Originally posted by Rage.Of.OlympusI still remember back in the days when I told you Marvel intended hulk to be superior to thor and you would always dismiss me an ignorant fool...would you care to apologize now?
They collided, Thor brought power to the table and Hulk ended up circling the planet.Pretty sure we even see Hulk flying away from the clash.
Originally posted by JakeTheBank
I don't think Thor: Vikings is canon. IIRC, it was a Marvel Knights mini. And personally, it's a good thing it wasn't canon due to Dr. Strange's rather OOC dialogue and the fact that like...hundreds of thousands of New Yorkers were murdered/tortured/raped in a grisly fashion by the Vikings.
I thought all Marvel Knights series were canon to 616.
Originally posted by JakeTheBankActually I was going by memory on the calculation. I never calculated the Lobo feat but rather how much force does it take to hit a 600lb being into orbit.
How did you get that number as far as Lobo is concerned?
I'll calculate a simplified low estimate of Superman getting Lobo into space. I'll ignore the fact Lobo went through the space ship afterwards and assume Lobo barely got into orbit.
Escape velocity is
Ve=11 201m/s
Assuming the distance Superman uses to accelerate his arm's center of mass is
d=.5m
Lobo's mass
m=1600lb = 725kg
Using the work kinetic energy theorem
and F*d=change in KE = .5 * m * Vf^2
which implies
1] F=m*Vf^2 since d=.5
Now to calculate Vf, or Superman's center of arm mass speed when his fists comes in contact with Lobo's face.
Assuming Superman's arm is 13.5lb (6% of body weight). He hit with both arms or ms= 27lb.
Assuming that collision was elastic then using the law of conservation of momentum
ms * Vf= m * Ve
or Vf = m/ms*Ve =59Ve
So from 1] F=725*(59*11201)^2 N
= 3.17E14 N or 35.6 billion tons of force
Now this is without air resistance. Adding in air resistance results in several orders of magnitude more force (but still under planetary force). Now add in the fact that once Lobo reached orbit, he still had enough velocity left to go through the thick highly durable metal walls of the alien spacecraft like a bullet through cardboard then this makes the feat planetary of several orders.
^ "planetary force"?
look at this weapon and scale it upwards to see how misguided you are:
"The gun was capable of hurling a 94 kilogram (210 lb) shell to a range of 130 kilometers (81 miles) and a maximum altitude of 40 kilometers (25 miles, 131,000 ft) — the greatest height reached by a human-made projectile until the first successful V-2 flight test in October 1942. At the start of its 170-second trajectory, each shell from the Paris Gun reached a speed of 1,600 meters per second (5,250 ft/s)."
Originally posted by psycho gundam
^ "planetary force"?look at this weapon and scale it upwards to see how misguided you are:
"The gun was capable of hurling a 94 kilogram (210 lb) shell to a range of 130 kilometers (81 miles) and a maximum altitude of 40 kilometers (25 miles, 131,000 ft) — the greatest height reached by a human-made projectile until the first successful V-2 flight test in October 1942. At the start of its 170-second trajectory, each shell from the Paris Gun reached a speed of 1,600 meters per second (5,250 ft/s)."
Several things.
1. The object didn't achieve low orbit or any orbit (must reach a minimum of 125miles with escape velocity to remain in orbit forever).
Escape velocity is 11 201 m/s and not 1 600 m/s. Big difference. Plus it has to have 11 201 m/s at 125 miles and not from initial launch. That means it must be launched at much higher than 11 201 m/s.
2. The distance of acceleration is much much larger than Superman's half a meter. That is to say, the distance of acceleration being much larger makes the force needed much less.
3. The object was only 210lb. I big difference from 1600lb.
4. Superman's lobo feat only becomes planetary because of Lobo going through the ship's walls like a bullet through paper AFTER entering space. Barely getting Lobo into space is not planetary at all.
5. The projectile was aerodynamic and capable with cutting through the air with minimal drag.
i know that, but:
paris gun____lobo punch_____________________"planetary" initial force
the feat is a lot closer to what that gun did than what your miscalculations make is seem.
look at it from the top down:
Let's calculate the energy of a 1 gram meteor. We must use mks units. So we set m = (1/1000) kg = 0.001 kg. v = 30 km/sec = 30,000 meters/sec. Plug into the equation:
E = (1/2) (.001) (30000)^2
= 450000 joules
(Every gram would have the energy of 100 grams of TNT.)
