Figuring out the yield of Apophis' supership's destruction

[Mr. Oragahn]

Remember the first episode Enemies, from the firth season.

There, both SG-1 and Apophis jump out of hyperspace 4 million light years away from intended plot point, because of the destruction of Vorash's sun, which propelled both ships farther than planned.

They end in a part of a galaxy controlled by Replicators. They control an alien ship which attacks Apophis' supership, a much bigger and much more powerful ha'tak.
That supership is basically an advanced, superior, ha'tak, something like 3.65 times bigger than a standard ha'tak.

We learn this:

JACOB: He's arming weapons.

JACK: Shields?

JACOB: Not responding. Not that they'd do much good anyway. That ship has weapons powerful enough to penatrate our shields at full power.

Note: all ha'taks have enough power to penetrate another ha'tak's shields, even at full power, over time.

This can only mean that the supership's weapons are far more powerful.

Apophis sets the self destruct on, and the Replicators can't reprogram the ship fast enough to stop it. SG-1, as they had boarded the ship, realize this, and try to escape as fast as possible.

Here's how things roll out, once they're back inside Cronos' ship:

INT—CORRIDOR, KRONOS' MOTHERSHIP

[Carter, O'Neill and Jacob are running down the corridor.]

CARTER
There's not enough time to fix the hyperdrive, is there?

JACOB
We're going to have to hope the sublight engines can get us far enough away.

CARTER
Of course, that's only going to matter if the replicators can't shut down the auto- destruct.

JACOB
They'd have to learn to reprogram a Goa'uld computer awfully quick.

[They run onto the bridge.]

CARTER
It's possible, if they get control of that ship.

O'NEILL
Hey! You guys are just being too negative!

JACOB
We're not going to make it!

DANIEL
Going as fast as we can.

EXT—SPACE

[SG-1's ship is flying away from Apophis' mothership as it explodes.]

INT—BRIDGE, KRONOS' MOTHERSHIP

JACOB
Right, shut down the engines, divert full power to the shields!

DANIEL
(stepping out of his way)
Be my guest.

[Jacob takes the controls.]

EXT—SPACE

[The orange glow of the shields appear just as the blast wave from Apophis' ship washes over them, leaving them undamaged.]

Two important things to know.

First, Apophis' ship was so far away that we couldn't see it just before it exploded.

Secondly, as the ha'tak got its engines shut off, and slowed down (inertia dampeners, same as the tech on puddle jumpers or Murio's cargo), the wave splashed against its shields, and this pushed the ship. Slightly, but noticeable up to the point where it moved a bit faster after being hit, and banked a little bit.

Thirdly, it was clearly a case of life or death, as shields were absolutely necessary. This means the ship could have not protected the crew without shields, yet it succesfully protected the crew for one hour when sitting close to a blue giant.

So we're talking about an important level of kinetic due to a flux of unknown particles, released from an explosion that occured far far away.

That has to be measurable, with mass, displacement over time and thus acceleration, or the levels of energy calced for ha'tak shields (what was it, several megatons per second, right?) and the inverse square law regarding the distance between the ha'tak and the supership.

What's the eye's maximum sight range, roughly?

If we need to estimate an acceleration, what's the most conservative size for a ha'tak? I had one which was about a 675.88 meters wide ha'tak, for a height of 255.9 meters. What amount of mass could we get from that, depending on the materials we'd consider?

[l33telboi]

First off, man i really need to re-watch the older SG episodes. It seems as if i've forgotten a ton of important stuff. Or things that could be quantified.

Secondly, as the ha'tak got its engines shut off, and slowed down (inertia dampeners, same as the tech on puddle jumpers or Murio's cargo), the wave splashed against its shields, and this pushed the ship. Slightly, but noticeable up to the point where it moved a bit faster after being hit, and banked a little bit.

Could you get the KE from that by using the KE formula? I.e. you first use the KE formula to get the initial, pre-impact, energy. Then compare it with the post-impact energy. The only thing changing in the formula would be the velocity, so that should be easy, if this approach is valid.

If we need to estimate an acceleration, what's the most conservative size for a ha'tak?

I've used the round figure of 500m for length. But i've only used that when trying to get extremely rough stuff. Though i have to point out that there are several episodes (including the latest one, Dominion) that seem to show the Ha'Taks being quite a lot larger then BC-304:s. So the size thing can very quite a lot.

What amount of mass could we get from that, depending on the materials we'd consider?

