I'm having a problem with the asteroid belt mining rules

E4MC

Banded Mongoose
So I'm going through the asteroid belt mining rules in High Guard and I'm trying to calculate the yield of a claim using the table on page 77. I rolled for the Size/Extent (Tons) of the asteroid and the result was "Small Planetoid". Then I rolled for the Resource Yield and the result was "4D %". The sum of the 4D was 14%. But here's where I'm having a problem: The "Small Planetoid" doesn't list its mass in tons so I don't have a number to multiply by 14% to calculate the yield. Am I missing something?

Also, what is the "object radius" column on the table used for?
 
The object radius tells you how big the object is. It is intended mostly as a description but it can be used to calculate the volume of the planetoid. Since "tons" in Traveller is almost always a measurement of volume, you can use the result as the tonnage for the yield calculation.

Note that planetoids are intended to be used to create habitats or ships and will yield an obscene amount of ore if mined. There is nothing in the rules prohibiting you from mining the planetoid and then selling it off as a ship hull as well.
 
E4MC said:
So I'm going through the asteroid belt mining rules in High Guard and I'm trying to calculate the yield of a claim using the table on page 77. I rolled for the Size/Extent (Tons) of the asteroid and the result was "Small Planetoid". Then I rolled for the Resource Yield and the result was "4D %". The sum of the 4D was 14%. But here's where I'm having a problem: The "Small Planetoid" doesn't list its mass in tons so I don't have a number to multiply by 14% to calculate the yield. Am I missing something?

Also, what is the "object radius" column on the table used for?

The object radius column tells you how big of an object it is. It can also tell you the volume of the planetoid. In your case (you rolled an 8 on the size), the average radius is 2D x 100 meters, or an average of 700 meter radius. That is a sphere with a volume of 1,436,760,000 cubic meters. Converted to displacement tons (divide by 13.5) you get 106,426,666.67 displacement tons on average. Of which 14% (or about 14.9 million tons) is your resource yield.

The smallest of the small planetoids on that table still comes in at almost 2.5 million displacement tons.
 
Jeraa said:
E4MC said:
So I'm going through the asteroid belt mining rules in High Guard and I'm trying to calculate the yield of a claim using the table on page 77. I rolled for the Size/Extent (Tons) of the asteroid and the result was "Small Planetoid". Then I rolled for the Resource Yield and the result was "4D %". The sum of the 4D was 14%. But here's where I'm having a problem: The "Small Planetoid" doesn't list its mass in tons so I don't have a number to multiply by 14% to calculate the yield. Am I missing something?

Also, what is the "object radius" column on the table used for?

The object radius column tells you how big of an object it is. It can also tell you the volume of the planetoid. In your case (you rolled an 8 on the size), the average radius is 2D x 100 meters, or an average of 700 meter radius. That is a sphere with a volume of 1,436,760,000 cubic meters. Converted to displacement tons (divide by 13.5) you get 106,426,666.67 displacement tons on average. Of which 14% (or about 14.9 million tons) is your resource yield.

The smallest of the small planetoids on that table still comes in at almost 2.5 million displacement tons.

Thanks, Dick and Jeraa.

Jeraa, I understand everything except where did the 13.5 come from?
 
E4MC said:
Thanks, Dick and Jeraa.

Jeraa, I understand everything except where did the 13.5 come from?

Number of cubic meters in a single displacement ton. At least in some versions of Traveller. Others use 14 cubic meters per displacement ton. Not sure which one MgT 2e uses.
 
Jeraa said:
E4MC said:
Thanks, Dick and Jeraa.

Jeraa, I understand everything except where did the 13.5 come from?

Number of cubic meters in a single displacement ton. At least in some versions of Traveller. Others use 14 cubic meters per displacement ton. Not sure which one MgT 2e uses.

Oh, okay, thanks! That seems really light to me when you consider that a cubic meter of aggregate can weigh between 1 and 3 tons depending on the material.
 
Core said:
A displacement ton displaces a volume equal to one ton of liquid hydrogen. One displacement ton is roughly equal to 14 cubic metres.

Only in MT is a Dt defined to be 13.5 m³ (to be exactly 2 map squares of 1.5 m × 1.5 m × 3 m).
 
E4MC said:
Oh, okay, thanks! That seems really light to me when you consider that a cubic meter of aggregate can weigh between 1 and 3 tons depending on the material.
Displacement tonnage is based on a volume of hydrogen, which tends to be a little bit lighter than rock.
 
The size 7 result is 1 million tons. adjust upwards fot your find.

The key importance for planetoids, as far as Travellers
are concerned, is their suitability as a stellar habitat.
Given the right Tech Level (TL8+), money and resources,
planetoids can be rendered habitable simply by
hollowing-out the inside, installing atmosphere generators
or sealed life support systems akin to those found on
ships, and then either installing gravitic generators or
increasing the planetoid’s spin artificially (with surfacemounted
thrusters, perhaps) to impart gravity. A body the
size of Ceres, with a diameter of 950 kilometres, could,
if rendered habitable in this way, support a population of
several millions in relative comfort.

