Ha'taks

[Emperor Tippy]

A little while ago there was an animation going around of an asteroid about 500KM in diameter hitting Earth. It basically liquefied the planet. Very cool animation. It even shows the gravitational tides from Earth heating up the Asteroid as it approaches. Anyway, when you mentioned "cracking continents" that is what I thought of. So I tried to calculate the yield such an asteroid would give (pretending that it was a kinetic warhead basically). I ended up getting a yield of 3 billion gigatons. ... wow. I don't know if I did that right, but if I did that is a freaking lot of energy 0_0. Kinetic Warheads are soooo the way to go if you want to destroy a planet.

Yeah. At .8C the energy released on impact is 100% of the impacting object. Anything over that and its more powerful than an equal mass of Anti-Matter.

At .95c a fist sized iron projectile releases 22 MT's of energy across the impact area (so about a 3 inch diameter circle). A continent killer is about the size of a house.

[gopher65]

Yeah. At .8C the energy released on impact is 100% of the impacting object. Anything over that and its more powerful than an equal mass of Anti-Matter.

At .95c a fist sized iron projectile releases 22 MT's of energy across the impact area (so about a 3 inch diameter circle). A continent killer is about the size of a house.

EDIT: I'm just going to edit the whole post. When I originally did this I accidentally forgot to convert grams to kilograms, so I ended up with 87 gigatons (heh). It seemed excessive, so I posted my whole process here so someone could point out to me what I did wrong. Eventually I figured it out on my own, and edited the post all over the place. It was really messy, so I'm just going go edit out my mistakes, and leave this here as an explanation. /EDIT
0_o so uh, how exactly are you getting that? I got 87 megatons. It is quite possible that I'm wrong, so I'll post what I did here with all work included so you all can figure out what I did wrong, if I did anything wrong.

I'm rounding here, but I left the full numbers in my calculator at each step.

1)First, I figured out how big the ball of iron was, and what its mass was:

r = (1.5in)*(2.54cm/in) = 3.81cm
V = (4/3)*(pi)*(r^3) = (4/3)*(pi)*(3.81cm^3) =231.67 cm^3.
I did a quick search, and the density of iron that came up was 8(g/cm)
M = Volume*Density = (231.6 cm^3)*(8 g/cm) = 1853.3 g
M(kg) = (M(g))/(1000(g/kg)) = (1853.3g)/(1000(g/kg)) = 1.8533kg

2)Next I used a rearranged version of the relativistic kinetic energy formula which you should be able to find on any number of websites (If I made a mistake, it is probably here):

m=mass(kg)
y=speed(%speed of light, decimal form)
KE(joules) = ((1/sqrt(1-(y^2)))-1)*(m)*(9*10^16) = ((1/sqrt(1-((0.95c)^2)))-1)*(1.8533kg)*(9*10^16) = 3.6739*10^17 joules

3)Next I converted to megatons:

mt = (KE(j))/(4.184*10^15 (j/mt)) = (3.6739*10^17j)/(4.184*10^15(j/mt)) = 87.81 megatons

There. That is the newly edited version. Sorry about that.

[Col. Shadow Quinn]

A little while ago there was an animation going around of an asteroid about 500KM in diameter hitting Earth. It basically liquefied the planet. Very cool animation. It even shows the gravitational tides from Earth heating up the Asteroid as it approaches. Anyway, when you mentioned "cracking continents" that is what I thought of. So I tried to calculate the yield such an asteroid would give (pretending that it was a kinetic warhead basically). I ended up getting a yield of 3 billion gigatons. ... wow. I don't know if I did that right, but if I did that is a freaking lot of energy 0_0. Kinetic Warheads are soooo the way to go if you want to destroy a planet.

You mean this one?

[Emperor Tippy]

Well just a quick glance tells me you came up with the mass in grams in step 1 and plugged it in in step 2 as kilograms.

So divide the final number by 1,000.

That brings it down to about ~88 MT's.

Using the various unit calculators available at http://www.unitconversion.org and this kinetic energy calculator http://www.csgnetwork.com/kineticenergycalc.html I got 17.983784 MT's (using your figure for the mass of iron).

I used 299,792,458 for c in m/s.

It is highly dependent on mass. A 1 kg object at .95c has 9.693195 MT of kinetic energy on impact.

EDIT:
A couple of other things. At the speeds and amount of energy we are discussing your impacter is almost guaranteed to just punch right through whatever it hits. This isn't so much of a problem if it impacts a shield (because the shield will most likely overload trying to block that much energy in that small an area) but it is a big concern after you punch through the shields. You will just punch a fist sized hole straight through the target. And unless you happen to his fairly important systems the ship will be ok.

So what you should do is create a shaped charge impactor that blows before it hits, spraying out millions of much BB sized or smaller impactors. It's essential a shotgun shell instead of a hard ball round.