No idea. I usually go for a simple "Compare with Nimitz-Class" approach, it's been used by others, but other then that i couldn't say if it's really valid. Though in this case, i guess it would, Ha'Taks are supposed to be massively armored, and Naquadah, which comprizes the hull, is usually made out to be quite a heavy metal (considering Stargates for instance). So i guess it would be ok when getting a minimum value for mass.

[Mr. Oragahn]

First off, man i really need to re-watch the older SG episodes. It seems as if i've forgotten a ton of important stuff. Or things that could be quantified.

I believe there are other things to quantify, but Enemies has been one of those rare episodes where you could get calcs while we were living during a period of non descriptive events typical to Stargate. It's only been lately that apparently Stargate gets very specific, generous and verifyable about yields.

Could you get the KE from that by using the KE formula? I.e. you first use the KE formula to get the initial, pre-impact, energy. Then compare it with the post-impact energy. The only thing changing in the formula would be the velocity, so that should be easy, if this approach is valid.

In a case of constant acceleration for a finite amount of time, I guess it would not be too hard. But it's not my favorite method. I'd rather go with the hull's protective factor and calcs done about ha'tak shielding.

I've used the round figure of 500m for length. But i've only used that when trying to get extremely rough stuff. Though i have to point out that there are several episodes (including the latest one, Dominion) that seem to show the Ha'Taks being quite a lot larger then BC-304:s. So the size thing can very quite a lot.

I think the right size for a 304 is around 450m in lenght, so a ha'tak could still be bigger than such a ship and still fit in the less than 1 km wide ballpark.

No idea. I usually go for a simple "Compare with Nimitz-Class" approach, it's been used by others, but other then that i couldn't say if it's really valid. Though in this case, i guess it would, Ha'Taks are supposed to be massively armored, and Naquadah, which comprizes the hull, is usually made out to be quite a heavy metal (considering Stargates for instance). So i guess it would be ok when getting a minimum value for mass.

It would probably even be too conservative.

I'll try to retrieve the ha'tak shield calcs from SB.com, and see what was the estimated wattage at the distance considered.

For example, if the calcs showed that a ha'tak's hull could protect the crew by withstanding 1 MT per second, over one hour, we could already estimate that the wave that hit the ship was at least worth 1 MT, if not 3600 MT, if we consider the total energy.

We'd just need the eye's acuity range, for a ship such as Apophis' one, remembering that it was rather dark than bright, for a ship that would be a around 2.5 km wide, though the "brightest" part of the ship, the pyramid core, would be slightly wider than a ha'tak (maybe 10% more at best).

This has to give us some info, or we'd have to make assumptions about the camera's resolution angle I suppose.

[l33telboi]

In a case of constant acceleration for a finite amount of time, I guess it would not be too hard. But it's not my favorite method. I'd rather go with the hull's protective factor and calcs done about ha'tak shielding.

Using the pre-established durability of the hull might be better yes, because we'd only get the KE by using the method above, so there would be a lot of other factors not taken into account.

I think the right size for a 304 is around 450m in lenght, so a ha'tak could still be bigger than such a ship and still fit in the less than 1 km wide ballpark.

Yes, but in episodes like the one where Baal is stealing Stargates, the Ha'Taks seem to be at least twice as wide as a BC is in length. But who knows, going for the minimum value would in any case be the right thing to do in this case.

It would probably even be too conservative.

That it undoubtedly would.

I'll try to retrieve the ha'tak shield calcs from SB.com, and see what was the estimated wattage at the distance considered.

I think it would be better to make a completly new one instead. I believe things like size and the lika were way off in those calcs. Though my memory could be a bit shoddy. In any case, Wong's put up a calculator for exactly these kinds of isntances, all you have to type in is the luminocity size of the Ha'Tak jadda jadda jadda, so it shouldn't be too hard.

I'll check back later today, too see if we can't get some specifics going.

[l33telboi]

Ok, it seems to be a slow Sunday today (not even the almighty Khorak has graced our IRC channel with his presence yet, although it's Khorakurday and all), so I thought I’d get started on the calc.

I’ll start with the blue star incident, since that will be used as a basis for how much the hull of the Ha’Tak can take.

I believe the star was referred to as a "Blue Giant", and as such, we would probably be talking about a B-class star, as the O-types are luminous to the point where they're almost completely white and are also usually referred to as supergiants.

So the Luminosity for the star would be 20 000 Lsun. And 1 Lsun (the luminosity of our sun) is 3.827*10^26 W. So we're talking about 7.654e30W for a blue giant (3.827*10^26 * 20 000).

