Resistence is futile!(and other invasions stories)

[Roondar]

Okay, after having watched this thread degenerate into a pissing contest, can either of you tell me what is the point of all of this? I mean, I've lost track as to why you two are arguing the insane minutiae of an ENT-era photonic torpedo's yield. What has this to the with the premise of the OP?
-Mike

Well, unless I've misread it, the key issue seems to be that they disagree on the correct yield of said torpedoes by 3 orders of magnitude or so.

But I could be wrong there ;)

[Mike DiCenso]

Yes, but to what end are they disagreeing on tthis for?
-Mike

[Mr. Oragahn]

Yes, but to what end are they disagreeing on tthis for?
-Mike

Perhaps to get a good interpretation of Reed's throwaway remark? :)

For one I'd really like to get my hands on the source Wong used to get his three formulas.

Besides, you want me to try inputing 15 megajoules. Why that number exactly? And above all, do you even know if the formulas you think I should use are meant to be used for such low figures?

Because 15 megajoules is about the yield of a small artillery shell. Now if your forumla is right and mine is wrong, then yours should provide an accurate result and mine a faulty one, yes?

This is what is generally called testing. We test if your forumla works with modern examples.

Oh please, cut the lecturing.

Again, what is the formula I am supposed to verify??
The only ones I could possibly deem "mine", and that's a very big stretch right now, are Wong's cratering ones, and that's only because you didn't use them... so they kinda became mine... but they're not.

Then there are the others, mentioned two or three pages ago, but which as you understood I left behind.
That said, I've read two days ago something about how much energy goes into cratering for a strictly vertical impact from a solid rock.
I thought this could be used in comparison to the amount of energy a nuke allows to cratering (much lower). So by using the cratering results obtained from formulas which had a kinetic energy backdrop in design, I could multiply the energy result by the adequate nuclear factor to obtain the yield of a nuke. This would, by a rough guess right now, probably lead me to get low megatons out of these mid to high hundreds of kilotons of cratering energy on asteroids.

But I digress.
If you could stop being facetious, that would help. Tell me formula you want me to use, that's all. I don't really get what's hard to understand here. It's really a simple request.
I've been asking this you over several posts and you're still refusing to give me the simple and clear answer that would promptly unlock this.

[Roondar]

Yes, but to what end are they disagreeing on tthis for?
-Mike

Perhaps to get a good interpretation of Reed's throwaway remark? :)

Reeds remark is troublesome because it's imprecise though. My remark about depth might not be to accurate, but even wide could be taken to mean both diameter (which is usually not used in physics equations) or radius (which usually is).

Nor is it clear just how big Mr. Reed envisions this asteroid to be and of what type (though asteroids as depicted in Star Trek tend to be rather massive affairs - they are definitely are not depicted as the scientific 'loose rubble' standard).

Strangest is that in my memory he called it a 3km crater in a moon. But that is entirely besides the point ;)

So by using the cratering results obtained from formulas which had a kinetic energy backdrop in design, I could multiply the energy result by the adequate nuclear factor to obtain the yield of a nuke. This would, by a rough guess right now, probably lead me to get low megatons out of these mid to high hundreds of kilotons of cratering energy on asteroids.

It would seem really quite odd to me if the yield gained that way is higher than when using a nuke 'proper'. After all, using a nuke will lead to vaporizing a significant portion of the material, which is -compared to a kinetic impact- a rather energy-wastefull way to do things.

Especially since it's pretty much established that ST (and SF in general) asteroids are not 'quite' the same as the real world ones any way (as I said, they tend to be depicted as if they're just a big hunk of rock/iron/whatever and not as a bunch of rubble waiting to disintegrate).

Then again, I do understand part of his issue - here on Earth a couple of kilotons are not going to get you anywhere near a 3km crater. Wide, deep, or circumference ;)

[Jedi Master Spock]

If you like, I could go back and split it off where it becomes strictly a technical discussion just about Star Trek technology.

[Mr. Oragahn]

If you like, I could go back and split it off where it becomes strictly a technical discussion just about Star Trek technology.

Could really turn out to be a hassle.
Perhaps in some near future, we shall make a thread entirely focused on craters on asteroids.

@ Roondar:

253 Mathilde looks like your typical SF rock, yet its bulk density is 1.3 g/cc. Don't judge an asteroid by its cov... its, err... surface?... ah?

[Roondar]

If you like, I could go back and split it off where it becomes strictly a technical discussion just about Star Trek technology.

Could really turn out to be a hassle.
Perhaps in some near future, we shall make a thread entirely focused on craters on asteroids.

@ Roondar:

253 Mathilde looks like your typical SF rock, yet its bulk density is 1.3 g/cc. Don't judge an asteroid by its cov... its, err... surface?... ah?

Interestingly enough, I found an asteroid impact calculator* which describes a 90 degree impact by a 17Km/sec 100m wide asteroid as follows on Earth (where gravity & density are higher) and on the Moon (where they are lower):

Earth impact: 54 MT, 3.0 km crater diameter, 0.65 km deep
Moon impact: 54 MT, 4.42 km crater diameter, 0.88 km deep

This suggests that gravity and density do play a part. The Moon has roughly 1/6th of Earth and its density is about 2/3rds of earths.

If this calculator is correct (and I see no reason why it shouldn't be) it doesn't make Reed's asteroid comment any better - if the asteroid is small enough that would make the yield, well, unrealistcally low (there is a lower limit here - if the torpedoes are too weak it makes little sense to use them over standard nukes. A 20th century nuke is able to do a megaton or or so in a surprisingly small package, especially if you add ST style propulsion instead of the huge rocket normally attached).

*) See: http://www.classzone.com/books/earth_sc ... page08.cfm

[Mr. Oragahn]

The problem lies in the fact that an impact even will lose a large portion of energy in cratering, while a mass-efficient nuke won't. That, as long as the impact is surfacic. A torpedo that rams an asteroid at an impressive speed would carry a lot of momentum and the reaction would occur while the torpedo's impact is moving matter.
What is important in crater formation is the amount of matter involved, notably the density of matter surrounding the origin of release of energy in the close surrounding environment.

That's why a buried nuke becomes akin to a non-mass-efficient nuke since it's like it's surrounded by a casing of some sort, in this case soil or rock. And then, logically, an impactor brings its own density into the mix.