That's interesting (the rolling a handful of D6s for larger ports) in its own right.

You'd think many A type ports, despite having larger berthing, may well have larger traffic patterns (and if there is a lot of shipping going on, even an A type might have you join a queue for some hours (that other product we discussed had that perspective).

So, having 11D of 0-modified dice should average to about 38-39 available pads. But does that mean you can fly to them without a long traffic/customs delay? Maybe not.

Anyway, you can fix it however you prefer. Maybe one of the authors could be located and asked but the point is small enough that you can just decide and likely nobody will quibble.

The challenge, one supposes, in designing sectors, is that populations would gradually develop (or reduce) and trade would be tied to that (to service the masses and flee dying areas where populations are crashing) and starports would follow trade (the more trade, the more port upgrades you need). To do this sort of modelling is a big project.

But if one used Gurps WTNs (assuming they don't require a Starport as part of WTN calculation in which case you'd have a recursion) to determine overall sector trade flows, you could also let that process largely drive port quality and sizes throughout your sector.

On Tue, Sep 29, 2020 at 1:27 PM Thomas RUX <xxxxxx@comcast.net> wrote:

On 09/28/2020 8:27 PM xxxxxx@gmail.com wrote:

Well, I don't disagree that it will, but the original author seems to think that the table should look (in this instance) linear.

(IMO: 10 landing pads is not even close to enough for a class A starport for a tens of billions of population planet. Maybe it should that number squared (100)).

When I recall looking at the Highport in Dragon #59 (Exonidas Starport), it has a lot of pads. And it fit my idea of a large port. 10 reminds me of what I'd expect at a C Class.

But the authors of this supplement seem to think a Starport upgrade are only worth 1 extra pad per level.

B->A is worth one extra landing pad over B just like B is over C, etc. all the way down according to the authors. I'm just trying for the fix that fits their progression.

The odd part is what if I have a type D port on a pop 2 planet. That's pop digit - 3. Does that mean you have -1 landing pads? (I'm assuming minimum one).

If we look at the one crazy setup in the Trojan Reaches where it's a type A on a pop 1 world, it'd only have 1 pad. The logic folks on the list suggested was 'lots of transient contract labourers plus millions of cargo handling bots'.

I can understand why they coupled population digit and landing pad count, but I think they're way out to lunch generally about the size and scope of larger population planets as far as the landing capacity they need. If you go with the 'heavy trade' model (vs. the 'bring in some antiquities, curiosities, and high value, low bulk goods only'), then you need a lot of pads.

So if I were fixing this inconsistency while maintaining the out of whack numbers the whole scheme generates, I'd still preserve their table and edit the max pad count, but that's totally a subjective thing.

--- but here's what I'd do in MyTU ---

The real world way pad capacities would tend to arise would be based on trade volume. So if you have a trade map and you know what your aggregate trade to a system is (and thus the main world Starport), then you could develop an equation.

It's likely want to be some form of exponential (or at least higher than linear) progression. WTN squared (World Trade Number for GT) would work at the top end, but near the bottom, too many pads for small outposts.

WTN-3 squared might work.

Powerhouse (WTN 15) : 144 pads
Industrial Center (WTN 12): 81 pads
Decent Frontier Capital (WTN 8): 25 pads
Small Mid-Pop Frontier system (WTN 6): 9 pads
Low Pop World: (WTN 3): 1 pad (minimum)

Or something like that. The bigger WTNs should get very large pad counts for their busy ports.

But that'd be MyTU.
My apologies for not including the complete instructions. Here is attempt 2 which I believe clears up some the the confusion I created.

Berthing: "The maximum number of landing bays available in a starport can be set by a simple calculation. The chart shows a basic number modified from the star system's UWP figure (P); thus, the number for a Class A port in a system with an UWP Population Code of A (10) is 11. This is the number of D6 thrown to determine available berths. The Berthing DM is applied as a modifier to each die rolled, with rolls reduced below ), should be 0, equalling 0."

The Starport Contents Table has the following entries:
Berths: Class A = P - 0, Class B = P - 1, Class C/F = P - 2, Class D/G = P - 3, and Class E/H = p - 4.
Berthing DM: Class A = +1, Class B/C/F = 0, and Class D/E/G/H = -1

Again according to the text a Class A port on a world with a Population UWP of A (10) has a total of 11 dice to roll. In order to get the 11 dice the simple calculation should be P + 1 not P - 0.

The Class A port also gets a +1 for each die rolled. The number of berths available based on the example using 11 dice ranges from, I think, die roll of 11 + DM of 11 = 22 to die roll 66 + DM of 11 = 77

Tom Rux

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