Resistence is futile!(and other invasions stories)

[Mr. Oragahn]

Why should I even follow those parameters, when the density of an asteroid averages at 2800kg/m^3, and not 1500kg/m^3?

Then use 2.8 g/cc if you want.

EDIT: just for the record, there are many asteroids, notably large ones, that fall way below that density. I provided some examples earlier on.

[l33telboi]

Then use 2.8 g/cc if you want.

EDIT: just for the record, there are many asteroids, notably large ones, that fall way below that density. I provided some examples earlier on.

Why? I've already told you that I don't quite buy what you're trying to get across here.

Also, there are plenty of asteroids that are more dense then this, hence why it's called an average.

[Mr. Oragahn]

Then use 2.8 g/cc if you want.

EDIT: just for the record, there are many asteroids, notably large ones, that fall way below that density. I provided some examples earlier on.

Why? I've already told you that I don't quite buy what you're trying to get across here.

It's much more simple, and my arguments are irrelevant to this.
I should have asked you this question from the beginning, and you're ought to prove that once your remove the gravity-related term, you obtain something close to the figure you defend.
The equation is from the blog you cited as evidence, to back up the 51 megatons figure. You should be able to prove that without gravity, you can obtain a figure that already reaches close to those 51 megatons.

Also, there are plenty of asteroids that are more dense then this, hence why it's called an average.

Yes, there are high ends, medium ends and low ends.
Of course, a conservative stance would have one take the low ends, which are still numerous enough not be treated as exceptional and unfair.

[l33telboi]

It's much more simple, and my arguments are irrelevant to this.
I should have asked you this question from the beginning, and you're ought to prove that once your remove the gravity-related term, you obtain something close to the figure you defend.

Then I would use the same formula squish used, in the same way he did.

The equation is from the blog you cited as evidence, to back up the 51 megatons figure. You should be able to prove that without gravity, you can obtain a figure that already reaches close to those 51 megatons.

Search function seems down right now, or at least I can't get it to work, but I'll dig up the specifics tomorrow.

Yes, there are high ends, medium ends and low ends.
Of course, a conservative stance would have one take the low ends, which are still numerous enough not be treated as exceptional and unfair.

Why be conservative when you can get a figure that's far more likely to be true then the conservative one?

[Mr. Oragahn]

It's much more simple, and my arguments are irrelevant to this.
I should have asked you this question from the beginning, and you're ought to prove that once your remove the gravity-related term, you obtain something close to the figure you defend.

Then I would use the same formula squish used, in the same way he did.

No, you would use the exact formula provided on the blog which is behind your argument. You'll notice that the "no gravity argument" came hand in hand with the equation provided on the same page the quotation was taken from. The argument belongs to the logic that spawned Nige's equation.
All we need to know is what you find for the first term mX, since the second one is left out because gravity would not matter.

The equation is from the blog you cited as evidence, to back up the 51 megatons figure. You should be able to prove that without gravity, you can obtain a figure that already reaches close to those 51 megatons.

Search function seems down right now, or at least I can't get it to work, but I'll dig up the specifics tomorrow.

May I point out that you don't need to search for anything other than what has already been presented at SBC and in this thread?
Here, all the links you need:
1 and 2.
No more, no less.

Yes, there are high ends, medium ends and low ends.
Of course, a conservative stance would have one take the low ends, which are still numerous enough not be treated as exceptional and unfair.

Why be conservative when you can get a figure that's far more likely to be true then the conservative one?

That you don't know.
I shouldn't have to tell you why you have to come with a conservative calculation first.
But feel free to proceed with higher parameters, we'll see what you get.

[l33telboi]

No, you would use the exact formula provided on the blog which is behind your argument.

My argument is behindthis formula (also see the next post after that).

From the paper, the generalized equation for surface blasts:

2.5x10^-3m per J^1/3.
(Edit: For a 3000 meter crater, that's 48.074 MT.)

You'll notice that the "no gravity argument" came hand in hand with the equation provided on the same page the quotation was taken from.

'No gravity' came hand in hand with the formula provided above, because it doesn't factor in gravity.

No more, no less.

