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Subject: The Mathematical Strength of MJOLNIR Mk VI Shields
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid

Starting with a base point: it takes 11 MA5C bullets to eliminate a Spartan-II's Mk.VI shields (Halo 3). If you hate maths, skip the below and get the answer at the bottom of this OP.

------------------WARNING, DIFFICULT MATHS TIME------------------

Here, we're calculating the force of an MA5C Assault Rifle round's collision at 50 metres away. Seems a fairly average distance.

Velocity of MA5C bullet = 905m/s (Halopedian) BUT we do not use this, as the bullet starts at 0m/s, amirite? So ignore that, just forget it. We do use it to work out the time though.

Need to work out acceleration of the bullet. We will have to also calculate the bullet's final velocity, just an instant before hitting the shields.

Distance (d) = 50.0 m, Time(t) = (50/905) = 0.055 seconds, Acceleration = a, Initial velocity(u) = 905 m/s and Final velocity = v

d = (u+v) / 2 x t
50 = (0 + v)/2x0.055
50 = 0.055 v/2
v = 2x50/0.055
v = 100/0.055
v = 1818.1818181818 (recurring)
v = 1818.18

v = u + (a x t)
v = 0 + (a x 0.055)
1818.18/0.055 = a

a = 33,057.8 m/s/s

Now that crap is out the way....

Force (Newtons) = Mass (Kg) x Acceleration (m/s/s)
We are very lucky that in Halo Lore, the bullet from an MA5C is actually a modern bullet we can take values from. It uses the 7.62x51mm Full Metal Jacket bullet.

Weight of bullet = 9.7 grams
= 0.0097 Kg.

Force = 0.0097 Kg x 33,057.8 m/s/s
Here we go.

The force of an MA5C-fired 7.62x51mm FMJ bullet = 320.66 Newtons.

I repeat, 320.66 Newtons per single bullet. Doesn't sound like much but it probably really bloody hurts, and remember, our tiny bullet only weighs 0.09506 Newtons. 320 Newtons of force for something of less than a tenth of a Newton in mass, isn't bad.

Now, back on topic, here's the easy bit. 11 bullets to lower shields, right? So 11 x 320 = our Spartan's shields' maximum amount of force it can take.

11 x 320 = 3,520 Newtons of Force.

-------MATHS TIEM IS OVER, YOU CAN COME OUT NOW--------

If you didn't read all that Maths, then the answer is 3,520 Newtons of force can be taken by the MJOLNIR Mk. VI's shields before they collapse.

Is that good or bad? I don't know myself.

[Edited on 12.17.2010 7:01 AM PST]

  • 12.17.2010 7:01 AM PDT
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid

Please do not let this go to waste.... :/

  • 12.17.2010 7:18 AM PDT

The maths/physics you used to get to your answer is not that hard, I mean I can even understand it. ;P

But that also means that a drop in an 500 kg armor, would yield a force of 500 * 9,81 (gravity of Earth) = 4905 Newtons. not taking in mind speed. So that is not very comforting I'd say. (<-- unless this calculation is wrong :P)

edit: and I didn't name a height, but I mean ~10m


[Edited on 12.17.2010 7:21 AM PST]

  • 12.17.2010 7:19 AM PDT
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid


Posted by: duck nl
The maths/physics you used to get to your answer is not that hard, I mean I can even understand it. ;P

But that also means that a drop in an 500 kg armor, would yield a force of 500 * 9,81 (gravity of Earth) = 4905 Newtons. not taking in mind speed. So that is not very comforting I'd say. (<-- unless this calculation is wrong :P)

edit: and I didn't name a height, but I mean ~10m

The shields on the soles of the armour are usually lowered/not there, because walking on shielded boots was described as being like black ice, i.e. very slippery. So generally, I'd guess that the Titanium alloy armour and hydrostatic gel would take the impact of a fall (at least, a fall where you land on your feet).

P.S. I know that the maths isn't that hard, there's just a nice chunk of it, and Reach forum-goers may come over here at some point.