Next up is the size of the star, Wiki says that a blue giant is about 7 times larger then our own sun, which means it should be 4.827*10^9m in radius.

Now then, to get Luminosity/Area we first need the area of the blue star, which should be 2.928*10^20m^2. Then to get the wattage involved, let's divide Luminosity with the Area of the sun, which should be 2.614*10^10 W/m^2.

Ok, so far everything is pretty much identical to the calc made on SB. However now comes the fun part. Using the inverse square law to see how much of this radiation would be splashing against the Ha'Tak.

For this, we first need to determine the distance from the surface of the star to the Ha'Tak. Now this will most likely involve a heavy dose of guesswork, since the distance was never said, except that the ship was inside the Corona. The only way I can think of figuring this out is by simply looking at the pic where we see both the star and the Ha'Tak and guessing how far it was from the surface. For now, I’ll just use the solar radius as distance, the real value is undoubtedly much lower then that, but I’d like to get some input, or perhaps some other way of gauging distance before doing this more detailed.

One radius from the surface means that 1m^2 on the Ha'Tak will get a fourth of the wattage 1m^2 of the surface of the sun gives out. 6GW/m^2 in other words, which is pretty much identical with what Wong's calculator says, in other words, I believe I did it right.

Now all that is left is to examine how much of the surface area of the Ha'Tak was directly exposed to the sun to be able to figure out how much radiation the whole ship took.

After this all we need is to get the distance between Apophis exploading mothership and the escaping Ha'Tak to be able to use the inverse square law to see how powerful the blast would have been at the source. Of course, this calc could also be a lot more refined then it currently is, so any suggestions are welcome.

SIDENOTE: Wong's calculator also gives the temperature (15436K), this is interesting, merely because of the CME incident with the BC-304. It was said that the heat the ship would be receiving would be the most problematic for the ship, meaning that we are probably talking about some serious heat coming from the plasma stream. This is interesting in the sense that many people were suggesting the heat of the stream would be about equal to the photosphere of the sun (5800K), which seems highly unlikely now. In any case, that was just a sidenote.

[Mr. Oragahn]

There's been thus far two attempts at calculating the blue giant event:

The first, from An Anicient: http://forum.spacebattles.com/showthread.php?t=84338

His calcs are nice in the sense that he provides a low, medium and high end intensity we can use.
His mitake, however, was to assume a 2 km wide ha'tak, which earned me a heated discussion with him and some nasty bits.

The second one is from Murazor: http://forums.spacebattles.com/showthread.php?t=83332

They're good, but he only provides the high end and stops at locating the ha'tak in the photosphere. Of course, since we're doubling the radius, we just have to divide his high ends by four, but I don't get why he didn't do it.

That said, your calcs fit with An Anicient's ones, before he used the wrong shield area. That's around two digits tons per second.

As for the surface of the ship, well, here's an old model of a ha'tak we can eventually use to know how much of the hull is facing the star.

We already know it will be 2.614*10^10 joules multiplied by the exposed hull area.

Actually, when we think about it, more hull was exposed to the sun radiations than was exposed to the wave.

Here's an old measurement I made a long time ago:

(I can't remember where I put the calcs I used to obtain the pel'tak trench height though).

Below is a pic showing an estimation of distance ship-photosphere I tried to obtain. I can't remember which focus angle I adopted, but I remember it closely matched a distance of one sun radius.
Unfortunately, this is a cropped version of the original PSD I've lost on a damaged HDD.

After this all we need is to get the distance between Apophis exploading mothership and the escaping Ha'Tak to be able to use the inverse square law to see how powerful the blast would have been at the source. Of course, this calc could also be a lot more refined then it currently is, so any suggestions are welcome.

We'll able to do so once we'll get the first numbers.

SIDENOTE: Wong's calculator also gives the temperature (15436K), this is interesting, merely because of the CME incident with the BC-304. It was said that the heat the ship would be receiving would be the most problematic for the ship, meaning that we are probably talking about some serious heat coming from the plasma stream. This is interesting in the sense that many people were suggesting the heat of the stream would be about equal to the photosphere of the sun (5800K), which seems highly unlikely now. In any case, that was just a sidenote.

Where is that calculator which incorporates the temperature?

[l33telboi]

(I can't remember where I put the calcs I used to obtain the pel'tak trench height though).

You always make the nicest scalings. In any case, i don't have the time now, but i think i can get on the whole measuring thing later on today.

