MT: RM Power Plant - Fusion

[snrdg082102]

I think I found something that may be errata for the TL 14 Fusion Power Plant.

I have a soft cover and an electronic copy of MT: RM both list the following characteristics for a TL 14 Fusion Power Plant.

Power Out = 3 MW
Weight = 3 tons
Price = 200,000 Cr
(Min) Volume = 0.100
Kl/Hour = 0.005
Scale Efficiency = 0.25 (I think this would be right)
1 unit

I checked with the latest Consolidated MT Errata document created by Don McKinney and found that there is a correction for kl/hour that changes 0.005 to 0.0045.

Power Out = 3 MW
Weight = 3 tons
Price = 200,000 Cr
(Min) Volume = 0.100
Kl/Hour = 0.0045
Scale Efficiency = 0.25
1 unit

I pulled out my copy of The Travellers' Digest #3 and Don McKinney's document for A MegaTraveller Starship Design Example. The Travellers' Digest article, on page 44, shows the following:
Base numbers:

Power Out = 3 MW
Weight = 3 tons
(Min) Volume = 1
Price = 200,000 Cr
Scale Efficiency = 3
100,000 units

Using the 100,000 units listed in the example changed the Scale Efficiency from 0.25 to 3.

Power Out = 3 MW
Weight = 3 tons
Price = 200,000 Cr
Volume = 0.1
Scale Efficiency = 3 (Based on power plant volume of 14+)
100,000 Units

However, there is a difference between the Travellers' Digest article which has Volume = 1 p. 44 and MT: RM p. 64 which shows the volume as 0.1. I checked again in the Consolidated MT Errata document to see if I missed anything stating that the Min Volume needed to be changed from 0.1 to 1. No luck so I ran the numbers anyway.

Power Out = 100,000 x 3 x 3 = 900,000 (a match to the article)
Weight = 100,000 x 3 = 300,000 tons (looking good)
Volume = 100,000 x 0.1 = 10,000 kl (nuts my number doesn't march)
Volume = 100,000 x 1 = 100,000 kl
Price = 100,000 x 200,000 = 20,000,000,000 CR (back on track)

Have I totally misunderstood what Min Volume means?

 

[laning]

definition of Minimum Volume

I misunderstood for years what it meant, but Don McKinney's errata set me straight when I finally got it. I know how you feel, trying to make sense of the editorial and copy-edit mistakes in MT books is not always possible, and in quite a few cases not even the outstanding job Don does with the errata is enough to correct things. Anyway, with that preamble out of the way...

Minimum volume means the smallest size power plant that can be constructed at that TL. You cannot construct a power plant smaller than the "minimum volume".

This is rarely relevant when building starships at MT's TL14 or TL15, but for small craft it can be very relevant. For military and civilian vehicles smaller than, say a tank, it can be even more relevant. One might be best advised to use batteries, fuel cells, or internal combustion engines rather than a fusion plant on a vehicle, for instance.

As for looking at the rest of your post on this and trying to help, I just don't have the brain power at the moment to concentrate on it. (Literally. I have a brain injury that makes concentration a very hit or miss proposition. A disability, but not crippling, thank the fates. I will try to remember to revisit this post this week when all my mental spark plugs are maybe firing better, and see if I can contribute. Something sounded funny about your number for Scale Efficiency, but I didn't look at the rules yet and memory definitely does not serve well enough to use as a reference.

Nice job on checking all the source material carefully before posting a question, and nice job on presenting the relevant material so responsibly and professionally. Just think, if everyone on the Intarwebs was as good a netizen as you, we might still have usenet newsgroups that work. That would be pretty neat. :->

--Laning, reader of entrails and prophet of a murky future.

 

[snrdg082102]

Late morning (PDT) laning,

I appreciate any and all help with my questions. Of course there are times when an answer comes in and I forgot all about the question.

Yep, Don McKinney has down a great service at pulling together one source document to help make a good system better. Unfortunately, I am the exception to the rule. After reading the errata I thought I had a clue about Min Volume. Then I checked the design article only to discover I was still clueless. Even the simple recap in your reply hasn't really helped

For the most part I agree that at most tech levels the Min Volume is rarely relevant for large space/star vessel hulls. A UCP 100 space/star vessel has the same characteristics as a small craft, which means that Min Volume is an important consideration.

