That is the value for the spin speed. Quoting from the paper, 4 paragraphs after the 2km/s you referenced:

> Probably the most signicant feature of the flow in the concept reactor is that it involves a compressible liquid. With flow velocities exceeding 2 km/s and pressures reaching 400 GPa, compressibility is unavoidable.

"Flow" refers to the spin speed, not the collapse.

Like I said earlier, I used the equation for a liquid mirror telescope -- h = 1/(2g) * (omega * r) ^ 2 -- for my '"back of the envelope" thing.' See http://en.wikipedia.org/wiki/Liquid_mirror_telescope for the full derivation.

Using h = 2.7 m, g = 9.8m/s, and r = 0.20m gives 40 Hz (or 2,400 rpm). Note though that this gives 20 cm at the top, and 0cm at the bottom. It's a paraboloid, so the center is probably 14-15 cm across. Thus, 40 Hz is a lower bound.

Using 40 Hz in (2 * pi * 1.5m) * 40/s implies a minimum equatorial speed of 380m/s, which is Mach 1. The equatorial pumps of course must be providing fluid at an even higher speed. The g-forces at the equator, 1.5 meters away, are even larger than the g-forces at the surface 20 cm away. a = r * omega^2 = 1.5m * 1600/s/s = 2500 G.

To get a more uniform evacuated center requires higher speeds still, but the formula I used (first discovered by Newton, btw), no longer holds. It will likely need to be several times faster. Flow speeds of 2km/s require only 6 times faster than the minimum possible speed, which sounds reasonable. Even if 2km/s doesn't sound reasonable.

I don't think General Fusion has done the engineering testing to show that they can actually construct one of these, much less use it to provide power.

Feel free to correct my calculations.

>> Quoting from the paper, 4 paragraphs after the 2km/s you referenced

The paper doesn't even discuss what you claim. This is the paragraph before the one you quoted:

"There are several important flows in the concept MTF reactor. However, this work focuses only on the compressible aspects. Consequently, issues such as the vortex formation and cycling lead through the reactor to generate steam are left to follow up work by General Fusion."

I do think you know you are misrepresenting the content of the paper.

Your calculations are derived from a mild parabola, I'm not going to bother looking up where the formula fails. We're talking about a globe, which must change this drastically. Which I frankly also think you're aware of.

AGAIN: 10+G in a centrifuge would certainly put any liquid against the wall. And already 1000 RPMs @ 20 cm is > 110G! (See reference in previous comment.)

After your misrepresentation of the paper above and forcing me to repeat trivial points repeatedly I'm not going to bother with the differing results from your math or how it might be applicable to a sphere.

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.

So your claim is that when "2 km/s" is referenced on page 3 of the reference, the first which obviously is about the implosion speed after the shock is not related to the second -- which is the necessary speed for the lead.

So you repeat the claim of 2km/s for a vortex in the GF reactor.

A few decimeters irregularity in the bottom of the vortex for the lower plasma injector invalidates that, which you certainly know.

I have no physics background (cs/chemistry), but considering e.g. your insisting on the exact definition of Asimov's prediction above (which I acknowledged in my first comment), I believe you're trolling.

So I'd really like to see other's arguments on this. I'm not going to spend time to get the cobwebs out of the part of my head where old math studies are stored for what likely is a troll. I have lost too much time as it is.

Edit: I don't think you really care, but cavity compression for a 1 m sphere was done 2012. Google for a pdf called "Update on Progress at General Fusion". ("Fusion Power Associates, 2012 Annual Meeting")

Addendum:

General Fusion has gotten tens of millions of dollars from serious investors. The idea that their design would need to accelerate many tons of molten lead to 2 km/s every pulse (i.e. once per second!) is just too ridiculous.

I need to be less polite/trollable.