"asteroid the size of San Francisco
Lets estimate 10 km x 10 km x 10 km = (10,000)^3 = 10^12 cubic meters. A cubic meter weighs about a ton, so the mass would be 10^12 tons = 10^6 megatons. The kinetic energy would be 100 times larger, or 10 million megatons = 10 million hydrogen bombs.
The US and Russia together had about 10,000 hydrogen bombs."
Originally posted by psycho gundamNo! The actual Lobo feat was planetary ONLY because of him going through the ship. Let's say the alien ship was on the ground 5ft away from Lobo. How much force would it take someone to hit Lobo through it like a bullet through paper? So now imagine Lobo reaching space (he would have slowed down a lot by then) and still having enough speed to go through the ship like a bullet through paper again.
i know that, but:paris gun____lobo punch_____________________"planetary" initial force
the feat is a lot closer to what that gun did than what your miscalculations make is seem.
But If we calculate the force needed just to barely get Lobo into orbit then it is nowhere near planetary.
Also your gun feat irrelevant. The gun releases a 210 aerodynamic object at 1600m/s. That's not even escape velocity. Try 11201 m/s. Also if you launch something at 11201m/s then it still would not make it to orbit. Why? Because of air resistance. The escape velocity 11201 m/s is calculated neglecting air resistance. Also the gun had a huge ass acceleration distance. This makes the force needed much less. A small force over a long distance will yield a very high speed.
If you still don't buy it then look at the same calculation I use for Lobo I will use for the gun.
Final velocity is
Vf=1 600 m/s
Assuming the distance the bullets accelerates through the barrel is
d=30 m (I used a ruler and scaled factored it against the man).
Bullet's mass
m=94kg
Using the work kinetic energy theorem
F*d=change in KE = .5 * m * Vf^2
which implies
1] F=.5*m*Vf^2/d
= .5*94*1600^2/30
=4.01E6 N or
=451 tons of force
Using the same calculations for the bullet being released at escape velocity 11 201 m/s yields 22000 tons of force. And doing the calculation again but with a shorter barrel of d=.5 m (shorter distance to accelerate) we get 1.33 million tons of force. This is all without air resistance. You see the difference?
Some mistakes here.look at it from the top down:
Let's calculate the energy of a 1 gram meteor. We must use mks units. So we set m = (1/1000) kg = 0.001 kg. v = 30 km/sec = 30,000 meters/sec. Plug into the equation:
E = (1/2) (.001) (30000)^2
= 450000 joules
(Every gram would have the energy of 100 grams of TNT.)
"asteroid the size of San Francisco
Lets estimate 10 km x 10 km x 10 km = (10,000)^3 = 10^12 cubic meters. A cubic meter weighs about a ton, so the mass would be 10^12 tons = 10^6 megatons. The kinetic energy would be 100 times larger, or 10 million megatons = 10 million hydrogen bombs.
The US and Russia together had about 10,000 hydrogen bombs."
1. 100 times 10^6 megatons =/= 10 million megatons. Try 100 million megatons.
2. All hydrogen bombs are more than 1 megaton each. Try 15-50 megatons each.
3. There is no point here. Clearly my argument is that Superman can knock someone into orbit. What does this stuff have to do with that?
Lastly, I made a mistake with my original calculation.
I used the wrong m or mass. I'll redo.
1] F=m*Vf^2 since d=.5 where m is the mass of Superman's arms.
Now to calculate Vf, or Superman's center of arm mass speed when his fists comes in contact with Lobo's face.
Assuming Superman's arm is 13.5lb (6% of body weight). He hit with both arms or ms= 27lb.
Assuming that collision was elastic then using the law of conservation of momentum
m * Vf= m_lobo * Ve
or Vf = m_lobo/ms*Ve =59Ve
So from 1] F=27*(59*11201)^2 N
= 1.18E13 N or 1.33 billion tons of force
