Two items, both stemming from the Campaign Operations Handbook-
On page 24, when calculating the amount of spare parts required for normal peacetime operations, the book reads "Monthly generic spare part consumption (by tonnage) is 0.1 percent of a 'Mech, fighter, vehicle, Dropship, Warship or battle armor unit's mass."; however, all the calculations in the examples are done using 0.01 percent, which I presume is what the calculations should be made with. Correct?
If I'm not slipping decimal places when I convert percentages to a decimal multiplier, 0.1% = 0.001. Double checking: 100% = 1, 10% = 0.1, 1% = 0.01, and 0.1% = 0.001, correct?
The first example with Arnold, who looks at 65- and 70-ton 'Mechs; 20- and 60-ton tanks; 50-ton fighters, and 6700-ton DropShips:
65 x 0.1% = 65 x 0.001 = 0.065 tons
70 x 0.1% = 70 x 0.001 = 0.070 tons
60 x 0.1% = 60 x 0.001 = 0.060 tons
50 x 0.1% = 50 x 0.001 = 0.050 tons
6700 x 0.1% = 6700 x 0.001 = 6.7 tons
Those are the numbers in the example on p. 26, both in the text and in the TO&E example sheet.
Arnold moves on to spare parts. The BattleMechs, at 65
and 70 tons each, use 0.065 and 0.070 tons of spare parts
per month (0.1 percent of their mass), respectively. ...
The Bulldogs each use 0.060 tons of spares per month,
while the Packrats use 0.020 tons...
The 50-ton Corsairs consume 0.050 tons per month at 15,000 CB per
ton of the pricey, high-tech aerospace components, while
the 6,700-ton Seeker uses 6.7 tons (100,500 CB) per month
and the Invader (at 0.01 percent of its mass) uses 15.2 tons
The Hannah example has 80-ton 'Mechs, 50-ton fighters, and a 3,600-ton DropShip.
80 x 0.1% = 80 x 0.001 = 0.080 tons
50 x 0.1% = 50 x 0.001 = 0.050 tons
3600 x 0.1% = 3600 x 0.001 = 3.6 tons
The example p. 27 TO&E sheet (not easily copy-n-pasted here) shows 0.080 tons, 0.050 tons, and 3.6 tons.
Jason's pirate force example is a bit lengthier, so I'll spot check the math on the 108 Scorpions, one Locust, and one Mule DropShip:
108 x 25 x 0.1% = 108 x 25 x 0.001 = 2.7 tons
20 x 0.1 % = 20 x 0.001 = 0.020 tons
11,200 x 0.1% = 11,200 x 0.001 = 11.2 tons
Which is what's shown on the pg. 28 TO&E.
Where can I find the correct chart to calculate the cost of ammunition per ton? (Addendium: I have been able to piece the information I needed together by using a combination of Tech Readouts, Sarna.net and a few other sources. I understand that there is a chart in the Tech Manual, but the only copy I have been able to purchase for less than a house payment is in German and does me no good. Having some information in the Campaign Operations handbook as to where to find said chart, or having it reprinted in the CO manual would have been helpful.)
If you can survive a .pdf over a hard copy, it's available at the CGL game store and DriveThruRPG for $15:
https://store.catalystgamelabs.com/products/battletech-techmanual-pdfhttp://www.drivethrurpg.com/product/24750/BattleTech-TechManual?manufacturers_id=2216&it=1However, ze non-German hard copies do seem a wee bit pricey on eBay. Amazon references some used ones for "36.95 and up" at the moment I'm typing this. Barnes & Noble heads back for the stratosphere, starting at $99.99.