[Edited on 12.17.2010 7:30 AM PST]

  • 12.17.2010 7:29 AM PDT

Yeah you're right, totally forget about that, but it still means that a fall from 10 meters would cost you your shields. But then again the gel layer combined a lot of layers of armor might soften the blow a little bit. :P

Another thing that also shot to mind, since gravity influences the force of the bullets a shield might take more bullets to drop on Reach than on Earth. And what about shooting in space, where the only thing generating Newtons is the acceleration and where gravity is neglectable?

  • 12.17.2010 7:48 AM PDT

I think considering that Covenant weapons will have roughly the same force if not slightly worse because they're weapons are slower, I suppose that's quite a good strength for the shields.

  • 12.17.2010 7:59 AM PDT

01000001 01101100 01101100 00100000 01111001 01101111 01110101 01110010 00100000 01100010 01100001 01110011 01100101 00100000 01100001 01110010 01100101 00100000 01100010 01100101 01101100 01101111 01101110 01100111 00100000 01110100 01101111 00100000 01110101 01110011

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.

  • 12.17.2010 7:59 AM PDT

I don´t get it, sorry.

  • 12.17.2010 8:13 AM PDT
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid


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

O I C

My equations are "dumb".

...okay, go devise some better ones in your own thread?

@Poster above me: This is to work out how much physical force the shields can take in terms of Newtons (using it as a unit of Force, not mass) before they collapse. The solution is that MJOLNIR Mk. VI can take approximately 3,520 Newtons of Force, a.k.a 11 bullets from an MA5C.

  • 12.17.2010 8:24 AM PDT

Et Eärello Endorenna utúlien.
Sinome maruvan ar Hildinyar.
tenn' Ambar-metta!

So according to the distance, it changes the numbers of bullets to take down your energy shield?

Plus what kind of weapon are we talking about? Remember the weight of the bullet changes according to the weapon.

How about the DMR? or the Pistol?

Update: MA5C Assault Rifle I solved my own question.

How about the DMR?

[Edited on 12.17.2010 8:43 AM PST]

  • 12.17.2010 8:41 AM PDT
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid


Posted by: MasterSin
1.So according to the distance, it changes the numbers of bullets to take down your energy shield?

2.Plus what kind of weapon are we talking about? Remember the weight of the bullet changes according to the weapon.

3.How about the DMR? or the Pistol?

Update: MA5C Assault Rifle I solved my own question.

4.How about the DMR?


I numbered your questions.

1. Yes. Over longer distances, drag will slow the bullet's velocity (deceleration) and thus less force will be projected onto the shield. But I chose 50m for an average range of combat for Halo's Multiplayer experience, to get a roughly approximate value of force-per-bullet so as to work with.

2. You answered this yourself.

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.

4. The answer to this is just half a rephrasing of question 3, really.

  • 12.17.2010 9:01 AM PDT

Your maths and physics are complete nonsense and there's no reason for me to look further into it. The very fact that you wanted to figure out a force in newtons invalidates everything you tried to do. Sorry, but your post sucks, hard. Nothing personal.

You did get the numbers we need, though. We have the mass, we have the velocity. E = 11/2 * 0.0097kg * (905 m/s)^2 = 43694.98375 J =~ 44 kJ of energy.



I've tried to decipher what you did, and this is what your post consists of:
You assume that the bullet starts at velocity 0, accelerates during the entire flytime, with the average velocity being the 905 m/s. Based on this ridiculous assumption (in real life, halo canon and halo gameplay, the entire acceleration of the bullet happens inside the weapon), you calculate the acceleration of the bullet correctly. Then you use this acceleration to calculate a force, again correctly. However, this force is the force that must be given to the bullet to make it accelerate, and has nothing to do with the final impact.

So, it appears that you did manage to calculate something proper: the required propulsion force output of a rocket-powered AR bullet. Too bad that has nothing to do with spartan shields.

[Edited on 12.17.2010 9:31 AM PST]

  • 12.17.2010 9:21 AM PDT
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid


Posted by: Mutoid Log
Your maths and physics are complete nonsense and there's no reason for me to look further into it. The very fact that you wanted to figure out a force in newtons invalidates everything you tried to do. Sorry, but your post sucks, hard. Nothing personal.

