HG2 Black Globe and jump drive capacitors

[McPerth]

In any case a little digging or following the process I laid out one can determine that Jump Drive EP = 0.01 x Hull Tons x Jn

This would be the same formula as for the PP EPs, and only half of what page 39 says (2 turns worth for a PP equal to the Jn, so 0.02 x hull tons x Jn). Aside from bing less than what do you need to break off by jumping, it wouldn't fill you capacitors. Only about 1/18 (5.55%) of them would be filled.

E.g.: a 1000 dton jump 5 ship, using your formula, has its JD producing 50 EP (1000 x 0.01 x 5), while it has 25 (1000 x 0.005 x 5) dtons of capacitors, with a maximum capacity of 900 EP.

 

[mike wightman]

900EP = 225GW (assuming Striker/MT conversion of EP to MW)

That's a pretty substantial chunk of energy...

and it's not the jump drive producing the 50EPs it is the power plant. The rest comes from the jump drive when it burns its hydrogen.

The fast burn fusion reactor in the jump drive is generating 850EP almost instantly to activate the jump drive.

The handwavium has to explain what becomes of this energy.

 

[mike wightman]

The EP required for jump is explicitly defined on pg 39 (no reference is needed on following pages - where else does any text reference another page explicitly?).


snipity

Enjoy!

The bit that is contentious is where it states you still need the full allocation of fuel to the jump drive to jump. It restates this in the black globe rules that the jump fuel is still required.

So I would think that if you filled your capacitors to full using the black globe and then tried to jump the extra energy from the jump drive would blow up the ship.

 

[McPerth]

900EP = 225GW (assuming Striker/MT conversion of EP to MW)

That's a pretty substantial chunk of energy...

Sure it is. Any jump transit in Traveller uses more energy than most today's nuclear weapons, and yet only a small quantity of the hydrogen needed would be used for it...

and it's not the jump drive producing the 50EPs it is the power plant. The rest comes from the jump drive when it burns its hydrogen.

I agree. I just used the formula Tom suggested to show that the energy it would give will not be enough for the jump.

 

[BytePro]

This would be the same formula as for the PP EPs, and only half of what page 39 says (2 turns worth for a PP equal to the Jn, so 0.02 x hull tons x Jn). Aside from bing less than what do you need to break off by jumping, it wouldn't fill you capacitors. Only about 1/18 (5.55%) of them would be filled.

E.g.: a 1000 dton jump 5 ship, using your formula, has its JD producing 50 EP (1000 x 0.01 x 5), while it has 25 (1000 x 0.005 x 5) dtons of capacitors, with a maximum capacity of 900 EP.

Nope - reread page 39 - assuming you are reading from HG 1980 edition (mine is from Classic Reprints).

It states 2 turns as effectively the max time for the caps to be charged to support jump but the stated requirement is still the same as for the PP - 'EP required = 0.1MJn.

I explained the rationale for this exhibited by the math - as PP is required for more than just the J-Drive. RW caps do not hold a charge indefinitely - this mimics that as well - so you can't just charge up (to required EP) over a period greater than 2 turns and expect to jump. There won't be enough charge left when you are ready.

NOTE that the text does not say 'per turn'. It just gives the formula required. Suspect are reading into it, especially given how it is presented.

I will try to restate what I have been stating regards discharge rate...

The max EP amount capacitors can store cannot be released in the time requirements for jump.​

Again, for a TL-9-TL-E 1000 dton Jn-1 (Pn 1) starship we have:

dTons for J-Drive = 20 dtons.
dTons of caps = 5 dtons
EP generated by equal rated PP = 10 EP.
EP required for jump (within 1-2 turns) = 10 EP.
Max EP dischargeable per turn = 10 EP.
Total EP storage (before failure) = 180 EP.​

So, 10 EP = 10 EP = 10 EP. Per turn. The capacitors, for purposes of the Jump Drive do not have any extra capacity as they need to be as big as they are to provide for the rate required by the Jump drives.

The fact that the capacitors can store more EP is irrelevant to the Jump Drive requirements - its simply a bonus when talking about Black Globes. If you can get enough EP within 2 turns you can jump (if you have the fuel).

