Well, total NP *Is* equal to the average of the NPs of the segments if you use a 4th-order average. I can spell it out with an example. To make it easy, lets say we have two segments, 1 and 2, and that each has just two samples:

NP1 = ((a^4 + b^4) / 2)^0.25

NP2 = ((c^4 + d^4) / 2)^0.25

Call the NP of the whole ride NP12, which would be:

NP12 = ((a^4 + b^4 + c^4 + d^4) / 4)^0.25

If we isolate the sum on the right hand side of the NP1 and NP2 equations, we get:

2 * NP1^4 = a^4 + b^4

2 * NP2^4 = d^4 + c^4

Substituting those into the equation for NP12 gives:

NP12 = ((2 * NP1^4 + 2 * NP2^4) / 4)^0.25

NP12 = ((NP1^4 + NP2^4) / 2)^0.25

NP12 = Avg^4(NP1, NP2)

If the two segments aren't of equal length, then the average needs to be weighted by their relative durations, but it works out the same.

rmur was right that the reason TSS isn't additive is b/c of the mismatch in the exponents between the NP formula (4th order) and the TSS formula (2nd order). TSS is IF^2 * T. Imagine instead that it was IF^4 * T. Then, TSS1 (using the same segment 1 as above) would be:

TSS1 = IF1 ^ 4 * T1

Substitute in IF = NP/FT:

TSS1 = NP1^4 / FT^4 * T1

Substitute in the definition of NP1 from above:

TSS1 = (((a^4 + b^4) / 2)^0.25)^4 / FT^4 * T1

Simplify the (...^0.25)^4:

TSS1 = ((a^4 + b^4) / 2) / FT^4 * T1

T1 = 2 if we assume a sample rate of 1hz (even if we don't it's just a constant difference), so:

TSS1 = (a^4 + b^4) / FT^4

By the same logic:

TSS2 = (c^4 + d^4) / FT^4

TSS12 = (a^4 + b^4 + c^4 + d^4) / FT^4

Summing TSS1 and TSS2 gives:

TSS1 + TSS2 = (a^4 + b^4) / FT^4 + (c^4 + d^4) / FT^4

TSS1 + TSS2 = (a^4 + b^4 + c^4 + d^4) / FT^4

But that's the definition of TSS12 above, so:

TSS1 + TSS2 = TSS12

Ergo, it's additive.

As for whether these equations are right or wrong or inflate or deflate, I've got no opinion. I'm just handy with the numbers