You ask me to prove the figure stated earlier, but for some reason I'm not allowed to actually use the formula used to obtain those figures? Come one, that's absurd. Have you noticed that your formula doesn't even factor in how hard it would be to crack a certain material? It literally only deals with density. But even a school-child could point out that it's easier to break up a handfull of sand into smaller pieces then a solid rock.

And out of curiosity, have you tried scaling your figures down to say 15 megajoules and see what that gets you?

That you don't know.

I do. It's statistitically provable that the average is far more likely to be closer to the real value then a conservative figure. This is simple logic. If you have a die with 10 sides, then the number 5 will most likely closer to the real figure then 1 whenever you throw it.

But feel free to proceed with higher parameters, we'll see what you get.

High parameters would have us assume we're talking about a nickel/iron asteroids, you know, with a density around 8000 - 7000 kg/m^3.

[Roondar]

Is there any reason why we are taking the 3km crater to mean a 3km wide one and not a 3km deep one?

After all, Reed does not actually specify either way which it is.

(Note, English is not my native tongue so I could be mistaken about the exact meaning behind the sentence. It just sounds to me it could mean both)

[Mr. Oragahn]

Is there any reason why we are taking the 3km crater to mean a 3km wide one and not a 3km deep one?

After all, Reed does not actually specify either way which it is.

(Note, English is not my native tongue so I could be mistaken about the exact meaning behind the sentence. It just sounds to me it could mean both)

Or he could have meant circumference. ;)

[l33telboi]

No one says 'a 3km crater' while talking about circumference. We're assuming it's that wide because that's what he most likely meant and the deep bit shoots the figures up in the region where it doesn't fit with the rest of the series. I suppose on might leave in a small cave-eat and point out that he might’ve been talking about depth, but that it’d be very unlikely.

[Mr. Oragahn]

No, you would use the exact formula provided on the blog which is behind your argument.

My argument is behindthis formula (also see the next post after that).

Not really. Remember, you talked about an equation that "let's you adjust what type of composition the asteroid is made of" and that while "it assumes zero gee", we couldn't "find a better suited formula because there quite simply is none", as "[it] fits every single parameter we need it to fit."

We're yet to see the one which allows you to input the composition parameter.
Considering that the simple one you used at SBC didn't have such a parameter, and that the only one which allowed to consider the material, via its density, was E = mX + mgh, although not used for asteroids, there were not many other choices than using this last one.

From the paper, the generalized equation for surface blasts:

2.5x10^-3m per J^1/3.
(Edit: For a 3000 meter crater, that's 48.074 MT.)

Now you have brought another formula, but seem a tad reluctant to use the blog's formula, minus the gravity-relative term (mgh), despite being directly associated to the quotation picked from the Glasstone blog.

You'll notice that the "no gravity argument" came hand in hand with the equation provided on the same page the quotation was taken from.

'No gravity' came hand in hand with the formula provided above, because it doesn't factor in gravity.

We don't know the parameters this other formula you brought, by citing Squishy, is allowed to be used for.
We've seen thus far that many crater formula for small yields do one single same thing: pick a given yield in kiloton, preferably a sub kiloton yield, divide by mass (defined by density and volume) of the corresponding crater, slap a small constant to that relative crater diameter, and that's it, you have an average energy per unit of volume to break or destroy it.
I still find this methodology quite poor if used for greater yields, because of the complete lack of care for the inverse square law, which I imagine could pass for small yields (<1 kT).

You ask me to prove the figure stated earlier, but for some reason I'm not allowed to actually use the formula used to obtain those figures? Come one, that's absurd. Have you noticed that your formula doesn't even factor in how hard it would be to crack a certain material?

It's not my formula, I didn't even know about it until you mentionned the blog. By all means, it's yours, since it's the one that comes with the blog you cited as part of your evidence.
You use all sorts of formulas, add new ones, but not the very one found on the blog. Why is that?

Also, why do you say it doesn't address what's necessary to crack a certain material? None of them do, and Nige's formula is by far one that has shown to contain the most parameters for the simple that it contains the mass, obtained with the material's density.

It literally only deals with density. But even a school-child could point out that it's easier to break up a handfull of sand into smaller pieces then a solid rock.

Yet it's a cratering formula which the author -- who couldn't understand how Glasstone and others super brains could miss the importance of gravity for so many years -- says can be used for all sorts of densities.