Oh and the size an object needs to be to totally destroy earth if traveling at .95c is 4,907,961 kilograms (roughly). Again I recommend the shotgun type blast unless you just want a hole through the planet.

[gopher65]

You mean this one?

Yeah that is the one Col. Shadow Quinn. I love that animation.

Well just a quick glance tells me you came up with the mass in grams in step 1 and plugged it in in step 2 as kilograms.

So divide the final number by 1,000.

That brings it down to about ~88 MT's.

Yeah, I caught that a few minutes ago. I can't believe I did that. Oh well.

Using the various unit calculators available at http://www.unitconversion.org and this kinetic energy calculator http://www.csgnetwork.com/kineticenergycalc.html I got 17.983784 MT's (using your figure for the mass of iron).

Well, you can't use a Newtonian equation to calculate a relativistic velocity. It won't take into account the increase in mass of the object as the velocity increases.

[Emperor Tippy]

Yeah that is the one Col. Shadow Quinn. I love that animation.

Yeah, I caught that a few minutes ago. I can't believe I did that. Oh well.

Well, you can't use a Newtonian equation to calculate a relativistic velocity. It won't take into account the increase in mass of the object as the velocity increases.

Yes, that is true.

It's still somewhere between 20 and 80 MT's. I saw teh figures for it a while back but I don't remember it exactly. And with slighty differning masses you get very different numbers.

[Mister Oragahn]

Well just a quick glance tells me you came up with the mass in grams in step 1 and plugged it in in step 2 as kilograms.

So divide the final number by 1,000.

That brings it down to about ~88 MT's.

Using the various unit calculators available at http://www.unitconversion.org and this kinetic energy calculator http://www.csgnetwork.com/kineticenergycalc.html I got 17.983784 MT's (using your figure for the mass of iron).

I used 299,792,458 for c in m/s.

It is highly dependent on mass. A 1 kg object at .95c has 9.693195 MT of kinetic energy on impact.

And 46.94 MT on relativistic kinetic energy.

EDIT:
A couple of other things. At the speeds and amount of energy we are discussing your impacter is almost guaranteed to just punch right through whatever it hits. This isn't so much of a problem if it impacts a shield (because the shield will most likely overload trying to block that much energy in that small an area) but it is a big concern after you punch through the shields. You will just punch a fist sized hole straight through the target. And unless you happen to his fairly important systems the ship will be ok.

I think that the more armour there would be behind that shield, the worst the effects would be. The more there's dense materials to try to stop an object, the more confrontation there will be. Friction, and lots of heat due to sudden stress, deformation, against the tensile properties of the materials that make up the superstructure of a vessel.

That's why in Andromeda, the structure is so special that it's made weak, in a sort of way, on purpose apparently, to actually let solid projectiles pass through while doing the lesser damage possible.

So what you should do is create a shaped charge impactor that blows before it hits, spraying out millions of much BB sized or smaller impactors. It's essential a shotgun shell instead of a hard ball round.

Well of course, there are issues with this. If you try to flatten your object, by keeping the same mass, but fiddling with the volume, you're going to get funky effect due to elasiticity.

Ideally, it would be better to fire a thin lens/pancake-looking round, rather than a bullet shaped projectile.
Assuming you're in vacuum, and that the object is accelerated thoroughly, it will hit more hull surface.
Like a big slap with the palm, rather than a seringe's sting, I think.
It won't go as deep, but what it does on the surface is largely more noticable, which is quite what matters here.

But trouble is that the more spread it is, the more subject to deformations it is. Avoiding this would require a perfectly homogenous coil gun.

[gopher65]

Yes, that is true.

It's still somewhere between 20 and 80 MT's. I saw teh figures for it a while back but I don't remember it exactly. And with slighty differning masses you get very different numbers.

It will scale linearly with the mass itself, because you are simply multiplying the mass into the rest of it. So, 10 times the mass equals 10 times the kinetic energy in a perfectly elastic collision with any velocity, assuming that the two velocities are the same.

[Emperor Tippy]

It will scale linearly with the mass itself, because you are simply multiplying the mass into the rest of it. So, 10 times the mass equals 10 times the kinetic energy in a perfectly elastic collision with any velocity, assuming that the two velocities are the same.

Oh, yeah it scales linearly but each .1 kg increases the energy (according to Mister Oragahn's figures) by ~5 MT.

[Jarnin]

Raw and unrefined are just the same.

The comma was there for grammar's sake. Example: A giant, blue hamster.

I suppose Naquadah Ore would be a better description.

Again, we'd probably need a page dedicated to naqahdah, filled a long list of quotes from the show.

Agreed. There was a thread I contributed to where I tried to explain how Naquadah could be used for bombs, building materials, super conductors, etc by using different allotropes of Naquadah.

