Ship's jump shadow

[Carlobrand]

I think we may have touched indirectly on this before, but my memory's fuzzy on details.

A planet has a region in which it influences the transition to jump space, making it risky to jump when too close: the 100-diameter limit. How risky varies from game variant to game variant, but the 100-diameter bit's pretty consistent, as far as I know.

That influence does not seem to be related to gravity. A tiny Size-1 world has a 100-diameter limit, a big Size-A has a 100 diameter limit, even though gravity out at the 100 diameter limit of a size A is about what you'd experience at - I think - about 62 diameters from the Size-1. A tiny planetoid exerts an influence out to 1000 kilometers, according to the venerable Traveller Handbook (presumably we're meaning a planetoid of 10 km diameter). A gas giant has a 100-diameter limit, even though the typical Sol system gas giant's density's about a quarter that of a terrestrial planet. Similarly, a star has a 100-diameter limit but, for sunlike stars, a density around a quarter that of the terrestrials.

So, the influence seems to be a spatial thing: a body's volume exerts a "shadow" into jump space, or perhaps into real space in some manner other than gravity, that influences the ability to transition to jump space (and that can prompt a precipitation out of jump space).

Can one presume ships likewise exert a "shadow"? If the little 10dT fighter's tenaciously matching course with you 50 meters to starboard, are you going to suffer the same penalty as if you were within 10 diameters of a world? Can you evade the pirate by jumping as he approaches to dock and board you, or are you risking a catastrophic misjump? Are there canonical descriptions of ships jumping with other ships close at hand, and do they discuss misjump?

 

[HG_B]

Can one presume ships likewise exert a "shadow"? If the little 10dT fighter's tenaciously matching course with you 50 meters to starboard, are you going to suffer the same penalty as if you were within 10 diameters of a world?

Depends. I use a lower limit of a 10km diameter solid for creating a "shadow" in jump space. Anything less massive doesn't impinge upon jump space. So, IMTU, ships don't matter... There MAY be a rule about this in T4. I don't recall.

 

[aramis]

For terrestrial bodies, it correlates VERY closely to tidal stress.

Peter Trevor's website has a great article on this. http://www.trisen.com/sol/default.asp?topic=10&page=29

 

[Carlobrand]

For terrestrial bodies, it correlates VERY closely to tidal stress.

Peter Trevor's website has a great article on this. http://www.trisen.com/sol/default.asp?topic=10&page=29

Okay, I like it a lot for an IMTU variant, and it even helps a lot with stellar jump shadow (see the other post) - except that it puts me inside the body of Betelgeuse: I'm getting a value of about 0.135 diameters if I follow his math right. The big M's are incredibly ... undense? Well, about as dense as space in close orbit ... but the idea of precipitating out in the middle of one is still vaguely disturbing.

However, given that the typical loaded ship runs a bit less dense than water, the formula's still generating some interesting jump shadows for ships and boats. Question remains: can the ships interfere with each other, whether OTU or via this alternate interpretation?

(Being somewhat weak in the physics department, I am obliged to ask a question that likely has a very simple answer: why does tidal force decrease with cube of distance?)

 

[esampson]

I think we may have touched indirectly on this before, but my memory's fuzzy on details.

A planet has a region in which it influences the transition to jump space, making it risky to jump when too close: the 100-diameter limit. How risky varies from game variant to game variant, but the 100-diameter bit's pretty consistent, as far as I know.

That influence does not seem to be related to gravity. A tiny Size-1 world has a 100-diameter limit, a big Size-A has a 100 diameter limit, even though gravity out at the 100 diameter limit of a size A is about what you'd experience at - I think - about 62 diameters from the Size-1. A tiny planetoid exerts an influence out to 1000 kilometers, according to the venerable Traveller Handbook (presumably we're meaning a planetoid of 10 km diameter). A gas giant has a 100-diameter limit, even though the typical Sol system gas giant's density's about a quarter that of a terrestrial planet. Similarly, a star has a 100-diameter limit but, for sunlike stars, a density around a quarter that of the terrestrials.

So, the influence seems to be a spatial thing: a body's volume exerts a "shadow" into jump space, or perhaps into real space in some manner other than gravity, that influences the ability to transition to jump space (and that can prompt a precipitation out of jump space).