You did get the numbers we need, though. We have the mass, we have the velocity. E = 11/2 * 0.0097kg * (905 m/s)^2 = 43694.98375 J =~ 44 kJ of energy.



I've tried to decipher what you did, and this is what your post consists of:
You assume that the bullet starts at velocity 0, accelerates during the entire flytime, with the average velocity being the 905 m/s. Based on this ridiculous assumption (in real life, halo canon and halo gameplay, the entire acceleration of the bullet happens inside the weapon), you calculate the acceleration of the bullet correctly. Then you use this acceleration to calculate a force, again correctly. However, this force is the force that must be given to the bullet to make it accelerate, and has nothing to do with the final impact.

So, it appears that you did manage to calculate something proper: the required propulsion force output of a rocket-powered AR bullet. Too bad that has nothing to do with spartan shields.


Sorry, but your solution isn't really much use. You could either a) Work out the applied energy required to take out a Spartan's shields or b) Work out the applied force required to take out a Spartan's shields.

I chose the latter. If you were approximating using plasma bolts (don't know why, using bullet values is easier?), you'd obviously chose to approximate using energy.

Here is an example (albeit, a silly one): Imagine some kind of uber-robot who can take out a Spartan's shields in a prod. It's going to need a certain balance of mass and acceleration to generate the force needed, i.e. APPROXIMATELY 3,520 Newtons of applied force. I know what you're thinking: "BUT WAIT! THAT'S THE FORCE REQUIRED!!!!1". Well, at a mere fifty metre range or a prod, factors such as drag aren't really going to do much to an approximated value, so the force in NEARLY equals force out. For the approximated value I achieved, it's negligible.

So, back onto the robot example thingbob. That robot's going to need one of the factors - either mass or acceleration - for his prodding finger to eliminate the Spartan's shields. It'll either have to have a damned heavy hand or a hugely fast prodding hand, assuming he's going to fully eliminate the shields with a considerable 3,520 Newtons of force. Technically, the robot could use a 1,000,000,000 Kg prodding finger moving at just 0.1mph and still inflict the same force - so long as the force was applied through a finger with a surface area that entirely made contact with the shields. (Would this mean one would have to take into account the surface area of the point of a bullet? Head is swimming.)

ANOTHER example to help what I'm explaining there ^ : You brush past someone and your shoulders barge at 5mph. Not much force applied, really. Then, at 5mph, you walk straight into someone. Bam, full force applied, captain.

I hope this clears things up if the OP was a bit rushed.

  • 12.17.2010 10:00 AM PDT


Posted by: Mega Fox
Sorry, but your solution isn't really much use.

Actually, his solution is far better than yours. It uses a reasonable model for shield function, whereas your compounding forces model is absolute nonsense. If 20N acts on an object for a second, and then a second goes by with nothing acting on the objects, and then 20N acts on it for another second, the object did NOT "receive 40N of force".

The rest of your physical model is nonsensical as well; why are your bullets accelerating during flight?

[Edited on 12.17.2010 10:12 AM PST]

  • 12.17.2010 10:12 AM PDT

Learn to accept people interest.

ITS OVER 9000

  • 12.17.2010 10:15 AM PDT
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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."

Posted by: AvengerTam1010
launch texas at the asteroid


Posted by: Tupolev
If 20N acts on an object for a second, and then a second goes by with nothing acting on the objects, and then 20N acts on it for another second, the object did NOT "receive 40N of force".

3,520 Newtons for one second of time. Sorry, I should have specified, that is my error.


Posted by: Tupolev
The rest of your physical model is nonsensical as well; why are your bullets accelerating during flight?

A bullet will accelerate after being propelled out of the rifle for a very short period of time before hitting it's terminal velocity. When considering a 50m range, which a bullet covers incredibly quickly, it is in fact quite easy to assume the bullet is still accelerating at this point, perhaps even up to the 80 metres mark.

