Base Delta Zero

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Post by Who is like God arbour » Tue Jun 19, 2007 10:33 am

Mr. Oragahn wrote:Mike's figure was the most generous, and that's based on generous exploitations of the 900 km DS9. Even the "tube", the conundrum between the DS2 and the moon, he made it bigger than what it is, by makign it a cylinder.
That shouldn't be a negative criticism at Mike's calculation.
I have understood, that he has deliberately choosen to take the most generous figures "for 'fairness' sake".

But now, it would - as it seems to me - be appropriate to take realistic figures to get a realistic result.

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Post by Mr. Oragahn » Tue Jun 19, 2007 10:36 am

Who is like God arbour wrote:
Mr. Oragahn wrote:Mike's figure was the most generous, and that's based on generous exploitations of the 900 km DS9. Even the "tube", the conundrum between the DS2 and the moon, he made it bigger than what it is, by makign it a cylinder.
That shouldn't be a negative criticism at Mike's calculation.
I have understood, that he has deliberately choosen to take the most generous figures "for 'fairness' sake".

But now, it would - as it seems to me - be appropriate to take realistic figures to get a realistic result.
Agreed.
The results will be funnay.

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Post by Mr. Oragahn » Tue Jun 19, 2007 3:24 pm

I'll use the values which Gari M. Sarli considered correct:

- DS2 orbit: 2130.9 km above surface.
- DS2 radius: 80 km (diameter: 160 km).

For further analysis:

- Endor radius: 5200 km (Saxton's figure).

Image

I consider that the shield's radius around the DS2 should be 1.5 larger than the station's radius. This is probably superior to the real value, and even extends beyond the range shown on the rebel hologram.
Also, I considered that the altitude between the DS2 and the surface of Endos was the distance between the ground and the tip of the battle station's south pole.
This will lead to a larger surface area for the shield.

Let's calculate the lateral surface area.

Image

Height: h
Radius of bases: r, R
Slant height: s
Lateral surface area: S

R = 120,000 m
r = 20,000 m
h = 2,210,900 m


s = sqrt([R-r]²+h²)
S = Pi(r+R) x s

S1 = Pi(r+R) x sqrt([R-r]²+h²)
S1 = Pi(140000) x sqrt(100000²+2210900²)
S1 = 973,398,769,019.74103328701168820348 m²

Surface area of the hemisphere (since I considered that the cone's largest base was located at the equator of the DS2):

S2 = (4 x Pi x R²)/2
S2 = 90,477.86842338604526772412943845 m²

S = S1 + S2
S = 973,398,859,497.60945667305695592753 m²


What about a shield around Endor?

- Endor radius: 5200 km (Saxton's figure).

A planet could not afford loosing its mesosphere, which is below the exosphere. It's important for natural protection against radiations. It has to be protected.
What about the exosphere?
Earth's mean radius is 6,372.797 km. The upper boundary of the exosphere extends at 10,000 km above the surface, and the lower boundary is located 500 km to 1,000 km above the surface.
So Endor's exosphere's upper boundary would probably be located at roughly 1.569 times Endor's radius, above the surface, or +8,159.68 km above the surface.

Though this is a great distance, it is true that since the light gases are found in the exosphere, it means that heavier gases and other particules will see their orbit drop over time.

There wouldn't be that many ways to cleanse a poisoned exosphere. Either torching it with absurd levels of energy, which would damage it anyway, or find a way to neutralize the poison via chemicals or radiations, and this would possibly create secondary ill effects. Not to say that the volume to cleanse is enormous.

If the planet has to be protected, it has to be protected entirely, and thus the shield will need to be slightly below 2 planetary diameters large.

So Endor's planetary shield's the surface area would be:

SE = 4 x Pi x (5200000 + 8159680)²
SE = 2,242,859,018,200,218.3711015620362272 m²

Divided by the [DS2-Endor]'s shield surface area, we obtain:

2,304.1520917312380994306188666987

So roughly 2,304 shield generators.

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Post by GStone » Wed Jun 20, 2007 1:31 am

Where did the 40 kms figure come from? I know I saw it somewhere before, but I can't remember.

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Post by Mr. Oragahn » Wed Jun 20, 2007 2:06 am

Invented.
It's far more than what the holo shows, in terms of proportions, which means that the shield's area would be smaller than what I obtained.
But we don't really know that much about how far the shield extends, and I'm not in the mood to go look at ROTJ to see if there's a clue about how far the Tydirium shuttle was landed from the base.
So I was generous.
It is roughly what I would expect if, for example, Hoth's shield was extruded upwards.

