Strategy for mining an asteroid, part III

Strategy for mining an asteroid, part II

In part 2,  I discussed how to begin construction of the torus.

How much would this thing weigh?  Since it is made of iron, the mass can be calculated based upon the dimensions.  Let's say we want 4 half thicknesses of iron, which should block 15/16ths of the gamma radiation.  The only radiation it wouldn't block at this point would be neutrons.  But there probably isn't any neutron source anywhere, except for the nuclear reactor, which should be shielded anyway.

At this thickness, we have two of the dimensions.  The only one left is how "wide" to make it.  This may not be set in stone, as it may be desirable to change it later.  I ran the numbers on the following dimensions:  2 kilometers in circumference, 10 cm thick, and 100 meters wide.  With these numbers, the calculations are

2000*100*.1 = 20000 cubic meters

Since there are 100 cm in a meter, then 100*100*100 = 1,000,000 cc in 1 cubic meter.

Iron's density is 7.87 g per cc, so 1 cubic meter equals 7.874 million grams.  The nickel, if it is kept, is slightly more dense at 8.91 g per cc.  If there's 90% iron and 10% nickel, the average is  7.97 gm/cc

So, if we were to make it a square torus (?), it would have 4 sides as a square and then connected to each other into a big torus structure.  To finish the structure then, we need 4 * 20000 cubic meters ( from above calculation ) to get 80,000 cubic meters.  Multiply that times 7.97 million and you get the mass of the torus--- 637,600,000,000 grams.  Divide by 1000 to get kilograms which makes it 637,600,000 kg.  Multiply that by 2.2 gives the mass in pounds  ----  1,402,720,000 lbs.  In tons, divide by 2000 which yields 701,360 tons.

Sounds like a lot, but the mass of the asteroid is something like 20 billion tons, so this barely scratches the surface.

Assuming that there's 100 parts per million of platinum, that's 1 pound in 10,000 pounds.  Or 140k pounds of platinum, which at a rough estimate of 1350 per ounce, would be worth---$189,367,200.  Or another calculation yields $245,476,000 based upon the percentage of the asteroid taken.

Would it all be worth it?  You would need a more economical way to mine the thing for the platinum to be worth it alone.  But you could pay the people for their work and for some of the materials and costs of transport, I would think.

A structure that size could house who knows how many people.  The value of that could be hard to calculate.

To compare, an Nimitz class aircraft carrier weighs about 100k tons.  Or about 15% the weight of this thing.  There may be over 5000 personnel onboard one of those vessels.  This is 7 times as massive, and that doesn't count the height inside the torus, which is probably more than the aircraft carrier.  Let's just say 35k people could be housed in the torus.

One hundred thousand dollars per head yields a value of $ 3,500,000,000.  More than the platinum.

This doesn't count the finishing on the inside of the torus.

You might get a nice Battlestar Galactica out of it.

Update:

There's some errors in the calculations.  Gasp.

Okay.  Let's just start with the work I did.  Most of it seems to check out in the first part.  But, 2000 meter circumference is too small of a torus ( if you can believe that ).  I forgot to multiply by pi ( 3.14 in order to get circumference of a circle ).  Let's set aside that error, because there's another.

The mass of the asteroid is more precisely 17.6 billion tons.  We're taking 701360 tons divided by 17.6 billion or .4 ten thousandth of the asteroid.  The $7 trillion dollars of platinum written about in John S. Lewis' book was based upon 1996 prices for platinum.  So, that's throwing the calculations off as well.  Prices have gone way up since then.  The asteroid could be worth much, much more.

I'm guesstimating the concentration of platinum as well.  At today's prices and assuming 100 parts per million in the calculations above, 140k pounds of platinum @ $1350 oz equals 140000 * 1350 * 16  equals $3,024,000,000.  That's assuming 100 parts per million of platinum in the asteroid.

Now, going back to the torus, if you make it as big as I anticipated, you can multiply that by 3.14.  Nearly 10 billion dollars worth of platinum.

Also, the thing could house 3.14 times as many people, or 100k people.  Over 10 billion in value for the torus.  A significant difference.

At the end of the day, you will have housing available for 100k people and enough asteroid left to build 20k more toruses just like the one just completed.  But that's getting ahead of ourselves.  We haven't outfitted the thing yet, so it isn't ready for occupancy.

Update:

A few adjustments still to be made.  It looks like the number should be 56 parts per million platinum, based upon the numbers I've seen.  That means if we use the first number, that is 2000 meter circumference, then the metal value should be about 1.5 billion dollars.  Maybe it will house up to 35 k people.  The mass would be 700k tons.  It could be spun up at less than 2 RPM to yield 1 g.

Another adjustment would be to shorten the height of the torus so that it isn't 100 meters tall inside.  Perhaps a quarter of that that will do.  That leaves substantial mass to work with to fit the insides of the torus for living compartments.  That would mean 3/16ths of the mass would be used to compartmentalize the interior.  You could possibly cut that even more, if needed.  The goal would be to get the mass as close to that 700k ton number as possible.   This is important for propulsion reasons.

I'll leave the discussion here for the next part of the series.