Can one presume ships likewise exert a "shadow"? If the little 10dT fighter's tenaciously matching course with you 50 meters to starboard, are you going to suffer the same penalty as if you were within 10 diameters of a world? Can you evade the pirate by jumping as he approaches to dock and board you, or are you risking a catastrophic misjump? Are there canonical descriptions of ships jumping with other ships close at hand, and do they discuss misjump?

In the CT rules when you are using vectored spaceship combat it works something like this:

Each turn you take the current vector of the ship. The ship can accelerate its entire G-rating in any direction it wishes to go. You modify the existing vector by the acceleration and that gives you a new vector. You then move the ship along that new vector.

This is not mathematically accurate, however. Mathematical accuracy would be to compute the new vector but then to move to a position equal to the old vector modified by only 1/2 of the change in vector.

As an example, if you were moving to the 'left' at '5 squares per turn' and you accelerated 'down' at '4 squares per turn' your new position at the end of the turn would be '5 squares left and 4 squares down' and your new vector would be '5 squares left and 4 squares down per turn'. A more mathematically accurate model would be that at the end of the turn your ship would move '5 squares left and 2 squares down' at the end of the turn even though you would still have the final vector of '5 squares left and 4 squares down'.

Ok, so why am I bringing this up when you are talking about jump shadows? Because I think it is a good illustration that Traveller is, at its core, a game. It is not a space simulation obeying physics as closely as possible. That means that in many cases rules are written in a manner that allows them to behave in a way that is 'correct enough' without getting bogged down with math more complex than it needs to be.

My own personal take is that the 100D limit is one of these rules. It is used as a simple and convenient rule to limit how close a ship can get to a large body without using math formulas that involve densities, division by numbers that are raised to the second or third power (or alternately square or cube roots, if working the other side of the equation), and possibly dealing with shapes that aren't very symmetrical.

If the reason for jump shadows does deal with gravitational forces it is unlikely that any kind of ship is going to generate the gravitational forces necessary to foul up jump simply through their own mass.

Of course that's my own interpretation and it isn't completely held up by some things that have been published, so YMMV.

All of that said, it would probably be incredibly difficult for a small ship to keep itself close enough to cause a problem (and most rules say that only the larger of two object creates a jump shadow on the other anyway, so a tiny ship couldn't prevent a larger ship from jumping simply through it's mass. All it could do is raise the likelihood of misjump).

 

[Vladika]

In the CT rules when you are using vectored spaceship combat it works something like this:

Each turn you take the current vector of the ship. The ship can accelerate its entire G-rating in any direction it wishes to go. You modify the existing vector by the acceleration and that gives you a new vector. You then move the ship along that new vector.

This is not mathematically accurate, however. Mathematical accuracy would be to compute the new vector but then to move to a position equal to the old vector modified by only 1/2 of the change in vector.

As an example, if you were moving to the 'left' at '5 squares per turn' and you accelerated 'down' at '4 squares per turn' your new position at the end of the turn would be '5 squares left and 4 squares down' and your new vector would be '5 squares left and 4 squares down per turn'. A more mathematically accurate model would be that at the end of the turn your ship would move '5 squares left and 2 squares down' at the end of the turn even though you would still have the final vector of '5 squares left and 4 squares down'.

Ok, so why am I bringing this up when you are talking about jump shadows? Because I think it is a good illustration that Traveller is, at its core, a game. It is not a space simulation obeying physics as closely as possible. That means that in many cases rules are written in a manner that allows them to behave in a way that is 'correct enough' without getting bogged down with math more complex than it needs to be.

My own personal take is that the 100D limit is one of these rules. It is used as a simple and convenient rule to limit how close a ship can get to a large body without using math formulas that involve densities, division by numbers that are raised to the second or third power (or alternately square or cube roots, if working the other side of the equation), and possibly dealing with shapes that aren't very symmetrical.

If the reason for jump shadows does deal with gravitational forces it is unlikely that any kind of ship is going to generate the gravitational forces necessary to foul up jump simply through their own mass.

Of course that's my own interpretation and it isn't completely held up by some things that have been published, so YMMV.

