Lets say we have a Dragon born with Volume=v (v=xyz), Mass=m. It’s Density would be d=m/v. We also know that it kick ass with a Force=F, where F=m*a, being a the acceleration of it’s arms while punching.
If our DB double it’s size in every dimension, it means that it’s Volume is: V’=2x2y2z > V’=8(xyz) > V’=8V.
If we assume it’s density keeps constant, then it means it gains some mass.
D=m/v, but now D=m’/v’ so
m/v=m’/v’ , mv=m/(8v) , m8v/v=m’ , m’=8m
If we assume that it will punch with the same acceleration as always, F=m*a
F’=m’a, F’=8ma F’=8(m*a) F’=8F
So, changing from medium to large would imply changing 8 times the strength. If this keeps up for every step, then we are talking about one extra change to get to huge, and then one to get to gargantuan . So 8^3 times the force. And if we assume a linear relation between force and damage, 8^3 times the damage…
Math time!!!
Lets say we have a Dragon born with Volume=v (v=xyz), Mass=m. It’s Density would be d=m/v. We also know that it kick ass with a Force=F, where F=m*a, being a the acceleration of it’s arms while punching.
If our DB double it’s size in every dimension, it means that it’s Volume is: V’=2x2y2z > V’=8(xyz) > V’=8V.
If we assume it’s density keeps constant, then it means it gains some mass.
D=m/v, but now D=m’/v’ so
m/v=m’/v’ , mv=m/(8v) , m8v/v=m’ , m’=8m
If we assume that it will punch with the same acceleration as always, F=m*a
F’=m’a, F’=8ma F’=8(m*a) F’=8F
So, changing from medium to large would imply changing 8 times the strength. If this keeps up for every step, then we are talking about one extra change to get to huge, and then one to get to gargantuan . So 8^3 times the force. And if we assume a linear relation between force and damage, 8^3 times the damage…
Yeah… A few extra d4s seems… Meh