The relationship between max available energy in a given discharge time vs a larger max safe capacity mimics (not to individual scale) how electrical capacitors work in the real world. Hence why this seems so obvious to me - and probably the original authors thought so as well... though it is counter-intuitive if you are not familiar with such.

The bit that is contentious is where it states you still need the full allocation of fuel to the jump drive to jump. It restates this in the black globe rules that the jump fuel is still required.

Yep - but how is that 'contentious'?

If your EP from your PP is being used for other things (like agility, weapons, etc) or is damaged below the normal rating required for charging the capacitors in 2 turns or less, but you have an operating BG that absorbs enough EP into the capacitors, then you can still jump. If you have the jump fuel to support the jump - of course.

What is wrong with that?

So I would think that if you filled your capacitors to full using the black globe and then tried to jump the extra energy from the jump drive would blow up the ship.

Er, if you fill your capacitors to full the ship is destroyed.

If you fill it to just barely, almost, but not quite full... then the J-Drive drains out its EP for the jump and you're free!


NOTE: None of this is about how I 'think' it should work - nor an IMTU thing - I am directly reading the text and applying the maths as I comprehend them. Not arguing a point, either - just trying to help others understand.

 

[mike wightman]

HG is pretty clear - 2 turns output from a power plant rated equal to the jump drive - or at least the jump being performed

It then goes on to put a formula in parenthesis that contradicts this

It must expend energy points equal to two turns output from a power plant whose number is equal to the jump being attempted (EP required =O.OlMJn).

2% of the hull tonnage per jump number worth of EP to initiate the jump according to the text, 1% of the hull tonnage per jump number worth of EP according to the formula.

They both can't be right.

The example that follows clarifies things it is the text that is correct and the formula that is wrong since a power plant 8 jump 5 takes 2 turns.

The formula should read EP required = 0.02MJn

Plus you need to then burn the fuel for the jump.

Marc's jumpspace article and the rules for drop tanks also show that the capacitors can discharge in less than a 20 minute combat turn.

And filling your capacitors does not destroy the ship - the rules state that you have to take more damage on top of having full capacitors

 

[snrdg082102]

Howdy McPerth

This would be the same formula as for the PP EPs, and only half of what page 39 says (2 turns worth for a PP equal to the Jn, so 0.02 x hull tons x Jn). Aside from bing less than what do you need to break off by jumping, it wouldn't fill you capacitors. Only about 1/18 (5.55%) of them would be filled.

I understand the formula to read that the energy points required to make a routine jump is equal to 0.01 times hull tons x the jump number. To make a jump under combat conditions (breaking off by jumping) requires twice the normal EP.

E.g.: a 1000 dton jump 5 ship, using your formula, has its JD producing 50 EP (1000 x 0.01 x 5), while it has 25 (1000 x 0.005 x 5) dtons of capacitors, with a maximum capacity of 900 EP.

I think I see where we differ. On HG page 27 the formula is E = .01MPn, where E =the energy points available (energy output, produced, power out) by the power for use by the computer, weapons, shields, and the maneuver drive. The formula EP required = 0.01MJn to me means that under normal jump conditions the jump drive requires (draws) from the power plant 0.01MJn EP which is calculated on the week traveling to the jump point around the 100 D limit. Breaking off by jumping in combat you have between 20 and 40 minutes to do the same calculations which means certain steps like the normal slow charge time is condensed by requiring 2 x the normal EP.

There has been a couple of issue here that delayed my replies the first was with our power. The front half of the house appeared to be getting half the needed power while the back half was fine.

The other issue is the pump in our water well gave up the ghost.

Two electrical issues while I'm discussing power plants and jump drives could Grandfather be messing with me?

 

[snrdg082102]

Hello BytePro

Nope - reread page 39 - assuming you are reading from HG 1980 edition (mine is from Classic Reprints).

I've got two HG2 1980 LBBs one is a 1st printing the other is the 15th. The one in my big floppy FFE 001 Classic Reprint is the 12th printing. All three are identical. Of course I also have a 1979 HG1 2nd printing which is totally different. I still wish they would have kept the 10-ton bays.

It states 2 turns as effectively the max time for the caps to be charged to support jump but the stated requirement is still the same as for the PP - 'EP required = 0.1MJn.

