Immortal Hulk vs. Yoda

Text-only Version: Click HERE to see this thread with all of the graphics, features, and links.



LordGod
Most thought Hulk would wreck Vader, so would Yoda do any better?

Galan007
A serious Yoda could most likely ragdoll Hulk.


https://i.ibb.co/QcTfnVk/Star-Wars-030-004.jpg https://i.ibb.co/7XtDLB7/Star-Wars-030-005.jpg https://i.ibb.co/CJVJSfQ/Star-Wars-030-006.jpg https://i.ibb.co/mHdhdMk/Star-Wars-030-007.jpg

StiltmanFTW
Hulk is greener.

Galan007
Hulk is the greenest there is. thumb up

Disturbedrogue
Immortal Hulk

Sin I AM
Has anyone tried to grabbing Hulk before?

Eon Blue
Yoda wins.

Damborgson
Originally posted by Sin I AM
Has anyone tried to grabbing Hulk before?

Carver tried to grab Hulk once...results were, mixed....

StiltmanFTW
laughing out loud laughing out loud laughing out loud

Magnon
Yoda could throw Hulk into space with such an acceleration that, by the time Hulk realized what was happening, the distance between him and the planet they were on would already be equal to several planetary radii.

Lets do some maths, based on Yoda's mountain-lifting feat. Assume the gravitational acceleration on that planet was 1 g, the mass of the mountain 2E14 kg, and the mass of Hulk 600 kg.

CALCULATION: (you can skip this, and go directly to RESULTS below)
The force F (actual physical force, not the Force) required to lift the mountain equals 1.96E15 newtons. Such a force when applied on Hulk will accelerate him to relativistic speeds in a fraction of a second, so a special-relativistic analysis is required. In relativity, the relation between (proper) force F and proper acceleration α is given by (where m = Hulk's mass) :

F = mα (eqn 1)

This resembles the famous Newton's second law. The proper acceleration is related to coordinate acceleration "a" by

α = γ^3 a

where γ is the Lorentz factor. Substituting this into (eqn 1) and integrating yields the speed v (of Hulk) at time t:

v = c / sqrt (eqn 2)

Here, c is the speed of light. Integration of (eqn 2) with respect to t yields the distance d traveled (by Hulk) in time t.

RESULTS:

In the table below, t is the time from the start of Yoda's force-push on Hulk, v is Hulk's speed at time t due to the force-push, and d is the distance traveled by Hulk at time t.

t (seconds) v (in units where c = 1.00) d (in units where Earth's diameter = 1.00)
-----------------------------------------------------------------------------------------
0 0 0
0.0001 0.737 0.001
0.001 0.996 0.022
0.01 0.99996 0.23
0.1 ~1 2.4
1 ~1 24

Within one second, Hulk would be 24 Earth's diameters away from the planet they were fighting on. Now that's BFR!

StiltmanFTW
Hello, h1a8

Proper_
This thread is laughable. Hulk blitzes and one shots

MrMind
Originally posted by Magnon
Yoda could throw Hulk into space with such an acceleration that, by the time Hulk realized what was happening, the distance between him and the planet they were on would already be equal to several planetary radii.

Lets do some maths, based on Yoda's mountain-lifting feat. Assume the gravitational acceleration on that planet was 1 g, the mass of the mountain 2E14 kg, and the mass of Hulk 600 kg.

CALCULATION: (you can skip this, and go directly to RESULTS below)
The force F (actual physical force, not the Force) required to lift the mountain equals 1.96E15 newtons. Such a force when applied on Hulk will accelerate him to relativistic speeds in a fraction of a second, so a special-relativistic analysis is required. In relativity, the relation between (proper) force F and proper acceleration α is given by (where m = Hulk's mass) :

F = mα (eqn 1)

This resembles the famous Newton's second law. The proper acceleration is related to coordinate acceleration "a" by

α = γ^3 a

where γ is the Lorentz factor. Substituting this into (eqn 1) and integrating yields the speed v (of Hulk) at time t:

v = c / sqrt (eqn 2)

Here, c is the speed of light. Integration of (eqn 2) with respect to t yields the distance d traveled (by Hulk) in time t.