I guess my issue is that the Power Plants table is geared toward large scale, with the exception of Price which is in Cr.

Thanks for the positive comment on the my details to support my question. I try to give as much, usually way too much, about my point of confusion or why something might be errata.

I apparently don't have the brain power either, since I have to ask for clarification. So take whatever time is needed I don't have anywhere to go anytime soon.

The Small Scale Efficiency numbers for the characteristics I entered are another thing I'm a bit hazy on.

Here is how I came up with the small scale efficiency for the TL Fusion plant. Min Volume, per the table is 0.1 kiloliters = 100 liters. Looking at the Scale Efficiency table's Small Plants side and using 100 liters:

Plant Volume from .75 kiloliters/750 liters to .501 kiloliters,501 liters reduces the base power output by 67%.

.5 kiloliters/500 liters to .251 kiloliters/251 liters power output is half of the standard output.

Anything from .25 kiloliters/250 liters or smaller generate 25% of a full size power plant.

Since .1 kiloliters/100 liters is smaller than .25 kiloliters/250 liters my conclusion is the Scale Efficiency is x .25.

Going the other way in scale is really simple any fusion plant over 14 kiloliters has a scale efficiency of 3.

Me professional, with an Alfred E. Newman grin (Mad magazine), after admitting to being a sailor. What have I done wrong.:rofl:

Again thanks for the positive comments and for all assistance.

 

[E.D.Quibell]

Min Vol is the smallest you can make that perticular power plant.

So for Anti Matter there isn't one, as there is none stated.

So if we want a 50 Mw power plant at TL17:

50 is 10% of the 500 Mw power plant listed, so the size of a 50 Mw one is 0.8 kl, it weights 0.6 tons and costs 50,000 Cr. It also needs 25 kl of fule for the year.

The minimum size fuel pod is 2 kl (from the errata) 2/25=0.08 so 0.08*365=29.2 or 29.2 days duration at max output.

There is no scale efficency with Anti Matter so we don't have to worry about it.

Does that make sence?

Best regards,

Ewan

 

[E.D.Quibell]

Ok so say we want to try and get a TL14 Fusion power plant that produces 1 Mw of output.

1 Mw is 33% output (33% of 3) of the listed 1 kl TL14 Fuison plant so we want one that is only 0.34 kl. This is ok becasue it's higher than the min volume of 0.1 kl.

However as it's a small scale fusion plant comming in at just under the 0.5 but not the 0.25 volume we have to reduce it's output by multipluing it by 0.5 giving just us 1*0.5=0.5 Mw output.

However this doesn't change the price weight or fuel consumption.

So our 0.5 Mw output fusion power plant still costs 20,000*0.34=6,800 Cr weights 3*0.34=1.02 tons, and uses 0.005*0.34=0.0017 kl/h of fuel.

Make sence?

To get a 1 Mw output power plant at TL14 you would probably need a 0.51 kl plant. This only hits the 0.75 scale efficency so the power out is *0.67. A 0.51 kl plant produces 3*0.51*0.67=1.0251 Mw of power output, but costs 20,000*0.51=10,200 cr, weighs 3*0.51=1.53 tons and consumes 0.005*051=0.00255 kl/h of fuel.

The scale efficencies only apply to their relative types of power plants, so there is no scale effiecncies for anti matter, batteries, fuel cells or sollar cells.

Turbine scale effiecncies cover both turbines, and Internal Combusion scale efficencies cover both combutions engines. Fusion covers all TL of Fusion plants from 9 to 16. Also scale effiecencies are per volume of power plant.

Hope this helps.

Best regards,

Ewan

 

[snrdg082102]

Min Vol is the smallest you can make that particular power plant.

So for Anti Matter there isn't one, as there is none stated.

So if we want a 50 MW power plant at TL17:

50 is 10% of the 500 MW power plant listed, so the size of a 50 MW one is 0.8 kl, it weights 0.6 tons and costs 50,000 Cr. It also needs 25 kl of flue for the year.

The minimum size fuel pod is 2 kl (from the errata) 2/25=0.08 so 0.08*365=29.2 or 29.2 days duration at max output.