And then there is:

16 Psyche
As an example of the value of M class asteroids, an
average earth-sized world of TL6 or 7 has an average
iron ore production value in excess of 1,000 million
metric tons. Compare this with an M class asteroid
with a mean diameter of just 1 kilometre: at this size
and density, a 1-kilometre M class asteroid could yield
in excess of 2,000 million metric tons of iron-nickel
ore, which is two to three times the annual production
requirements for the planet. In our own asteroid belt,
the asteroid 16 Psyche is believed to contain 1.7 ×
1019 kilograms of iron-nickel, enough to satisfy present
iron ore production requirements on Earth for several
million years.


So you have enough material to satify a planet for years.
 
Okay. So I started over from the beginning and found a 1,000,000 Ton asteroid composed of Dense Materials (including metals). The Yield is 32%, so 1,000,000 Tons x 32% = 320,000 Tons. I'm assuming Type D material is Dense Metals on the commodity table (I'm not really sure because the rules don't really explain it.) worth Cr50,000 per ton. So the value of the yield is 320,000 Tons x Cr50,000 = Cr16,000,000,000. But because it is unrefined ore, the value of the claim is only 10% of that, or Cr1,600,000,000. So I open a channel with a corporation and they offer 14% of the value. I'm assuming that's 14% of the value of the unrefined ore and not the value of the yield (The rules aren't clear on this) for an immediate cash offer of Cr1,600,000,000 x 14% = Cr224,000,000. Did I do this right or am I missing something?
 
I would argue for a different version.

1 000 000 tons: 32 percent dense metals so 320 000 tons dense metals worth 50 000 a ton or 16 Billion Credits
pg 76. the remaining material of the asteroid is 680 000 tons: 50 % is Common Ore @1000 /ton or 340 million Credits
30% Uncommon ore or 204 000 tons @5000 / ton for 1.02 Billion Credits
And 136 000 tons of waste material of no value that you might be able to melt and form into a slag ship and create your own rock fleet. :p

Total Value of Asteroid is 17.36 Billion Credits.
pg 79 10% offer of total yield of claim, so the offer on the table would be 1.736 Billion Credits.

How a claim is sold depends on the relationship the belter has with those who would buy the material. Also the asteroid’s yet-to-be-mined ore is of a lesser value
to fully processed material, so the base price in any negotiation is 10% the value of the commodity. With an Average (8+) Admin check (1D days, SOC) the
belter can establish a sales channel with a nominated corporation. The belter must provide the estimated market value of the yield and the corporation will
conduct a cursory check of the detailed survey analysis. It then offers the belter 1D + 7% (that is, 8% - 13%, with an average of 10%) of the claim’s value as an immediate cash realisation. This is increased by +1% for every level the belter has in the Broker skill.

So every 1 rank in Broker is worth 173 million extra credits. It pays to study in your off time.

Moral of the story, asteroid mining will create the worlds first trillionaires. :p
Now your example is a bit of a fluke in that you scored a 12+ on the 2D roll to get a D asteroid.

HMMMM... wait a second. As I doublechecked this I saw something odd. How did you determine that the Dense material was actually Dense worth 50 000 and not metallic nickel iron worth 400? There seems to be a gap in the rules.
pg 74
M Class: Metallic-based. Usually nickel-iron, but with the
potential to yield precious and heavy metals depending
on the location within the belt. M class asteroids are
reasonably dense in composition, containing high-grade
sources of metal, along with exotic elements such as
radioactives and super-dense metals. This makes them
valuable for manufacturing industries and larger examples
can be turned into starship hulls at any orbital A class
starport or similar grade of dockyard. M class asteroids
are most common in the N zone of the belt and are
targets for most prospectors.

Scanning on pg 75 shows dense objects including metals, Crystalline, including ice, and exotics on a roll of 12.
Detailed survey andm logistics on pg 75 shows metallic asteroids, Carbon asteroids or ice.

To be honest I do not see how you get Crystals at 20 000 / ton, dense at 50 000 or radioactives.

With a nickel iron find the value of the 32% drops to 128 million and 1.488 Billion total. This would be a 148 million offer for the claim.
 
Thanks, PsiTraveller. How did you come up with the composition of the remaining material of the asteroid?

"Now your example is a bit of a fluke in that you scored a 12+ on the 2D roll to get a D asteroid."

Yea, I got lucky and rolled a 12 on the Asteroid Composition Table. In fact, I got lucky with pretty much all the rolls that mattered, so I know this is a rare situation.

"How did you determine that the Dense material was actually Dense worth 50 000 and not metallic nickel iron worth 400?"

