Starcraft vs Halo

[Sapper007]

again reading comp 101
try figuring for carbon steel not iron...
like what i used...

[Stark]

I think lecturing people on reading comprehension when you haven't posted a complete sentence in this thread instead of addressing the eight-magnitude error is probably a bit of a stretch there sailor.

[Thanas]

again reading comp 101
try figuring for carbon steel not iron...
like what i used...

So you are saying carbon steel is worth eight orders of magnitute? Source, please.

Also, what is your source for BCs being build of carbon steel?

[Agent Sorchus]

Ok, I am dealing with an ignorant buffoon. I can deal with this.

First, you never tell what type of carbon steel you are pretending to use. There is no way for me to look up something that is totally not defined. Whereas Iron is easy to look up and should work as a way to determine order of magnitude. See carbon steel is a name that can be used for all sorts of different alloys, including 99% Iron 1% Carbon by mass. Using Iron as an estimate for 99% Iron is perfectly valid.

Second you are so ignorant of energy magnitude that is used to vaporize stuff that I weep for humanity. Look here. This chart lists off how large 10^14 joules is, falling roughly halfway between the energy of a one Megaton bomb and the fat man, or about 100kilotons. You honestly expect me to believe that we can produce a material that with only two Kilograms can resist a nuke? Cause if we could guess what? We would have fucking Orion Drives already and I would live on the moon.

Your calcs are based on bullshit.

Thanas > when doing this sort of analysis one must make a certain number of assumptions, carbon steel would be good enough of a replacement for "neosteel," except that the term carbon steel is too broad to use without actually naming the grade. Even if he names the grade it is inferior, because of how hard it is getting exact data on differing grades of steel. That is why I simplify the matter by using raw Iron. No confusion on terms there.

PS: yes Sapper007, it is important to remember sig figs because the extra figures are not worth the time to not convert to scientific notation. But then it would be easier to see where your numbers are bullshit.

[Sapper007]

Fine you got me...
I used AISI Steel Alloy 4340 for its use in armor in modern combat vehicles...(which classifies as carbon steel)
its density is 7.83g/cm^3
specific heat is 490
boiling point 1282 (C)
I used 0 as the starting temp (instead of -272C)
used E = mcp(T2-T1) to figure the required energy to reach heat of vaporization...
and
E = mhv
to figure the the total energy expended upon complete vaporization...

where did i go wrong?

[Sapper007]

You honestly expect me to believe that we can produce a material that with only two Kilograms can resist a nuke? Cause if we could guess what? We would have fucking Orion Drives already and I would live on the moon.

Ground zero - Hiroshima

The supports for the dome on the bank was made of iron...

[Lancer]

You honestly expect me to believe that we can produce a material that with only two Kilograms can resist a nuke? Cause if we could guess what? We would have fucking Orion Drives already and I would live on the moon.

Ground zero - Hiroshima

The supports for the dome on the bank was made of iron...
http://farm1.static.flickr.com/22/31790 ... fc120b.jpg

You're discounting the fact that Little Boy detonated half a kilometer above ground level. All that damage you see is from the blast, not the fireball from the detonation itself.

[Stark]

If only the metal struts survive in a 40s Japanese town... maybe the building was destroyed by fire and not vaporised.

Just saying.

[Agent Sorchus]

I am going to have to retract some of what I said above. You did not make an error in the MJ/kg required, your method differs from mine enough that it was hard for me to see that.

What you did make an error in was in squaring the mass. E=mcp? and then plugging in E=mhv?, why did you do that. It meant that you multiplied the mass by the mass. All you had to do was simple dimensional analysis. Take the specific heat capacity for 4000 series alloy steel, which is easier to find than 4340 and is .475j/(g*K), close enough to your .490j/(g*K) for our work. Then you cancel grams based on the weight of the armor that is vaporized (1.17e12 grams), by the change in Temperature (1300K is close enough for our math). You now have the total energy (7.25e14 joules, or roughly in the 100kt-500kt range).

But in yours you went a step further and multiplied by mass again. None of this would have happened if you A) kept good scientific notation and B) didn't drop units. B is more important and would have told you that you didn't need to go on to your second step.

As for Hiroshima, the metal did not suffer more than a very small fraction of the blast thanks to the, uh, Inverse square law I believe. It did not suffer every joule of energy that the Nuclear bomb put out.

An alternative method of determining the Yamato blast is to assume that the fireball that is generated prior to the beam is the entirety of the Nuke's fireball and compare that to known fireballs. Doing so from the clip I posted in my first post you can see that the beams diameter is almost equal to the hammerhead section, but not quite. Looking around that would be slightly less than the radius of a 500kt nuke, about 330meters in minimum radius. An 150Kt nuke would have a ground burst fireball of 330m radius. Since both of these easily fall within the range I determined for vaporization of 4000 series steel, it is additional proof that I am within the actual range.

