Ramming Speed!

[Botcho]

Hello all,

Long time lurker here with a question on how to calculate the damage of a collision between small craft.


I am running my first small craft space combat using the Mongoose core rules. It's been a blast! Heh. Things are comming to a close nicely. The bad guys have been run off and all major ordinance has been expended, the remaining problem is that a 20dt hull fragment is on an intercept course for a private orbital station with no means to maneuver out of the way. The impact will happen next turn. All ships are within close range of one another.

There are three scenarios that may play out next turn.

A 60dt tugboat could ram the hull fragment at maneuver 2

A 6dt midge fighter could ram it at maneuver 5

If those actions fail or prove insufficient the 20dt hull fragment will impact with the orbital station at maneuver 2

The small craft doing the ramming are comming at the fragment head on.

My maths are insufficient to the task of injecting any realism into the situation. Do any of you fine Travellers have thoughts on how to resolve the ships damage in these scenarios?

Thank you for your interest and creative thought.

 

[flykiller]

don't know what you mean by "at maneuver 2" or "at maneuver 6". maneuver 2 (for example) means acceration at 64ft/sec/sec, so the impact speed will be based on how many seconds the acceleration has been underway. for example at m6 after 20 seconds of acceleration the vehicle will be moving at 3840 ft/sec - that's well over half a mile a second. of course that's also half a mile a second relative speed of the fragment against the hull of the incoming vehicle - at 20dtons at half a mile a second the damage to the incoming vehicle will be total, no calculation necessary.

now there is no need for the incoming ship to actually ram the fragment. it very easily can accelerate towards it, then decelerate to match vectors, then gently nudge or bump the fragment to a different course.

 

[Botcho]

There are three scenarios that may play out next turn.

A 60dt tugboat could ram the hull fragment at maneuver 2

A 6dt midge fighter could ram it at maneuver 5

If those actions fail or prove insufficient the 20dt hull fragment will impact with the orbital station at maneuver 2

don't know what you mean by "at maneuver 2" or "at maneuver 6". maneuver 2 (for example) means acceration at 64ft/sec/sec...

Thanks for your input flykiller.

Let's call it like this:

The 20dt ton hull fragment is moving at a steady pace, broke off of devastatingly foiled, rapidly decelerating boarding attempt. It is moving roughly 22784 ft per second. And will intercept the key join on a distributed orbital structure of 1200dt if not dealt with immediately.

The small ships on hand have spent most of their maneuverability getting to close range with the fragment from extreme tangential directions to the fragments course (basically head on) As you may find by their relative speeds, I suspect there is little they can do to other than make minor adjustment to ensure a forcible impact with the fragment.

A 60dt tugboat could ram the hull fragment, head on, at maneuver 2 for 360 seconds (again my maths are pretty weak I'm not sure what to do with 64ft/sec/sec)

A 6dt midge fighter could ram the hull fragment, head on, at maneuver 5 for 60 seconds.

Does that clarify the problem somewhat?

 

[pendragonman]

Let's go completely nerdy:

Figure the joules (or kj, or mj) of energy on impact. Then decide how much damage per joule.

Kinetic energy E= (1/2)m v^2 where m is mass and v is velocity, in Joules.

Two issues: dtons are a volume not a mass and how to figure damage per joule.

 

[Botcho]

Let's go completely nerdy:

Figure the joules (or kj, or mj) of energy on impact. Then decide how much damage per joule.

Kinetic energy E= (1/2)m v^2 where m is mass and v is velocity, in Joules.

Two issues: dtons are a volume not a mass and how to figure damage per joule.

I'll go down that rabbit hole with you as long as you keep supplying the maths. So we need the mass eh? TL-10 all around, only the Midge Fighter has a point of armor. Everything else is industrial grade. Anyone have a trusty conversion of DT to Mass?

 

[mike wightman]

A rough rule of thumb is multiply by 10 if only lightly armoured, by15 if heavily armoured.

 

[Garyius2003]

I don't have my book in front of me, but isn't the Mongoose tug set up for 12G with push pads in the front connected through to the drive pads?

That would seem to be the best bet, make it a very difficult pilot roll to hit the fragment square with it rolling, and difficult to keep it balanced enough for the push to take it clear of the station's cross section.

 

[McPerth]

how to figure damage per joule.

For this it may be useful to remember a nuke does 2d6 damage. IIRC, some time ago I calculated that CT:HG nukes are between 7 and 13 kt warheads (I'll have to digg de exact information in this same board...)

