Nicely put, LA.
I can´t tell if the factors reflect reality but the logic and math behind your formula does make sense except for one case. I´ll explain that below.
I don´t think we need long explanation for our rules and so I propose to use just the results.
Misc weight (wm) to be added for (x = machinery weight)
a) Turbines, electric drive
wm = x*0,1
b) Diesels, geared drive
wm = x*0,5
c) Diesels, electric drive
wm = x*0,65 [= x(1,5*1,1-1) = x(1,5*1,1)-x = (x*1,5*1,1)-x]
d) Mixed propulsion, geared drive
wm = x*0,5*Pc/Pf where Pc = Power/cruisespeed and Pf = Power/fullspeed
e) Mixed propulsion, electric drive
wm = x*0,1 + x*0,5*Pc/Pf*1,1
(-> I´ll explain what I did in the post scriptum.)
I´m aware of the fact that SS handles misc weigth differently from machinery weight but LAs rules are the best I´ve seen so far to handle other machinery - even if we need to assume that poweroutput scales linear (otherwise we couldn´t use Pc/Pf).
What I´m missing here is what effect those drives will have on range and damage resistance. To assume electric drive allows better subdivision and thus such a design could gain 10% resistance sounds okay to me. On the other hand I´ve read about several occasions where turbines on US ships still ran even while submerged due to flooding - something a diesel never would do. Further more diesels are higher, need more room than turbines and thus I would expect damage reduction to be somewhat less than on a ship using turbines....
What does the board think?
Regards,
HoOmAn
PS: Now it´s time to explain why I added that factor above.
If we look at the formula Bernhard uses for diesels, electric drive we can see that he does not only add a 10% penalty to the original machinery weight but also on the additional 50%.
He uses x*1,5*1,1 meaning that the original weight is increased by 50% and then another 10% are added. This sums up to a total 65% penality on diesel driven ships with electrical transmission.
wm = x*0,65 [= x(1,5*1,1-1) = x(1,5*1,1)-x = (x*1,5*1,1)-x]
Later he explains how one has to calculate misc weight for hybrid propulsion plants/electrical drive. He uses
wm = x*0,1 + x*0,5*Pc/Pf
If you look at it closely you can see that he now adds 10% on the original machinery weight but _not_ on the diesel weight. Thus the result does not compare to what he got before using the other rule. It becomes more obvious when using his remark. He states that Pc/Pf=1 if one uses pure diesels. In other words Pc/Pf=1 means that you´re simply using diesel, electric drive as explained above.
The problem now is that if you use Pc/Pf=1 the misc weight you have to add is only
wm = x*0,1 + x*0,5 = x*0,6
and not
wm = x*0,65 !!!
To fix that error we have to make sure the 10% penalty for electric drive does also apply to the diesel component when modelling hybrid drives. That´s why I added the factor 1,1.
Then we get
wm = x*0,1 + x*0,5*1*1,1 = x*0,65
if Pc/Pf=1 and everything´s fine again.