[Agent Sorchus]

I would also like to point out that while 4340 steel is called carbon steel, it is not a carbon steel proper. It is a Molybdenum steel. A proper carbon steel is 1060 plain carbon steel and any other 1xxx series steel.

It is not a major point, but is one that I thought you should know.

[Sapper007]

I used the 1st formula to find the energy required just to reach heat of vaporization...
the second one is used to determine the energy required to completely vaporize the mass...

[Stark]

That's neat, since you've so conclusively established that all the material was vapourised.

[Sapper007]

That's neat, since you've so conclusively established that all the material was vapourised.

since the beam goes clean through in a fraction of the second... and all i was calculating for was the initial armor plate... not the second or third it goes through... yes

[Stark]

I'm sorry, was that evidence? I didn't even notice a capital letter.

What do you think a spaceship with a path vapourised through would look like? HINT - what happens when metal is rapidly evaporated and has nowhere to go?

[Agent Sorchus]

The first formula is all you need. When you did the second your units became joules*kilograms, not an energy value.

Look sapper I am going to post my math and that is all.



Stark normally I would ask you to stop helping, but you know what? This guy is too much of an idiot, so please be my guest.

[Sapper007]

The first formula is all you need. When you did the second your units became joules*kilograms, not an energy value.

Look sapper I am going to post my math and that is all.

Actually both are required.
The first calculation is to discover the heat of vaporization of the total mass of steel. The point at which you get the entire mass of steel to boil. The next calculation is determine the continuous application of energy to vaporize the mass.

For water its much simpler, because we know the certain temp's and energy output required, and have the applications for this.
For example we know boiling water vaporizes at 2260 J/g or 2260000 J/Kg.
It means 1g of water absorbs 2260 J of energy to vaporize at its boiling point. So with the latent heat of vaporization of water, we can calculate how much energy is required to vaporize any amount of already boiling water.
*note - This does not include the energy expended to get the water to boil.

So we will use the exact mass of steel from my picture. And compare the results:

E(required)=Mass*Heat of Vaporization(the boiling point)
so...
1,195,523,550*2260000=2.70x10^15

I wonder why your number for energy required to reach the boiling point of steel and then vaporizing it is less than just vaporizing already boiling water...?

[Guardsman Bass]

Well, at any rate, it means that Terran game mechanics may be a lot closer to canon than we otherwise have imagined.

Which makes you wonder about their manufacturing capabilities. Even discounting the obvious game mechanics, they still seem to have some pretty good compact manufacturing, if Raynor's Foundry Guy on the Hyperion was any indication.

[Formless]

I wonder why your number for energy required to reach the boiling point of steel and then vaporizing it is less than just vaporizing already boiling water...?

Are you seriously suggesting that it takes a 100 KILOTON NUCLEAR BOMB just to vaporize water?

[Xess]

Actually both are required.
The first calculation is to discover the heat of vaporization of the total mass of steel. The point at which you get the entire mass of steel to boil. The next calculation is determine the continuous application of energy to vaporize the mass.

For water its much simpler, because we know the certain temp's and energy output required, and have the applications for this.
For example we know boiling water vaporizes at 2260 J/g or 2260000 J/Kg.
It means 1g of water absorbs 2260 J of energy to vaporize at its boiling point. So with the latent heat of vaporization of water, we can calculate how much energy is required to vaporize any amount of already boiling water.
*note - This does not include the energy expended to get the water to boil.

So we will use the exact mass of steel from my picture. And compare the results:

E(required)=Mass*Heat of Vaporization(the boiling point)
so...
1,195,523,550*2260000=2.70x10^15

I wonder why your number for energy required to reach the boiling point of steel and then vaporizing it is less than just vaporizing already boiling water...?

Using 1.2e11 g as your mass and the heat of vapourization of iron of 19 kj/g as I couldn't find one for steel; we have an energy required to vapourize the iron after raising it to boiling of 2.28e15 Joules. Adding that to the energy required to raise the steel to boiling found by Agent Sorchus we get the approximate total energy required to vapourize the armour as 7.25e14 Joules + 2.28e15 Joules = 3.01e15 Joules. That translates to about 719 kilotons.

[Sapper007]

I wonder why your number for energy required to reach the boiling point of steel and then vaporizing it is less than just vaporizing already boiling water...?

Are you seriously suggesting that it takes a 100 KILOTON NUCLEAR BOMB just to vaporize water?

If the mass of said water is over a billion kg's.... Yes...

Using 1.2e11 g as your mass and the heat of vapourization of iron of 19 kj/g as I couldn't find one for steel; we have an energy required to vapourize the iron after raising it to boiling of 2.28e15 Joules. Adding that to the energy required to raise the steel to boiling found by Agent Sorchus we get the approximate total energy required to vapourize the armour as 7.25e14 Joules + 2.28e15 Joules = 3.01e15 Joules. That translates to about 719 kilotons.

except the formula calls for the product of the two numbers... not the sum...

