Ship Maintenance Theory

[SpikeTheHobbitMage]

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If a ship has any engineering spaces, the AFR is equal to [0.04%/engineering tonnage percentage*total tonnage].
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This is the same as:  (Total HS ^ 2) / (Eng Tons)

(Keeping in mind that 1 HS = 50 tons.)

In your example, 9995Tons = 199.9HS.
So: 199.9^2 / 100 = 399.6001

Yup. There's a bunch of ways to show the math. I figured the formula that I used was the most useful in normal circumstances, but if there's something specific you want to look at, you can and should play with the math here.

The major takeaway for me is that failure rate is proportional to the square of ship size, but inversely proportional to the (non-squared) engineering tonnage.
Meaning that if you double the ship size, you have to quadruple the engineering tonnage to keep the same AFR.

As annoying as it is, I can see why Steve did it that way:  If you combine two 10,000 ton ships part-for-part into a 20,000 ton ship, the combined AFR doesn't change.

[Alsadius (OP)]

The major takeaway for me is that failure rate is proportional to the square of ship size, but inversely proportional to the (non-squared) engineering tonnage.
Meaning that if you double the ship size, you have to quadruple the engineering tonnage to keep the same AFR.

Sure, but the same is true of doubling the number of ships. Two ships with 100% AFRs have a total of a 200% AFR. Increased failure numbers with increased tonnage is the expected behaviour, not a particular function of a big-ship design philosophy.

For me, what this really drove home is that AFR targets should never be fixed numbers - they need to be tonnage-based.