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This topic has moved here: Subject: The Mathematical Strength of MJOLNIR Mk VI Shields
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Subject: The Mathematical Strength of MJOLNIR Mk VI Shields
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"Time was your ally human. But now it has abandoned you. The Forerunners....have returned. And this tomb... is now yours". - The Didact

And why is this based on game mechanics?

  • 12.18.2010 2:52 AM PDT

i can only count to jagermeister.

Posted by: drummer0702
I'm sorry, but none of your math makes sense. Realistically, the bullet will have reached its max velocity near the end of the barrel so all of those calculations are moot. Also, your force equation has nothing to with the force each bullet delivers. Rather, it is telling you the force that moved the bullet in the first place (assuming you use the proper bounds). Forces due to impacts is calculated a completely different way... F= (mv1-mv2)/(time of impact). My other other problem is you're neglecting air resistance.

Another way to look at this problem is to find the proper height that it takes to pop one's shield.
Ok dude this has already been brought up. Out of anyone here so far the OP is closest to the answer. And if anyone here knew anything about ballistics in real life, closest is the best you can get. Air resistance means nothing when a shot is fired under 50m. OP already said that. And I believe I explained fairly clearly why a bullet does not reach max velocity at the end of the barrel. That said, it's impossible to know without extensive study exactly how a bullet accelerates out of a certain gun.

I'm not saying I know more about math or that I'm smarter than anyone else, but I do know a thing or two about guns.

  • 12.20.2010 8:27 AM PDT

On hiding dead bodies:
Posted by: Psuedo
Posted by: teh Chaz
Inside another dead body. It's the last place they'll look
A corpse within a corpse.
CORPSEPTION.
Win.

You appear to have the round accelerating whilst in flight. This is not possible, as the moment it exits the barrel it will begin to decelerate, since the force used the propel it, the pressure of expanding gases within the barrel and chamber, no longer applies and air resistance and bullet drop now begin to take effects. It is also somehow able to accelerate at almost 4000G.
At a distance of 50m (especially from 905m/s), deceleration will be negligible. It probably won't be much under 890m/s.

However - this is my theory on how Energy Shields work, probably ought to be a help.

[Edited on 12.20.2010 8:53 AM PST]

  • 12.20.2010 8:42 AM PDT

We're never what we invent or intend.


Posted by: duck nl
Yeah you're right, totally forget about that, but it still means that a fall from 10 meters would cost you your shields.
Don't forget, either, that 10 meters is almost 31 feet. That'd be fatal, if not simply put you completely out of action, to almost anyone.

SPARTANs in full MK. V armor survived a fall from the upper atmosphere, albeit with some damage. Still, that's a good 15k fall. IMO, I think the shields actively adapt. The baseline projection level is the output that can take 11 MA5C bullets. In the case of falling, they can overcharge their shields/overpressurize the hydrostatic gel, and thus take a lot more punishment.

  • 12.20.2010 9:38 AM PDT

On hiding dead bodies:
Posted by: Psuedo
Posted by: teh Chaz
Inside another dead body. It's the last place they'll look
A corpse within a corpse.
CORPSEPTION.
Win.

Posted by: Adragalus

Posted by: duck nl
Yeah you're right, totally forget about that, but it still means that a fall from 10 meters would cost you your shields.
Don't forget, either, that 10 meters is almost 31 feet. That'd be fatal, if not simply put you completely out of action, to almost anyone.

SPARTANs in full MK. V armor survived a fall from the upper atmosphere, albeit with some damage. Still, that's a good 15k fall. IMO, I think the shields actively adapt. The baseline projection level is the output that can take 11 MA5C bullets. In the case of falling, they can overcharge their shields/overpressurize the hydrostatic gel, and thus take a lot more punishment.
Generally, a regular human can withstand a 50ft drop, but be on the borderline of living and dying.

SPARTANS with their augmentations can withstand higher jumps - in TFOR John looks down an elevator shaft and thinks "a thirty-metre plunge into darkness. His bones wouldn't break, but there'd be internal damage." So SPARTANS without MJOLNIR can take at minimum a near-100ft drop and still be able to walk or hobble away.
MJOLNIR is armoured so will take some impact out anyway, but will take out a very heavy factor due to its cushioning gel layer. I'd say that MJOLNIR minus shields could probably take a 100m+ fall before the SPARTAN inside became too wounded. Total guess, BTW.