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Post by Mike DiCenso » Wed Jun 20, 2007 10:44 pm

Well, one way to calculate out the radius of that area would be to try and figure out how many kilometers a day the Rebel commando team could have reasonably covered in about a day's time, still maintain some level of stealth, and account for the stop over in the Ewok village.
-Mike

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Post by Mr. Oragahn » Wed Aug 08, 2007 1:13 pm

Mike DiCenso wrote:Well, one way to calculate out the radius of that area would be to try and figure out how many kilometers a day the Rebel commando team could have reasonably covered in about a day's time, still maintain some level of stealth, and account for the stop over in the Ewok village.
-Mike
Yes. Any idea on how to do that?

More on the initial subject. I'd like to adress the Nar Shaddaa BDZ.

While reading an old thread here, I clicked on a link that led me to an old thread on Spacebattles.
There, a member tried to obtain firepower figures from the events of The Hutt Gambit.
Here are the calcs:
Heh. I shouldn’t really be here but I heard of this from a friend and couldn’t resist.

This new information fits in quite well (although we don’t yet know how big these quad cannons are, they might be as large as ISD HTL’s), it supports the planned 15 minute Nar Shaddaa BDZ using 3 dreadnoughts and 4 Bulkcruisers (not exactly top of the line ships). That particular operation had it not been ordered to fail would have entailed wiping out 72-95 billion inhabitants and "razing the entire moon", the surface as described by Page 207 of the SW encyclopaedia:

"Nar Shaddaa - the ungoverned "smugglers' moon", it is completely covered by interlocking spaceport facilities and kilometers high docking towers reaching into orbit. Nar Shaddaa's "vertical city" was built over thousands of years and is protected by often-malfunctioning shields."

20 Psi blasts do not guarantee destruction of modern concrete buildings (5.6.2 of the FAQ), 100 Psi would probably be required to flatten all-steel construction buildings if we were to try building "multi Km" structures today (refer to DasBastard’s comments in the "tale of two mikes thread").

From the NukeFAQ:

http://www.fas.org/nuke/hew/Nwfaq/Nfaq5.html

r=Y^0.33*const-bl

Where this constant is ~2.2*P^-0.7 (P in Psi, refer to 5.6.2 of the FAQ)
I.e. for 100 Psi 0.088

Using:
a) Surface area of Mars (diameter 3394Km)
b) Surface area of Earth (diameter 6250Km)

And assuming the ships in squadron (primarily the dreadnoughts as the bulk cruisers apparently don’t even have turbolasers) between them are capable of the same number
of shots as a single WEG "60TL" ISD (seem reasonable 1 ISD=3 dreadnoughts?), in 15 minutes of continuous firing they could rattle off 54,000 shots.

In case a) that has to cover an area of 1.45e8 Km^2, i.e. blasts of radius ~29 Km. requiring a yield of 37 GT per shot.

In case b) that has to cover an area of 4.91e8 Km^2, i.e. blasts of radius ~54 Km. requiring a yield of 232 GT per shot.

Even if we said the squadron was equivalent to 3 WEG ISD's and thus capable of 162,000 shots the figures are as follows:

a) Mars sized- 17Km blasts, 7 GT per shot
b) Earth sized- 32Km blasts, 45 GT per shot.
I wanted to know how many ships were involved there, how long the bombardment would have lasted, and just how big Nar Shaddaa was.
As such, I landed on Wong's mail exchange with Edam:

http://www.stardestroyer.net/Empire/Hat ... -2b-3.html
Hutt Gambit Lies
[Defending use of "Hutt Gambit" as BDZ] Because they ordered a BDZ - it shows what a BDZ is, and how some survivors are acceptable. It is also supported by Cronus' attack on Khomm, and the three ISDs attacking Dankayo, the attack on Emberlene (VotF) - there are some survivors after the initial attack.
As usual, you are fabricating facts. They did not execute a BDZ because they weren't even trying! Read the quotes from p249:
    • "Greelanx had memorized the message, and that was easy, for it had been short.

      The message had read: "Admiral Winstel Greelanx, eyes only, destroy after reading. Regarding Nar Shaddaa/Nal Hutta ENGAGEMENT. You are advised for the good of your Empire to engage the enemy and suffer a strategic defeat. Minimize Imperial losses, and withdraw in good order.