There is no scale efficiency with Anti Matter so we don't have to worry about it.

Does that make sense?

Best regards,

Ewan

How did an answer to my Antimatter Fuel pod post at http://www.travellerrpg.com/CotI/Discuss/showthread.php?t=25349 get over here? I mean really the web gremlins are adding to my very confused state.

Ewan thanks for the post which I'll read over several more times.

 

[snrdg082102]

Ewan thanks for showing me the steps and I may be seeing some light at the end of the tunnel.

I think one of the stumbling blocks started with the 100,000 units detailed in the article A MegaTraveller Starship Design Example from The Travellers' Digest #13 on p. 45. To me the word unit means 1 item, in case of 1 unit of power plant appears = 1 kl of volume.

Using the definition of 1 unit = 1 kl means, at least in my fuzzy mind, creates a TL 14 fusion power plant possible provides a power output of 3 MW, a Volume = 1 kl, weight = 3 metric tons, with a price of 200,000 Cr. Fractional units just hard for me to wrap my head around.

Another area which has helped in losing my understanding is reverse engineering from the power draw to the power output.

I'll have to read through an ponder the post responding to my Min Volume and antimatter fuel pod questions.

Thanks again for the reply.

 

[E.D.Quibell]

From the example Starship design they are looking for 1,637,121 Mw of power out.

At TL14 each kl gives 3 Mw, so in the first instance we would be looking at 545,707 kl of power plant (1,637,121/3).

The example says 545,728 kl which is overpowering it a little, but that's fine.

So scale efficency says every fusion plant over 14 kl of volume has 3 times the output, so we divide our power plant size by 3 giving 181,910 kl of power plant (545,728/3).

Working the other way. Any fusion plant at TL14 over 14 kl of volume gives 9 Mw of power per kl (3 MW *3 scale efficency). A 14 kl power plant provides 126 Mw of power. A 181,910 kl power plant provides 1,637,190 w of power.

Make sence?

Best regards,

Ewan

 

[snrdg082102]

From the example Starship design they are looking for 1,637,121 MW of power out.

At TL14 each kl gives 3 MW, so in the first instance we would be looking at 545,707 kl of power plant (1,637,121/3).

The example says 545,728 kl which is overpowering it a little, but that's fine.

So scale efficiency says every fusion plant over 14 kl of volume has 3 times the output, so we divide our power plant size by 3 giving 181,910 kl of power plant (545,728/3).

Working the other way. Any fusion plant at TL14 over 14 kl of volume gives 9 MW of power per kl (3 MW *3 scale efficiency). A 14 kl power plant provides 126 MW of power. A 181,910 kl power plant provides 1,637,190 w of power.

Make sense?

Best regards,

Ewan

The light is getting stronger at the end of the tunnel, I'm just not out of the tunnel yet.

I am too used to designing by entering the number of units installed to get the resulting output. The power plant step in the design example goes from the output to the number of units. Basically, I'm an old dog being shown a new trick and my mind, what little I have, just is not absorbing the method.

Hopefully all the help I'm receiving will finally get my mind wrapped around the concept.

Thanks for the numbers, I may even catch on and like the method in another 5 or 10 centuries.

 

[E.D.Quibell]

I understand (I think ).

In MT you have to work from the Mw needed to power the craft.

Basically you add all the components power together and come up with a figure in Mw. From that figure you then use the power out and scale efficenses to work out how much power plant you need and then knowing the size you work out the fuel needed.

When designing ships you need to have an understanding of the size of your Power plant before adding all the maneuver, wepaons etc otherwise you run out of space.

For the envirment you need about 1Mw per displacemnt ton (13.5 kl volume) this includes everything (basic to inertial comp). You need to know how much your maneuver drive is going to take, and what weapons you are likely to add.

These will give you a rough idea of the size of plant. Wack that rough size in. Work everything else out and then adjust the size of the power plant to fit the exact figure.

Best regards,

Ewan

 

[snrdg082102]

Well, Ewan I was just about finished making a reply your latest post when something I did sent me back to the forum and lost my work.

I'm confused by this line of text from the latest reply

When designing ships you need to have an understanding of the size of your Power plant before adding all the maneuver, weapons etc otherwise you run out of space.

which appears to be in conflict with this line.