Below is a pic showing an estimation of distance ship-photosphere I tried to obtain. I can't remember which focus angle I adopted, but I remember it closely matched a distance of one sun radius.
Unfortunately, this is a cropped version of the original PSD I've lost on a damaged HDD.

The Ha'Tak was a whole solar radius from the blue giant? Are we talking our suns solar radius or the blue giants solar radius? If it is one blue giant radius, then the figures i gave should be quite spot-on.

Where is that calculator which incorporates the temperature?

Here. It doesn't incorporate the temperature though, it merely shows us what the equlibrium temperature at that distance would be. Though i wonder if he's accounted for that whole "the further from the photosphere, the hotter and less dense it gets" thingy. In any case, it's in the box labeled "Equlibrium Temperature".

Question, when we're measuring the radiation (as in the luminosity), the heat coming from the sun doesn't get taken into account, right? So would it be possible to convert the temperature to Watts, and then incorporate that value as W/m^2, adding this to the already existing radiation figure?

[Mr. Oragahn]

You always make the nicest scalings. In any case, i don't have the time now, but i think i can get on the whole measuring thing later on today.

You mean the exposed surface? What kind of primitives could be used?
A tetrahedron and some kind of elipsoid capsule?
Self masking surfaces and other voids should be considered as well (the dark scaffolding is full of such gaps).

The Ha'Tak was a whole solar radius from the blue giant? Are we talking our suns solar radius or the blue giants solar radius? If it is one blue giant radius, then the figures i gave should be quite spot-on.

It was close to one radius from that blue giant.

Here. It doesn't incorporate the temperature though, it merely shows us what the equlibrium temperature at that distance would be. Though i wonder if he's accounted for that whole "the further from the photosphere, the hotter and less dense it gets" thingy. In any case, it's in the box labeled "Equlibrium Temperature".

Question, when we're measuring the radiation (as in the luminosity), the heat coming from the sun doesn't get taken into account, right? So would it be possible to convert the temperature to Watts, and then incorporate that value as W/m^2, adding this to the already existing radiation figure?

Well, since luminosity is the amount of energy radiated by a black body I think (with the side effect of being more or less bright, as luminous), whatever figure we obtain from luminosty, as power, is all we can get from a star I'd say. At least, anything of relevance. Heat will turn into EM radiations as energy is lost by bodies, so it seems to pretty much stop there.

As a side note, remember Robert's page on ISd acceleration and weapon range.
http://www.st-v-sw.net/STSWwarsrange.html#II-B-1

For an object 885 meters wide, occupying ten pixels on the screen (and with an assumed 30° angle of view), he already got a distance of more than 135 km.

Just imagine what we'll get with a ship 2.65 times bigger than a ha'tak (so that would be 1.77 km for a 675 m wide ha'tak, though we could focus on the supership's core only, which itself is still slightly bigger than a ha'tak), and yet it is not even visible!

See the following gif:



Frames (zoomed in):

1. Vanilla still, last frame before explosion.
2. Vanilla still, first frame of explosion.
3. Explosion with contrast enhanced. Crosshair.
4. Former frame, with enhanced contrast. Crosshair.
5. Former frame, with enhanced contrast.

[l33telboi]

You mean the exposed surface? What kind of primitives could be used?
A tetrahedron and some kind of elipsoid capsule?
Self masking surfaces and other voids should be considered as well (the dark scaffolding is full of such gaps).

Wouldn't it be better to instead just use a plain old 2D model instead of 3D one? Both the radiation and the shockwave will hit anything that is within their line of sight. Wouldn't this mean that one can take your image and then simply map out the 2D area of the ship it provides? Naturally there are going to be discrepancies, like if the ship moves a greater area will be exposed and the large 'light-source' is going to mean that every nook and cranny of the ship is going to get lighted.

But this would give us a workable base, no? That is, if we need this figure at all. I mean, a W/m^2 figure should be enough for us to determine the source strength, right? What would be more interesting is how much of Wattage the should could have taken all at once to the hull.

As a side note, remember Robert's page on ISd acceleration and weapon range.
http://www.st-v-sw.net/STSWwarsrange.html#II-B-1

Good! I've actually been looking for a way to calc that very same thing. And here the formula comes landing right on my doorstep, as it were.

In any case, i'm going to do a quick calc to see what distance this will yield.

For the purpose of this calc i'm going to assume the ship was 1 pixel wide and a 30 degree angle, like RSA did (i have no idea whether that would be the right thing to do though). The ship, as you said, would be 1770m long. The image i have of the incident is 640 pixels in width and the ship would be 1, giving us a theta value of 1.5625e-3.

