Technology doesn't work that way - SamuelA's Pit Thread

[Tripler]

@SamuelA

Now my motivations were different than your guesses. (1) I did read the Washington Post article first, I didn’t lie about that. But that lobby group’s article, that I only skimmed, seemed more credible.

I don’t care.

@SamuelA

(2) Sorry about any personal attacks.

I don’t care. I’ve been called far worse by far better men.

@SamuelA

(3) Yep, my bad

I don’t care.

@SamuelA

(4) **“Eventually, the brass adopted the idea that, when it came to nuclear matters, they were at the beck and call of the president. It was not generals’ responsibility to make the order; it was their responsibility to carry it out.” **

Isn’t this exactly the point I was making? Like, wasn’t this the entire topic of discussion? Not sure why you put this here.

It was to illustrate to you that even Flag Grade military officers acknowledged they couldn’t ‘go rogue’ and had to answer the President’s orders.

@SamuelA

(5) It’s the War Powers Act of 1973?

It goes both ways.

@SamuelA

Wait. So the original topic of discussion was that you said that Congress was *helpless *to stop the President from killing us all. Something that is a real, though small risk.

I don’t care. This is a redirect, and it belongs in the other thread.

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SamuelA I will now raise the questions on the “assumptions” you have made for your ‘estimates.’ I’ll be “nice”. I will point out where your assumptions are either lacking, or do not cut the mustard due to politics or law. This is not “nitpicking”; imagine you’re trying to sell this to a panel of Engineers (you are).

@SamuelA

Asteroid is 1.4×10^12 kilograms, a number I googled for a 1 kilometer asteroid.

There are gaps here: Based on what composite material? And you’re not clear on 1 km diameter or radius.

@SamuelA

In order to push it’s impact probability down acceptably, a 1 cm/second change in velocity is needed. (I read that in a wired magazine article as being enough)

There are gaps here: What is it’s distance from Earth? And what is it’s current velocity?

Link to that article please. That change in velocity is only valid at certain ranges from the Earth. The closer it gets, the more of a velocity change on the asteroid you need. Swapping the frame of reference (and point of view) for a second to the asteroid (a puck) because we can’t move the Earth (you); if I take an accurate slapshot at you from the far end of the rink, you only need to move 0.5 m/sec until that puck crosses the ice for you to avoid it. If I’m 3 meters away from you with the same accurate slapshot, you need to move a kilofuck quicker to get out of the way. But, since we can’t move the Earth, we have to move the asteroid. You need a megafuckton more energy the closer it gets to move it aside. So, you need to define how far away the asteroid is before you assume a 1/cm change in velocity.

You need to assume a distance and velocity (ETA: I do this later for you), because this will A) tell you if you need a higher change in velocity, and B) will give you an estimated time to impact which will drive other factors (e.g. spacelift) later. . .

@SamuelA

Nuclear pulse charge + small thrusters for guidance is assumed to be 300 kilograms, with a 150 kiloton yield. ISP is assumed to be 7500. (10k-20k from a paper on Project Orion, reducing it because you need a guidance system)

There are gaps here: What warhead are you using? Are you going to design a warhead from scratch? [Hint: no you’re not, not even in 10 years]. The Project Orion Drive was abandoned as wildly impractical [Hint: you will not detonate nuclear weapons over American soil, nor can you detonate them in the atmosphere due to the PNTBT as you pass through the atmosphere on the way to space.] Spacelift will come later. You will also need to tell me the blast overpressure of your chosen weapon’s effect in a vacuum when it is fuzed for a PD (or point detonation)–it will affect the force you impart on your asteroid. Why do you assume a 150 kT yield, and how does that correlate to blast overpressure in a vacuum?

@SamuelA

So how many nukes do you need?

(1.4×10^12)(0.01) = (7500*9.8)(300)(n)

n = 634. So 300 kilograms * 634 = 190,200 kilograms must reach the asteroid.

This will be recomputed when you fill the earlier gaps. You have a faulty assumption: I do not understand why you are claiming a 300 kg nuclear-tipped projectile will impart 300 kg of force onto this asteroid. You don’t need nuclear weapons for this, you need giant fucking 300 kg stones. On detonation, there may or may not be 300 kg of force imparted by the nuclear explosive. You assume a 150 kT device–again, how much force will that impart on detonation in a vacuum?

@SamuelA

. . . I used the yields of actual devices for a Fermi estimate. Let’s test your background : which warheads hit 3 kilograms/kiloton? Do optimized nukes exist that are that good or not?

I do not understand your 3 kg/kT correlation. Are you talking about primaries, secondaries, tertiaries, or total system mass? [Hint: there is no correlation between system mass and nuclear yield]. What ‘actual device’ did you use, and please link a citation for it.

@SamuelA

Falcon heavy stated payload to mars is 16,800. So you need 11 launches to do it.

16,800 what? Also, of course, you need to actually develop those nuclear pulse charges. Maybe you can’t get a really efficient one in the time you have and you need 10 times as many inefficient ones. Not all the rockets will make it.

You do not understand how difficult it is to build, test, and deploy a nuclear weapon with respect to politics or engineering. And, why won’t all of the rockets ‘make it’?

@SamuelA

Bigger asteroids - 10 kilometer spheres - yeah, it’s not good. 1000 times the mass, this problem becomes 1000 times harder. Now you would need to hope you can eject craters in it, which in turn means it better be made of the right kind of material for that.

You haven’t done any sort of assumption of what the asteroid is made of…

Here’s how a legitimate estimate should be done (to start):

A 1km diameter asteroid, of Lead (the densest material–worst case scenario) is headed towards Earth. It’s volume is 0.523 km^3 (or 5.23x10^14 cm^3). With lead at 11.34 gm/cm^3, you’re kind of fucked with 5.93*10^12 kg of mass coming at you. For arguments’ sake, we shall assume we have spotted it 10 years before it crosses the point of our orbit (and hits us directly–acceleration or deviation due to gravity notwithstanding). This is a perfect, solid sphere tumbling (rotating about all three axes, but I am simplifying the rotational dynamics for you, otherwise targeting and imparting moments of inertia due to blast cannot be speculated unless we had a real-world shape to discuss) in space, zipping along at 10,000 km/sec, magically on orbital plane. 10,000 km/sec * 31,536,000 sec/year * 10 years gives you a range of 3.15x10^12 km from Earth. We will assume, for academics, that 1 cm/sec adjustment imparted orthogonally to it’s trajectory (our orbital plane) shall be sufficient to “lift” it above the Earth. This range and forces involved will come into play once you have a warhead identified. This is where we stop for now.

Do you now see how complex of a problem this is, and that you cannot just gloss over it with handwaving? You really screwed the pooch by not identifying a weapon system. And here is no correlation between system weight and yield.

I am no longer interested in participating in your science fair project. I simply highlight where you’ve passed over some very important details.

Tripler
Mr. Fermi, I hope I do you justice.