      Repeat: you are to LOSE, Admiral. Do not attempt to confirm these orders. Do not discuss them with anyone. If you fail to comply, no excuses will be accepted."
And how about the following quote from p282?
    • "The admiral sighed. His battle plan called for the entire engagement to be over in less than fifteen minutes. He had better get busy, figuring out how he was going to manage to lose."
Get it? They were ordered to fail! It was apparently a fake order, but Greelanx believed it was real, and acted accordingly. Therefore, it is ridiculous to use their limited effectiveness as proof of BDZ limitations! Their misson timeframe was only 15 minutes, so their attack was designed to do token damage during that 15 minutes! Stop using this bullshit red herring, Edam. Yet again, you have distorted the facts beyond recognition and hidden important information. See the following quotes:
    • "Suddenly scrutiny from the Empire brought al normal life on Nar Shaddaa to a screeching halt. Moff Sarn Shild proclaimed the Hutts' lawless territory would benefit greatly from stricter Imperial control. As a public-relations stunt, Shild was authorized to blockade Nal Hutta and turn the smuggler's moon into molten slag."- Essential Chronology p29.

      "Han watched as the Princess drifted closer and closer to the large moon. Nar Shaddaa was actually the size of a small planet, almost a third the size of Nal Hutta."- Hutt Gambit p56.

      "Han tensed, but made himself stay calm. He could tell Greenlax was really tempted. "Sir, what are your orders?" he asked. "Perhaps we can think of something that will benefit us both, and yet leave you free of any charge of wrongdoing."

      Greelanx laughed bitterly, a short, bitten-off laugh. "Hardly, young man. My orders are to enter the Hutt system, execute Base Delta Zero upon the smuggler's moon Nar Shaddaa, and then blockade Nal Hutta and Nar Hekka until the Hutts agree to allow full customs inspections and a complete military presence on their worlds. The Moff doesn't want to cripple the Hutts too badly, but he wants Nar Shaddaa reduced to rubble.

      Han swallowed, his mouth dry. Base Delta Zero was an order that called for the decimation of a world- all life, all vessels, all systems- even droids were to be captured and destroyed. His worst nightmare come true."- Hutt Gambit p223/224.

      "Nar Shaddaa- the ungoverned "smugglers' moon", it is completely covered by interlocking spaceport facilities and kilometers-high docking towers reaching into orbit. Nar Shaddaa's "vertical city" was built over thousands of years and is protected by often-malfunctioning shields... most of the moon's seventy two to ninety five billion inhabitants live in the highest levels of the spaceport-cities."- SWE, p207.
Get it now? Nar Shaddaa is not freakishly small, it is covered with multi-kilometre urban agglomeration, and a BDZ does exterminate "all life, all vessels, all systems". But what am I telling you for; you are the one who keeps bringing up "Hutt Gambit" even though if you have read it, you know perfectly well that it is not the proof you need. You are shamelessly misrepresenting it, taking events out of context and witholding key information (such as that minor detail about how the Imps were ordered to fail!)
It fits all the stated requirements of a BDZ. The only reason it shouldn't be considered is if we think there is more to a BDZ than is actually stated. Obviously, you do (otherwise you wouldn't be lying about the SWTJ description of a BDZ on your imperial weapons page), but we can't simply go round saying "that source isn't right, it's far worse than that" for no reason other than we want to
Repeating the lie again.
[Downgrading Nar Shaddaa's size] Nar Shadda was a moon of an inhabitable planet (moon, not sister planet), which means it wasn't particularly large. infact, you'd be lucky if it was Mars sized. If it was half the size of the Earth you'd ahve quarter the surface area. Oh, look, my figures are higher than that. My examples gave the first hour of the attack - it actually takes much longer (a matter of hours, not one single hour)

[Ignoring multi-kilometre tall buildings on Nar Shaddaa] And the majoriy of buildings in Star Wars are NOT multi-kilometre tall buildings. Those that are will take additional firepower, but an "idealised" BDZ will never happen. Everyone will be different, because every planet is different. SW buildings in general don't seem any different to modern buildings. There's no reason for them to be built better, so why assume they are?
See above quotes. You continue to lie.
The SWTJ says an ISD can slag a planet, but nothing anytime anywhere says a BDZ requires slagging of a planet, so anyone claiming that it does is lying. a BDZ (not requiring the entire slagging of a planet) takes a few hours for an unknown number of ships. Obviously, if you want to slag the planet to an arbitrary depth you'll need more ships and more time.
...
The Nar Shadda attack was designed to fulfill every requirement of a BDZ. That it was not completed does not invalidated the stated goals, and the fact they were acceptable as a Base Delta Zero operation
See above quotes. You continue to lie. And your energy estimates are completely inadequate regardless of whether we slag the planet or pound its surface to rubble.
An interesting exchange, on which there is much to say.

