On the other hand, I just realized that how f*cking scary TIR shielded missiles would be.
Shepard Quest Mk VI, Technological Revolution
- Thread starter Hoyr
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I think that's getting too close to time-traveling munitions, since we're talking about adding FTL quality barriers and eezo cores to missiles, and we have @Hoyr asking us to please stop. For the children.
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Do we actually have a counter-missile? That sounds like something that would be relatively useful. As soon as TIR shields come online, the only thing that would reliably be able to kill a missile or a fighter would be a Kinetic Impact.
I don't know if the MFTM-1 "Super Driver" is considered canon, but something similar to the Block 1 Standard Missile is probably something that would be sorely needed in the future.
True, true. But I'm not talking FTL munitions (which won't be time travelling somehow, given that they aren't in canon). As I noted above, the power and eezo amounts required to generate a ME field scale exponentially with mass affected and with the "grade" of the field (how strong it is). Barriers don't actually (normally) affect much of anything - the field is projected to near the protected object. This lowers the amount of eezo needed greatly. Hence why Normandy's shields don't cost nearly as much as Normandy's drive core. The same with fully developed TIR shielding - it'll cover vacuum near the missile, but not the missile itself.
Although yeah, this is getting into very scary areas in general. Space combat in ME, especially with our additions, is incredibly asymmetric. Essentially, the one to attack first wins, and there's little that can be done to protect stationary targets, the only real safety measures being either complete and permanent shielding or simply not being hit in the first place.
Though gravity sensors would mitigate this a lot.
Well the Tiger is already using Sagitta missiles to counter Piliums and other heavy weapons. Our Improved Fire Control VI seems able to shoot other weapons out of the sky fairly accurately as-is.
Or we start using SR4 rules, in which case the missiles suddenly stop being able to hit anything because reasons. I have an omake idea for that: the universe decides to play dice with people at random intervals (the Illusive Man fails to open a door and face-plants; Joker crit-fails a Piloting check in a simulator and crashes; Gaver Dor fails a strength check and fails to lift his desk), and Revy tries unsuccessfully to harness this power FOR SCIENCE! Don't have time to whip my weak and uncooperative muse into gear, though, as I'll be out of the country for most of the month starting Sunday and I need to pack.
Please tell me you are not going to make me differentiate between stations in LPO and deep space ones... that's going to make things more complicated...I guess you could have somethings not have much shielding... just don't move it ever.
That said you still need some of the the stuff. Having trouble finding any numbers though
I think for rad shielding that we just need to straight-up detail how we're dealing with cosmic rays for each station on an individual basis, separate from the cost of construction. The three we'll be relying on:
- LPO stations: no special cosmic ray shielding needed.
- Hanging out behind a large asteroid. Requires the asteroid, plus a thruster array for station-keeping.
- TIR-quality shields.
Lesser companies, who don't have access to our awesome Hyper-modularity tech, have to pay those exponentially-increasing costs that you outlined... or call our construction company.
Actually, the reasons why the Normandy SR 2 can use both Cyclonic Barrier tech and the Thanix Cannon together is because of its oversized drive core plus the fact that its closer to a cruiser and thus has more space to store the needed equipment, according to the codexes about the Normandy upgrades.
In the War Assets/Alien listing, there is an Asari Cruiser that had a dreadnought class drive core installed by accident and it could generate dreadnought class kinetic barriers because it.
I think its more like the drive core pumps out dark energy/makes dark matter appear, they pull the dark energy/matter away and shunt it into whatever system needs it.
The bigger the drive core, the more dark energy/matter you have, thus stronger shields or more powerful guns or both if you built your ship to accomodate it.
@TheEyes Your forgetting Galactic Cosmic Rays, solar origin stuff is only part of the equation.
Minimal shielding. Still needs some though, covered in build cost.
Actually you still need shielding for those Galactic Cosmic Rays, sure it blocks the solar stuff, but the other stuff... needs more shielding.
Only covers photonic radiation. Cosmic rays are particles. (Naming error)
Sound like a reasonable starting point, so maybe something like that? Let me think about it a tad more.
*Mumble, Mumble* Bioware writers... The Drive core can't be the primary source of the shields. Shield use Heavy* Dark Energy, the drives system uses Lightening Dark Energy. You need both when flying around at FTL, otherwise you ship melts to interstellar plasma/dust. A bit of eezo only makes one at a time. Well unless you want to retard functionality, make a warp or something like that. I can understand using the drive core to support the shields if your not using it, that's at least consistent. Or some sort of subdivided core... which could mean the Asari have something like a Multi-core system.
