TNE Only: Dimensions for a 100 dTon ship in FF&S

[atpollard]

Taking a shot at trying to get the most basic data correct (Length, Diameter, Volume and Surface Area) for two simple and common shapes … a SPHERE and a CYLINDER using the RULES AS WRITTEN. If this basic data is incorrect, then everything else will be wrong.

GIVEN: [FF&S pg 11]

• 100 dTon ship
• Volume (Vol) = 100 x 14 = 1400 cubic meters
• Material Volume (MV) = 6 cubic meters
• Length of Hull (L) = 14 meters

SPHERICAL HULL (because nothing is simpler than a sphere):

• Diameter (D) = L x LM = 14 x 1 = 14 meters
• 1 cm Hull Material Volume (HMV) = MV x MVM = 6 x 1 = 6 cubic meters
• Surface Area (SA) = HMV x 100 = 6 x 100 = 600 square meters.

CHECK SPHERE USING BASIC GEOMETRY:

• Volume (V) = 4/3 x 3.14 x (D/2)^3 = 4.186 x 7^3 = 1436 cubic meters [close enough]
  • [Actual Diameter for 1400 cubic meters = 13.88 meters]
• Surface Area (SA) = 4 x 3.14 x (D/2)^2 = 12.56 x 7^2 = 615.44 square meters [close enough]
  • [Actual Surface Area for 1400 cubic meters = 605.21 meters]

CYLINDRICAL HULL (classic 'Rocket' shape):

• Length (Lcyl) = L x LM = 14 x 2 = 28 meters
• 1 cm Hull Material Volume (HMV) = MV x MVM = 6 x 1.1 = 6.6 cubic meters
• Surface Area (SA) = HMV x 100 = 6.6 x 100 = 660 square meters.
• Diameter (D) = unknown from FF&S … 14 m from the original table???

CHECK CYLINDER USING BASIC GEOMETRY:

• Diameter (D) = 7.98 meters (Radius = 3.99 meters)
• Volume (V) = 3.14 x Lcyl x (D/2)^2 = 3.14 x 28 x (3.99)^2 = 1400 cubic meters
• Surface Area (SA) = 2 x 3.14 x (D/2) x (Lcyl + D/2) = 802 square meters

Hull Weight is based on Surface Area and LOTS of components are based on Surface Area … so 660 sq. m. vs 802 sq.m. is sort of a big deal.

• Is this FF&S Errata ...
• or did I misread something ...
• or did I make a math error?

What are the correct dimensions (Length, Diameter, Surface Area) ​

for a 100 dTon (1400 cu.m.) cylinder in FF&S?​

 

[atpollard]

I played around with one of those TNE Starship Spreadsheets (by someone else) and they, too came up with Surface Areas of 600+ sq.m. rather than 800+ sq.m ... so I didn't make a mistake with FF&S.

I played with an online "Cylinder Calculator" to find the most "efficient" 1400 cubic meter cylinder (Height = Diameter). At H = D = 12.14 meters, the Volume (V) = 1405 cu.m. and the Surface Area (SA) = 695 sq.m.

Any other Cylinder (longer and narrower or shorter and fatter) will have a LARGER surface area for a 1400 cu.m. volume.
Thus the FF&S LM and MVM is broken for a Cylinder craft and requires errata.

Does anyone have any idea what the correct data or ratios should be for a Traveller Cylinder craft?

 

[infojunky]

Hum... Depends on the streamlining as well.

Historically I did the geometry for surface area with FF&S.

 

[AnotherDilbert]

SPHERICAL HULL (because nothing is simpler than a sphere):

• Diameter (D) = L x LM = 14 x 1 = 14 meters
• 1 cm Hull Material Volume (HMV) = MV x MVM = 6 x 1 = 6 cubic meters
• Surface Area (SA) = HMV x 100 = 6 x 100 = 600 square meters.