So looking at wankypedia, I wanted to know more about The Hutt Gambit.

Now, there are several key factors not to forget.

1. Chances are that when a building, even a modern one, has its upper levels sqashed by thermonuclear level blasts, they will crumble on themselves, or fall down on the neighbouring buildings. In a urban agglomeration such as the one on Nar Shaddaa, we'd be expecting a domino cascade.
Especially if the buildings, much like the planetary shields, are not pristine.

2. Wong liked to insist on how the BDZ was to leave no life, no vessels, no systems. He didn't, however, insist on the idea that droids could be captured. Once again. The wookipedia page also revealed that troops were landed on the moon, to find survivors. No mention of the magmatroopers.

3. We can see, once again, a contradiction. Page 29 of the Essential Chronology says that the moon was to be turned into molten slag. Well, this could surely count as hyperbole. Even more, because there's precisely no mention of the surface.
Taken literally, truly literally - and not cherry picking - that would mean all solid materials need to be melted, at least. That means melting the whole crust. And of course, the solid part of the core and mantle.
Pointless and overkill. And of course, this would make it impossible, and equally pointless, to send mop up teams for further inspection of, ahem, survivors.

Taking the comment literally is absurd. Just limiting ourselves to the terms molten slag, but add their our sauce, by assuming that it only meant the surface, and of course the multi-kilometer high buildings, is fallacious.
That said, in the light of parsimony, and strategical sense, it should mean the surface. It should be impliedly clear.

It also highlights a major problem I've seen, a question of philosophy maybe; What takes precedence, between a guide, and a story?
This is a subject for another thread.

4. As a continuation of the point above, Greelanx is rather very clear about the goal of the operation:

"My orders are to enter the Hutt system, execute Base Delta Zero upon the smuggler's moon Nar Shaddaa, [...]. The Moff doesn't want to cripple the Hutts too badly, but he wants Nar Shaddaa reduced to rubble.

As we can see, there's a difference between turning a small planet into molten slag, and have it reduced to rubble. Both are hyperboles of their own kind.
We likely realize that in fact, they're talking about the surface.

5. All life has to perish. But please, let's remember that the EU largely points out how poisonous and radioactive turbolasers are. Coupled to raging fires spread over a planet's whole surface, resulting from thousands of thermonuclear blasts spread evenly over the entire surface, how could anything survive here?
This will be made easier, since "most of the moon's seventy two to ninety five billion inhabitants live in the highest levels of the spaceport-cities".

They're living on the points which are easiest to shoot, but most importantly, easiest to destroy, since you just need to topple the buildings to kill them. The crash of the buildings on themselves will leave no survivor, since all the surface is about tightly packed buildings. Coupled to the turbolaser bolts.

6. How many ships were used here, precisely? The sectorial fleet was moved to Nar Shaddaa. According to wookiepedia, there were sixteen Guardian-class light cruisers. There were a couple of capital ships there as well. The flagship was a Dreadnaught-class heavy cruiser, named Imperial Destiny. So I suppose the other capital ships also were heavy cruisers of the same class.
SB.com's member Evil S'Tan assumed only 3 heavy cruisers, and 4 light cruisers, instead of 16.

7. The book describes Nar Shaddaa as "a small planet, almost a third the size of Nal Hutta."
Nonetheless, Evil S'Tan uses Earth and Mars as models. Maybe Mars is a good starting point, but using Earth is outright wrong.
Is it an error, or a dishonest move to reach an upper limit?
Well, we don't know, but it's wrong from boot.

8. In his calc, he already ignores the light cruisers, since they apparently don't have turbolasers - oh there, another idea of theirs, wherein laser cannons are peashooters in the EU.

He assumes that 3 dreadnoughts = 1 ISD. Based on West End Games's (horrible) 60 turbolasers figure, by 900 seconds, he gets his 54,000 shots.
Nevermind if by WEG's rules, capital ship turbolasers are often rated at a firepower of 4-5D, while laser cannons have 2D.
Yet, it's okay to ignore them. More for the bigger guns I suppose. :|
So there's like 16 ships which are ignored.