.5 A / D = tan (.5 (theta))
.5 (1770m) / D = tan (.5 (1.5625e-3))
D = 885m / tan (7.8125e-4)
D = 64,904,659.03m

So yeah, it's quite far away indeed. If i did things right, which i in no way guarantee. I never was good with numbers.

So now to see what this gives us if we factor in the inverse square law and take the already established W/m^2 value. The W/m^2 value is a wee bit low though, it would be better to figure out how much of that wattage the hull could take in one single blow, as that would give us a more accurate figure.

S = I*4*pi*r^2
S = 6000000000*4*pi*64904659^2 = 3.18e26W (or 76PT/s)

But like i already said, i was never any good with numbers so this should all be checked. Consider this a post that can be edited and corrected.

[Mr. Oragahn]

Wouldn't it be better to instead just use a plain old 2D model instead of 3D one? Both the radiation and the shockwave will hit anything that is within their line of sight. Wouldn't this mean that one can take your image and then simply map out the 2D area of the ship it provides? Naturally there are going to be discrepancies, like if the ship moves a greater area will be exposed and the large 'light-source' is going to mean that every nook and cranny of the ship is going to get lighted.

But this would give us a workable base, no? That is, if we need this figure at all. I mean, a W/m^2 figure should be enough for us to determine the source strength, right? What would be more interesting is how much of Wattage the should could have taken all at once to the hull.

Hm, yes. After all, we know that the ship could not withstand a wave of that power. Since the calcs regarding the ha'tak's shield were based on a solar intensity at the distance where the ha'tak was sitting, from the sun's core, we don't need to calculate the hull surface in fact. The intensity is all we need.

However, 6 GW/m² was what the hull could withstand per second. There'd be no urge to raise the shields if the intensity was that. We learn that the hull could resist for a full hour before failing.

That's 21,600 GW/m². Technically, I think that's the real intensity the ha'tak will face when hit by the explosion's wave.

There's also the fact that the figure is based on an intensity from a low end blue giant.

Good! I've actually been looking for a way to calc that very same thing. And here the formula comes landing right on my doorstep, as it were.

In any case, i'm going to do a quick calc to see what distance this will yield.

For the purpose of this calc i'm going to assume the ship was 1 pixel wide and a 30 degree angle, like RSA did (i have no idea whether that would be the right thing to do though). The ship, as you said, would be 1770m long. The image i have of the incident is 640 pixels in width and the ship would be 1, giving us a theta value of 1.5625e-3.

.5 A / D = tan (.5 (theta))
.5 (1770m) / D = tan (.5 (1.5625e-3))
D = 885m / tan (7.8125e-4)
D = 64,904,659.03m

So yeah, it's quite far away indeed. If i did things right, which i in no way guarantee. I never was good with numbers.

So now to see what this gives us if we factor in the inverse square law and take the already established W/m^2 value. The W/m^2 value is a wee bit low though, it would be better to figure out how much of that wattage the hull could take in one single blow, as that would give us a more accurate figure.

S = I*4*pi*r^2
S = 6000000000*4*pi*64904659^2 = 3.18e26W (or 76PT/s)

But like i already said, i was never any good with numbers so this should all be checked. Consider this a post that can be edited and corrected.

It seems that you forgot to multiply the FOV angle to obtain theta, which is an angle.

With your numbers, I get theta = 0.046875°, or 4.6875e-3°.

Thus a distance of 2,163,488.5137408317764773305948765 m.

Now, other things that I'd change.
I'd fit with the parameters which the still was taken with. I took it from the video as it came on my screen. Its dimensions are 1024 x 576. So that's 1024 pixels wide instead of 640.

This would give me a theta of 2.9296875e-3°.

Thus a distance of 3,461,581.7396416590983962392815957 m.

1 pixel is rather pretty generous considering that the ship is literally non existent on screen. 1 pixel will be, in our case, the low end, and any smaller figure would be a high end (like 0.1 px).

Thus a distance of 34,615,818.143081742773493318130835 m

There's also the fact that the figure is based on an intensity from a low end blue giant. Plus my ha'tak dimensions may be undersized, and they correspond to the older models from the earliest seasons, while the newer ones seemed slightly bigger. Cronos' ha'tak was of the newer generation (but pre-Anubis).

Basically, I'd make the calc with a conservative distance of 3,461,581.7396416590983962392815957 m, coupled to a threshold intensity of 21.6 TW/m².

