# (DCEU) Aquaman Half-Million Ton Strength Debunk Thread

• 94 results
DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

Sorry guys . . . Just felt like an ass today! So here I go. I'm pretty sure you all know which feat I am referring to, and honestly I'm to lazy to put in all the details. If you want more details as to how a certain calculation is done, or where I got a certain number, or why a certain formula is being used then just ask. Also, if you think I'm wrong then please give reasons for your answer so I can improve the post. Without further ado . . .

I might accidentally start a shit-storm of a thread!

The submarine was an Akula class submarine which can weigh in from 8140 to 13800 tons. Let's high ball and say that it was 20000 tons!

Initial Vertical Velocity of the Sub as it broke the water = x m/s

Final Vertical Velocity of the Sub after it broke the water = 0 m/s

Video Time Tag --> [0:49]

Height that the sub went into the air < 30 meters. Let's High ball and say it was 50!

(V^2) - (x^2) = 2as

(x^2) = -2as

x = (-2as)^(1/2)

x = (-2*-9.81*50)^(1/2) ==> 981^(1/2) ==> 31 m/s

Acceleration = Velocity/time ==> x/t

t = [0:48] - [0:17] ==> 31 seconds :o

Acceleration = 31/31 == 1 m/s^2

F = ma

F = (20000*1000)*1/10000 ==> 2000 tons!

We also need to add in drag force

Time tag --> [0:25] Arthur begins 'bracing' the submarine. Feat turns from strength to durability from here on outwards.

Time of Acceleration = [0:25] - [0:17] ==> 8 seconds

Drag Force calculator = Co-Eff * Density * Reference Area * 0.5 * Velocity^2

Co-eff = 0.47

Density = 1000 kg/m^3

Ref-Area = 2709 m^2 [High-Balled]

Velocity = Acceleration*time ==> 1*8 ==> 8 m/s

Drag Force = 0.47 * 1000 * 2709 *0.5 * 64/ 10000 ==> 4074 tons!

Total Force = 4074 + 2000 = 6074 tons [High-Balled]

Any questions?

Eri_Joni

13159

Forum Posts

213

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

Here we go.

deactivated-60957cbcbe0f1

7362

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

This gon’ be good..

Michael Jackson popcorn meme.

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

#5  Edited By DarkDementor101
deactivated-60758db60e021

9525

Forum Posts

0

Wiki Points

0

Followers

Reviews: 1

User Lists: 0

Nicely done. I'm sick to death of people high-balling Arthur. I mean, he's powerful, but 500,000 tons? Lmfao, no

chuggachugga170

1459

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

bump

The_Hajduk

13849

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

6000 tons sounds right for Aquaman. Then there’s room for Superman to be about 10 times stronger.

Supermanforever

11193

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

#9  Edited By Supermanforever

not bad, but there is some misscalculation. Forexample that the velocity was not 0 as the sub jumped out of the water almost. But alright. There is also things like density etc that need to be considered

deactivated-627d8daf1de25

16791

Forum Posts

3038

Wiki Points

0

Followers

Reviews: 0

User Lists: 7

8k - 13k Tons seems a bit light for a submarine that big.

Where did you get this figure from? @darkdementor101

Shinne

20952

Forum Posts

294

Wiki Points

0

Followers

Reviews: 0

User Lists: 1

Makes more sense, I guess.

Mrnoital

9043

Forum Posts

3547

Wiki Points

0

Followers

Reviews: 3

User Lists: 0

#12  Edited By Mrnoital

so you say the sub is being high balled and weights 20 000 tons but it only takes 2000 tons to move it then send it flying into the air, then most the force came from overcoming drag (true)

it seems like you miscalculated at some point

godzilla44

8625

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 3

@mrnoital said:

so you say the sub is being high balled and weights 20 000 tons but it only takes 2000 tons to move it then send it flying into the air, then most the force came from overcoming drag (true)

it seems like you miscalculated at some point

^^^

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

@mrnoital said:

so you say the sub is being high balled and weights 20 000 tons but it only takes 2000 tons to move it then send it flying into the air, then most the force came from overcoming drag (true)

it seems like you miscalculated at some point

The sub, while being submerged in the water, would not have any weight of its own due to buoyancy. To accelerate it to the speed that I gave, and over come the drag that it would experience due to the water, it would only need the amount of force that I gave up top.

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

not bad, but there is some misscalculation. Forexample that the velocity was not 0 as the sub jumped out of the water almost. But alright. There is also things like density etc that need to be considered

I was referring to when the Sub was mid-air and was in a temporary state of rest, not when it broke the water!

The velocity value that I have to find (X) is what I calculated.

