from - every one of your pen marks will be moving away from it - choose a different pen make and every other one is moving away from that one too.

Point your telescope in any direction from the Earth and you observe distant galaxies moving away from us - conclusion your observatory is the point of the original singularity.
Now for the hard bit.
Imagine in a distant galaxy five billion light years away, imagine in an insignificant planet in some out of the way system , imagine the telescopes in the observatories of that world. When they look into the sky what do they observe? That distant galaxies are expanding away from them and that they are the point of the singularity.

 

[Hal]

There is no singularity point anymore - that's the whole point - there is no centre of the universe because the spacetime that makes up the universe is expanding in every direction, hence the balloon skin model; the centre of expansion of the balloon skin is effectively whichever point you decide to measure from - every one of your pen marks will be moving away from it - choose a different pen make and every other one is moving away from that one too.

Point your telescope in any direction from the Earth and you observe distant galaxies moving away from us - conclusion your observatory is the point of the original singularity.
Now for the hard bit.
Imagine in a distant galaxy five billion light years away, imagine in an insignificant planet in some out of the way system , imagine the telescopes in the observatories of that world. When they look into the sky what do they observe? That distant galaxies are expanding away from them and that they are the point of the singularity.

A couple of points, and I'll lay this to rest (having made my points specifically).

1) in order to make observations of where a given thing is relative to earth, some astronomers would measure the angle of their observed target at a given time on earth, and one that was 1/2 year later (knowing the diameter of the earth's orbit helping with the calculations.

If all things seem to be moving away from earth, it may be that we need more TIME and a known distance to move (time measured in lifespans greater than a single man, possibly even greater than man's entire existence to date) before we can get a sufficient reference point to accurately measure where those galaxies are moving relative to not only Earth, but to galaxy we're inhabiting.

2) even if we accept that there is no center of existence (which I don't have a problem going with for purposes of this discussion), you still have not answered the point I made with the math I used... If the Earth's sun moves at a speed of 220/km per second as a function of its orbital speed within this galaxy we're in, a change of 1% of 168 hours duration in Jump space, assuming that time moves at the same rate on both the jump space inhabiting starship and the normal space it is returning to - results in a potential miss by 1.33 million kilometers irrespective of where the planet itself will be in the time where it has to travel (which in the case of 1% of 168 hours, is 6,048 seconds. How far will the planet itself move in that time? It depends on its orbital velocity around its own sun for one, but it also has to include the velocity of the sun itself at the very least.

In short? Because CT jump rules don't really handle the realism I spoke of, we can either ignore certain issues because it makes the game more complex, or we can attempt to address those issues today, because they weren't addressed 30 or 40 years ago. <shrug>

 

[mike wightman]

I understand where you are coming from - my solution is to assume the jump space dimensions maintain some sort of relative motion to our galactic central supermassive black hole.

So all you need to do is consider the relative motion of your departure system and your destination system. This can be done in two ways:

assume the maneuver drive is used to match the relative velocity of your destination system - or body within your destination system you wish to jump close to

or

assume the jump drive somehow compensates for the relative velocities of the departure and arrival systems as part of the parameters of the hyperdimensional n body problem you have to solve to initiate the jump and that whatever normal space vector you had pre-jump is carried over to the new system.

 

[whartung]

Whether we're orbiting The Great Black Hole, whether the Big Bang happened "in the middle", or we're all simply marbles in God's Great Game, all of the objects in space are moving, and the movement can be measured relative to any other point that can be measured.

At one point, the Sun orbited the Earth, as it was the center of the Everything. We now have decided that the Earth orbits the Sun. But we also have decided that the Sun is moving relative to other Suns, and have seen images of galaxies that imply that a bunch of their contents orbit some galactic barycenter. What the Galaxies themselves are orbiting, I didn't see mentioned the Hitchhikers Guide to the Galaxy -- the seminal reference.

Canon states that when you leave Jump space, you maintain your vector.

And the question is "Vector relative to what?".

Answer: It's not important. Whatever the vector is, its "maintained". Everything moves relative to each other. Let's say you have a vector of "0" relative to the Sun. When you jump, you STILL have a vector of "0" to the Sun. The detail is that you're now someplace else. And now the important question is what are the vectors of everything ELSE, specifically things near you, relative to the Sun.

I can have a vector of "0" in relation to my bed. And I may be "ported" on to a busy street, where I still have a vector of "0" relative to my bed (and for discussion, the street itself). However, the Bus that is fast approaching, does not have a vector of "0" to my bed or the street and, ergo, has a non-zero vector in relation to me. The problem now is not my vector of "0", its that in my current location, having a vector of "0" in relation to the bed does not do me as much good as it was in the old location given the non-zero vector of the bus, of which I suddenly appeared in front of. Normally, I would not be at this location with this vector.

Survey data documents the relative motion of the major objects in the Galaxy. We "know" how Star A is moving relative to Star B. We also know how the planets around Star A move, as well as the planets around Star B. So, it's straightforward to understand that if I jump out of System A with a vector of "0" relative to the Star, when I arrive at Star B, I will have a relative vector to that Star identical to the relative motion of Star A to Star B.

So given an arbitrary start vector relative to your current location, you can see how the astrogator can predict what your relative vector to the major system objects will be in the new location.

Similarly, you can take advantage of this effect. The accelerating ship problem is simply "shooting the gun" and then moving the bullet to hit the target after the fact. The question is the accuracy of the bullet given the know limitations of jump (notably jump arrival point and time of arrival).

