erik818
Sorry to contradict, but your assumptions are not totally correct.
1) Correct, since 2 ^3 = 8, ie your scale factor, f, is f^3.
2) Incorrect. There are many formulae that are involved. However they can be summarised as EHP(old) = EHP(new)x f^3.5
3) Then you need to establish same speeds V = V(old)xsqrt(f)
4) Don't understand what you mean by "water resistance", in what context?
5) Not strictly correctly, see above. Since the EHP is different at the same Fn.
As for:
"..My problem now is that I don't know how to scale for increased displacement.."
Assuming that all is changing, is the displacement, it is pro-rata, thus:
EHP(new) = EHP(old) x Displacement(new)/displacement(old)
for small-ish increases in displacement.
I can't comment on Rick's "fog" worded explanation!
PS.
A delightful comment
"..Your personal attacks are getting creepy.."
Well, if you had a backbone to debate this publicly to explain yourself, would be useful. But that just exposes your own ignorance and prejudices...oh well. I don't think you would know what a personal attack is, compared to a factual verifiable justification if it hit you on the head. Still..ignorance is bliss....
That's 3 neg hits today....oh of course, it's hallowev'en, silly me. What else have you guys got to do....