Mrnoital

9043

Forum Posts

3547

Wiki Points

0

Followers

Reviews: 3

User Lists: 0

@mrnoital said:

so you say the sub is being high balled and weights 20 000 tons but it only takes 2000 tons to move it then send it flying into the air, then most the force came from overcoming drag (true)

it seems like you miscalculated at some point

The sub, while being submerged in the water, would not have any weight of its own due to buoyancy. To accelerate it to the speed that I gave, and over come the drag that it would experience due to the water, it would only need the amount of force that I gave up top.

so you're just ignoring that you still have to overcome the inertia of that mass? it's weight doesn't disappear, subarines use systems to rise and sink and it would take an undetermined amount of force to overcome those systems, then proceed to move that weight above where it's buoyancy wants it to be

pretty sure you're ignoring plenty of variables

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

dude, the force that I calculated was due to that inertia. Inertia is the resistive force by an object due to a change in momentum, which in this case was the acceleration that I calculated out. Don't know how that is a difficult concept to understand, but if you want I could make diagram explaining it.

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

thanks for asking about the weight! If you go search up Aquaman on Wikipedia you find the submarine class listed under the plot section. It was listed as an Akula class sub which can be anywhere between 8k to 13k tons depending upon the model. Regardless of that, I still high balled the weight anyways, having assumed that the sub was actually 20k tons rather than the weights already given

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

Also, bouyancy on an object, within a fluid, is the same regardless of the height that it is placed in. Bouyancy only really cares about the volume of the fluid displaced.

Mrnoital

9043

Forum Posts

3547

Wiki Points

0

Followers

Reviews: 3

User Lists: 0

you'd think buoyancy on an object would be the same, except this is a submarine, a machine that changes its buoyancy so it can rise and sink to different depths with different pressures, the fact that you can't calculate the power it takes to overpower these systems you have definitely skipped some variables

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

alright, you definitely are not understanding what I'm saying. I'll make a diagram tommorow to explain this in detail. Going to go play MC right now.

Mrnoital

9043

Forum Posts

3547

Wiki Points

0

Followers

Reviews: 3

User Lists: 0

#22  Edited By Mrnoital

@darkdementor101: make all the diagrams you want, that seems to prove you don't know how submarines work, cause the buoyancy needed to say float on top of the water, and sink to and stay at 100 ft, and to sink and stay at 500 ft are all different levels of buoyancy, and a submarine has the ability to change the depth it wants to stay at, lifting it any higher and you are lifting it's weight, much like lifting a weight under water

deactivated-627d8daf1de25

16791

Forum Posts

3038

Wiki Points

0

Followers

Reviews: 0

User Lists: 7

thanks for asking about the weight! If you go search up Aquaman on Wikipedia you find the submarine class listed under the plot section. It was listed as an Akula class sub which can be anywhere between 8k to 13k tons depending upon the model. Regardless of that, I still high balled the weight anyways, having assumed that the sub was actually 20k tons rather than the weights already given

Yeah I get that it's an Akula class sub but where is 8k-13k coming from?

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

Also Wikipedia

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

@mrnoital said:

@darkdementor101: make all the diagrams you want, that seems to prove you don't know how submarines work, cause the buoyancy needed to say float on top of the water, and sink to and stay at 100 ft, and to sink and stay at 500 ft are all different levels of buoyancy, and a submarine has the ability to change the depth it wants to stay at, lifting it any higher and you are lifting it's weight, much like lifting a weight under water

Okay, I'm not really sure who taught you about buoyancy, because that is not how it works at all! You could say the absolute pressure experienced by the body will change at different depths inside a fluid, but not the net upthrust that is applied to it.

It's just simple math really:

ρg = Density*Gravitational_Constant = ρg = C
Height of Object = K [Constant]
Upthrust[F1] = ρgΔh

Scenario A

F1 = C*[K-(K-H)]
F1 = C*[K-K+H] = C*H

Scenario B

F2 = C*[(K+X)-([K+X]-H)]
F2 = C*[K+X-K-X+H] = C*H

See, no difference in pressure! Honestly, I shouldn't have to prove this because it is already a well documented fact, but I posted it regardless.

However, while I was writing this, I thought of another way that you may have been trying to write your question.

Since Submarines have systems that ensure that they remain at a certain depth, then the submarine would try to re-adjust its depth by taking on water, as Aquaman was lifting it, to ensure that it remained at a certain height underwater. I think this may be what you were originally asking, but I don't know. You kind of phrased your question a bit weirdly! Regardless, if this is what you were originally trying to state, then this would not work either, as to the end of the clip we actually see the Submarine floating on water, not sinking back down as it should have. This would mean that the Submarine actually dumped the water that it originally took on, only leading to it having a net upward force acting on the submarine, which makes the Aquaman feat less impressive, which I'm sure is not what you originally intended.