More, the dreadnoughts are actually say to sport "10 turbolaser cannons, 20 quad turbolasers, and 10 turbolaser batteries" (wookiepedia).
The turbolaser cannons are likely the heavier guns. Probably two barrels per turret.
The Quad turbolasers are likely similar pieces, but with four barrels per turret. Maybe a slightly lower firepower, for a higher rate of fire.
The turbolaser batteries are likely racks of lighter turbolasers. We just don't know how many cannons a battery has.

The WEG. guide begs to differ on this.
It gives:
- 10 laser cannon, 2D.
- 20 quad laser cannons, 4D.
- 10 turbolaser batteries, 7D.

For the note, WEG says an ISD's 60 TLs are all the same, and all dealing 5D of damage. It's likely, at best, an averaged value.

We see that laser cannons can be as powerful as turbolaser cannons.
Ultimately, a dreadnought is more like 1/2 or 2/3 of an ISD (but is said to lack good shields and good sublight engines).

As for the 16 light cruisers, this means that they could have anything from 2D cannons to 4D cannons.
Even if each ship only had one 2D cannon, it would still be equivalent to a total of 8 4D cannons.
Of course, I suppose that light cruisers are likely to have more cannons than that.
For example, a Carrack-class light cruiser is said to have 10 HTLs (7D) and 20 laser cannons (2D), which puts them at 2/3 each of a heavy cruiser.
So they are hardly to be ignored.

Of course, the use of WEG's figures is most dubious, especially since they fail at giving ISDs any heavy turbolasers, despite the large turrets.
Wookiepedia lists 73 TLs, all calibers mixed (but a precise repartition is also provided).

But gloablly, he seems to underestimate the power of the 3 heavy cruisers, and completely dismisses the presence of 16 other warships.

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Post by Mike DiCenso » Thu Aug 09, 2007 12:40 am

Mike DiCenso wrote:
Well, one way to calculate out the radius of that area would be to try and figure out how many kilometers a day the Rebel commando team could have reasonably covered in about a day's time, still maintain some level of stealth, and account for the stop over in the Ewok village.
-Mike
Mr. Oragahn wrote:

Yes. Any idea on how to do that?
The problem is in getting an idea of how fast a squad of soldiers could manage at least a day's time. We know that the capture by the Ewoks cost time for Luke, Han, and the others, while the other part of the team went on to rendezvous at the bunker. But let us cut out the hours and the night spent at the Ewok village. Let us say the team went for 16 hours during a day, which, assuming the team made a couple kilometers per hour, would be 32 km, and then came to a stop to rest the night , and then reach the bunker in the morning (Han tells the remaining squad that they'll meet at the bunker at "0300 hours").

So I'am roughly estimating at most 40 some kilometers, also assuming no rest stops are made, except for nightfall.
-Mike

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Post by Mr. Oragahn » Thu Aug 09, 2007 3:18 pm

Mike DiCenso wrote: The problem is in getting an idea of how fast a squad of soldiers could manage at least a day's time. We know that the capture by the Ewoks cost time for Luke, Han, and the others, while the other part of the team went on to rendezvous at the bunker. But let us cut out the hours and the night spent at the Ewok village. Let us say the team went for 16 hours during a day, which, assuming the team made a couple kilometers per hour, would be 32 km, and then came to a stop to rest the night , and then reach the bunker in the morning (Han tells the remaining squad that they'll meet at the bunker at "0300 hours").

So I'am roughly estimating at most 40 some kilometers, also assuming no rest stops are made, except for nightfall.
-Mike
That would make the radius at least 40 km wide, thus the conundrum's base would be like 80 km wide.
Twice my take.

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Post by Who is like God arbour » Thu Aug 09, 2007 4:07 pm

I think, it's too far.

They had to move through a forest and they had to be extremly cautious to not alert other scouts or to activate potential positioned sensors.

Under these condition, they could move only very slow.

On the other side, they weren't en-route much time:
  • They landed and an indeterminably time later they have met the first scouts.
    Then the trip of Luke and Leia was following, while the task force with Han and Chewie has waited.
    Luke had to run the whole way back.
    • And as fast as they have flown, that was a not so little distance.
    Then he, Han, Chewie, C3PO and R2D2 have to walk the whole distance again to find the site of Leias crash.

    Now the whole Ewok-village-events were following.

    In all that time, they haven't approached the base - unless the Ewok village was per coincidence on the route to the base.

    Altogether they haven't had much time in which they were approaching the base.