Basically you add all the components power together and come up with a figure in MW. From that figure you then use the power out and scale efficiencies to work out how much power plant you need and then knowing the size you work out the fuel needed.

Don't you need to know the power draw of the major systems of Locomotion (J and M Drives), Communications, Sensors, Weapons, Screens, and Environmental?


Basically the required power supply size is reversed engineered from the total power draw of the major systems installed on board.

Number of Units/Volume of power plant = Vessel Power Requirement ÷ Plant Power output ÷ Scale Efficiency.

I think I'll just have to figure away to add a section to do the reverse calculations.

Thanks Ewan and all who have been trying to get me through the tunnel.

 

[McPerth]

Well, Ewan I was just about finished making a reply your latest post when something I did sent me back to the forum and lost my work.

I'm confused by this line of text from the latest reply

which appears to be in conflict with this line.

Don't you need to know the power draw of the major systems of Locomotion (J and M Drives), Communications, Sensors, Weapons, Screens, and Environmental?

Basically the required power supply size is reversed engineered from the total power draw of the major systems installed on board.

Number of Units/Volume of power plant = Vessel Power Requirement ÷ Plant Power output ÷ Scale Efficiency.

I think I'll just have to figure away to add a section to do the reverse calculations.

Thanks Ewan and all who have been trying to get me through the tunnel.

While you're right on the need to know your power needs before adding your PP, Ewan is also right that you must take your power needs in head when designing the power hungry systems.

Even if intuitively, if you don't have your power needs when adding weaponry and maneover to your design, you're heading to find you cannot fill the PP needed (or the fuel for a decent endurance) into your ship, and so having to begin anew. I gues this intuition comes with practice (at least it happened to me, and I had to redesign quite a ship before I learned how to count the power needs before thinking about the PP).

 

[McPerth]

For the envirment you need about 1Mw per displacemnt ton (13.5 kl volume) this includes everything (basic to inertial comp). You need to know how much your maneuver drive is going to take, and what weapons you are likely to add.

Of course, this only applies if you're designing a spaceship/starship. For other vehicles that are thought for planetary use, you usually don't need artificial gravity nor inertial compensors, so cutting environmental needs by more than 80%...

And that brings me another (perhaps unrelated) question: can a ship have inertial compensator if it has no artificial gravity?

Following the letter of the rules, it can, but, as I undersand them, inertial compensators are a fine tuning and upgrading of the grav plates to nullify the inertia from the acceleration you may give to your ship in any direction...

 

[snrdg082102]

While you're right on the need to know your power needs before adding your PP, Ewan is also right that you must take your power needs in head when designing the power hungry systems.

Even if intuitively, if you don't have your power needs when adding weaponry and maneuver to your design, you're heading to find you cannot fill the PP needed (or the fuel for a decent endurance) into your ship, and so having to begin anew. I guess this intuition comes with practice (at least it happened to me, and I had to redesign quite a ship before I learned how to count the power needs before thinking about the PP).

Morning McPerth,

The list that I put together for power using systems came directly form the Design Example in The Travellers Digest #13 and Don McKinney's errata document. My brain wasn't up to thinking at the time.

In the real world there have been a few designs that were flawed because of the power plant. The prototype P-51 Mustang was kind of underpowered before installing the Rolls engine. A lot of the early jets biggest short comings was the power plant/propulsion system.

Naval vessels have also had problems with not enough power to run everything.

Thanks for the reply.

 

[snrdg082102]

Of course, this only applies if you're designing a spaceship/starship. For other vehicles that are thought for planetary use, you usually don't need artificial gravity nor inertial compressors, so cutting environmental needs by more than 80%...

And that brings me another (perhaps unrelated) question: can a ship have inertial compensator if it has no artificial gravity?

Following the letter of the rules, it can, but, as I under sand them, inertial compensator's are a fine tuning and upgrading of the grav plates to nullify the inertia from the acceleration you may give to your ship in any direction...

Gremlins are out enforce, they timed out the server and I lost my post. So here goes a second shot.

Nice question which I don't have an answer for and I haven't been able to locate the rules for inertial compensator's. Can you or someone else provide the page numbers, that way I can stop crying.:rofl: