OT?: Goodbye to an era

Tell us how a spheroid in space is going to get pushed harder by a nuke if there is an atmosphere surrounding it or if there is a vacuum surrounding it. Which one experiences a greater push in the desired direction?

As a vacuum chamber is nearing emptiness (full vacuum) do the last few million molecules all leave hand in hand, strung together, or are they all completely unbound?

Can you 'pull' liquid through a straw? A gas through a hose? Are you sure about the forces in place in those circumstances?

In the milliseconds that comprise a nuclear blast, the presence of an atmosphere or not will make little difference as to how much force gets applied against the spheroid mass.

The omnidirectional blast, in both cases, puts the same amount of energy against the spheroid.

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Can you say "shock wave"? How about "over pressure"?

WTF are you yammering on about?

No.

Pull? No.

Yes, but obviously you're not.

Clueless.

Absolutely clueless.

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Ah, a higher pressure vacuum. To fill the bag back out, do you add more vacuum?

You still do not get it. We are talking about net displacement of a large spheroid by a nuke.

Short of building a mile wide parabolic dish to set it off in, the net result of total force against the spheroid mass from a nuke blown at a

This is why a teardrop is the right shape for traversing through air with the smallest cd. The air moves out of the way.

Wouldn't matter even if it doesn't. The net pressure against the sphere will be the same.

In the dish, it would result in a slightly higher perpendicular force.

Not a whole lot, and not worth finding a crater to blast within, but it would, nonetheless, be slightly higher. The flat spheroid 'air' blast would be the same air or no air.

and

Whay you actually just did here is prove my case that there will be no difference.

The nuke blast is far too big for an air layer to matter when the job is moving the entire planetoid/spheroid.

It would be akin to the bb passing through a layer of Saran wrap, before it hits the 4 foot marble. Zero change as to the impact energy on the ball.

You guys need some perspective.

Geez, you don't understand anything about electronics.

You don't understand anything about physics.

You don't understand anything about people.

Do you understand anything?

John

and

Yup. You can buy tanks of it.

John

Propellant? I don't follow you there. If you want to nudge an asteroid or a comet, xrays and gammas do the work. That's mostly what comes out of an h-bomb already.

John

Terrorist implications? Like from a James Bond movie, Smersh with their own rockets and astronauts in black space suits?

John

You lose.

I'd beat your ass in billiards as well. I also know how those rocks clack together and they are equal weight objects tied to a flat, level plane in/by the gravitational field of a planet.

You think there was sand over 200 feet deep and it got compacted to glass without being ejected?

Glass is about twice as dense as sand. To compact it by 200 feet, that would mean melting a 400 foot layer of sand into a 200 foot glass layer. That sounds like both doing that within a few seconds and leaving it there is a very tall order, much easier to blow most of it out. And there is enough energy available to vaporize that much.

Then again, where is there 400 foot deep sand? Things usually tend to get rocky below a few to several meters below the top surface of the sand or topsoil, or below the low spots in sand dune areas. (Barrier islands can get to 10's of meters deep in sand, but an island in the middle of the Pacific is not a barrier island.)

No. John does. He says 50 feet. I say a quarter inch. you need to grasp what is being talked about before you interject again.

It does not get 'compacted to glass'. Ity gets 'melted from afar' from the heat of the fireball. Long before any blast wave arrives.

He thinks that 50 feet of material gets brought "up to temp" and I say that far far less gets 'affected' to that 'degree'.

Hahaha... techno comedy.

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The lower energy but timely orbits come at the rock nearly perpendicularly to its path. Aim becomes critical under those conditions. Directly anti-pathwards flight paths are short time but very energy intensive. Orbital mechanics is mean stuff, and i use power tools to obtain mediocre results if at all.

I said many times here that my for-sake-of-argument figure for the .1 mg sand grain is 1 km/sec. I did not dig to see why that was not quoted back to me.