RESULTS:

In the table below, t is the time from the start of Yoda's force-push on Hulk, v is Hulk's speed at time t due to the force-push, and d is the distance traveled by Hulk at time t.

t (seconds) v (in units where c = 1.00) d (in units where Earth's diameter = 1.00)
-----------------------------------------------------------------------------------------
0 0 0
0.0001 0.737 0.001
0.001 0.996 0.022
0.01 0.99996 0.23
0.1 ~1 2.4
1 ~1 24

Within one second, Hulk would be 24 Earth's diameters away from the planet they were fighting on. Now that's BFR!

I love it

spetznaz
Originally posted by Magnon
Yoda could throw Hulk into space with such an acceleration that, by the time Hulk realized what was happening, the distance between him and the planet they were on would already be equal to several planetary radii.

Lets do some maths, based on Yoda's mountain-lifting feat. Assume the gravitational acceleration on that planet was 1 g, the mass of the mountain 2E14 kg, and the mass of Hulk 600 kg.

CALCULATION: (you can skip this, and go directly to RESULTS below)
The force F (actual physical force, not the Force) required to lift the mountain equals 1.96E15 newtons. Such a force when applied on Hulk will accelerate him to relativistic speeds in a fraction of a second, so a special-relativistic analysis is required. In relativity, the relation between (proper) force F and proper acceleration α is given by (where m = Hulk's mass) :

F = mα (eqn 1)

This resembles the famous Newton's second law. The proper acceleration is related to coordinate acceleration "a" by

α = γ^3 a

where γ is the Lorentz factor. Substituting this into (eqn 1) and integrating yields the speed v (of Hulk) at time t:

v = c / sqrt (eqn 2)

Here, c is the speed of light. Integration of (eqn 2) with respect to t yields the distance d traveled (by Hulk) in time t.

RESULTS:

In the table below, t is the time from the start of Yoda's force-push on Hulk, v is Hulk's speed at time t due to the force-push, and d is the distance traveled by Hulk at time t.

t (seconds) v (in units where c = 1.00) d (in units where Earth's diameter = 1.00)
-----------------------------------------------------------------------------------------
0 0 0
0.0001 0.737 0.001
0.001 0.996 0.022
0.01 0.99996 0.23
0.1 ~1 2.4
1 ~1 24

Within one second, Hulk would be 24 Earth's diameters away from the planet they were fighting on. Now that's BFR!


The Force is strooooong in you!

cdtm
Originally posted by LordGod
Most thought Hulk would wreck Vader, so would Yoda do any better?


By most, you mean two?


No one who really counted thought this. smile

LordGod
Well I mean there's no one arguing for Hulk here. Is Yoda really that far beyond Darth vader?

DarkSaint85
Originally posted by Sin I AM
Has anyone tried to grabbing Hulk before?

Internally? I once asked that in a Mate Grey thread, and Bran went on a rant against me lol.

I don't think he can fight against someone grabbing his heart and lifting it, tbh.

StiltmanFTW
Was that in the Nate vs. Hulk thread?

Star Wars characters were never meant to be taking on threats such as the Hulk.

Galan007
Originally posted by LordGod
Is Yoda really that far beyond Darth vader? TK-wise, Vader can likely approach Yoda/Palpatine-level, but they're still superior.

...There's also the fact that Vader doesn't have a TK feat on par with hurling a mountain-sized creature to the ground, while also holding back an entire group of force-sensitives.

Originally posted by StiltmanFTW
Star Wars characters were never meant to be taking on threats such as the Hulk. And rightfully so. That kind of BS would ruin the franchise for me.

StiltmanFTW
thumb up

DarkSaint85
Originally posted by StiltmanFTW
Was that in the Nate vs. Hulk thread?

Star Wars characters were never meant to be taking on threats such as the Hulk.

Indeed it was.

StiltmanFTW
I remember.

To be fair now, most of the Hulk hate happens only because of carver.

Ten years ago, it was like that for Wolverine when he was fangirling him.

Text-only Version: Click HERE to see this thread with all of the graphics, features, and links.