*The English language seems to lacks a good word for "makes heavier"... encumbering?
Ah, its not the eezo core that allows that ship to have Dreadnought shields!
"The Cybaen is an asari cruiser-class warship. A mix-up during the cruiser's construction left it with an unusually large drive core and engineers redesigned the ship around the excessive power source. The Cybaen's unusually robust engines provide enough energy to generate kinetic barriers normally reserved for dreadnoughts."
From the text the engineers got an XL sized core and though hey, why don't we upgrade everything else to match. It should be noted here that the drive core also seems to include the fusion (or fission) core as its noted as a power source (Okay...? I guess that makes ship terminology really complicated.). Anyway the upgrade to the power systems is what give her Dreadnought barriers, not the size of her eezo core. I'm going to assume that by engine they mean power generating engine not thrusters as that would be extra silly. Or in short XL Drive Core -> Power Upgrade = Better Shields not Bigger Core=Better Shields.
I'm slightly less disappointed in Bioware now.
Minimal shielding. Still needs some though, covered in build cost.
Actually you still need shielding for those Galactic Cosmic Rays, sure it blocks the solar stuff, but the other stuff... needs more shielding.
Only covers photonic radiation. Cosmic rays are particles. (Naming error)
Sound like a reasonable starting point, so maybe something like that? Let me think about it a tad more.
*Mumble, Mumble* Bioware writers... The Drive core can't be the primary source of the shields. Shield use Heavy* Dark Energy, the drives system uses Lightening Dark Energy. You need both when flying around at FTL, otherwise you ship melts to interstellar plasma/dust. A bit of eezo only makes one at a time. Well unless you want to retard functionality, make a warp or something like that. I can understand using the drive core to support the shields if your not using it, that's at least consistent. Or some sort of subdivided core... which could mean the Asari have something like a Multi-core system.
*The English language seems to lacks a good word for "makes heavier"... encumbering?
Ah, its not the eezo core that allows that ship to have Dreadnought shields!
"The Cybaen is an asari cruiser-class warship. A mix-up during the cruiser's construction left it with an unusually large drive core and engineers redesigned the ship around the excessive power source. The Cybaen's unusually robust engines provide enough energy to generate kinetic barriers normally reserved for dreadnoughts."
From the text the engineers got an XL sized core and though hey, why don't we upgrade everything else to match. It should be noted here that the drive core also seems to include the fusion (or fission) core as its noted as a power source (Okay...? I guess that makes ship terminology really complicated.). Anyway the upgrade to the power systems is what give her Dreadnought barriers, not the size of her eezo core. I'm going to assume that by engine they mean power generating engine not thrusters as that would be extra silly. Or in short XL Drive Core -> Power Upgrade = Better Shields not Bigger Core=Better Shields.
I'm slightly less disappointed in Bioware now.
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It could be possible that they use eezo to create gravitational forces to create the conditions needed for net gain fusion, but it requires the eezo to be almost wrapped around said reaction, thus 'drive core'.
It would certainly make some sense about plasma being vented from the drive core if you do not get all of the upgrades in ME 2.
Although I do not understand why they do not use pure hydrogen to hydrogen fusion, as it would allow you to produce helium four along with berylium eight which then can be fused together to create carbon twelve which is a perfectly safe carbon isotope that can be converted into diamonds...
Maybe Bioware did not realise that they handed everyone the ability to get the holy grail of fusion power?
Unless of course, you need a dreadnought scale one, which would explain the insane power output from a dreadnoughts main gun...
Also about the whole 'heavy' and 'light' dark energy....that is not what happens, they use dark energy to generate mass effect fields.
It is the two different mass effect fields that cannot exist in the same location at the same time, heck, trying to layer two of same field causes both to break down, quite possibly back into dark energy.
It would explain the Normandys drive core giving off both blue and purple light.
It would certainly make some sense about plasma being vented from the drive core if you do not get all of the upgrades in ME 2.
Although I do not understand why they do not use pure hydrogen to hydrogen fusion, as it would allow you to produce helium four along with berylium eight which then can be fused together to create carbon twelve which is a perfectly safe carbon isotope that can be converted into diamonds...
Maybe Bioware did not realise that they handed everyone the ability to get the holy grail of fusion power?
Unless of course, you need a dreadnought scale one, which would explain the insane power output from a dreadnoughts main gun...
Also about the whole 'heavy' and 'light' dark energy....that is not what happens, they use dark energy to generate mass effect fields.
It is the two different mass effect fields that cannot exist in the same location at the same time, heck, trying to layer two of same field causes both to break down, quite possibly back into dark energy.
It would explain the Normandys drive core giving off both blue and purple light.
Funny, yes. Still, fluff-wise, this all leads to the only possible long-term conclusion: abandoning planets and living in ships / habitats with shields and active sensors. Essentially Culture. And I am ok with that.
Well, protium to protium is harder than pretty much anything else. It requires more density and higher "temperatures".
It might also be that eezo degrades into normal matter under barrage of ionizing particle radiation, if slowly.
Also, pedantically, 8Be doesn't exist. It breaks down into two alpha particles in something like 10-16 seconds.
The issue of dark energy and how it's something like phlogiston is... Yeah, it's complicated, and never much touched upon.
Huh, so drive core stands for both the FTL generation system and the primary power generation system? That... makes sense, somewhat. Interesting to know.
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You I like.
edit: Also, new headcannon for arc reactors later on, for perpetual ones that don't need refueling.
That part was deliberate. It was accidentally a dreadnought core into a cruiser's construction dock. Probably just a paperwork screw up that they decided to capitalize on rather than try to fix.
It depends. It's one interpretation of Virmire kilotons at least. In my paper I go with "you get the same, as MC^2 remains constant, resulting in changes in C with changes in M", but there are, obviously, other interpretations too.
I think that @Hoyr is going with MC^2 is constant.
Here is a thought that occurred to me. Flawless Blackboxing means the tech is a literal blackbox right? So anyone who gets a copy of our Repulsor* once we've completed it won't be able to do stuff like mess with the Repulsor's programming right?
If so I imagine making the software such that it requires the main computer send the Repulsor a password and token to activate would make salvaging the thing practically impossible.
There would be a couple passwords programmed in by the device's users which would likely have to be done at a secure PI facility with a PI only interface device. But the real security would be in the tokens.
I'm picturing Tokens that only function for a preset time. The user receives theirs before a mission and it's pre-programmed to only last the expected length of the mission plus a reasonable margin of error.
At least that is my initial idea of how to secure Repulsors to a level that would satisfy the SA.
*I am assuming that we automatically upgrade tech to our highest level of blackboxing since there isn't really a reason not to and it limits the number of easier to crack copies out there.
If you read the text on eezo it pretty clear that a positive current increases mass and a negative current decreases it. (This is also what leads to the idea that eezo is a terminal, being a resistor would lead to inflow and outflow I think the nets 0, but I'm not an electrician). Since the effect of the field on mass is determined by the current being used the "type" of field must be determined at the time the dark energy is created (or concentrated or what ever it is eezo really does) and the dark energy needs to carry the typing. Thus a core can only create one of the two field types at a time. Ergo my use of the terms for 'heavy' and 'light' dark energy as it's typed at generation by the current used.
Current +/- and Eezo -> Dark Energy +/- -> Mass Effect Field +/-
I guess you could suggested that the current bit means something else... but I doubt it.
I like my universes mostly intact. Or at least like having breaking them take effort.
Yes. They can can still test around the black box, but you can always do that as long as the thing works.
Any newly made items use the highest level of blackboxing. Old ones you still own? Eh... part of me really thinks that should still take work/money. You may or may not need to rebuild the thing to integrate the new security or remove old bits. IDK
Well that's paranoid. IDK that maybe more than you need.
"Hey could you not spread your tech around?"
"Well, we kinda planned on using it... How about if we make it so that it used limited and require check-ins for it to be to refresh uses. I'm sure that it won't run out at the wrong time because our people couldn't make check in."
"And you could make it explode if it hasn't been checked in for too long." "Great idea! That too!"
"That was a joke."It's hard to find a compromise between don't spread tech around and wanting to use it.
It is. That is actually the original version I thought up last night. Not quite sure why I posted it...
Anyway I changed my mind on the timer thing because the more I thought about it the more impractical it became. Instead I'm thinking of just having it so there is a button on the token to erase the keys inside it
Still solves the problem the timer was created for (if a Gladius with pilot was captured) but has significantly less issues.
With the erase button someone would have to acquire a Repulsor, steal the matching Token without it being cleared, and torture/bribe the password out of the assigned user.
The system is vulnerable to one of our people selling out and just giving someone the device, Token, and password but I'm hoping we pay our people enough and inspire enough loyalty to avoid that since it's hard to design a system that is traitor proof.
Be-8 has a half life of 10^-17 seconds, and decays back to two He-4 nuclei, which basically gives you nothing. The only place where you can do anything with it is the center of a red giant star, where it reacts to form C-12 near-instantly; you can't do it in a controlled environment.
Be-6, the product of two He-3 nuclei fusing, has just as short a half-life; the difference is that the decay products are one He-4 nuclei and two protons, all of which can be caught in magnetic fields and be used to make electricity.
Well, given that the FTL eezo core is probably the biggest power hog on the ship, other than maybe the main gun, it makes sense.
"Hey, buddy, you're the one telling me to withhold the galaxy-changing technological revolution in order to save lives, and then deploying it to kill people. I'm just trying to match the level of crazy in the room.""Hey could you not spread your tech around?"
"Well, we kinda planned on using it... How about if we make it so that it used limited and require check-ins for it to be to refresh uses. I'm sure that it won't run out at the wrong time because our people couldn't make check in."
"Great idea! That too!"
Well, thinking more about it, saying that E=MC^2 is constant doesn't let one to preserve the explosive yields completely. Because yeah, sure, you have the same release of energy in and outside of the ME field. But then you have to take blueshift into account, and if you are dealing with annihilation, then it just gets nasty, as the total energy observed by the observer outside the field changes...
Still, it allows one to easily derive why C changes with the changes of mass, and how. So I'm fine with it.
You are good a physics. I can bearly follow (only through google). But as I had written I did not knew how it scales and run with the assumption that it scales linear. I know that that is unrealistic but I had no better idea.
As you pointed out if the emitters scale linear to the area it leaves the power as the second factor.
Emitters m³ = Thickness * Area * Scalierung ^ Logx(Stärke)
Thickness of the equipment.
Area protected
Scalierung ^ Logx(Stärke) => Building Emitters becomes harder as the projected power increases
Thickness of the equipment.
Area protected
Scalierung ^ Logx(Stärke) => Building Emitters becomes harder as the projected power increases
Power generation and Disturbing m³ = Scale1 * Root( Strength) + Scale2 * Volumen
Scale1 and Scale 2 are shameless scaleable integer. I am open for improvements.
Root(Strength) takes the assumption that bigger thinks have it easier.
Volumen of the shield as indicator for how hard it is to power all emiters.
Scale1 and Scale 2 are shameless scaleable integer. I am open for improvements.
Root(Strength) takes the assumption that bigger thinks have it easier.
Volumen of the shield as indicator for how hard it is to power all emiters.
But how to calculate them?
Lets take the assumption that a readnought is similar protected against its own weapon as an soldier.
The spinal MA is rated with 38 kilotonnes TNT per shoot. That are 158,992,000,000,000 Jules of energy.
I dont know how strong infantery weapons in ME are but scince they seem to work similar enough to "modern" weapons with recoil and all I am going to take them as a base.
Caliber 5,56 × 45 mm NATO rounds sports 1,800 Jules of energy.
There is a massive difference in zeros...
If your common soldier has a defence rating of 1 a dreadnaught would have a rating of 88,328,888,889.
What we "know" as well is how much equipment they are using. We even have 3 date points.
Soldier= Def:1 ; Equipment:0,01m³ (rougly 10 litre for hardsuit)
Hulk Buster Idea= Defreadnaught(88,328,888,889) ; Equipment:240m³ (as stated)
Dreadnaught= Def:88,328,888,889 ; Equipment: rougly 10% (aproximate 392,699,082m³ size point to 39,269,908.2m³ for the shield)
The next step had nothing to do with math but lots of testing until I got the answers I wanted
Emitters m³ = Thickness * Area * Scalierung ^ Logx(Stärke)
Emitters m³ = 0,0004* Area * 1,25^ Log1,75(Stärke)
Power generation and Disturbing m³ = Scale1 * Root( Strength) + Scale2 * Volumen
Power generation and Disturbing m³ = 0,000009 * SquareRoot( Strength) + 0,0015 * Volumen
The spinal MA is rated with 38 kilotonnes TNT per shoot. That are 158,992,000,000,000 Jules of energy.
I dont know how strong infantery weapons in ME are but scince they seem to work similar enough to "modern" weapons with recoil and all I am going to take them as a base.
Caliber 5,56 × 45 mm NATO rounds sports 1,800 Jules of energy.
There is a massive difference in zeros...
If your common soldier has a defence rating of 1 a dreadnaught would have a rating of 88,328,888,889.
What we "know" as well is how much equipment they are using. We even have 3 date points.
Soldier= Def:1 ; Equipment:0,01m³ (rougly 10 litre for hardsuit)
Hulk Buster Idea= Def
readnaught(88,328,888,889) ; Equipment:240m³ (as stated)Dreadnaught= Def:88,328,888,889 ; Equipment: rougly 10% (aproximate 392,699,082m³ size point to 39,269,908.2m³ for the shield)
The next step had nothing to do with math but lots of testing until I got the answers I wanted
Emitters m³ = Thickness * Area * Scalierung ^ Logx(Stärke)
Emitters m³ = 0,0004* Area * 1,25^ Log1,75(Stärke)
Power generation and Disturbing m³ = Scale1 * Root( Strength) + Scale2 * Volumen
Power generation and Disturbing m³ = 0,000009 * SquareRoot( Strength) + 0,0015 * Volumen
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I have punched some numbers in.
Let us assume that a Legionary (with an radius of 1,2meter) has five times as much space (50 litre or 0,05m³) as an military hardsuit for shields. Using my formula it would have a rating of 80.
If we assume that the Hulk Buster variant (with an radius of 1,6meter) has again five times as much space (250 litre or 0,25m³ or the space one legionary takes) as a legionary its rating would be 1,385.
That sounds extreme but I want to point out again that some stutends you can save in the academy in ME3 build with similar space a shield that Cerberus could not overcome with infantery weapons.
All of that is only true if my formula has at least some merit.
Let us assume that a Legionary (with an radius of 1,2meter) has five times as much space (50 litre or 0,05m³) as an military hardsuit for shields. Using my formula it would have a rating of 80.
If we assume that the Hulk Buster variant (with an radius of 1,6meter) has again five times as much space (250 litre or 0,25m³ or the space one legionary takes) as a legionary its rating would be 1,385.
That sounds extreme but I want to point out again that some stutends you can save in the academy in ME3 build with similar space a shield that Cerberus could not overcome with infantery weapons.
All of that is only true if my formula has at least some merit.
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Please explain your notation, it's very hard to follow, sorry.You are good a physics. I can bearly follow (only through google). But as I had written I did not knew how it scales and run with the assumption that it scales linear. I know that that is unrealistic but I had no better idea.
As you pointed out if the emitters scale linear to the area it leaves the power as the second factor.
That leaves the power generation for the shield.Emitters m³ = Thickness * Area * Scalierung ^ Logx(Stärke)
Thickness of the equipment.
Area protected
Scalierung ^ Logx(Stärke) => Building Emitters becomes harder as the projected power increases
I am not going into cost and focus on the size of the equipment.Power generation and Disturbing m³ = Scale1 * Root( Strength) + Scale2 * Volumen
Scale1 and Scale 2 are shameless scaleable integer. I am open for improvements.
Root(Strength) takes the assumption that bigger thinks have it easier.
Volumen of the shield as indicator for how hard it is to power all emiters.
But how to calculate them?
Does this help?Lets take the assumption that a readnought is similar protected against its own weapon as an soldier.
The spinal MA is rated with 38 kilotonnes TNT per shoot. That are 158,992,000,000,000 Jules of energy.
I dont know how strong infantery weapons in ME are but scince they seem to work similar enough to "modern" weapons with recoil and all I am going to take them as a base.
Caliber 5,56 × 45 mm NATO rounds sports 1,800 Jules of energy.
There is a massive difference in zeros...
If your common soldier has a defence rating of 1 a dreadnaught would have a rating of 88,328,888,889.
What we "know" as well is how much equipment they are using. We even have 3 date points.
Soldier= Def:1 ; Equipment:0,01m³ (rougly 10 litre for hardsuit)
Hulk Buster Idea= Def
Dreadnaught= Def:88,328,888,889 ; Equipment: rougly 10% (aproximate 392,699,082m³ size point to 39,269,908.2m³ for the shield)
The next step had nothing to do with math but lots of testing until I got the answers I wanted
Emitters m³ = Thickness * Area * Scalierung ^ Logx(Stärke)
Emitters m³ = 0,0004* Area * 1,25^ Log1,75(Stärke)
Power generation and Disturbing m³ = Scale1 * Root( Strength) + Scale2 * Volumen
Power generation and Disturbing m³ = 0,000009 * SquareRoot( Strength) + 0,0015 * Volumen
Some other notes:
Why? If each emitter covers a given area, and uses a given amount of power, then the power would also scale linearly with surface area.
Yes. The emitters for the area protection scale linear but have no input to the strength of the shield. My headcanon is that the emitters have to become more and more complex/big as the strength of the shield they project increases.