That was one of my questions too because the rules don't say how to determine it. On Page 75, on the Scan Potential Table, I rolled a three and the result was D. The table says that "D = Dense materials (including metals)." So I just assumed that would be Dense Metals on the Commodity Prices Table on page 79 but I wasn't sure. So, yea, I agree, Mongoose should have clarified it in the rules. All it would take is for them to match the results on the Scan Potential Table with the results on the Commodity Prices Table.

So to summarize, in just this one section of rules on Belt Mining, it appears the following necessary rules are missing:

1) How to calculate the mass of a planetoid. It would be nice if the rules listed the calculation for establishing the volume of a planetoid as well as the 13.5 divisor to establish displacement tons.

2) How to determine the composition of the remaining material of the asteroid.

3) How to determine the type of material actually present for commodity pricing.
 
I've always thought the mining rules were broken. The problem is the game system doesn't take into account economics. If, using Psitravellers example of 16 Psyche, were to be lassoed and returned to orbit, the price of platinum, nickel and any other salvageable material would plummet across the world for, well, a very long time. It would be a total bust for all planetary mining interests. Even if they didn't DO anything with the minerals to start, the markets would factor in that they would have to start pretty soon to pay off the costs of bringing it, that terrestial production would collapse.

The existing system is simply too generous. It's Monty Haul in space (for those that remember that moniker). Granted one could do a hand wavium on this and simply move on with extremely rich players, but it kind of sucks when the rules don't have much applicability without a generous hand wave. Spica publishing put out a book on their Katringa system. The mining rules in there seem better suited from a PC perspective than what's in the MGT core rules. Or at least the profitablity goes down (and expenses go up). It has a pretty generous write-up of mining and the ups and down of it. Something similar could easily be applied elsewhere.
 
Composition of asteroid is on page 76 under asteroid size and yield.

Roll 2D twice on the Asteroid Size and Yield table: once
for the object’s size, and again for the object’s Resource
Yield. The Resource Yield indicates the percentage
of the object’s mass that has been determined in the
Resource Presence table. Remember to apply any DMs
for Dense materials or Exotics. Any remaining mass is
split into 50% Common Ore and 30% Uncommon Ore,
with valueless waste taking the rest.

So that is where I get the common ore and Uncommon Ore numbers.

And Just to add some confusion into the mix, the numbers for station processing are different, (and WAY more profitable if you look at them)
Page 61 Mineral refinery
Once asteroids are delivered to the station, they must
be crushed, the ores and other by-products sorted, and
waste released back into space. The produce is split
between 50% Common Ore, 30% Uncommon Ore,
15% Crystals & Gems and 5% Precious Metals (as
defined on page 220 of the Traveller Core Rulebook).
So, for example, for every 100 tons of produce from
the refinery, 50 tons will be Basic Ore, 30 tons will be
Uncommon Ore, 15 tons Crystals & Gems, and 5 tons
Precious Metals.

So 100 tons of processed asteroid material gives you 100 tons of usable goods on the back end. 50 tons common ore, 30 tons Uncommon Ore, 15 crystals and gems and 5 tons of the precious metals.
Now I can understand this to some degree. With cheap fusion power you can melt things down to component molecules and fashion whatever you need. Silicate rock can be broken down into oxygen and the various elements the silicates are composed of. But look at the cash involved.

So yeah, the numbers get silly. And the million ton rock is the average find.
 
"Roll 2D twice on the Asteroid Size and Yield table: once
for the object’s size, and again for the object’s Resource
Yield. The Resource Yield indicates the percentage
of the object’s mass that has been determined in the
Resource Presence table. Remember to apply any DMs
for Dense materials or Exotics. Any remaining mass is
split into 50% Common Ore and 30% Uncommon Ore,
with valueless waste taking the rest.

So that is where I get the common ore and Uncommon Ore numbers."

My 2e High Guard doesn't say "Any remaining mass is split into 50% Common Ore and 30% Uncommon Ore, with valueless waste taking the rest."
 
I think the problem with the system is allowing players to take the easy way with a big score and sell the whole thing off to soon live the good life. I noticed I got such results too and it's not much fun.

Belters, like miners of old, should be spending their lives searching for the mythic motherlode but actually spending most of the time chipping out enough to fill that tiny hold hopefully to pay bills and have a little left over for a long weekend at the asteroid startown or buy some upgraded or new equipment to find better ores and mine it a little quicker including ship upgrades. The experience shouldn't be just Find & Mine. There should be encounters or 'exciting' finds in the field and adventures during R&R time. Good time to bring out Jumpcussers, pirates and just desperate free traders looking for easy money.
 
I think you need a licence and permit.

The legality in light of the Imperium ten diameter orbit tradition may seem dubious, but you can have cheap drones landing on rocks all over the place, planting flags and claiming sovereignty.
 
Just a quickie, chaps - we have been monitoring your comments, and we will be returning to mining in the future!
 
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