Aelryinth RPG Superstar 2012 Top 16 |
This is probably a calc question, but I've been out of school too long and can't figure out how to arrange the math.
Here's the set-up...I want to find out how fast the Hulk really is, and that's simply by normal jumping.
We're going to ignore the fact that he can jump into orbit. Truly. We are. Because jumping to escape velocity is more Supermanish.
No, no, I just want to know the following:
Assuming he jumps off at 45 degrees for maximum efficiency, how fast is he going if he jumps 1 mile?
If he jumps three miles? I think three miles was his 'old limit'.
The physics of it are wonky because his height going UP slows at the speed of gravity, while his speed ACROSS stays at the same speed as his initial velocity (cause we is ignoring comic book air drag).
a .22 rifle bullet usually goes about a mile. A 30/.06 bullet can hit three miles.
So...how fast is The Hulk? I just want to know... :)
If you're a bit of a math/physics whiz who can figure out the right equation, that'd be fun to see. It will also give people an idea just how fast you have to toss something to have it go a mile.
==Aelryinth
Bardarok |
Well you don't need calculus for this if you neglect air resistance, it is just algebra. I am going to give this a shot since I am bored right now but I am a little rusty so watch for mistakes.
If initial leap velocity is Vo and the angle is 45° the initial velocity can be defined as two components a horizontal components Vx and a vertical component Vy
In Magnitude: Vx =Vy=Vo 2^(-1/2)
We now have two equations
Horizontally the acceleration is zero therefore we have a simple equation: distance traveled = velocity*time
3 miles = T*Vx
or 0 = T*Vx - 3 miles
where T is the time of flight, an unknown
Vertically the acceleration is constant at g and we will use a similar equation but add an acceleration term
0 = T*Vy - (1/2)g T^2
zero (because we are returning to the ground) and - 0.5 g T2 because gravity is accelerating downwards at a constant rate. That it is (1/2) g T^2 instead of simply gT^2 comes from calculus (its the integral of the velocity relation) but this is a well documented equation that you can test yourself without any need for calculus.
Remember Vx = Vy so now we have two equations and two unknowns (V and T) So we solve this using standard algebra techniques, the units can be hard but its useful to know that the acceleration of gravity is 78999 miles per square hour.
So set equations equal to each other
T*Vx - 3 = T*Vy - 39500 T^2
3 = 39500 T^2
T = 0.00871 hrs or 31.374 s
plug back into the second equation
0 = 0.00871*Vy - 39500 0.00871^2
Vy = 344.2 mph but this is only the vertical (or horizontal) component of the velocity the total velocity is higher
Vo = 486.83 mph
Buuuuut we neglected air resistance so we will need to be a little bit faster so lets say
500 mph ish.
Note that is significantly slower than a bullet however the bullet is not going up it is only going horizontal. If the Hulk was doing a near horizontal leap (like the trajectory or a bullet) the required speed would be the same as long as their total fall distance was the same since gravity affects all objects on earth the same way.
If you want to include air resistance than its probably best to write a code in MATLAB or similar.
sunbeam |
Artillery does a bit better than that (though the Hulk is not as aerodynamic as an artillery shell).
Some artillery has a range of 40 to 50 kilometers (though I'm betting this is due to the rounds they are using). 1000 m/s is a common exit velocity for the shells with modern artillery, so about a kilometer per second.
Modern artillery uses different kinds of shells, including rocket shells so it's not a similar comparison.
From some casual googling though, this may have been the all time record holder for "conventional" shells.
"The Paris Gun - properly called the Kaiser Wilhelm Geschutz - was so-named for its sole purpose of shelling Paris from extreme distances starting from March 1918. A behemoth, the Paris Gun - regarded by many as the ancestor to the German V3 - was capable of firing shells into the stratosphere from locations as far as 131km from Paris.
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Designed and operated by the German Navy and manufactured by the German munitions firm of Krupp, some seven 210mm guns were made using bored-out 380mm naval guns, each fitted with special 40 metre long inserted barrels. However with only two railway gun mountings actually available just three of the guns were ever in use at any one time, fired from the Forest of Coucy.
Such was the rapid wear and tear of firing its 120kg shells, each requiring a 180kg powder charge, towards Paris - the aim was often wild - that the gun's lining required reboring after approximately 20 shots. Indeed, after every firing the succeeding shell needed to be of slightly greater width.