  • 12.20.2010 9:52 AM PDT

Posted by: Painbow 6
Posted by: drummer0702
I'm sorry, but none of your math makes sense. Realistically, the bullet will have reached its max velocity near the end of the barrel so all of those calculations are moot. Also, your force equation has nothing to with the force each bullet delivers. Rather, it is telling you the force that moved the bullet in the first place (assuming you use the proper bounds). Forces due to impacts is calculated a completely different way... F= (mv1-mv2)/(time of impact). My other other problem is you're neglecting air resistance.

Another way to look at this problem is to find the proper height that it takes to pop one's shield.
Ok dude this has already been brought up. Out of anyone here so far the OP is closest to the answer. And if anyone here knew anything about ballistics in real life, closest is the best you can get. Air resistance means nothing when a shot is fired under 50m. OP already said that. And I believe I explained fairly clearly why a bullet does not reach max velocity at the end of the barrel. That said, it's impossible to know without extensive study exactly how a bullet accelerates out of a certain gun.

I'm not saying I know more about math or that I'm smarter than anyone else, but I do know a thing or two about guns.
The OP isn't anywhere near the solution. Nobody's ever mentioned air resistance. Taking the post-barrel acceleration you mentioned into account isn't possible, since we only know the muzzle velocity, not the peak velocity. But it definitely doesn't increase the velocity of the bullet very much, and it most definitely does not lead to anything resembling the OP's calculations.

Me and AYF have the best answers one can get with the muzzle velocity and mass. Simple, and sensible. The OP isn't even trying.

  • 12.20.2010 10:53 AM PDT

seventh coloumn pride

ahh!!! the maths it burns

  • 12.20.2010 1:28 PM PDT

On hiding dead bodies:
Posted by: Psuedo
Posted by: teh Chaz
Inside another dead body. It's the last place they'll look
A corpse within a corpse.
CORPSEPTION.
Win.

Posted by: Destroyer0921
ahh!!! the maths it burns
It should do, it's wrong.

  • 12.20.2010 3:28 PM PDT
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  • Exalted Legendary Member
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I take it this means Sir Issac Newton is the deadliest son-of-a--blam!- in space.

  • 12.20.2010 3:38 PM PDT


Posted by: AYF 001
F = D(m*v)/Dt= (.0097*905)/(.028/905)= 283,773 N
Times 11 rounds = 3,121,070 N

And there's your answer
Yup, definitely this.

  • 12.20.2010 3:53 PM PDT
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  • Exalted Legendary Member

UWG

My jokes, so I don't lose them (ignore this):
ZedFish's Opinion on Sgt. Foley.
ZedFish's Forerunner Rickroll.

Posted by: Mega Fox
3. Maybe in the future I might take the time to work the DMR and Magnum out, but I doubt it, since there are no given values for their bullet weights, which is crucial to the calculation. However, whilst reading about the BR55 Battle Rifle, I did notice why the BR has more damage per bullet (only 9 bullets to lower shields); because it has a higher muzzle velocity due to the bullets having an additional propellant. So yes it depends on the weapon and the bullet very much, obviously.
Just throwing this in, I think the DMR fires 7.62x51mm NATO, and the Magnum fires a 12.7x40mm. Try finding weights of those calibres.

The Magnum's would be similar to the .500 S&W Magnum (.50 calibre; varying weights), and the DMR fires a NATO round (9.7g FMJ).

  • 12.20.2010 9:01 PM PDT

On hiding dead bodies:
Posted by: Psuedo
Posted by: teh Chaz
Inside another dead body. It's the last place they'll look
A corpse within a corpse.
CORPSEPTION.
Win.

Posted by: AngrydoG
Posted by: JobeTheConqueror
Your math cannot be trusted.

After all, by simply adding an N to your name, we get one of the dumbest chicks on the planet.

Your equations may be dumb, but they look awesome in a bikini. So that's something.


Bring it to a physics teacher. it looks right to me.
It's not.

OP has it so that upon exiting the barrel, over a 50m distance the round can double its velocity from 900 to 1800m/s, which is faster than a sniper rifle round. He thinks it can accelerate when the one and only force that could provide that acceleration has no more effect.

  • 12.20.2010 11:45 PM PDT

i can only count to jagermeister.

Nobody is reading my posts. OP may have figured the acceleration wrong, but other than that they are correct. You guys need to do a little research on real life ballistics. I know lots of guys who reload and they all use mass x velocity squared. That's for kinetic energy right? Ok done.

EDIT- here lookwutifound. 10 seconds via google.

[Edited on 12.21.2010 9:35 AM PST]

  • 12.21.2010 9:28 AM PDT

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