I am going to calculate a minimum of the water needed to cover the Earth to 8km above sea level. I am using 8km as it minimizes the volume required. Anything that lowers the radii involved lowers the over all volume needed.
Pay attention to the methodology.
dV = the minimum volume required to flood the Earth.
v2 = the volume at the minimum radius + 8km. Computed from r2, where r2 = r1 + 8km.
v1 = the volume at the minimum radius. Computed from the volume of a sphere for r1, where r1 is the minimum radius to the surface of the Earth.
vDisp = the volume displaced by topology.
For this exercise we are going to use the minimum radius of the Earth at its surface for r1. Why the minimum radius you ask? By using the minimum radius we can compute a lower bound of the volume. As volume increases at the cube of the radius (r^3), dV is minimized when r1 decreases. This lower bound will be independent of the actual variance of the surface.
In the egg drawing sub provided it would be equivalent to:
Using the radius of the lower (bigger) sphere inside the egg in the above image. This still computes a valid lower bound for this ovoid. The minimum is not the actual ammount needed, the actual value needed will be greater than this volume. But this model produces a value the would be the barest minimum needed if the longer radius of the oviod was only infinitesimally longer that the short one. As that variances increases in deviation from this minimum radius positively the over all volume required increases.
It is a valid lower bound. If you do not understand why this is a lower bound, let me know and we will fix the gaps in your education and/or reasoning.
The formula is dV = v2 - v1 - vDisp.
For vDisp we are going to maximize the impact of the topology of the Earth. How you may ask? By saying it is all at 8km above sea level. By using all of it at 8km above sea level we are maximizing the impact of the topology which in turn minimizes the volume of dV.
We are also going to say that this topology covers 30% of the surface of the planet.
So to calculate vDisp we use:
vDisp = .3(v2 - v1) if you don't understand why this is, just ask and we will walk you through the geometry and math.
so
dV = v2 -v1 -.3v2 - .3v1 = .7v2 - .7v1 = .7(v2 - v1)
Now lets plug in the numbers.
r1 = 6,353 km again remember that the smaller this number is the better as it minimizes dV.
v1 = 1074051671475 km^3
r2 = 6361 km
v2 =1078114273973 km^3
dV = .7(v2 - v1) = 2,843,821,748.6 km^3.
additional cubic kilometers of water as a lower boundary independent of the variances in the radius.
2,843,821,748.6 additional cubic kilometers of water as a lower boundary independent of the variances in the radius.
The USGS calculates the combined total of all water resources on the planet as 1,386,000,000 km^3 of water.
Again dV is a minimum bound of water required needed to raise the oceans an additional 8km, This is a minimum that grossly overestimates the displacement contribution of land by saying all the land is at 8km of elevation.
You need at a minimum 3 times the current volume of water on the planet to flood the planet to 8km above what we now have. Deal with it.
It is better to be a warrior in a garden, than a gardener at war.
Some of us, on the other hand, actually prefer a religion that includes some type of correlation with reality.
~Bill Hamblin