Also, sorry for the long delay, but I was travelling.

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

What do you guys think?

Also, bump!

deactivated-60957cbcbe0f1

7362

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I think you’re going to get people angry.

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I think you’re going to get people angry.

I was honestly expecting to garner a much larger response, but this is not too far off from what I predicted. The moment a feat is properly analyzed and given its proper due course of analysis, the two extreme ends of CV, High-ballers and Low-ballers, just disappear off the face of the vine. I still remember when people were using that faulty submarine calculation to justify the *Half-A-Million Ton* Aquaman feat. Not a single person who has been using that feat has actually come onto the thread to defend themselves. At that point you could only help but ponder over whether or not they truly know, deep down, that they were truly wrong and are now too afraid to admit it!

deactivated-60957cbcbe0f1

7362

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I just think you’re late to the party. The movie came out in December and we have already argued this to death.

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I just think you’re late to the party. The movie came out in December and we have already argued this to death.

Better late than never!

Heatforce

9461

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

So you are essentially debunking the below video? I would actually like to see you debunk this particular video if that wasn't your original goal. I like the channel and I am a dceu stan but I like accuracy. Also, for us poor math pleebs is there a way to have a third party confirm your calc?

deactivated-5e37510e25a10

2710

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

Bump

MonsterStomp

37581

Forum Posts

361

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

MetalJimmor

6962

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I never really cared what exact number the feat was at so long as it was acknowledged as a viable feat. The scene was clearly put there to showcase Aquaman's immense strength and shouldn't just be tossed aside.

4-6,000 tons seems plenty reasonable. It paints him as a powerhouse that can hold his own against beings like Steppenwolf and tank hits from a creature as big as the Karathen.

So congrats on your math. I am sure that was a lot of effort and you seem to have done a solid job.

ashrym

2003

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I see drag but I don't see the weight of the water above the sub that also needs overcome due to it's viscosity.

I consider the 500000 ton number debunked because it's obviously cinematic licence physics instead of real physics. It didn't take any math to make that judgement call. ;-)

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

for the third party calc confirmation I would recommend asking this question on physics stack exchange. They would give an even better result than I did, and I would recommend that people do not use my calc until they have a second level of confirmation from someone with actual certified credentials.

As for the video debunk, the guys used some bs about 'lifting the weight of the water above the sub', which is absolutely nonsensical. There is such a thing called buoyancy which makes a non-sinking submerged object act virtually weightless. It still has mass, but you are at no point working against its weight

Have more to say, but have to go right now!

deactivated-5e14500e3bd2c

770

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

4K-6k is more reasonable than the half a million ton imo

Heatforce

9461

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

Bump!

death4bunnies

19169

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

#43 death4bunnies  Moderator  Online

I have questions about the subs buoyancy.

——

Why did it float after Aquaman got it to the surface?

——

Would it of came to the surface (slower) if Aquaman didn’t lift it at all, based on how it floated?

DarkDementor101

1562

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

#45  Edited By DarkDementor101

Yeah sure, just go ahead and ask. You should have just posted the question tbh, as I would have answered it by now!

On to the second question, there are two possibilities. Since we know the sub is floating, on the surface, at the end of this entire ordeal, it means two things must have happened. Either the water bearing systems that allow the sub to sink and float must have dropped their loads to prevent damage due to a sudden change in pressure, or possibly some other reason (Scenario A). That or the buoyant forces that should act on the sub simply returned once a majority of the sub got resubmerged (Scenario B less likely). So scenario A involves the water bearing systems to drop their load, whereas scenario B requires nothing to happen as the buoyant forces of the sub were already shown to negate the subs weight, but are less likely to allow the sub to float, as it would require the sub to fully submerge in water. My calc assumes that Scenario B took place (less likely of the two scenarios) to show that even when the feat is high-balled, it is a long shot from the half-a-million ton feat people claim it is

Sorry, but I dont understand your third question

chuggachugga170

1459

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

b

deactivated-6349385499256

14104

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

6074 tons actually makes sense.

580k tonner Aquaman is hard to comprehend and also seemed an outlier.

Ccbm2208

5041

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

I knew something is wrong with that half a million ton calc . Several thousand tons seem more reasonable . At no point did Aquaman ever feel like a casual 1 million tonner.

Darkthunder

16601

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

6074 tons actually makes sense.

580k tonner Aquaman is hard to comprehend and also seemed an outlier.

its not just hard to comprehend. Its impossible beacause no way a sub would weigh half a milion ton