But I don't think they've made a distinctive difference between refined naqahdah and weapon grade naqahdah.

Well, here's a quote where they use refined Naquadah and weapons grade Naquadah interchangeably.

(Some scientists are working on a bomb. Above, in the observation room, Carter is working. Bauer enters the observation room.)

BAUER: At ease. How are things progressing, Major?

CARTER: Well, the hardware's almost ready. But the problem is the ratio of Naquadah to Lithium Hydride. According to my calculations, we don't have enough weapons-grade material to complete the bomb.

BAUER: Then we'll just have to get some more.

CARTER: Sir, we're talking about refined Naquadah. It's extremely rare.

It doesn't make sense that refined and weapons-grade would be the same thing. Writer's mistake?

Why, exactly? It's just an even more powerful variant of naqahdah. Used for shield, hyperspace, sublight drives and weapons, if the Goa'uld can control such a power, then fine, really.
I don't see any problem with it.

Here's the problem:

First, we don't know how a Naquadah reactor works. If it's a simple fission reactor, then neutron emission from weapons grade material would start a chain reaction (if Naquadah is analogous to Uranium).

Second, with weapons-grade Naquadah, you'd want the chain reaction. With refined Naqudah, used for building materials and superconductors and such, you wouldn't want a chain reaction.
If you had a ship built from weapons-grade Naquadah, neutron radiation would cause it to go into chain reaction mode and explode.

Yes. There's a couple of other ships that do as well. I bet the sort of ion thrusters on al'keshs are just there for extra boost... but really, I don't see the point, considering the accelerations ha'taks can achieve.
They just added them because glowing engines is cool.

Maybe the Al'kesh uses the same inertia drive system as a Tel'tak, and the boosters are there to provide extra thrust to overcome those targets?

Huh, I tried, but didn't get any. Not only wiki search engines are the devious fruit of spelling nazis, and lack flexibility.

I haven't had problems using the search engines in a wiki. You just have to filter through the crap and use Ctrl+F a lot.

Huh, I pretty much dismissed the idea of triangle based ha'taks landing on classical pyramids since a long time.

Me too, but it's still canon that they're supposed to be able to land on pyramids. That means they're mostly hollow.

You got to wonder why they never had the gall to show landing pads especially made for the regular ha'taks, really.

Because they'd be tetrahedrons, not square pyramids, and fans would notice the difference between those and say, the great pyramid at Giza.

The fans haven't really notice the Ha'taks being tetrahedrons because of the angles they're shown, and we've only see one or two landing on a pyramid. Most people don't even notice things like that.

Not sure about what you mean.

I'm just saying that we don't have a firm number for the Mark-9. Saying it's 812 GT is like saying the sky disperses light at exactly 490 nm, when it's actually a pretty big range.

All I'm saying is that number should be presented as an approximation, not a well tested figure.

That said, with the evolution of gatebusters, it seems fairly and stupidly easy to increase the yields. If you'd fill an entire space rocket with naqahdria, you'd get enough energy to crack entire continents and entirely set ablaze the atmosphere of a planet, vapourizing all oceans within a couple of hours, turning your target into a dead world.

Too bad the Ori have control over the only source of Naquadria. I wonder how many Mark-9s are left in the SGCs arsenal?

We should probably try to steer this back on topic, since this thread is supposed to be about Ha'taks, not Naquadah

[gopher65]

Ore: Raw Naquadah, dug from the ground
Unrefined: The ore has been heated, and the Naquadah has been melted out, separating it from the useless rock.

^^ that is undeniable. That's just what those two words mean. Where the speculation comes in is what the difference is between Refined Naquadah and Weapons grade Naquadah.

If it is anything like Uranium (and every other element), then it exists in multiple isotopes while in both its Raw and Unrefined form. With Uranium only one Isotope is unstable enough to be capable of achieving a fission chain reaction. So to "refine" Uranium you separate out the two Isotopes from each other. Reactor grade refined Uranium is about 5% U235 (the good stuff). They need the 95% of U238 to act as a neutron buffer in order to keep the reaction from going supercritical (well, it can anyway, but they can control it). Weapons grade refined Uranium is normally 95% U235, or, basically, as pure as they can reliably make it. If you stick that stuff in a reactor it explodes.

Whether or not Naquadah is the same way is up for question. I've seen no evidence thus far in the series to suggest that there is such a thing as Reactor Grade Naquadah.

[Lord batchi ball]

Those numbers are way over my head
How do the gliders get back in the Ha'tak after they are launched?

[Emperor Tippy]

Those numbers are way over my head
How do the gliders get back in the Ha'tak after they are launched?

Magnets

[Lord batchi ball]

If you had magnets every glider would just fly towads it then you have a mess.

[gopher65]

The comma was there for grammar's sake. Example: A giant, blue hamster.

I would prefer "A giant, blue gopher." personally.