S = 21600000000000*4*pi*3461581.7396416590983962392815957^2 = 3,252,466,242,023,299,023,418,483,760.5449 W (~3.252e27 W).

Ergo, 776,838,215,826,717,068,74426381975351 tons of TNT. Rounded to 776.838 PT/s.

Now, it's per second, but considering that there's no way to know if it's a complete exposure (as duration) that was a lethal a danger, or just a fraction of that time, the figure will either be 776.838 PT spread over the duration of the whole wave, or 776.838 PT, first divided by the amount of time necessary to damage the ha'tak (for example 776.838 PT per 0.2 seconds), and then multiplied by the whole wave's duration, which lasted a couple of seconds.

So that figure comes as the bottom-of-the-barrel-scratching-anal-retentive figure, because we use one pixel, because we use the smallest ha'tak measurement, because we use the weakest blue giant, and because we assume that the ship would have been destroyed after a full elapsed exposure to the wave.







Extra info:

I have the PAL (25 full frames /s), widescreen version 16:9.

• Frame #1: Shield becomes visible and immediately fizzles over its whole surface.
• Frame #45: The blue wave hits the ship.
• Frame #86: Shield is still under strain, but intensity slightly disminishes.
• Frames #97-98: The shield flickers. Not from null to full, but from full intensity to a lower yield. Yet, it's still active as a whole.
• Frame #115: Last frame for which we can see the shield active. The intensity has been slowly decreasing until this frame.

[l33telboi]

Disregard this post.

[Mr. Oragahn]

Using An Anicient's medium calcs (before shield area being inputed), and dividing the intensity by four (since his numbers are based on the photosphere surface), we get 9.225e10 W/m².

Mixed to the hull threshold, we get 33.21e12 W/m², and using the same formula as above (with distance still obtained from a one pixel wide ship):

S = 33.21e12 * 4 * pi * 3461581.7396416590983962392815957^2 = 5e27 W

Or 1,194,388,756,833,577,493.1943056228712 T/s, rounded to 1.194 ET/s.

For a low to medium end, it's funny how this would bring us close to the level of power generation (1 )described in McCay's first book, for the Eye of Ra (Ra's battleship), though we're far from being close to those levels of firepower.

Well, it was a supership. If we consider the teraton level crash on Delmak, and associate the alt-figure of 200 MT per shot for a ha'tak, we can estimate that a goa'uld mothership could - at that time - make a cannon spit out e-6 of its total energy potential.

Of course, it's likely that the crash itself represents a yield of possibly two-digit TT, and that goa'uld standard motherships can fire salvos up in the low gigaton.

Anyway, with such a ratio, considering the power output found above, that would be around 1.197 teratons per salvo for the supership, possibly one order of magnitude more, which would fit relatively well with ha'tak shields estimations (high triple digit gigaton to single digit teraton), since Jacob said that the supership had weapons powerful enough to penatrate their shields at full power.

[Mr. Oragahn]

What's important to notice is that since the ship is not visible, we'll always have to use the ratio of one pixel against the maximum width of the video medium, because obviously, the clearer and more refined video quality has to be used to spot the ship.
If the episode was to be released as a new format with a higher definition, say HD whatever, we'd have to use that newer definition, and thus the numbers would grow even more.
That basis tells us that even using one pixel for the ship is a very special kind of low end in that case, one that is actually under the real low end.

[Mr. Oragahn]

Just some sidenote about the ha'tak's sublight acceleration, based on this event.

The self destruct was one minute away from blowing the ship up when Jacob called in Jack and Sam on the radio, still onboard Apophis' ship. Then they came rushing down the corridor with replicators on their six, Jacob, Jack and Sam run down the corridor and enter the ring room, ring out, get on the ha'tak and say to Daniel to go.
That has to be more than ten seconds, but let's say ten.
So when Daniel pushes the engines to max thrust, they have 50 seconds before the ship blows up.

We've seen earlier on that they crossed more than thousands of kilometers in that time. As usual, say that half that distance was used for acceleration, and half for deceleration.
So we have 3460 km within 50 seconds, that gives us 1730 km within 25 seconds, or a mean velocity of 69.2 km/s.
This value could be considerably higher, by one order of magnitude, considering that the distance I worked from was itself obtained by assuming that the super ha'tak was 1 pixel wide, which of course it wasn't as it was literally not visible at all, which implies a much greater distance (the one obtained by working from a width of 0.1 pixel was ten times higher).

This suggests an acceleration of a couple of a couple hundreds G.