Hello all,

Long time lurker here with a question on how to calculate the damage of a collision between small craft.


I am running my first small craft space combat using the Mongoose core rules. It's been a blast! Heh. Things are comming to a close nicely. The bad guys have been run off and all major ordinance has been expended, the remaining problem is that a 20dt hull fragment is on an intercept course for a private orbital station with no means to maneuver out of the way. The impact will happen next turn. All ships are within close range of one another.

There are three scenarios that may play out next turn.

A 60dt tugboat could ram the hull fragment at maneuver 2

A 6dt midge fighter could ram it at maneuver 5

If those actions fail or prove insufficient the 20dt hull fragment will impact with the orbital station at maneuver 2

The small craft doing the ramming are comming at the fragment head on.

My maths are insufficient to the task of injecting any realism into the situation. Do any of you fine Travellers have thoughts on how to resolve the ships damage in these scenarios?

Thank you for your interest and creative thought.

As I guess the true reason for this collision is to avoid the hull fragment to hit the orbital station, you should take into account that aside from the damage, the collision will put some acceleration to the fragment, so aletring the vector, hopely enough to make the job...

 

[flykiller]

Does that clarify the problem somewhat?

not really. verbal descriptions are difficult to employ properly.

if the tug and fighter are located near the orbital structure then meaningful interception will be difficult. even ramming will produce debris, some with a high remaining original vector towards the orbital structure.

but you seem to say that the boats are already alongside the fragment. if that is the case then (assuming ct) nudging the fragment to a new course should be a trivial matter. I have no idea what mongoose's vector movement rules are, but even so 6dtons/5g/60sec and 60dtons/2g/360sec, whatever exactly that means and whatever the game rules involved, seem more than enough to alter the fragment's trajectory.

 

[McPerth]

not really. verbal descriptions are difficult to employ properly.

if the tug and fighter are located near the orbital structure then meaningful interception will be difficult. even ramming will produce debris, some with a high remaining original vector towards the orbital structure.

but you seem to say that the boats are already alongside the fragment. if that is the case then (assuming ct) nudging the fragment to a new course should be a trivial matter. I have no idea what mongoose's vector movement rules are, but even so 6dtons/5g/60sec and 60dtons/2g/360sec, whatever exactly that means and whatever the game rules involved, seem more than enough to alter the fragment's trajectory.

As I understand the OP, there's no time to match vectors for such delicate curse change. Only in this case I guess the ramming option will be considered.

 

[flykiller]

As I understand the OP

The small craft doing the ramming are comming at the fragment head on.

sometimes I'm just stupid ....

yeah, in this case vector matching would involve moving out towards the incoming fragment, then reversing vector to match it. but 22784fsec combined with the descriptor of "next turn" leaves little opportunity for effective action. 20dtons/22784fsec is quite significant.

let's see. for the fragment, 20dtons x 22784fsec = 455680 "footsec dtons".

for the fighter, 6dtons x m5 x 60 seconds = 57600 "footsec dtons".

for the tug, 60dtons x m2 x 360 seconds = 1382400 "footsec dtons".

it is clear the ramming fighter will not have a major impact on the fragment's kinetic energy vector. the ramming tug however might, as it will have triple the energy of the fragment. however the fragment, a hatch, is quite compact, while the tug's structure will be much more dispersed - the fragment may barrel through the tug like an artillery shell and continue on its way. if the tug is lined up such that the fragment impacts all of its major structures such as maneuver drive and power plant and as many bulkheads as possible, and if the impact is timed such that the fragment hits the tug face-on rather than edge-on, then it could have a major effect on the fragment's kinetic energy vector.

 

[Vladika]

sometimes I'm just stupid ....

yeah, in this case vector matching would involve moving out towards the incoming fragment, then reversing vector to match it. but 22784fsec combined with the descriptor of "next turn" leaves little opportunity for effective action. 20dtons/22784fsec is quite significant.

let's see. for the fragment, 20dtons x 22784fsec = 455680 "footsec dtons".

for the fighter, 6dtons x m5 x 60 seconds = 57600 "footsec dtons".

for the tug, 60dtons x m2 x 360 seconds = 1382400 "footsec dtons".

it is clear the ramming fighter will not have a major impact on the fragment's kinetic energy vector. the ramming tug however might, as it will have triple the energy of the fragment. however the fragment, a hatch, is quite compact, while the tug's structure will be much more dispersed - the fragment may barrel through the tug like an artillery shell and continue on its way. if the tug is lined up such that the fragment impacts all of its major structures such as maneuver drive and power plant and as many bulkheads as possible, and if the impact is timed such that the fragment hits the tug face-on rather than edge-on, then it could have a major effect on the fragment's kinetic energy vector.