Equally, a school-child could point out that it's easier to break and crush a porous object made of a given material, than brake another object made of the same material, but which is not porous.
This can be verified in your very garden, with the soil mounds left by moles. Smashing the dry soil with the first barely leaves a mark, but if you'd move the escavated mass and carefully put it into a pot (in order to mimic the presence of that volume of earth like it had never left the ground), and tried to smash it, your fist would penetrate it easily.

All these nuke equations you presented are fine and dandy but none take into account that factor. Not even Wong's, as I said earlier on. A table on the blog shows that that a dry coral crater for a surface 1kT bomb is 1.1739 times bigger than for dry soil. That would mean that if we considered that the 1500 m radius for the crater was due to it being dry coral, the same device detonated on a dry soil surface would leave a crater with a radius of 1278 m. That alone with one of your formulas would return 32 megatons.

I also noticed, while observing the origins of the old formulas based on W (in kT) or J (in joules) that small differences creeped up, but didn't matter for small yields. However, when one would use those these formulas for higher yields, the differences would be multiplied in much noticeable ways.

Let's also rewind things a bit.
Glasstone's old formula worked this way, and was fairly simple in its construction:

For a dry soil, he estimated that a 1 kT device would leave a 120 feet wide crater (radius = 18.288 m). So to obtain the radius of a crater in feet, you had to do the following:
R = 60 x W^0.3

We may infer that the equation is almost primitive in a way. It's blunt. Small bomb makes small hole of dat size, urgh, bigger bomb makes same hole times bigger boom.
Wow.
Then things changed, casing density entered the stuff, and after more changes, came the introduction of gravity, for a final formula in 1994. Yet I don't see where they ever took care of the ISL. :/

Now, when I look at your first formula, the one picked from SBC which gave you 51 megatons (51.62 MT more specifically), what do we see? Well, first, since you used in a zero gee environment, we can understand that it means g either is not present, or is but the term it belongs to returns 1 when p = 9.8 m/s².

So we see that it's a very close cousin of Glasstone's formula:

Diameter = 5 e-3 * J⅓ , with D in meters.

Let's see what this formula gives us for 1 kT.
1 kT = 4.184 e12 J. Therfore the diameter is:

D = 5e-3 * (4.184 e12)⅓
D = 80.57

Our 1 kT device laves a crater in dry soil that is 80.57 meters wide, for a radius of 40.285 meters.
One of the blog's tables says that a 1 kT surface detonation on such a soil would leave a crater with a radius of only 18.4 m. Considering that we're talking of low yields, the difference is monstruous.

A Russian test the 26th of november 1962, with a yield of 1.1 kT, detonated half a meter above the ground (which is the equivalent of putting the large bomb on the ground, with the core being 0.5 m above the surface), left a crater radius of 13.5 meters, and that possibly with one of these bigger casing devices (Glasstone often points them as being low yield/high mass devices), and as such, making a larger crater than an "efficient" device.
Again, the difference is baffling.

We could compare this with Johnnie Boy (0.5 kT), again, an old device, detonated 0.585 m above the ground.
By the formula, we obtain D = 63.95, so the radius is 31.975 m.
The real test begs to differ, with R = 19 m.

Bufallo-2, 1.5 kT and detonated 0.2 m above the ground (almost peanuts) left a crater radius of 18.6 meters.
The formula argues that the radius should be 46.1 meters.

Now, can the detonations' heights of 0.5 meters matter? Possibly, but likely to a very minor degree.
To get a good bracketing idea, let's compare the formula with underground explosions then.

The Uncle test in 1951 is very interesting there. A 1.2 kT device, buried 5.2 meters below the surface, left a 78 m wide crater (R = 39 m). Adding to the casing the amplification of cratering, we also have depth, which matters a lot.
But what does the formula return for 1.2 kT?
A crater 85.62 meters wide!
It manages to make a bigger crater with an assumed more efficient device and, above all, a surface burst.

What about Sedan (1962)?
Yield, 104 kt. Buried 194 meters below the surface. Made a fantastic clean conical crater, with a radius of 186 meters.
The formula for the same yield, but a surfacic burst of a modern device?
R = 189.44 m

Finally, perhaps we can make a comparison between the estimated 184 m wide crater for a surface efficient 10 MT burst, from G&D's 1984 formula, and yours?
R = 867.9 m
When you look at the summary of the evolution of the formula, on the Glasstone blog it says the introduction of gravity brought the radius from 145 m to 92 m. We're nowhere close to the 867.9 m here.