  • 12.17.2010 10:23 AM PDT


Posted by: Mega Fox
3,520 Newtons for one second of time. Sorry, I should have specified, that is my error.

Now your model is becoming even more garbled. Are we assuming, now, that the eleven bullets strike the spartan simultaneously and that each apply 320N over the course of a whole second (strange bullets...)? And since we're using force over a time, are we now assuming that it's not force that the shield can take a certain amount of, but rather a change in momentum!?


Posted by: Tupolev
The rest of your physical model is nonsensical as well; why are your bullets accelerating during flight?

A bullet will accelerate after being propelled out of the rifle for a very short period of time before hitting it's terminal velocity. When considering a 50m range, which a bullet covers incredibly quickly, it is in fact quite easy to assume the bullet is still accelerating at this point, perhaps even up to the 80 metres mark.

No, bullets are not rockets. Bullets undergo their entire acceleration within the barrel of a gun, via the low-explosive expansion of gunpowder as a gas.

...Hence why it's possible to state a muzzle velocity.




Look, your model is inconsistant and founded on wrong assumptions.

[Edited on 12.17.2010 10:37 AM PST]

  • 12.17.2010 10:32 AM PDT
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Why are we assuming that the force required to shoot the bullet of the gun is the same as the force required to stop the bullet when it hits the shield? Wouldn't the force be the mass of the bullet times the acceleration as it goes from 900 m/s to 0 over however much time when the bullet hits the shield?

  • 12.17.2010 10:41 AM PDT

Posted by: opogjijijp
Why are we assuming that the force required to shoot the bullet of the gun is the same as the force required to stop the bullet when it hits the shield? Wouldn't the force be the mass of the bullet times the acceleration as it goes from 900 m/s to 0 over however much time when the bullet hits the shield?
And because it's incredibly hard to estimate the time of the deceleration, you never want to use force in impact scenarios with solid objects. The quantities that work are energy and momentum.

  • 12.17.2010 10:49 AM PDT

i can only count to jagermeister.

Posted by: Tupolev

Posted by: Mega Fox
3,520 Newtons for one second of time. Sorry, I should have specified, that is my error.

Now your model is becoming even more garbled. Are we assuming, now, that the eleven bullets strike the spartan simultaneously and that each apply 320N over the course of a whole second (strange bullets...)? And since we're using force over a time, are we now assuming that it's not force that the shield can take a certain amount of, but rather a change in momentum!?


Posted by: Tupolev
The rest of your physical model is nonsensical as well; why are your bullets accelerating during flight?

A bullet will accelerate after being propelled out of the rifle for a very short period of time before hitting it's terminal velocity. When considering a 50m range, which a bullet covers incredibly quickly, it is in fact quite easy to assume the bullet is still accelerating at this point, perhaps even up to the 80 metres mark.

No, bullets are not rockets. Bullets undergo their entire acceleration within the barrel of a gun, via the low-explosive expansion of gunpowder as a gas.

...Hence why it's possible to state a muzzle velocity.




Look, your model is inconsistant and founded on wrong assumptions.
Ah not true. And I has a picture from wikipedia to prove it lol. Muzzle velocity is only the speed at which the bullet exits the muzzle. That graph shows the velocity/distance/time/pressure of a 5.56mm NATO round fired from a 20" barrel. As you can see, the velocity is still steadily increasing by the time the bullet leaves the barrel. Once it does, and the gas is free to expand in all directions, the force propelling the bullet drops but absolutely does not disappear. There's a whole science dedicated to this called Transitional Ballistics.

  • 12.17.2010 12:26 PM PDT

Error 404:
-Error not found.

The armor shielding can take the weight of about three horses, is the result I find.

Three horses can sit on a normal 2010 day human and might not kill him.

Damn, Spartans are a bunch of pansies!

[Edited on 12.17.2010 2:47 PM PST]

  • 12.17.2010 2:46 PM PDT

Surrender? I've never heard of such. Defeat? Not a possibility! Why...why I will fight and fight and fight, until I am victorious, or I am dead? Because that is who I am, and that is why I fight; everyone will see that, when I am proclaimed greatest among men.

The Flood AdviceCenter- Where you can talk about stuff people usually want to, but can't, in The Flood.

A better way to solve the problem would be to use the impulse-momentum theorem, since a more accurate assumption is that the bullets travel at a certain velocity with effectively 0 acceleration.

You find the time it takes for a bullet to impact as the times it takes to travel its own length, then divide the change in velocity by that time to find the force ( D = change in)

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

And there's your answer

  • 12.17.2010 4:28 PM PDT

Error 404:
-Error not found.

Posted by: AYF 001
A better way to solve the problem would be to use the impulse-momentum theorem, since a more accurate assumption is that the bullets travel at a certain velocity with effectively 0 acceleration.

You find the time it takes for a bullet to impact as the times it takes to travel its own length, then divide the change in velocity by that time to find the force ( D = change in)

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

And there's your answer

That makes more sense..
Three hundred horses...

That seems right.

  • 12.17.2010 4:33 PM PDT

01000001 01101100 01101100 00100000 01111001 01101111 01110101 01110010 00100000 01100010 01100001 01110011 01100101 00100000 01100001 01110010 01100101 00100000 01100010 01100101 01101100 01101111 01101110 01100111 00100000 01110100 01101111 00100000 01110101 01110011


Posted by: Mega Fox

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

O I C

My equations are "dumb".

...okay, go devise some better ones in your own thread?

@Poster above me: This is to work out how much physical force the shields can take in terms of Newtons (using it as a unit of Force, not mass) before they collapse. The solution is that MJOLNIR Mk. VI can take approximately 3,520 Newtons of Force, a.k.a 11 bullets from an MA5C.


Actually, I wasn't even attacking your "math". That kinda thing didn't need my help or the other posters, it kinda broke itself, you know?

It was merely some light-hearted poking fun at your name.

  • 12.17.2010 6:34 PM PDT
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Posted by: Mega Fox
------------------WARNING, DIFFICULT MATHS TIME------------------

Here, we're calculating the force of an MA5C Assault Rifle round's collision at 50 metres away. Seems a fairly average distance.

Velocity of MA5C bullet = 905m/s (Halopedian) BUT we do not use this, as the bullet starts at 0m/s, amirite? So ignore that, just forget it. We do use it to work out the time though.

Need to work out acceleration of the bullet. We will have to also calculate the bullet's final velocity, just an instant before hitting the shields.

Distance (d) = 50.0 m, Time(t) = (50/905) = 0.055 seconds, Acceleration = a, Initial velocity(u) = 905 m/s and Final velocity = v

d = (u+v) / 2 x t
50 = (0 + v)/2x0.055
50 = 0.055 v/2
v = 2x50/0.055
v = 100/0.055
v = 1818.1818181818 (recurring)
v = 1818.18

v = u + (a x t)
v = 0 + (a x 0.055)
1818.18/0.055 = a

a = 33,057.8 m/s/s

Now that crap is out the way....

Force (Newtons) = Mass (Kg) x Acceleration (m/s/s)
We are very lucky that in Halo Lore, the bullet from an MA5C is actually a modern bullet we can take values from. It uses the 7.62x51mm Full Metal Jacket bullet.

Weight of bullet = 9.7 grams
= 0.0097 Kg.

Force = 0.0097 Kg x 33,057.8 m/s/s
Here we go.

The force of an MA5C-fired 7.62x51mm FMJ bullet = 320.66 Newtons.

I repeat, 320.66 Newtons per single bullet. Doesn't sound like much but it probably really bloody hurts, and remember, our tiny bullet only weighs 0.09506 Newtons. 320 Newtons of force for something of less than a tenth of a Newton in mass, isn't bad.

Now, back on topic, here's the easy bit. 11 bullets to lower shields, right? So 11 x 320 = our Spartan's shields' maximum amount of force it can take.

11 x 320 = 3,520 Newtons of Force.

-------MATHS TIEM IS OVER, YOU CAN COME OUT NOW--------



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.

[Edited on 12.18.2010 2:42 AM PST]

  • 12.18.2010 2:41 AM PDT

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