Heck no....I'm not putting my fish up against Bruce's.....his HBG would likely mop the floor with mine! 
"Forget pounds and ounces, I'm figuring displacement!"
If we accept that: MBG(+)FGSF(=)HBG(F1)
And we surmise that: BG(>)HBG(F1) while GSF(<)HBG(F1)
Would it hold true that: HBG(F1)(+)AM500(x)q.d.(=)1.5lbGRWT?
PB answer: It depends.
| 