A-HEM would be an understatement!
Can we trust an equation that is obviously so wrong even for low yields? Not really, no. The problem is, what equation to use then?

And out of curiosity, have you tried scaling your figures down to say 15 megajoules and see what that gets you?

What figures? And why 15 megajoules?

That you don't know.

I do. It's statistitically provable that the average is far more likely to be closer to the real value then a conservative figure. This is simple logic. If you have a die with 10 sides, then the number 5 will most likely closer to the real figure then 1 whenever you throw it.

Still, googling average density returned you a nice number, but this number is pretty much totally useless and false.
First, the reason you don't know that 2.8 kg/m³ is a more probable scenario is because above all things, you need an asteroid that can have a crater of 3 km.
It means you'd need to work with asteroids being at the very least that wide, and wider. It completely reduces the whole sample to a new one, and we don't know what is the average density and composition of asteroids of that size.
Secondly, the exposé assumes that the average asteroid if 10 km wide, despite establishing "that there are approximately 1 million asteroids of diameter 1 km or larger". That's to obtain an nice value that doesn't look completely bullshit, but it's just as good as the premise is: totally assumed and arbitrary.
Then the author and uses a spherical volume formula to obtain the average volume. Nevermind if asteroids of that size hardly meet the sphere criterium. Of course that would increase density, but it tells you just how accurage that exercise attempted to be: very little.
Now, the rubble pile model is very well favoured, and the 2007 Encyclopedia of the Solar System uses Mathilde as a reference.

Finally, if we look at wikipedia, for what it's worh, we see that our knowledge of the Kuiper Belt is still very limited. Still, observations made prove the existence of methane ice and water ice in the make up of several large objects.
Also, the overall mass of the belt varies a lot: 1/10th, 1/30th or 66 times Earth's mass, yet all are above the figure from that edu website you found, making it pointless as a reference.
Notice, by the way, that what you used only was the so called average mass of an asteroid of the Kuiper Belt.

But feel free to proceed with higher parameters, we'll see what you get.

High parameters would have us assume we're talking about a nickel/iron asteroids, you know, with a density around 8000 - 7000 kg/m^3.

I meant higher than the ones I suggested earlier on, below the 2.8 kg/m³ density. Use this one in E = mX if you want.
Hey, if you have some spare time, you may even plug the heavier densities, just for the kicks! :)

Next, the part about the famous "torpedo glow".

[Mr. Oragahn]

Besides, what makes the glow is there is no force field involved?

Propulsion, sensory systems, or some form of sensory jamming system, to name a few possibilities that are much more plausible then shields, because those would actually explain why the torpedoes glow the way they do.

You have hardly shown why any of these suggestions would make the glow. Throwing terms like there's no tomorrow is not a good methodology.
Propulsion? I don't even know where you're going with that.
Sensory systems? What's the link?
Jamming? Not only flaring like a star would be a stupid thing to do to mask the weapon, but torpedoes are locked if needed and plain to see on the most shitiest sensor screen ever.

See, that's why parsimony is useful. For example, what would you need to start arguing about to say the glow is related to propulsion would obviously be terribly complicated and largely made up, baseless. It would make no sense for an entire vehicle to glow just because its engines are one, while it makes sense for the same device to do so if there's a force field of some kind at play there.
We can go on that way but I suppose you get it.

Ultimately you're trying to dodge the issue though. You're the one claiming they are shielded, which means that you're going to have to fork over the evidence for it. Simply sitting there and saying “what else could it be?” is not evidence, especially when it could plausibly be a billion and one different things.

That's why the principle of parsimony is of interest, because there could be plenty of other explanations, but their validity is only as good as the elements they're composed of.
On the other hand, we have quite a lot of natural phenomena which show all sorts of fields, be they based on gravity or EM, which act as protection and can destroy matter and deflect particles, and violent fields burst in a way that is quite "solar", which would fit with what we see about torps.

But so far you've presented none. Your sole argument seems to be that because shields exist in the TNG era, they must also exist in the ENT era,...