I explained the rationale for this exhibited by the math - as PP is required for more than just the J-Drive. RW caps do not hold a charge indefinitely - this mimics that as well - so you can't just charge up (to required EP) over a period greater than 2 turns and expect to jump. There won't be enough charge left when you are ready.

NOTE that the text does not say 'per turn'. It just gives the formula required. Suspect are reading into it, especially given how it is presented.

I will try to restate what I have been stating regards discharge rate...

The max EP amount capacitors can store cannot be released in the time requirements for jump.​

​

The formula on page, as I understand the wording, is the EP required (from the power plant) equal to 0.01 x Hull tons x the desired jump range under normal jump conditions.

A 60,000 ton hull making a jump 1 under normal conditions requires the power plant to provide 0.01 x 60,000 x 1 = 600 x1 = 600 EP to charge the capacitors for the jump. The computer calculates the time and amounts dumped into the lanthanum coils and out to the jump network of wiring on the hull which is how I understand the JUmp Space article in JTAS 24.

Again, for a TL-9-TL-E 1000 dton Jn-1 (Pn 1) starship we have:

dTons for J-Drive = 20 dtons.
dTons of caps = 5 dtons
EP generated by equal rated PP = 10 EP.
EP required for jump (within 1-2 turns) = 10 EP.
Max EP dischargeable per turn = 10 EP.
Total EP storage (before failure) = 180 EP.
​

So, 10 EP = 10 EP = 10 EP. Per turn. The capacitors, for purposes of the Jump Drive do not have any extra capacity as they need to be as big as they are to provide for the rate required by the Jump drives.

The fact that the capacitors can store more EP is irrelevant to the Jump Drive requirements - its simply a bonus when talking about Black Globes. If you can get enough EP within 2 turns you can jump (if you have the fuel).

The relationship between max available energy in a given discharge time vs a larger max safe capacity mimics (not to individual scale) how electrical capacitors work in the real world. Hence why this seems so obvious to me - and probably the original authors thought so as well... though it is counter-intuitive if you are not familiar with such.

I'm trying to figure out what is bothering me about this section and how eventually make a solid comment. Of course the light bulb, okay a candle maybe, my illuminate at some point.

Yep - but how is that 'contentious'?

If your EP from your PP is being used for other things (like agility, weapons, etc) or is damaged below the normal rating required for charging the capacitors in 2 turns or less, but you have an operating BG that absorbs enough EP into the capacitors, then you can still jump. If you have the jump fuel to support the jump - of course.

What is wrong with that?

I think that Mike Wightman might think that the jump drive uses the entire fuel load allocated for a maximum jump. A 75,000 ton hul with a J-5 drive requires a full fuel load of 30,000 dtons. From the wording in his quote I get the impression that the 75,000 ton ship has 10,000 tons of jump fuel. During the pre-combat step the ship's skipper declares the ship is breaking off by making a jump of 1 parsec. To complete a normal J-1 trip the ship consumes 7,500 tons of fuel, however the wording could be considered as saying that the jump requires the entire 10,000 tons of fuel. Of course I'm probably out to lunch.

Er, if you fill your capacitors to full the ship is destroyed.

If you fill it to just barely, almost, but not quite full... then the J-Drive drains out its EP for the jump and you're free!

I'll agree with this one.

NOTE: None of this is about how I 'think' it should work - nor an IMTU thing - I am directly reading the text and applying the maths as I comprehend them. Not arguing a point, either - just trying to help others understand.

I too have directly read the text and applied the math as I understand them which has resulted in in some differences. Which is what got the thread started. I do really appreciate the points and view point you have provided. thanks for the help.

 

[Carlobrand]

Hello all,

With the references to the Jump Space article in JTAS 24 and the Annic Nova adventure I've read through both. This is my third attempt at trying to post my comments about them.

Jump Space first since, if I get it right, the article appears to explain why the Annic Nova doesn't use fuel.

Per Jump Space there are several basic components needed for a jump drive to operate listing them in the order found on JTAS 24 pages 35-36:

A. Power source: A jump uses large amounts of energy to open a passage between normal and jump space. The standard Traveller Fusion power plant is best a supplying the energy demand. Alternate methods are Solar power generators which take longer to accumulate the required energy, and anti-matter power systems which are rare, very high tech, and expensive.