In a perfectly inelastic dead-on collision, the 2,500,000,000.1 mg combo of the 2.5 metric ton ball and the .1 mg sand grain has the same momentum that the sand grain alone had, assuming a frame of reference where the 2.5 metric ton ball was stationary before the collision.

.1 / 2,500,000,000.1 * 1 km/sec times (1000 m / 1 km) (which is unity)

• (31,557,600 seconds / 1 year) (which is unity) = 1.26 meter/year.

And you said above 2 mm? You either did not do the math, or otherwise both of us did and one of us made a boo-boo of 3 orders of magnitude.

I have already explained in this thread how nuking rocks a km or 2 in diameter can get bigger numbers.

What if we improve the detection system and the armament? What if we double to triple the advance warning, reduce the delivery time slightly, and have handy 1 or 2 hundred megatons to blast with?

Can you work out numbers according to the mathematical laws of physics even from reasonable-sounding assumptions with lightweight/featherweight support? I did that in this thread, as applicable to rocks several hundred meters to 2 km in diameter. Do you have to-same-extent-supportable numbers to plug into the equations and then can you post yourself doing the math?

My math-based arguments earlier in this thread were in part on ratio of magnitude-of-vector-impulse to magnitude-of-vector-momentum along lines of the like of 2 km per 2 years / 30 km/sec to deflect a little over half of problematic asteroids having such average speed in their solar orbits from colliding with Earth or entering problematically low layers of Earth's atmosphere.

This is ratio of 1 in 946 million, if I did not screw up the math.

Translating to angle - I see 1.06 nanoradian, while 1 second of arc is

I doubt that. Medical gamma sources are shielded by modest amounts of lead. The gammas from an h-bomb denonated in the atmosphere are mostly converted to blast energy by being absorbed by air relatively close to the bomb. A few meters of rock is denser than the entire column of the atmosphere.

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Ken, if the rock is only a few kilometers across, you want to be rather close (Say 10 to 100 meters) to use as much of the yield as possible. Being kilometers away would waste most of the available energy.

Even if nuclear explosives are only 1% as efficient at translating stored energy to momentum (efficiency of which improves with quantity of mass ejected from impact point[s] with ejecting more mass at lower radial velocity until mass*velocity product declines), I seem to think that nukes ejecting at average-radial-velocity of hundreds of m/sec 100's of kilotons of ejected-from-crater material would do better than rocket engines ejecting 100's of tons or a few kilotons of rocket engine exhaust at several (under 10) km/sec.

Most firecrackers have less stored chemical energy than most "bottle rockets" have. Most firecrackers have their chemical reaction products less gaseous at temperatures typical of "bottle rocket exhaust" and at atmospheric pressure than is achieved by "bottle rockets".

While the total launch mass of a Space Shuttle including SRBs including mass other than fuel is about .002 megaton, and a Saturn 5 rocket has mass of about .003 megaton including other than fuel. In comparison to nukes of around or over a megaton, let alone nukes of 10's of megatons having mass of 10's of tons even in the early 1960's and late 1950's, good payloads for launch vehicles no bigger than Saturn 5. At this rate, more-routine launch vehicles that are cost-effective to launch in multitudes can deliver to even almost-Mars-orbit-level (even with the added difficulty of the likely-unnecessary circular solar orbit) such multitudes of warheads around a couple hundred kilotons.

If a detonation produces a blowout of mass from a surface target, then the directional pattern is ideally hemispherical for a zero altitude burst. In an ideal hemispherical mass ejection, the ratio of sum of perpendicular-to-surface components of ejected-particle momentums to what that would be if all mass was ejected perpendicularly is .5 if I figure correctly. (I do remember posting before calculations based on

In a vacuum, and adjacent nuclear fireball makes a mortar out of the being-formed crater.

At this rate, reduced efficiency even by over an order of magnitude times .5 times 2.5-3 orders of magnitude more energy is probably more.

-- - Don Klipstein ( snipped-for-privacy@misty.com)