[Formless]

I wonder why your number for energy required to reach the boiling point of steel and then vaporizing it is less than just vaporizing already boiling water...?

Are you seriously suggesting that it takes a 100 KILOTON NUCLEAR BOMB just to vaporize water?

If the mass of said water is over a billion kg's.... Yes...

Where the fuck did you get that number? Please, show us your math so we can laugh at you again.

Also, I would like to echo Stark in asking for evidence of vaporization being involved.

except the formula calls for the product of the two numbers... not the sum...

Except that if you knew anything about stoichiometry you would know that when you multiply those two numbers together you get a nonsense unit. AS already told you this, and you flat out ignored the point. You have no idea what these equations mean, do you?

First you need energy to raise it to a boil. Then you need additional energy (ADDITION, not multiplication) to vaporize it. You NEVER multiply energy units. Anyone who has ever taken a high school chemistry or physics class should know this (assuming you payed any attention in school, which I find doubtful).

[Nieztchean Uber-Amoeba]

Well, at any rate, it means that Terran game mechanics may be a lot closer to canon than we otherwise have imagined.

Which makes you wonder about their manufacturing capabilities. Even discounting the obvious game mechanics, they still seem to have some pretty good compact manufacturing, if Raynor's Foundry Guy on the Hyperion was any indication.

Indeed. The Terrans appear to have the sort of self-sufficient industrial bootstrapping capabilities that nanotechnologists dream of.

Hell, look at the picture of Korhal on the mission selection screen (or the community website) - it looks like a goddamn city-planet. If its population is 'only' 6.3 billion, all those lights must mean something. Or New Folsom - I could be wrong, but the latticework covering the entire planet - with each thread large enough to be visible from an astronomical distance - must require some pretty extraordinary industry. I'd mention the re-Terraforming of Mar Sara in 4 years, but we don't know how thorough Tassadar was, aside from the fact that it clearly wasn't enough to entirely eliminate all the Zerg spores there.

EDIT: You know, I think the Thor might be the most impressive example. Swann not only had the facilities to manufacture a war machine on the Hyperion, but he was able to design an entirely new mech, and not only create a functioning prototype, but start up full industrial production on this new model. And he did all of this in no more than a few weeks, and probably within days. On the Hyperion. If this doesn't say some mind-boggling things about the mobile industry of the Terrans... I dunno, man. That shit is wacky.

[DrStrangelove]

Just using MW's figures off the main pages asteroid calcs, he gets a figure of 7.63MJ to vaporize a kilo of iron.
Thus a billion kilos of iron would require 7.63e15J to vaporize or less than 2 megatons.
You should really use some common sense and try to figure out why your calc to vaporize an astronomically insignificant amount of iron and carbon, is anywhere near the amount of energy required to melt a 1m thick planetary sized shell of rock

[Raxmei]

Isn't there a latent heat missing? Starting from a solid you'd need mass times specific heat times change in temperature + mass times heat of fusion + mass times heat of vaporization.

And while it is reasonable not to set the starting temperature at one degree above absolute zero, why would you see the need to devote space to saying that isn't what you're doing?

[Xess]

except the formula calls for the product of the two numbers... not the sum...

Well to find the energy required to boil a substance, we use E(boiling) = mass (kg) * heat capacity (J/kg*K) * change in temperature (K) where the final temperature is the boiling point and the final unit is Joules. To be more precise we first need to raise the temperature of the solid material to melting, then find the energy to raise the molten material to boiling, but we'll just be simple for now and go straight from solid to boiling. Then we need to find the energy required to overcome the latent heat of vapourization (and of fusion in the case of melting but again being simple). That equation is E(vape) = mass (kg) * latent heat of vapourization (Joules/kg) final unit is again Joules. Thus total energy is E(total) = E(boiling) + E(vape) in order to keep the final units in Joules. No multiplication of the two equations.

In your math what you did was find E(boiling) and then substitute that number in for the latent heat of vapourization. That gave you E = E(boiling) * mass, which gives you a unit of J*kg which is not a real unit.

Anyway, I did the math for vapourizing 1.2e9 kg of pure iron again using the correct way by first melting then vapourizing it and got E(melt) = (1.2e9 kg * 450 J/kg*K * 1538K) + (1.2e9kg * 247.3 kJ/kg) = 1.13e15 Joules and got E(vapourization) = (1.2e9 kg * 611 J/kg*K *1324 K) + (1.2e9 * 6088 kJ/kg) = 8.28e15 Joules. This gives E(total) = 1.13e14 J + 8.28e15 Joules = 9.41e15 Joules = 2.25 megatons.

Since steel will not vary so much as to increase the energy required by more than an order of magnitude we have a range of 1 megaton < yield < 100 megatons for the Yamato cannon.

EDIT: Grr should have used 1.2e9 kg for mass but I used 1.2e8 originally. I fixed it.