Since it's incomplete, what you think you're quoting out of me is not my argument. The argument you just invented is like me saying Romans should have iPods because we have them now.

My argument would be that if Romans had device which looked exactly like iPods, it would be rather safe to assume they were iPods like ours.
Then, the other issue being linking shields to glow, but considering that there are not many solutions, and that those you have proposed don't fit, we're left with the one which has always been very simple and always assumed by so many: glow = force field.
In lack of a better idea, that's what I go with.

... and then you cite the glow on the torpedoes as evidence, even though you can't even prove that there's a connection between the glow and the shields.

First, we know the Klingons did have shields back in ENT.
Secondly, my point, if needs to be repeated, is not linked to TNG in particular, but to the fact that all torps glow like that, from the ships of the TOS era to the ones from DS9/VOY, and we know they had shields.

The sad fact is that the evidence is clearly against this. Starfleet during this period don't have shields, let alone stable force-field technology.

Yet humans didn't invent the torpedo technology, they got it handed from the Vissians, as I said earlier on, and I'm all open to evidence that humans are even capable of producing their own torpedoes.
You assume it, perhaps with a good reason, but I'd like to see this evidence.

Now, what torps do? They glow. Perhaps one would say like a sun. And sun burns, and its projections can deflect particles with enough force.
Could it be that the force field, in this case, is not stable, and actually not required to be, so the EM radiations and other particle emissions could also deflect and destroy matter on the torpedo's path, as well as provide a moderate defense against small caliber phaser-like weapons?

You're bonkers if you really think wild speculation like that has any merit in a debate.

It's still better than your wide speculation that fails at linking the glow to anything you listed as possibilities, without even bothering to establish the link.
The naive would be the one thinking you presented an argument.

I might as well start pondering the possibility of the torpedoes being inhabited by glowing pixies and pretend I've come up with a plausible idea supported by canon.

That wouldn't be serious, contrary to my suggestion.

Also, we’re shown how unstable force-fields work in the previously quoted episode, and oddly enough an unstable force-field doesn’t glow. It just fizzles out and disappears.

Fizzling out is precisely what a leaking shield should be: it can't retain its particles within the force field. Interestingly, I assume that by saying it disappears, it means we saw it fizzle out. Now let's imagine a form of force field that is precisely meant to leak out constantly, by bursting like a star until its reserves are depleted.

Ultimately, this whole argument is not even really useful there. As I said, it only mattered if we considered that the torp could get inside the asteroid like some torps in TNG+ could smash into matter and burrow or some such. But I did not consider this. I only evoked the significant burrowing as a possibility, yet ignored it for the calcs.

That said, the burrowing argument would matter quite a lot for torps of the TNG era and beyond.

[l33telboi]

Not really. Remember, you talked about an equation that "let's you adjust what type of composition the asteroid is made of" and that while "it assumes zero gee", we couldn't "find a better suited formula because there quite simply is none", as "[it] fits every single parameter we need it to fit."

That is the above one, yes.

Now you have brought another formula, but seem a tad reluctant to use the blog's formula, minus the gravity-relative term (mgh), despite being directly associated to the quotation picked from the Glasstone blog.

That is the formula I've been talking about all along. Unless you noticed, "the formula in the NX vs. Whitestar thread" has been mentioned rather often. And you probably also missed the fact that I've been talking about these formulas you've been using as something I have no interest in.

the parameters this other formula you brought, by citing Squishy, is allowed to be used for.

I know, because I was along for the ride when he and DF argued about it on the IRC. The first number there is a constant assiged depending on the material you're supposed to crack. It factors in how how much energy it takes to break up said material, and presumebly densities as well.

It's not my formula, I didn't even know about it until you mentionned the blog. By all means, it's yours, since it's the one that comes with the blog you cited as part of your evidence.
You use all sorts of formulas, add new ones, but not the very one found on the blog. Why is that?

I've been behind Squishy's forumla since post one. Anyone can see that. You're the one mixing and matching with whatever formulas you randomly find lying around.

Also, why do you say it doesn't address what's necessary to crack a certain material? None of them do

Squishy's formula does. And I'm right, ain't I? It's a little silly for a formula like this not to factor in hardness of a material. Indeed that's probably why you've been talking about dirt all along. But dirt is a lot easier to crack then actual rock.