B. Energy Storage nodes are usually capacitors or large fast-discharge batteries that hold the generated power until the instant a jump is made.

C. The ship's hull is constructed to withstand the rigors encountered in both normal and jump space. Additionally, the hull contains the network of wiring that maintains the jump field around the ship.

D. A computer is needed to control the precise power requirements and the needed high level of accuracy needed in the calculations for a jump.

E. Lanthanum Jump coils channel the ship's energy and control the jump energy being sent to the jump network of wiring on the hull.

A jump follows a process which is described in JTAS 24 on pages 36-37.

1. During the transit of the ship to the jump point past the 100 D limit calculations for the jump are being refined.

2. The decision to jump is made and the required amount of fuel is fed into the power plant generating the necessary energy for the jump which charges the jump drive capacitors to the required energy capacity.

3. The computer feeds the energy to appropriate jump drive coil sections beginning the jump. The jump duration is fixed based on the specific jump space entered, drive input energy, and other factors.

In one of the posts it was mentioned that the jump drive might be a specialized fusion power plant. From my reading I don't think the jump drive is a fusion power plant, unfortunately I haven't a guess on what type of device it is.
...

The idea of the jump drive containing a high-speed fusion furnace comes out of the Megatraveller Starship Operator's Manual, a Digest Group Publications product but nonetheless considered the canon source for details of starship operations in MegaTraveller. It is also the source of that bit about the jump field standing a meter out from the hull and of people going crazy when they get too close to the jump field.

The basic idea is that it needs a heck of a lot more power to trigger the jump than the typical power plant puts out, and apparently it's cost-effective to have a second power plant designed to burn a whole lot of fuel quickly and inefficiently rather than to carry the size of plant that would be needed to do it more efficiently. Because of the level of energy produced in such a short time, a lot of the fuel ends up going as coolant rather than getting burned, so it's not really possible to draw conclusions about power output and such, but at a guess the power needed would require a traditional plant so big that you couldn't fit it in the jump field produced, ergo a smaller and vastly more inefficient plant with much higher fuel consumption.

There are, not surprisingly, some ill-considered ramifications to the idea. For example, one would think that the introduction of antimatter technology at TL 17 would have led to a radical revision in the way jump drives work. However, they continue to be the clunky things that burn hydrogen - albeit with increasing improvement in fuel efficiency - right through TL 21.

I don't know if the concept's occurred elsewhere before that. Me, until this idea came out, I assumed the hydrogen was being compressed to create a singularity that was in some manner being used to open the way into jumpspace. This evolved from some earlier ideas about naked singularities, but I think the science has left some of those ideas behind.

add: Ooh, there's a new thingie on my banner, 1k! I so proud!

 

[rancke]

Because of the level of energy produced in such a short time, a lot of the fuel ends up going as coolant rather than getting burned, so it's not really possible to draw conclusions about power output and such...

The problem with that explanation is that hydrogen is not a very efficient coolant. With volume being the crucial factor in jump ship design, it would be much more efficient to carry only as much hydrogen as you plan to use to produce energy and water for coolant.

Or so I've been told by people who know more about physics than I do.

I don't know if the concept's occurred elsewhere before that. Me, until this idea came out, I assumed the hydrogen was being compressed to create a singularity that was in some manner being used to open the way into jumpspace. This evolved from some earlier ideas about naked singularities, but I think the science has left some of those ideas behind.

The explanation I came up with a good many years ago was that jumpspace is denser than real space. Getting trust across the dimensional barrier into unmodified jumpspace would be like hitting water at high speed: the water does not have time to get out of your way and it's like hitting rock.

A jump begins with opening an aperture into jumpspace and pumping in a lot of hydrogen, creating a small region[*] of jumpspace that is thin enough for a ship to enter unharmed. Then you activate the jump before the area has had time to dissipate (max 40 minutes, of course ).

[*] I originally referred to this as a jump bubble, but that tends to be confused with the bubble of normal space (the unjump bubble, as it were ) created by the jump grid around a ship. Unfortunately, I can't think of a good synonym for 'bubble'.​

Higher levels of jumpspace are denser and requires more hydrogen to create the less dense area.