All of that said, it would probably be incredibly difficult for a small ship to keep itself close enough to cause a problem (and most rules say that only the larger of two object creates a jump shadow on the other anyway, so a tiny ship couldn't prevent a larger ship from jumping simply through it's mass. All it could do is raise the likelihood of misjump).

Your vector example i.e. " one half " is totally wrong. Vectors, graphically, are summed by adding each and every current and new vector head to tail.

Try it simplified:

4 to the left (current vector) plus 3 down (new vector) = position 4 left & 3 down with a MAGNITUDE of 5 (resultant vector) Note: Review the Pythagorean Theorem as it was used in this example.

Every vector has exactly TWO parts; MAGNITUDE and DIRECTION

Add them head to tail if you can't, or won't, do it mathematically. You'll never go wrong.

There is not an Engineer, or a Physicist, here, or anywhere else that is going to argue the above.

 

[mike wightman]

The thing is you can't magical produce the 3 down. At the end of the turn your vector is 3 down, but during the turn of acceleration you only travel 1.5 down. On the turn of the acceleration the vector arrow should only be half of the drive rating - it's pretty basic physics.

 

[Vladika]

The thing is you can't magical produce the 3 down. At the end of the turn your vector is 3 down, but during the turn of acceleration you only travel 1.5 down. On the turn of the acceleration the vector arrow should only be half of the drive rating - it's pretty basic physics.

From a Newtonian frame of reference my math and science are absolutely correct.

I think you are confusing thrust/force acting on a body with gravitation?

On looking at your original post I may have misunderstood your intent. If you are considering a planetary or stellar gravitational field influencing a passing ship there is merit in what you are/were saying. From a purely thrust of ship point of view my example is correct. That pesky Distance x Time Squared thing is a pain isn't it?

 

[Carlobrand]

...Ok, so why am I bringing this up when you are talking about jump shadows? Because I think it is a good illustration that Traveller is, at its core, a game. ...

Yes, it's a role playing game, and if I can think of it, then the players can think of it - and usually at an inconvenient moment. I'm as capable of a fiat explanation as anyone else, but I prefer to be able to throw out something a bit more science-fiction.

Gravity is fine for an IMTU explanation but it imposes some changes to canon - small worlds depart from the 100 diameter bit. Trevor's idea (presented by Aramis) corrects by doing a cube root thing instead of a square root thing, which seems to work rather well. Question remains: is there a lower limit, and why? When your player tells you he's going to keep the target from jumping away by hugging close in his fighter - and then points out that a 1G free trader can't change vector rapidly enough to put him at risk in his 6G 1-seater at a range of 30 meters, least of all when he has a computer to assist him - do we allow his clever plan to work, respond with some reasoned explanation why only a mass of size X can interfere with jump, or pull fiat out of the air?

 

[tjoneslo]

(Being somewhat weak in the physics department, I am obliged to ask a question that likely has a very simple answer: why does tidal force decrease with cube of distance?)

This Page has a good explanation. Short answer: Tidal forces measure the differential force of gravity across the body. That is if your space ship is orbiting a planet the side further away from the planet experiences a different gravitational force than the side closer to the planet. So calculus says the differential of 1/r^2 is 1/r^3.

 

[HG_B]

do we allow his clever plan to work, respond with some reasoned explanation why only a mass of size X can interfere with jump, or pull fiat out of the air?

I simply say that a mass of <X does not curve normal space deeply enough to influence J-space.

 

[McPerth]

Can one presume ships likewise exert a "shadow"? If the little 10dT fighter's tenaciously matching course with you 50 meters to starboard, are you going to suffer the same penalty as if you were within 10 diameters of a world? Can you evade the pirate by jumping as he approaches to dock and board you, or are you risking a catastrophic misjump? Are there canonical descriptions of ships jumping with other ships close at hand, and do they discuss misjump?

IITR having read somewhre (probably in this same board) that to exert such a shadow, the body must be larger (or at least a significant portion of the mass) than the ship.

So, this little fighter will not affect the ship, but a tanker (usually quite large ships) will, and a scout with a cruiser near will be so hindered, while the cruiser will not.

IDK how canon this might be, but I find it logical enough...