CHECK SPHERE USING BASIC GEOMETRY:

• Volume (V) = 4/3 x 3.14 x (D/2)^3 = 4.186 x 7^3 = 1436 cubic meters [close enough]
  • [Actual Diameter for 1400 cubic meters = 13.88 meters]
• Surface Area (SA) = 4 x 3.14 x (D/2)^2 = 12.56 x 7^2 = 615.44 square meters [close enough]
  • [Actual Surface Area for 1400 cubic meters = 605.21 meters]

Using TL-12 armour I get:

I cheat and calculate the dimensions of the sphere directly, rather than using the Hull Size table.
Like you I get 13.9 m and 605 m² rounded numbers.

CYLINDRICAL HULL (classic 'Rocket' shape):

• Length (Lcyl) = L x LM = 14 x 2 = 28 meters
• 1 cm Hull Material Volume (HMV) = MV x MVM = 6 x 1.1 = 6.6 cubic meters
• Surface Area (SA) = HMV x 100 = 6.6 x 100 = 660 square meters.
• Diameter (D) = unknown from FF&S … 14 m from the original table???

CHECK CYLINDER USING BASIC GEOMETRY:

• Diameter (D) = 7.98 meters (Radius = 3.99 meters)
• Volume (V) = 3.14 x Lcyl x (D/2)^2 = 3.14 x 28 x (3.99)^2 = 1400 cubic meters
• Surface Area (SA) = 2 x 3.14 x (D/2) x (Lcyl + D/2) = 802 square meters

Even FF&S is simplified... It does not calculate the area of the cylinder, just applies a mod to the sphere numbers.

I get:

• Is this FF&S Errata ...
• or did I misread something ...
• or did I make a math error?

I don't think so.

What are the correct dimensions (Length, Diameter, Surface Area) ​

for a 100 dTon (1400 cu.m.) cylinder in FF&S?​

Undefined. It should be ~28 m somewhere, but I take that more as a "characteristic length" than an absolute length. It could be a diagonal length through the cylinder.
I would draw deck plans as I saw fit, rather than be limited to a single specific cylinder.

 

[atpollard]

If we take the FF&S dimensions (28m long x 8m diameter) as the correct "design intent" since the LM seems reasonable compared to the other FF&S values. We can add a NOTE to FF&S that the Diameter of a cylinder is the Modified Length (28m in this case) divided by 3.5 (for Starship deckplan design).

We KNOW the actual SA is 802 sq. meters. Working the rules backwards 802 / 100 = 8.02, so 8.0 cubic meters is the correct HMV. Since HMV = MV x MVM and the MV for a 100 dTon hull is 6, then 8 = 6 x MVM and MVM = 1.3333 [rather than 1.1] ... use 1.3

So repeating the FF&S Cylinder calculations with MVM = 1.3 ...

• Length (Lcyl) = L x LM = 14 x 2 = 28 meters
• 1 cm Hull Material Volume (HMV) = MV x MVM = 6 x 1.3 = 7.8 cubic meters
• Surface Area (SA) = HMV x 100 = 7.8 x 100 = 780 square meters.
• Diameter (D) = Lcyl / 3.5 = 28 / 3.5 = 8

The Hull area is still slightly underestimated for THIS geometry (780 sq.m vs 802 sq.m) but not below the 695 sq.m. minimum possible area for a 1400 cu.m. cylinder.

If we try again with MVM = 1.4 ...

• 1 cm Hull Material Volume (HMV) = MV x MVM = 6 x 1.4 = 8.4 cubic meters
• Surface Area (SA) = HMV x 100 = 8.4 x 100 = 840 square meters.

The FF&S SA using MVM=1.3 (780 sq.m = -2.7%) was closer to the actual SA (802 sq.m) than the FF&S SA using MVM=1.4 (840 sq.m = +4.7%)

It looks like MVM for a Cylinder needs to be changed from 1.1 to 1.3 in FF&S.
Does anyone see anything that I missed or any other problems this creates?