Hans

 

[snrdg082102]

Evening Carlobrand,

The Jump Space article in JTAS 24 on pages 34 to 38 is the source I use for CT. Now if I hadn't been directed to the article I would being using MT. My impression from the article the Jump Drive is not a fusion furance, unfortunately I'm not sure on how the magic happens.

I started to re-create the flow charts in the Operator's Manual unfortunately my version of Visio, MS bought them out, doesn't play well with System 7.

The idea of the jump drive containing a high-speed fusion furnace comes out of the Megatraveller Starship Operator's Manual, a Digest Group Publications product but nonetheless considered the canon source for details of starship operations in MegaTraveller. It is also the source of that bit about the jump field standing a meter out from the hull and of people going crazy when they get too close to the jump field.

The basic idea is that it needs a heck of a lot more power to trigger the jump than the typical power plant puts out, and apparently it's cost-effective to have a second power plant designed to burn a whole lot of fuel quickly and inefficiently rather than to carry the size of plant that would be needed to do it more efficiently. Because of the level of energy produced in such a short time, a lot of the fuel ends up going as coolant rather than getting burned, so it's not really possible to draw conclusions about power output and such, but at a guess the power needed would require a traditional plant so big that you couldn't fit it in the jump field produced, ergo a smaller and vastly more inefficient plant with much higher fuel consumption.

There are, not surprisingly, some ill-considered ramifications to the idea. For example, one would think that the introduction of antimatter technology at TL 17 would have led to a radical revision in the way jump drives work. However, they continue to be the clunky things that burn hydrogen - albeit with increasing improvement in fuel efficiency - right through TL 21.

I don't know if the concept's occurred elsewhere before that. Me, until this idea came out, I assumed the hydrogen was being compressed to create a singularity that was in some manner being used to open the way into jumpspace. This evolved from some earlier ideas about naked singularities, but I think the science has left some of those ideas behind.

add: Ooh, there's a new thingie on my banner, 1k! I so proud!

 

[BytePro]

HG is pretty clear ...

As mud perhaps!

...
The example that follows clarifies things it is the text that is correct and the formula that is wrong since a power plant 8 jump 5 takes 2 turns.

The formula should read EP required = 0.02MJn
...
Marc's jumpspace article and the rules for drop tanks also show that the capacitors can discharge in less than a 20 minute combat turn.

And the Black Globe text has max caps discharging at the rate the PP supports... which would agree with the 0.01MJn formula. I didn't really pay attention to the example - nor that for lower model computers no EP would be expended if one weren't firing weapons or maneuvering.

Given the wording and rules that have jump occurring within a turn elsewhere (I'll take your word on that) - I'd tend to what snrdg082102 says about EP requirements being doubled during combat. However, even if the formula is 2x the EP for normal jumps - it still does not change...

The max EP amount capacitors can store cannot be released in the time requirements for jump.​

So having a greater EP storage available than needed for Jump is still not an issue because it is the discharge rate, not the total quantity per dton, that defines how many dtons of capacitors are required.

And filling your capacitors does not destroy the ship - the rules state that you have to take more damage on top of having full capacitors

Oops - you are absolutely correct!

Again though, doesn't change what I was responding to - Jumping wouldn't cause a problem as it would just reduce EP from caps.

...I don't know if the concept's occurred elsewhere before that. Me, until this idea came out, I assumed the hydrogen was being compressed to create a singularity that was in some manner being used to open the way into jumpspace. This evolved from some earlier ideas about naked singularities, but I think the science has left some of those ideas behind.

Sorta likewise - though I presumed the hydrogen was being fed into an evaporating artificial singularity throughout the time of the jump in order to violate the normal laws of physics captain! <waves hands frantically>

...A jump begins with opening an aperture into jumpspace and pumping in a lot of hydrogen, creating a small region[*] of jumpspace that is thin enough for a ship to enter unharmed. ...

Ah, a jump bubble - but for the opposite reason MgT stipulates.

The kink with the jump bubble concept for me is the issue of how this happens with huge volumes of LHyd - for example the 60 k-dtons of a 100 k J-6 starship. Though using all the hydrogen before/at the time of jump does fit with misjumps that can span more than a week...