 

[tjoneslo]

Question remains: is there a lower limit, and why? When your player tells you he's going to keep the target from jumping away by hugging close in his fighter - and then points out that a 1G free trader can't change vector rapidly enough to put him at risk in his 6G 1-seater at a range of 30 meters, least of all when he has a computer to assist him - do we allow his clever plan to work, respond with some reasoned explanation why only a mass of size X can interfere with jump, or pull fiat out of the air?

There is for the tidal model, and Peter's article mentions it, but isn't explicitly discussed.

From the article 1pS (picostress) is the where the safe jump distance is defined to be. If your object (say the fighter) doesn't generate that much gravity, it won't matter how close you get.


Starting from 1ps = 2 G M m / r^3 we set r = 1 (if the other mass is inside the jump field it comes along) and m = 1 (for simplicity) and solve for M

pS = 2 G M
ps / 2G = M
3.05E-4 / 2 * 6.67E-11 = M

or M = 2,286,356 kg. or 2,200 metric tons. Not a whole lot. Based upon the MT design system most craft end up at around 10 metric tons per displacement ton, so a 200Dton craft might be sufficient. But that craft has to stay 1 meter or less away. And if you read the lower Jump Difficulty chart, you can still make a Difficult roll to jump safely for ships to 1000 times that mass, or 220kt. For a well armored craft, call it 100,000 dTons. And that's assuming a 1 meter distance. At 1km, the stress even from the destroyer is so small as no longer impact your ability to jump.

Preventing a jump with a ship is difficult but possible.

Note this would make an interesting idea for an asteroid base defense. If you read the original paper on a Dyson sphere, the idea was to have an large number of small orbiting stations. Enough to capture all the sunlight. So take an asteroid (where you base is) and put several hundred 1Mton block of rock or ice in Dyson sphere orbit around it. This would push out the safe jump distance from the base asteroid by (hopefully) sufficient distance to make an early warning when the Vargr raiders come jumping in. You can now also have the cool star wars like flight through the close packed asteroids, playing hide and seek.

The more serious problem with the 6G fighter glued to your hull would be what happens if you do engage the jump drive anyways. Either the jump field encloses the fighter, but you no longer have sufficient jump fuel and or jump drive strength for the combined ship+fighter. Or the field generated by the jump shears off the fighter somewhere in the middle, and now your jump field is contaminated with the remains of the fighter. Neither bodes well for a good outcome.

 

[Carlobrand]

...
Starting from 1ps = 2 G M m / r^3 we set r = 1 (if the other mass is inside the jump field it comes along) and m = 1 (for simplicity) and solve for M

pS = 2 G M
ps / 2G = M
3.05E-4 / 2 * 6.67E-11 = M
...

Okay, lost again.

Article presents the formula F = N M / D^3.
F is the stress, set to 1: the minimum stress to start influencing jump.
M is the mass, which the article is describing in Earth masses.
D is distance in kilometers, same as your r.
I don't know what N is, appears to be a constant: 1,973,990.

How does that relate to your formula: 1ps = 2 G M m / r^3? Where does the 6.67E-11 come in? Sounds familiar, but I'm not thinking well today (three days walking kids around for Halloween events and then Renaissance Festival today).

From the article formula, I'm getting something like dreadnought-size objects to influence something at a meter, which for all practical purposes means ships aren't influencing each other.

 

[tjoneslo]

Okay, lost again.

Article presents the formula F = N M / D^3.
F is the stress, set to 1: the minimum stress to start influencing jump.
M is the mass, which the article is describing in Earth masses.
D is distance in kilometers, same as your r.
I don't know what N is, appears to be a constant: 1,973,990.

How does that relate to your formula: 1ps = 2 G M m / r^3? Where does the 6.67E-11 come in? Sounds familiar, but I'm not thinking well today (three days walking kids around for Halloween events and then Renaissance Festival today).

From the article formula, I'm getting something like dreadnought-size objects to influence something at a meter, which for all practical purposes means ships aren't influencing each other.

G is the universal gravitational constant. The formula F = 2 G M m / r^3 is the original formula. In my formula r is in meters.

The N in the formula from the article seems to be G * (Vland Mass in Kg) / (Vland Radius in Km) in order to derive the values for the tables in planetary radius and masses.

I didn't used the formula in the article, but revered to the base formula because I wasn't looking for planetary scale masses or distances.