This is why when jumping from Planet A of Star A to Planet B of Star B, there will likely be departure lanes, since most any ship leaving Planet A for Planet B will want to have a favorable vector on arrival, and since they have to accelerate out to 100D of the source planet, they may as well leverage that time to set up their arrival vector relative to the destination planet to be as optimal as possible. So, in simple terms, they'll want to arrive "in front of" the destination planet (i.e. in the leading orbit), while flying toward the planet, so that they can leverage the velocity of the planet to lower their arrival times once in system. It's better to maneuver toward a planet that's racing toward you than to arrive trailing the planet and have to catch it. Since most ships would want to optimize this, you'll see several ships leaving on similar vectors from the originating planet, thus creating traffic lanes.

You can see how using this technique one can make an assumption about where a ship is heading. "All ships take that vector out if they're heading for Regina this time of year.". It's not fact, but a good presumption.

Similarly, you can see that if a ship captain wants to optimize their transit to jump, they simply need to accelerate away from the source planet, so that they get the planets velocity away from them "for free". However, this may not be the optimal arrival vector for their destination. So, that's part of the whole astrogation task to try and manage the flight times more efficiently. Of course, when you're on the run from Space Command, you're arrival vector may not be your primary concern. "We're running, burn baby burn -- 105% on the reactor."

 

[warwizard]

Now to open another worm container, I do not believe in the big bang, rather the scientists have missed a physical property of light and misinterpreted it as the red shift property.
Light is affected by gravity, it bends it and also as it climbs out of a gravity well, it is red shifted, and as it falls into a well, it is blue shifted.
I propose that light also loses energy as it interacts with matter or dark matter as it passes through the billions of light years, on a scale of about 20 billion years that light will degrade and eventually decompose into some as yet unidentified particle, which aggregates and forms normal matter, giving rise to matter in intergalactic space that powers the continual formation of galaxies.

As to the fractional C ship issue. Even if you have a reactionless drive such that you do not need to throw matter out the back, you still need to generate energy equal to your E=M(.nC squared) for your acceleration. Fusion reaction only converts a tiny fraction of the mass of H1 or D2 into energy, giving you only a few grams of energy per ton of LH2, ergo you cannot obtain such velocities due to the limitation of your power sources. If you had 25% of the ship's mass is antimatter and you got a 100 % conversion matter to energy and energy to acceleration, you could reach .5 C. Forty five% Antimatter would be needed to get to 90% C... Just land the damn ship and lose containment on the antimatter, that's a lot easier.

 

[Thot]

But OTU M-Drives don't obey the E=1/2m*v² rule. Though maybe they just should?

Assuming 250 MW per dton of fusion reactor, and taking a regular Free Trader with 4 dtons for the reactor (so an energy output of 1 GJ per second, or 1 GW), the best we could achieve with a 10m/s² (1 G) acceleration would be an increase of speed by 10 m/s that costs no more than our reactor can at most produce. Also, let us use an approximation for the ship's mass derived from MegaTraveller rules, which means our 200 dton ship should have about 2000 tons of mass. E=1/2m*v² applies, which means that your acceleration with a given amount of joules per second (=watts) decreases as your speed increases.

Filling these considerations into a spreadsheet, we get speeds and accelerations like this:

Second	Speed(m/s)	Energy spent this second
0	0,00	0
1	10,00	100.000.000
2	20,00	300.000.000
3	30,00	500.000.000
4	40,00	700.000.000
5	50,00	900.000.000
6	59,16	1.000.000.000
7	67,08	1.000.000.000
8	74,16	1.000.000.000
9	80,62	1.000.000.000
10	86,60	1.000.000.000
...
100	312,25	1.000.000.000
...
200	444,41	1.000.000.000
...
500	705,34	1.000.000.000
...
1,000	998,75	1.000.000.000
...
5,000	2235,51	1.000.000.000
...
10,000	3161,88	1.000.000.000
...
50000	7070,89	1.000.000.000
50001	7070,96	1.000.000.000
etc.

 

[Hal]

Probably the better thing to do is simply place a cap on maximum velocity that can be attained by the maneuver drive. It won't solve all of the problems of course, but it will at least place a limit on the whole thing.

Time to reach a given destination would be based on the distance displaced trying to reach max velocity, the cruising time at max velocity, and the turn over time to decelerate to orbital velocity of the planet you're trying to match velocities with when the time comes.

I think that GURPS SPACESHIPS might have that formula somewhere...

Bugger. The formulas in GURPS SPACESHIPS deal with delta V rather than straight acceleration. Ah well. I'm sure there would be a reasonably easy method for computing the process.

 

[Thot]

Probably the better thing to do is simply place a cap on maximum velocity that can be attained by the maneuver drive. It won't solve all of the problems of course, but it will at least place a limit on the whole thing.

That does indeed seem like a very gameable solution.

[...]I'm sure there would be a reasonably easy method for computing the process.

For all practical purposes, the size of the M-Drive wouldn't matter, only the power from the reactor - and the acceleration from that would decease significantly. But an easy, simplified solution would probably be to take the M-Drive rating, multiply it by a factor, and use that as a cruise speed. A factor of 1000 should do, but then we'll have to adjust the jump distance if we want the travel times to remain roughly the same..

 

[Hal]

For all practical purposes, the size of the M-Drive wouldn't matter, only the power from the reactor - and the acceleration from that would decease significantly. But an easy, simplified solution would probably be to take the M-Drive rating, multiply it by a factor, and use that as a cruise speed. A factor of 1000 should do, but then we'll have to adjust the jump distance if we want the travel times to remain roughly the same..

I don't think travel time will be all that much impacted by capping max velocity at 10% light speed.

Just back of envelope calculations using values for velocity of light found on the net (186282.3971 miles per second) and values of 1 G acceleration as being 32.1741 feet per second - I get roughly 35 days acceleration to reach 10% light speed at 1G. That's not going to impact on travel time to reach 100 diameters.

What it will impact upon is intra-system travel between planets when not micro-jumping between worlds with large distances to travel. So "boats" will see the worst impact overall...