P.S. snrdg082102 - I recommend http://www.yworks.com/en/products_yed_about.html in place of Visio. I use it on multiple platforms and have found it very stable for a free product (not to mention feature rich).

 

[BytePro]

There has been a couple of issue here that delayed my replies the first was with our power. The front half of the house appeared to be getting half the needed power while the back half was fine.

The other issue is the pump in our water well gave up the ghost.

Ouch. Hopefully they are related and resolving power or fixing/replacing well pump resolves (in U.S. the ones I've come across are 220 - so good candidate to cause partial power loss).

Whatever the case - make sure the water system is properly disinfected if it is opened (with attention to aerators and valves to avoid clogging issues). I constantly come across folks who think that rotten egg smell coming from their taps is okay...

 

[Carlobrand]

The problem with that explanation is that hydrogen is not a very efficient coolant. With volume being the crucial factor in jump ship design, it would be much more efficient to carry only as much hydrogen as you plan to use to produce energy and water for coolant.

Or so I've been told by people who know more about physics than I do.

They are most likely right. Certainly I don't have the physics to evaluate the argument, but my guess is a denser medium would be better at carrying away heat, and water isn't all that hard to come by. Nonetheless, such is the story that Starship Operator's Manual offers, their handwave for how the jump drive can consume so much hydrogen while the power plant consumes so little.

The explanation I came up with a good many years ago was that jumpspace is denser than real space. Getting trust across the dimensional barrier into unmodified jumpspace would be like hitting water at high speed: the water does not have time to get out of your way and it's like hitting rock.

A jump begins with opening an aperture into jumpspace and pumping in a lot of hydrogen, creating a small region[*] of jumpspace that is thin enough for a ship to enter unharmed. Then you activate the jump before the area has had time to dissipate (max 40 minutes, of course ).

[*] I originally referred to this as a jump bubble, but that tends to be confused with the bubble of normal space (the unjump bubble, as it were ) created by the jump grid around a ship. Unfortunately, I can't think of a good synonym for 'bubble'.​

Higher levels of jumpspace are denser and requires more hydrogen to create the less dense area.

It's as good an explanation as any and likely a good deal better than a fusion furnace that consumes fuel at 20,000 times the usual rate.

 

[snrdg082102]

Evening from the Pacific Northwest BytePro

As mud perhaps!


And the Black Globe text has max caps discharging at the rate the PP supports... which would agree with the 0.01MJn formula. I didn't really pay attention to the example - nor that for lower model computers no EP would be expended if one weren't firing weapons or maneuvering.

Given the wording and rules that have jump occurring within a turn elsewhere (I'll take your word on that) - I'd tend to what snrdg082102 says about EP requirements being doubled during combat. However, even if the formula is 2x the EP for normal jumps - it still does not change...

The max EP amount capacitors can store cannot be released in the time requirements for jump.
​

So having a greater EP storage available than needed for Jump is still not an issue because it is the discharge rate, not the total quantity per dton, that defines how many dtons of capacitors are required.


Oops - you are absolutely correct!

Again though, doesn't change what I was responding to - Jumping wouldn't cause a problem as it would just reduce EP from caps.


Sorta likewise - though I presumed the hydrogen was being fed into an evaporating artificial singularity throughout the time of the jump in order to violate the normal laws of physics captain! <waves hands frantically>


Ah, a jump bubble - but for the opposite reason MgT stipulates.

The kink with the jump bubble concept for me is the issue of how this happens with huge volumes of LHyd - for example the 60 k-dtons of a 100 k J-6 starship. Though using all the hydrogen before/at the time of jump does fit with misjumps that can span more than a week...

I'm still pondering the above information before throwing in my thoughts.

P.S. snrdg082102 - I recommend http://www.yworks.com/en/products_yed_about.html in place of Visio. I use it on multiple platforms and have found it very stable for a free product (not to mention feature rich).

Thanks for the link. I picked up Visio because the IT Department I used to work for using the application.