In my headcanon as well is the generator the central part of the emitter array that holds the power (capacitor) and transfer it to the emitters that need it to repell an attack. Following that allows the same shield strength everywhere until the central capacitors/shield generator runs out of juice.
I don't know what I can do to explain the notations better but let me try:
Is the notations now better?You are good a physics. I can bearly follow (only through google). But as I had written I did not knew how it scales and run with the assumption that it scales linear. I know that that is unrealistic but I had no better idea.
As you pointed out if the emitters scale linear to the area it leaves the power of the shield as the second factor.
That leaves the power generation for the shield.To project a shield for a given area a given amount of equipment is needed. Scince it fits into infantery hardsuits it can be fairly thin. Lets call it Thickness for the moment.
How many of the shield emitter you need is linear to the protected Area.
So far we only have the ability to create a shield. Its strength is unknown but most likely the emitters will get bigger if the strength of the shield increases. That means that we need to Scale the needed equipment based on the planned strength of the shield.
The weakest shield fit into hardsuits and can handle infantery weapons but the strongest are build into dreadnaughts that hit with the strength of atome bombs. I doubt that the scaling is linear to the strength so I will use a function based on logarithm to influence it. Log(Stärke)
At this time I do not know wich logarithm to use so I am going to use X as a placeholder.
That leaves the following function to describe how much space is needed for the shield emitters:
Equipment needed for Emitters = Thickness * Area * Scale ^ )Logx(Stärke)
I am not going into cost and focus on the size of the equipment.The central part of a shield is he generator. It is most likely a glorified capacitor or an array of those. I can not explain how those work or if they are still working the same way in 2170+. To make it easy (and most likely false) asume that all the other parts that determine the strength of the shield scale in efficenty exportential to their size. That way we can describe it as the SquareRoot to Strength: Root(Strength)
How big is that? I don't know. I only know that it can fit into an infantery hardsuit. For the moment place the variable Scale that lineary influence Root(Strength).
For the shield strength we have now:
Equipment needed for shield strength = Scale * Root(Strength)
Now we still need the logistic of distributing the energy. That will most likely Scale lineary to the size of the systeme projecting the shield. How big that is can be guessed on the Volume of the projected shield.
For the distribution of energy we have the following function:
Equipment needed for Distribution of energy = Scale * Volume
But how to calculate them?
Lets take the assumption that a readnought is similar protected against its own weapon as an soldier is protected against infantry weapons.
The spinal MA is rated with 38 kilotonnes TNT per shoot. That are 158,992,000,000,000 Jules of energy.
I dont know how strong infantery weapons in ME are but scince they seem to work similar enough to "modern" weapons with recoil and all I am going to take those as a base.
Caliber 5,56 × 45 mm NATO rounds sports 1,800 Jules of energy.
There is a massive difference in zeros...
If your common soldier has a defence rating of 1 a dreadnaught would have a rating of 88,328,888,889.
What we "know" as well is how much equipment they are using. We even have 3 date points.
Soldier= Def:1 ; Equipment:0,01m³ (rougly 10 litre for hardsuit)
Hulk Buster Idea= Dreadnaught(88,328,888,889) ; Equipment:240m³ (as stated by Hoyr the volume taken by two tanks)
Dreadnaught= Def:88,328,888,889 ; Equipment: rougly 10% of its room (aproximate 392,699,082m³ total size point to 39,269,908.2m³ for the shield)
Lets put in the numbers we know for a soldier:
0,01m³ Equipment = Thickness * Area * Scale1 ^ Logx(Strength) + Scale2 * Root(Strength) + Scale3 * Volume
We know already Strength (rated 1) and we can calculate Area and Volume if we handle the soldiers shieldbubble as a ball with radius of 1 meter.
How does it looks now:
0,01m³ Equipment = Thickness * 12.57m² * Scale1 ^ Logx( 1 ) + Scale2 * Root( 1 ) + Scale3 * 3.14m³
And now we have five variables that are unknown.
The next step had nothing to do with math but lots of testing in Excel until the results came close to the three date points above. Here what came as close as I have managed:
Thickness = 0.004
Scale1 = 1.25
x = 1.75
Scale2 = 0.000009
Scale3 = 0.0015
The complete formular can than be solved to avaiable space for equipment, planned shield strength or radius of shielded area.
Equipment = 0.0004 * Area * 1.25 ^ Log1.75(strength) + 0.000009 * Root(strength) + 0.0015 * Volume