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Post by Mike DiCenso » Thu Aug 09, 2007 9:59 pm

Mr. Oragahn wrote:
Mike DiCenso wrote: The problem is in getting an idea of how fast a squad of soldiers could manage at least a day's time. We know that the capture by the Ewoks cost time for Luke, Han, and the others, while the other part of the team went on to rendezvous at the bunker. But let us cut out the hours and the night spent at the Ewok village. Let us say the team went for 16 hours during a day, which, assuming the team made a couple kilometers per hour, would be 32 km, and then came to a stop to rest the night , and then reach the bunker in the morning (Han tells the remaining squad that they'll meet at the bunker at "0300 hours").

So I'am roughly estimating at most 40 some kilometers, also assuming no rest stops are made, except for nightfall.
-Mike
That would make the radius at least 40 km wide, thus the conundrum's base would be like 80 km wide.
Twice my take.

I would like to note that this is likely an uppermost limit. The presumption was that the Rebel commandos did not stop to rest at all until night fall, nor did they have to take any time to hide or check for sensors/traps on their way to the bunker.
-Mike

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Post by Mr. Oragahn » Sun Feb 10, 2008 7:13 pm

Just for the record, you now have at Spacebattles one big thread, which we can safely claim to be one of the most detailed BDZ related threads made in the recent years, as far as I've been able to check.

The thread was closed upon extremely disputable criteria, but it was complete enough to have a good look at all the data relative to the BDZ order.

While it started as a broad SW weapons' strenght, it ultimately ended into a complete and exhaustive analysis of the Base Delta Zero order, so that's why I put it the link here.
It also took a look at other sources which aren't related to the BDZ order, but still serve the point just as well.

It ends with another very short lived follow up, where among other things, Poe was asked to defend his many and erroneous claims, and notably present the evidence of the huge numbers predating the AOTC:ICS.

We already have a thread about that one here:

The Lost Scrolls: Uber Firepower calculated for Star Wars

I'll come up with a summary of the discussion I had with Vympel/Leo1, but we can clearly see where the errors creep in, and identify the fallacious standards they've been using, which in the end point out how Saxton's inprint on the ICS has made these two books be the sole EU sources to ever claim such over the top firepower capabilities.

More than one week later, Vympel didn't post his rebuttal, despite his further participation to the discussion, nor did he try to post it in the second thread I opened, trying to obtain a delay by noticing the mods of his plan to post said rebuttal of his own.

Thus far, since his arguments were more or less all of SDN's in a nutshell, his lack of response was considered a whole concession. If he really wanted to, he could have asked the mods to reopen the thread, for him to post his reply, or PM me, but he did not.

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Post by Cock_Knocker » Sun Feb 10, 2008 7:52 pm

Mr. Oragahn wrote:It ends with another very short lived follow up, where among other things, Poe was asked to defend his many and erroneous claims, and notably present the evidence of the huge numbers predating the AOTC:ICS.
Which has been presented for quite a few years now, and which you've proved you've known about: The Mike January calcs. But that doesn't stop from your, "Poe can't prove those numbers were around before the ICS" rhetoric, does it?

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Post by Mr. Oragahn » Sun Feb 10, 2008 9:38 pm

Cock_Knocker wrote:
Mr. Oragahn wrote:It ends with another very short lived follow up, where among other things, Poe was asked to defend his many and erroneous claims, and notably present the evidence of the huge numbers predating the AOTC:ICS.
Which has been presented for quite a few years now, and which you've proved you've known about: The Mike January calcs. But that doesn't stop from your, "Poe can't prove those numbers were around before the ICS" rhetoric, does it?
It doesn't stop me doing so for simple reasons.

When asked to provide the calcs, you didn't comply.

You can complain all you want that I aware about those faulty calcs beforehand, but first, Spacebattles is not just about me you know, there's like plenty of members on that board, many of which don't even know about those calcs and would have liked to hear about them, and secondly, the rules forbid linking to other discussion boards.
Thirdly, I really wanted to know if the calcs you mentionned were the exact same.

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Post by Cock_Knocker » Sun Feb 10, 2008 9:46 pm

Mr. Oragahn wrote:It doesn't stop me doing so for simple reasons.

When asked to provide the calcs, you didn't comply.

You can complain all you want that I aware about those faulty calcs beforehand, but first, Spacebattles is not just about me you know, there's like plenty of members on that board, many of which don't even know about those calcs and would have liked to hear about them, and secondly, the rules forbid linking to other discussion boards.
Thirdly, I really wanted to know if the calcs you mentionned were the exact same.
Now you know, and actually, have known, whether you could link someone else or not.

So can I expect to see no more posts with, "Poe can't prove those calcs were around before ICS?"

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