 

[McPerth]

HG is pretty clear - 2 turns output from a power plant rated equal to the jump drive - or at least the jump being performed

It then goes on to put a formula in parenthesis that contradicts this

It must expend energy points equal to two turns output from a power plant whose number is equal to the jump being attempted (EP required =O.OlMJn).

2% of the hull tonnage per jump number worth of EP to initiate the jump according to the text, 1% of the hull tonnage per jump number worth of EP according to the formula.

Nope - reread page 39 - assuming you are reading from HG 1980 edition (mine is from Classic Reprints).

It states 2 turns as effectively the max time for the caps to be charged to support jump but the stated requirement is still the same as for the PP - 'EP required = 0.1MJn.

I explained the rationale for this exhibited by the math - as PP is required for more than just the J-Drive. RW caps do not hold a charge indefinitely - this mimics that as well - so you can't just charge up (to required EP) over a period greater than 2 turns and expect to jump. There won't be enough charge left when you are ready.

NOTE that the text does not say 'per turn'. It just gives the formula required. Suspect are reading into it, especially given how it is presented.

I agree that the formula should explicit that this is the EP required per turn, but I always understood as such, so giving ships with largr PP an advantage, as they can jump in one turn.

I will try to restate what I have been stating regards discharge rate...

The max EP amount capacitors can store cannot be released in the time requirements for jump.​

Again, for a TL-9-TL-E 1000 dton Jn-1 (Pn 1) starship we have:

dTons for J-Drive = 20 dtons.
dTons of caps = 5 dtons
EP generated by equal rated PP = 10 EP.
EP required for jump (within 1-2 turns) = 10 EP.
Max EP dischargeable per turn = 10 EP.
Total EP storage (before failure) = 180 EP.​

So, 10 EP = 10 EP = 10 EP. Per turn. The capacitors, for purposes of the Jump Drive do not have any extra capacity as they need to be as big as they are to provide for the rate required by the Jump drives.

I don't know too much about RW capacitors, but the few things I've read about them they are precisely to store power for quick release.

In this way, one can assume that the limit about discharging them at the same output as the PP allows may be to avoid overloading ship's circuits, not because they cannot be discharged in less time. The JD coils may be thought just to accept this very quick release, so allowing the capaictors to be fully discharged into them in a single turn, and thus sending the ship into jump space.

The fact that the capacitors can store more EP is irrelevant to the Jump Drive requirements - its simply a bonus when talking about Black Globes. If you can get enough EP within 2 turns you can jump (if you have the fuel).

The fact that you mount capacitors with 9 or 18 times the capacity needed, at 4 MCr per Dton of capacitors doesn't seem me irrelevant, when you won't use them and they raise your ship's price and reduce its cargo capacity. Unless you have a BG to absorb energy, they are just a waste.

And those are just the capacitors for your jump drive, regardless if you have a BG or not. That's what makes me think you need more power stored in them than just the 0.01*M*Jn (or 0.02*M*Jn) shown in the formula, and so that the jump drive needs quite more power to initiate transit.

 

[mike wightman]

As mud perhaps!

I think I put my motie in the wrong place - yup, clear as mud.

And the Black Globe text has max caps discharging at the rate the PP supports... which would agree with the 0.01MJn formula. I didn't really pay attention to the example - nor that for lower model computers no EP would be expended if one weren't firing weapons or maneuvering.

Ahhh - you are talking about discharging the capacitor back into the ship to power its systems.

The discharge rate into the jump coils is very much faster - Marc's jumpspace article says so

Given the wording and rules that have jump occurring within a turn elsewhere (I'll take your word on that) - I'd tend to what snrdg082102 says about EP requirements being doubled during combat. However, even if the formula is 2x the EP for normal jumps - it still does not change...

The max EP amount capacitors can store cannot be released in the time requirements for jump.​

The wording is for using the capacitors to power ship systems normally designed to be powered by the power plant.

The jump capacitors are actually meant to feed a very very large amount of energy very quickly into the jump coils.

So having a greater EP storage available than needed for Jump is still not an issue because it is the discharge rate, not the total quantity per dton, that defines how many dtons of capacitors are required.

The discharge rate of energy to the ship systems is limited to the ability of the power plant operated systems to consume power, we are not told, other than in Marc's jump space article, about the discharge rate of the capacitors through the jump coils.