 

[atpollard]

@AnotherDilbert
In post #2 ... it was YOUR spreadsheet that I used to compare the design with. It confirmed that I didn't make a mistake* with the rules ... your spreadsheet 100dTon Cylinder came out with an impossibly small surface area of 666 sq. m. [just like FF&S 660 sq.m.].

(*... or we both made the same mistake)

 

[AnotherDilbert]

(... or we both made the same mistake)

Using FF&S without any mistake or misunderstanding seems unlikely, but in this simple case I don't think so.

 

[AnotherDilbert]

Does anyone see anything that I missed or any other problems this creates?

You make the Cylinder less desirable for naval vessels, as armour would mass considerably more, requiring larger drives. MVM = 1.3 would make the armour as expensive as the needle that has a superior length for better spinals.

Naval vessels would be those actually worrying about surface area the most.


You now probably need to recalculate all the MVMs...

 

[AnotherDilbert]

Original:



MVM = 1.3:

(I had to go to mass-based drives, driving up the need for fuel so that I don't really need mass-based drives. Still volume-based drives don't work.)


Cost increased from MCr 320 to MCr 360.
Mass decreased due to more required fuel.
Free space decreased from 4600 m³ to 2500 m³.

I guess more ships will use spherical config?

 

[AnotherDilbert]

Perhaps a fairer comparison:
MVM = 1.1, max armour (115) with volume-based drives:



MVM = 1.1, max armour (95) with volume-based drives:


Same cost and volume, considerably less armour value.

 

[atpollard]

You make the Cylinder less desirable for naval vessels, as armour would mass considerably more, requiring larger drives. MVM = 1.3 would make the armour as expensive as the needle that has a superior length for better spinals.

Naval vessels would be those actually worrying about surface area the most.


You now probably need to recalculate all the MVMs...

That's the sort of "unintended consequences" that I was concerned about.

ALL of the ships might be broken for calculating Surface Area, but changing the MVM on all of the ships will have dramatic impacts on a mass-sensitive MD system.

The MVM is really just the ratio of the SA of the shape to the SA of a sphere of equal volume, right?
That means that there really are MINIMUM MVM ratios inherent in each basic geometry ... pyramid, cylinder, cube ... each has a different SA to V ratio ...

... it is late and I have a lot to think about.

 

[atpollard]

One quick FF&S question ...

What difference does LENGTH make?
How important is it to other systems?

 

[AnotherDilbert]

The MVM is really just the ratio of the SA of the shape to the SA of a sphere of equal volume, right?
That means that there really are MINIMUM MVM ratios inherent in each basic geometry ... pyramid, cylinder, cube ... each has a different SA to V ratio ...

Yes.

FF&S applies a heavy dose of simplification...

 

[AnotherDilbert]

One quick FF&S question ...

What difference does LENGTH make?
How important is it to other systems?

Spinals are better.

But no-one needs spinals, as high powered lasers trumps everything. So armour is more important...

 

[atpollard]

Volume (V) = 1400 cubic meters
Surface Area (SA) =

• 16.2 m Pyramid SA = 849 square meters = MVM = 849/605 = 1.403
• 14.1 m Wedge SA = 877 square meters = MVM = 877/605 = 1.450
• 12.1 m High x 12.1 m Dia. Cylinder SA = 693 square meters = MVM = 693/605 = 1.145
• 11.2 m Cube (Box) SA = 753 square meters = MVM = 753/605 = 1.245
• 13.9 m Dia Sphere SA = 605 square meters = MVM = 605/605 = 1.000
• 17.5 m Dia Dome SA = 720 square meters = MVM = 720/605 = 1.190

I don't know exactly what "simple geometric shape" the other choices [Open Frame, Close Structure, Slab] are.
THESE are the absolute minimum MVM values for these shapes. Less than this and there isn't enough hull to create that shape at that volume in any configuration. All these are compact forms (close to sphere or cube proportions) to minimize the MVM.