 

[aramis]

IITR having read somewhre (probably in this same board) that to exert such a shadow, the body must be larger (or at least a significant portion of the mass) than the ship.

So, this little fighter will not affect the ship, but a tanker (usually quite large ships) will, and a scout with a cruiser near will be so hindered, while the cruiser will not.

IDK how canon this might be, but I find it logical enough...

It's in T5. It's canon.

 

[Carlobrand]

G is the universal gravitational constant. The formula F = 2 G M m / r^3 is the original formula. In my formula r is in meters.

Ancient rusty memories triggering - that's the derivative of the standard F=GMm/R^2. For a ship 100 diameters from Earth, that's 2*(6.674×10−11)*(5.97219*10^24)/(1274100000^3), in which F comes out to 3.85 something something x10-13. Question then is what mass generates the same F when r=1. And the answer is ... absurd. Why are my answers coming out wrong??? Seems to be doing right for planets, seems to deal with density differences the same, seems to give the results the gentleman's getting in that link when I play with planetary and solar masses. Maybe its the business of not considering the jumping ship's mass, in which case McPerth has the right of it.

Another example of why I gave up physics and went into psychology in college. I love the stuff, but I don't seem to have the right circuits upstairs. :nonono:

 

[Carlobrand]

I just had the wildest thought: build a dyson sphere around a star, a brobdingnagian jump drive for it, then using the star for power to jump the star itself. It'd have to be an Annic Nova style drive - can't see a way to draw that much hydrogen from the star for a jump without destabilizing the star, nor any way to get enough hydrogen anywhere but the star itself. But, boy howdy, that would be one to make Grandfather look like a primitive.

 

[tjoneslo]

Ancient rusty memories triggering - that's the derivative of the standard F=GMm/R^2. For a ship 100 diameters from Earth, that's 2*(6.674×10−11)*(5.97219*10^24)/(1274100000^3), in which F comes out to 3.85 something something x10-13. Question then is what mass generates the same F when r=1. And the answer is ... absurd. Why are my answers coming out wrong???

I get 3.83E-13 N/m, which matches your answer. At 1 m I get 0.002 kg (or 2 grams about the size of a large paint chip), which says my original calculation is wrong. Probably because I assumed the initial 1pS value was correct (it's not for this calculation, the units are wrong).

This gives an large question of how much of your craft has to be affected by the tidal forces. When dealing with planetary scale masses, it's pretty obvious the whole of your ship either is or is not affected. But for smaller objects the question is less obvious. Does the 2 gram paint chip at 0.01m prevent your ship from jumping safely? If so, it presents an interesting idea for a jump prevention weapon.

I wouldn't think so, the jump drive is capable of compensating for little changes like this over the whole of the jump field. (within limits).

A 100dTon sphere is a little less than 14 meters across. Assume we need the the 3.83E-13 N/m at the 14m mark from the object (i.e 14m radius) so the gravitational distortion encompass the entire ship. This calculates out to about 8kg (7.87kg).

This may be part of the reason for the 100dTon lower limit on a starship size. The Vilani caution about not building star ships smaller than this because it's too likely that some piece of space debris would wander too close to your ship just as it went to jump and causes instability in the field. 8kg is large enough to be easily spotted.

A 100dTon starship is about 1,000,000 kg, with a minimum safe distance of 700M. Which may sound like a lot, but space is big.

There are two simplifying assumptions here: the central mass is a single point mass, which for a planet at 100D is an reasonable assumption. But a 100dTon starship at 70m is no longer true, and may have significant effects of the calculation.

The second is we're only dealing with two objects (The central mass and the star ship with the jump drive). If you accept the idea that calculation of the tidal forces is a simplification of using General Relativity and calculating the Tensor for space curvature, the interaction of the gravity field of two objects can, as some points, become flattened enough to allow jumping safely.

 

[Rabbit 3]

I just had the wildest thought: build a dyson sphere around a star, a brobdingnagian jump drive for it, then using the star for power to jump the star itself. It'd have to be an Annic Nova style drive - can't see a way to draw that much hydrogen from the star for a jump without destabilizing the star, nor any way to get enough hydrogen anywhere but the star itself. But, boy howdy, that would be one to make Grandfather look like a primitive.

But then again, just how do you go about creating a pocket universe anyway?