You are assuming the maximum rate of jump coil discharge flow (wow - good technobabble roll) is equal to power plant energy distribution ability, while my argument is that the extra large storage capacity of the capacitors and Marc's article show that the capacitors can actually dump that energy into the jump drive but not back into ship systems.

 

[BytePro]

I'm using the math and text that are in the pages we have been quoting (aside from my own mistakes!)...

We have EP required for jump as = 0.02MJn worst case, or 0.01MJn as explicitly stated for a total (and inline with other sources) in a combat turn.

This is a rate.

Further - for energy 'leaving the ship', the rate of disposing energy from caps is given on pg 42 (which could be in addition to the rate of use of power for the ship - but, that was errata'd away according to a prior post). That max rate is the power plant output per turn (0.01MPn EP).

As to how that rate applies to having more EP capacity than required for jump, I'll try an analogy: I require an electric vehicle that can accelerate to 200 kph in 10 seconds.
It requires power to do so (battery, generator, capacitor, dancing hamsters creating static...).
If the power source can provide for longer than 10 seconds - that is fine, but irrelevant to my requirement. So, if the nature of the power source is that the only practical way I have for the power source to provide enough power within 10 seconds (the required rate) is to be big enough that it can actually provide power for three minutes (18 times that needed )- that's fine, regardless of being over-rated for my time requirements.

Again - max rate of discharge and total storage capacity are two different things for a single given system.

The scalar quality can be the same or different (i.e. 10 EP/1000 seconds and 10 EP total, or 10 EP/1000 seconds and 180 EP total).

In the case of Jump drives, there may be other aspects of the discharge rate that are crucial to why the caps are bigger in EP capacity than they need to be based simply on the JD requirements. In the RW, capacitor systems can be setup exactly this way. The rationale for doing such in the Traveller system could involve thermal requirements and the characteristics of the the non-linear discharge rate. Doesn't really matter exactly - it is what the rules predicate (intended or not).

As to plausibility - those disagreeing have all stated a level of unfamiliarity with capacitors. Capacitors are used for a large number of other applications than simply storing up a charge for quick release. Such as conditioning, filtering, DC blocking, coupling, etc. They have characteristics like self resonance (based on size and material), thermal limits, hysteresis and dielectric leakage, polarity, etc., etc., that need to be accounted for. (It gets real hairy sometimes!) I've scratch built capacitors (extremely large plate); I replace them nearly every month in consumer electronic equipment; and, I spec and install them on rare occasions in industrial environs. So, the concepts presented in Traveller seem natural to me - but explaining them is not.

 

[snrdg082102]

Hello McPerth,

The fact that you mount capacitors with 9 or 18 times the capacity needed, at 4 MCr per Dton of capacitors doesn't seem me irrelevant, when you won't use them and they raise your ship's price and reduce its cargo capacity. Unless you have a BG to absorb energy, they are just a waste.

And those are just the capacitors for your jump drive, regardless if you have a BG or not. That's what makes me think you need more power stored in them than just the 0.01*M*Jn (or 0.02*M*Jn) shown in the formula, and so that the jump drive needs quite more power to initiate transit.

I may be confused by the above so my comment may be out to lunch. My understanding is that the jump drive has the capacitors installed during the time the component is built. The tons and cost of the capacitors included as part of the total drive cost.

Adding 10 tons of jump capacitors as part of a black globe installation, at least to me, is a separate installation from the jump drive.

 

[mike wightman]

We are not talking about real world capacitors though

We are talking about fictional jump drive capacitors whatever they may be and however they may work.

We know they require 0.02MJn EPs to charge in preparation for a jump and that this can be provided in 20 mins.

We also know that the next step varies by rules edition

(In CT'77/MT the jump drive contains a fast burn fusion reactor that feeds a massive burst of energy into the capacitor which then discharges into the jump coil/grid. In Marc's article the ship's power plant throttles up and uses the jump fuel in a few minutes to fully charge the capacitor which then discharges through the jump coil.)

We know that energy may be taken out of the capacitor to power the ship at the same rate the power plant can produce and use energy within the ship.

We do not know how fast the capacitors feed energy into the jump coil - we just know that Marc says it is fast and only takes a few minutes.