Looking at the CYLINDER again, it looks like they used the compact Length to Diameter ratio and just rounded down to 1.1 ... That is where the "error" comes from. The tall slender cylinder of the FF&S Modified Length (3.5:1 ratio) is less "efficient" and uses more hull area than the compact (1:1 ratio) form.

It is just a disconnect between the simplified rules for LENGTH and the simplified rules for SURFACE AREA.
I don't know that we really want to change the Length modifiers to create short fat Starships.
At this point, I am content to say: "Yeah, L and SA don't agree." and just move on with life.

 

[Condottiere]

Flying bricks are probably the easiest to calculate, planetoids would be lost in fractal geometry.

Armour with spheres, assuming consistent thickness, are basically a larger sphere Matryoshkaing a smaller inhabitable sphere.

 

[mike wightman]

I find it best to use simple geometry and a sketch of the ship I want to design reduced to simple geometric shapes. Still not completely accurate but better results.
Designing the hull and armouring it could take a couple of hours (I don't use spread sheets and I am too old to try and learn).

 

[Rupert]

I was wandering through the forums, and I saw this old thread and thought "What is the 'cylinder' is actually a cylinder with hemisperical ends?"

After some mucking around with a calculator and spreadsheet, I found that such a shape, when 28m long overall (as FF&S says it should be) would have a radius of 5.205m and an area of 740 m^2, giving MVM = 1.22. Better than the conventional cylinder but still too high.

So I checked what a cylinder with hemispherical ends and MVM = 1.1 would look like. I got overall length = 21m, radius = 5.025m, area = 663m, and volume = 1400 m^3. So such a shape has a length multiplier of 1.5. Not too bad, I think. A pity that all the other shapes end up really short too.

 

[GypsyComet]

One quick FF&S question ...

What difference does LENGTH make?
How important is it to other systems?

It sets the length available for spinal mounts, which scale with installed length.

What are the correct dimensions (Length, Diameter, Surface Area) ​

for a 100 dTon (1400 cu.m.) cylinder in FF&S?​

Only Length is fixed and "correct" for a particular hull Configuration, due to the constraint on spinal weaponry. Surface area will *always* be an approximation for simplicity. Any other physical dimension is in the details. A cylinder could have an oval or hexagonal cross-section, or a complex and/or variable cross-section. Each produces a different value for diameter, and, because it doesn't matter, all are technically correct.

The standard equation for a cylinder, if strictly adhered to, will produce pre-Zho-contact Vargr soup can hulls and nothing else. No room for the Type R's wing root engine mounts, as one obvious example.

 

[atpollard]

Only Length is fixed and "correct" for a particular hull Configuration, due to the constraint on spinal weaponry. Surface area will *always* be an approximation for simplicity. Any other physical dimension is in the details. A cylinder could have an oval or hexagonal cross-section, or a complex and/or variable cross-section. Each produces a different value for diameter, and, because it doesn't matter, all are technically correct.

The standard equation for a cylinder, if strictly adhered to, will produce pre-Zho-contact Vargr soup can hulls and nothing else. No room for the Type R's wing root engine mounts, as one obvious example.

Actually, what I was pointing out is that 660 sq. m. is not enough surface area to enclose 1400 cubic meters in any cylinder ...

I played with an online "Cylinder Calculator" to find the most "efficient" 1400 cubic meter cylinder (Height = Diameter). At H = D = 12.14 meters, the Volume (V) = 1405 cu.m. and the Surface Area (SA) = 695 sq.m.

Any other Cylinder (longer and narrower or shorter and fatter) will have a LARGER surface area for a 1400 cu.m. volume.
Thus the FF&S LM and MVM is broken for a Cylinder craft and requires errata.

Ultimately, I decided that FF&S was "broken" for game balance reasons and greater accuracy in surface area would destroy other areas of the rules (like Armor). It is simply not suitable for simulating greater reality, it is just a system with greater detail and complexity. "Precision without Accuracy" is how I would describe it in Stormwater Modeling at work.