Based on your responses, I can only conclude that you have neither an engineering nor physics background.

I quote from the abstract:

> An Eulerian compressible flow solver suitable for simulating liquid-lead flows
involving fluid-structure interaction, cavitation and free surfaces was developed and applied to investigation of a magnetized target fusion reactor concept. The numerical methods used and results of common test cases are presented. Simulations were then performed to assess the smoothing properties of interacting mechanically generated shocks in liquid lead, as well as the early-time collapse behavior of cavities due to free surface reflection of such shocks.

That's three parts. 1) fluid-structure interaction, 2) effects of shocks, and 3) collapse behavior. I'm discussing part #1, which mentions 2 km/s fluid flows. There's also part #3, which includes jets up to 6 km/s.

It's a compressible flow solver because liquid lead compresses at those pressures. That's why the author talks about "Probably the most significant feature of the flow in the concept reactor is that it involves a compressible liquid" and why the author later goes on to develop the equation of state for liquid lead, including a term for cavitation effects.

You might have confused it with compressing the plasma. But plasma compression isn't covered until "cavity collapse" on page 71.

You wrote "Your calculations are derived from a mild parabola."

Yes, of course it's a parabola. The free surface of all bulk liquids in a system with constant rotation forms a paraboloid. That's what the physics says. Why do you think otherwise?

In this case it's also a "mild" parabola. That's why it's the minimum bound for a solution.

The fact that this is a rotating sphere instead of cylinder doesn't change anything. The free surface is a consequence of the centripetal force balancing out potential gravitational energy, not the shape of the container it's in.

For some reason you think using a sphere "must change this drastically", but you offer no explanation. The counter-proof to your supposition is very simple. For all but thin-films, the surface has no way to know the shape of the container. It can't tell if it's in a cylinder or a sphere. Indeed, if you replaced the lead that's more than 30cm away from the rotation axis with a solid, then the free surface wouldn't change, and now it's spinning in a cylinder.

You also believe that 10g "would certainly put any liquid against the wall." This is true, but the actual question is how high will it push the liquid? You offer your conclusion without explanation. The counter-example is easy to show.

Let's suppose that 10g is enough to be near-vertical. We know the angle of the free surface, relative to vertical, must be arctan(1g / 10g) = 5.7 degrees. Let's say the vortex surface is 2 meters long. In that case, if the top is 20 cm from the rotation axis then the bottom - descending at 5.7 degrees - must be tan(5.7 degrees) = d/(2 meters). Since tan(5.7) = 1/10, then d = 0.2 meters. But there isn't enough room for 200cm of inclination!

Obviously, with 100 g then the displacement across 2m will be about 20cm. This means the 20cm vortex surface must have well over 100 g in order to maintain a near-vertical surface. Yes, this requires more than 1,000 rpm.

I've now worked this out in two different ways, and presented the math. I get basically the same result each way. Those also seem to agree with the engineering analysis you point out.

All you've said is that I'm "misrepresenting" things, that you're "not going to bother looking up where the formula fails", and give seat-of-the-pants answers that are easily shown to be invalid.

Since you had to look up a method to compute the G formula for RPMs - something covered in introductory physics for physics/engineering majors - I therefore conclude that you don't actually understand the engineering physics involved.