Now then, you asked me to verify the figures - I have. And I'm still not interested in your jury-rigged equations.

What figures? And why 15 megajoules?

Try it, get the crater radius for a 15 megajoule explosion and see what results you get. Then try the same with the equation I provided just now. Notice something funny?

Still, googling average density returned you a nice number, but this number is pretty much totally useless and false.
First, the reason you don't know that 2.8 kg/m³ is a more probable scenario is because above all things, you need an asteroid that can have a crater of 3 km.

I can quote a number of other sources that laugh a the idea that this asteroid should be around 1500kg/m^3 in density. We go by the average density - that's final.

I meant higher than the ones I suggested earlier on, below the 2.8 kg/m³ density.

No, if you insist on calling this higher, I will insist on using a density of 8000kg/m^3 and a material hardness that of iron.

[Mr. Oragahn]

Not really. Remember, you talked about an equation that "let's you adjust what type of composition the asteroid is made of" and that while "it assumes zero gee", we couldn't "find a better suited formula because there quite simply is none", as "[it] fits every single parameter we need it to fit."

That is the above one, yes.

What "above one"?
The last one you brought contains nothing like that, not even a note about density, and there's no formula that you have actually brought that specifically refers to the "composition of an asteroid".

Now you have brought another formula, but seem a tad reluctant to use the blog's formula, minus the gravity-relative term (mgh), despite being directly associated to the quotation picked from the Glasstone blog.

That is the formula I've been talking about all along. Unless you noticed, "the formula in the NX vs. Whitestar thread" has been mentioned rather often.

I'm afraid that you'll have to be much clearer when you use the word "formula" on its own without caring to identify it properly, because you lost me here.

And you probably also missed the fact that I've been talking about these formulas you've been using as something I have no interest in.

I remember in the past that you had no issue refering to craters on Earth to argue for those 50 megatons.
Anyway, I have not asked you to use them either, and that it's been several posts ago since I last mentioned them.

the parameters this other formula you brought, by citing Squishy, is allowed to be used for.

I know, because I was along for the ride when he and DF argued about it on the IRC. The first number there is a constant assiged depending on the material you're supposed to crack. It factors in how how much energy it takes to break up said material, and presumebly densities as well.

Well, at least we finally get some information. :)
Now, what material would that constant correspond to, by chance? How do you know the constant in that latest formula would correspond to an asteroid?

It's not my formula, I didn't even know about it until you mentionned the blog. By all means, it's yours, since it's the one that comes with the blog you cited as part of your evidence.
You use all sorts of formulas, add new ones, but not the very one found on the blog. Why is that?

I've been behind Squishy's forumla since post one. Anyone can see that.

Hey? The first time you pointed at a formula, you were citing Isynic, and exchanging with higbvuyb.
Yet later on it apparently became Squish's and DF's formula or some such. This is getting messy and ultimately irrelevant.

You're the one mixing and matching with whatever formulas you randomly find lying around.

Yeah well whatever, I guess I can add that to the list of baseless accusations of dishonesty for non-disclosure of information, also having a -Trek agenda, along being bonkers and other things.

Now let's resume. higbvuyb quoted the blog to defend the formula, but the quoted material was presented by Nige to explain the E = mX + mgh formula, yet for some reason, after several posts asking you what X is about (although this is not necessary anymore, I'm just telling you that, since you don't seem to have figured that out) and asking you then to show me that mX matches something around 51 megatons, you have managed to fill several posts, yet all devoid of anything remotely approaching a result.
Just why, pray tell?

Besides, as I have shown, that formula you first posted completely fails to match any logged real nuclear tests of 0.3~1 kT and above, and that's on terrains which rate at densities around 1,700 kg/m³; it's averaged, as the Russian test involved a density of 1.85 g/cc and the soil for Nevada tests had a density of 1.6 g/cc.

So it's absolutely worthless. And it's not even a question of me supposedly having an "nerf Trek" agenda by all means, because the implication of what I'm saying is that the yields would even be greater!

However, the yield would become so huge that it would turn out to be a problem on its own. You already saw what claiming 50 megatons did, after all.

Now, let's also check out Squishy's other equation, for the same kind of yields.