In any of the given cases enough deflection may very well occur. This is easily solved using vector mechanics. A glancing blow is all it will take as ALL vectors are additive.

Establish ANY arbitrary reference grid, reduce the occurring vectors into their axial components, sum those components, establish the new resultant vector.

Think of a pool table; a ball striking another at any angle not through the enter of gravity will cause BOTH balls to change vectors relative to one another. Sometimes you put a ball into a pocket, sometimes you don't.

 

[flykiller]

In any of the given cases enough deflection may very well occur.

true. but since 1) the size of the orbital station is not given and 2) time to intercept is not given then it is not possible to say.

in addition the final closing speed between the hatch (22784fsec) and the tug (23040fsec) would be 45824fsec or 8.7 miles per second. achieving an off-angle intercept could not be done "by eye" and would require very serious navigation skills. the only way to guarantee impact would be a straight-on approach, meaning a full face-to-face impact with no deflection.

 

[Garyius2003]

Not trying to derail your train of the game, but Mongoose tugs are 30Dt M14s.

 

[Botcho]

Fantastic reading folks thanks for all the input. I found this http://www.rocketpunk-manifesto.com/2009/08/space-warfare-vi-kinetics-part-1.html useful when wrapping my head around the problem.

Kinetics! I always suspected the devastating power of throwing rocks at one another at speed, but to have it spelled out has been an eye opening experience.

The exact damage caused ended up being moot. The tug, non standard, was a modular ship. In a spectacular display of piloting prowess it's operator slung everything but the main drive component at the fragment significantly changing its course and saving the day. The resulting debris field was still troublesome but nothing a little luck and the various point defense systems couldn't handle.

Thanks for walking me through the various complexities of this problem. I am still interested in a megajoule to "number of dice of damage" conversion rule of thumb. This "Boom Chart" could be a good place to start. http://www.projectrho.com/public_html/rocket/spacegunconvent.php#boom

 

[McPerth]

Fantastic reading folks thanks for all the input. I found this http://www.rocketpunk-manifesto.com/2009/08/space-warfare-vi-kinetics-part-1.html useful when wrapping my head around the problem.

Kinetics! I always suspected the devastating power of throwing rocks at one another at speed, but to have it spelled out has been an eye opening experience.

The exact damage caused ended up being moot. The tug, non standard, was a modular ship. In a spectacular display of piloting prowess it's operator slung everything but the main drive component at the fragment significantly changing its course and saving the day. The resulting debris field was still troublesome but nothing a little luck and the various point defense systems couldn't handle.

Thanks for walking me through the various complexities of this problem. I am still interested in a megajoule to "number of dice of damage" conversion rule of thumb. This "Boom Chart" could be a good place to start. http://www.projectrho.com/public_html/rocket/spacegunconvent.php#boom

Maybe those numbers may help you to convert raw energy (MJoules) to damage tables:

As I've seen this question in several threads already, let me givve you a resume of the calculations I've done, based on the energy absorbtion of the Black Globe capatiors:

In the Black Globe section of combat rules, it’s specified that nuclear missiles inflict 25000 Mw per factohat means a TL 7-12 single missile inflicts 25000 Mw, and one TL 13-20 inflicts 50000 Mw. As Mw are power and we are talking about energy released, and assuming the black globe rules are in Mw turn, they should inflict about 8333 (TL 12-) or 16666 (TL13+) mw hour.-

1 w = 1 joule/second

So, 1 wat hour= 3600 joules

1 kton= 4.184 x 10^12 joules = 1.16222 x 10^9 wat hour=1162 Mw/hour

That puts the TL 12- missile at the 7 kton range, and the TL 13+ on the 14 kton range.

This is calculated in MT numbers, but HG numbers (keeping the 1ep=250 Mw given in several sources) give more or les the same results.

In MT there are no differences among turet and bay missiles, just bays launch more of them . In HG, AFAIK, the missiles themselves are not talked about (no volume nor price is given), so it's difficult to know if bays launch larger missiles or just more of them.

As calculating the kinetic energy is relatively easy, I hope that could help you (and, of course, you're free to ignore it if it doesn't )