From the paper, the generalized equation for surface blasts:

2.5x10^-3m per J^1/3.
(Edit: For a 3000 meter crater, that's 48.074 MT.)

Moving aside from the fact that once again, there's a constant which we don't know what it corresponds to, what this formula says is basically, 0.0025 meters per J^(1/3).
It's nothing more than the former one, safe that the constant is divided by two.

If it gives the diameter (in meters), then:

D = 2.5e-3 * E^(1/3)
E = (D * e3 /2.5)³
E = 1,728 e15 J.

413 megatons.

If it's the radius, then it's nothing more than the equation you first showed me in your first link to SBC, and still returns 216 PJ, or 51.62 megatons.
I don't know where Squishy got his 48.074 MT from, perhaps a wrong button somewhere.

That, and the fact that if the formula provides the crater's diameter, I don't even need to redo the comparisons, since as the yield is far far greater than if it was meant for the radius in that 3 km wide crater case, it obviously will fail even more to match the real tests, because what we needed is an equation returning considerably lower yields.

Also, why do you say it doesn't address what's necessary to crack a certain material? None of them do

Squishy's formula does. And I'm right, ain't I? It's a little silly for a formula like this not to factor in hardness of a material.

Smells like conjecture, guessing if it should be right or not. It's not more silly than saying a cratering equation, described as to calculate nuclear cratering (Glasstone blog's formula) would be less likely to have the part about what's necessary to crack the material.

Or perhaps, are you asserting that the blog's equation is erroneous?

Indeed that's probably why you've been talking about dirt all along. But dirt is a lot easier to crack then actual rock.

Now then, you asked me to verify the figures - I have. And I'm still not interested in your jury-rigged equations.

I don't know if citing the same figure several times, and dropping names like Squish, DF (??) matches the definition of "verification".
What I really wonder is how you could check any figure considering that you largely used appeals to authority, without actually being certain of what you were doing, quoted an erroneous figure, which shall be enough to prove that contrary to your claim, you checked neither the figures you hung to as unmatched in their accuracy, nor the blog's content, and merely repeated Squish's erroneous result. Add to that how you tried to recall vague memories, and it's only now that we start to talk about that the constants are the parts that are related to density, for some reason, and that the values differ.
We're yet to know if we used the right values though. Which means getting the right light of constant for each formula.

Besides, the only one term formula I'm talking about since several posts is one that has nothing to do with the ones I mentioned much early on about asteroid cratering, but is the blog's one. That's all.

What figures? And why 15 megajoules?

Try it, get the crater radius for a 15 megajoule explosion and see what results you get. Then try the same with the equation I provided just now. Notice something funny?

I'm asking you what figures for a reason, because I need to know what formulas you're talking about. You may have not noticed, but there's been quite a good number of figures presented.
So please be specific. Tell me what formula you used, and tell me what formulas I shall use to make a comparison.

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?

15 MJ nuclear firecrackers don't really exist atm, and the only explosives that could provide that firepower and being small enough would be very different. You know rather well, for having argued it countless times here and at SBC, that it's only around high yields that the differences in the structure of explosives (without considering casing or directed bursts that is) don't really matter. The reverse should be equally true.

Still, googling average density returned you a nice number, but this number is pretty much totally useless and false.
First, the reason you don't know that 2.8 kg/m³ is a more probable scenario is because above all things, you need an asteroid that can have a crater of 3 km.

I can quote a number of other sources that laugh a the idea that this asteroid should be around 1500kg/m^3 in density. We go by the average density - that's final.

There's nothing "final" until you provide quotes.

I meant higher than the ones I suggested earlier on, below the 2.8 kg/m³ density.

No, if you insist on calling this higher, I will insist on using a density of 8000kg/m^3 and a material hardness that of iron.

Geez... I said 2.8 kg/m³ was higher than the 1.xx ones I mentionned. You want to call 2.8 kg/m³ average or conservative, sure, by my guest. I call them higher. Wait. Ûber Higher now. There.
We both know nickel-iron is not conservative anyway.
As I said, feel free to use that 2.8 kg/m³ density of yours, and even play with nickel-iron density if you wish.
You know the formula.

[l33telboi]

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.

[Mike DiCenso]

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