An undoubted sensation when first deployed (at 7.18 on the morning of 21 March 1918) the appearance of heavy shells in Paris caused initial and widespread alarm among its inhabitants which nevertheless quickly subsided. Once fired a shell took 170 seconds to reach Paris, rising as high as 40 km above the earth."
So say 132 km/3 mins = 44 km/min or 2640 km/hour. So 2640/1.6 = 1650 miles per hour.
Mach is 761 miles/hour at STP. So let's just say the Hulk can duplicate the performance of this WWI artillery piece and travel just over Mach 2.
Interestingly he would also jump into the stratosphere. Not to mention create a sonic boom whenever he leaped in this manner.
Strangely enough this sounds eerily similar to pre-Crisis Wonder Woman as well, when she would jump up there and catch a ride on the jet stream (when she wasn't tooling in the invisible jet).
Aelryinth RPG Superstar 2012 Top 16 |
Yeah, WW used to 'glide on air currents', not fly. They finally did away with the limitation and just gave her flight in the 90's reboot.
When I gave the bullet ranges, that's shooting for distance, not straight, btw. A .22 bullet will hit the ground a few hundred yards away if fired level at shoulder height. It's only got 5 feet to fall, after all, so it would only be airborne for 1/3rd of a second or less.
I'm suprised the Hulk is only moving 500 mphish for a 3 mile jump. Wouldn't Vy be the 'average' velocity during the jump, and the starting be higher? Just kind of surprised here, but I guessing air resistance would have a huge effect on rifles.
At a 45 degree angle, the maximum height the hulk would reach would be 1.5 miles on a 3 mile jump. I don't think that's quite the stratosphere (since assuming gravity is a constant, he'd rise until vertical velocity is 0, then come back down at the same rate as he started, and forward momentum remain constant).
And hey, when they found that German Battleship, it started firing 15 miles away! :)
It would probably have been simpler to just say he jumps 1.5 mile into the air at 45 degrees, how fast did he need to go, aye?
I remember a show where they were talking about the stupidly large giant shark movie, and the scene where it does indeed jump into the sky to try to grab a passenger plane. I think at 10,000 feet they said it had to reach a launch speed of Mach 1.
==Aelryinth
sunbeam |
Oh, I was assuming the Hulk could equal the performance of the vintage WWI artillery piece I referenced in that excerpt.
Using that the Hulk is going to leave the ground at a much higher velocity than that the other poster posited, which is simply based on assuming the Hulk can leap 3 miles.
I'm assuming he can jump as far as an artillery shell can be launched.
Actually I think it would be impossible for him to do that though, if you use anything like real world physics.
I used some figures I thought I remembered from Handbook of the Marvel Universe one (and the opening blurb at the top of comic books that used to be there up till the 80s).
"Seven feet tall, 1000 pounds of jade jawed fury," or something with Stan Lee hyperbole.
The problem with the hulk is that he doesn't have an "extended" time to maintain force.
Assume the Hulk is two meters tall (close to 7 feet). Say his legs are half that and about a meter in length. No matter the force he can exert he can only do it as long as his feet are in touch with the ground (and still extending).
As soon as he has exerted enough force to lift the center of mass of his 1000 pounds more than 1 meter off the ground, he is no longer in contact with the ground and cannot apply more force.
It sounds wonky, but just playing around a little bit I didn't think I could solve it with algebra (though maybe related rates). You get a very simple differential equation (or you could just call it a calculus problem at that point).
What I'm saying is I don't think it is possible for anything to leap like the Hulk does (yeah I know he is pretty much impossible for a number of reasons), because of ... well action and reaction for one thing. He is accelerating his body upwards from the beginning; when he has enough impulse to rise further from the ground than his legs are long, he can't generate more.
You can speculate that his legs flex really quickly (super speed quickly), but the laws of action and reaction still apply the same as always (this is actually the part where you run into calculus).
But when you consider that the Hulk can gain and lose about 850 pounds every time he transforms, and somehow turns his pants into incredibly elastic purple ones every time he transforms, well...
Aelryinth RPG Superstar 2012 Top 16 |
The hulk stands at a minimum of 7 feet tall, ranging up over 8 feet depending on incarnation, and with a minimum weight of 1000 lbs, ranging up to 2400 lbs (the World of Hulk incarnation is 8'8 and 2400 lbs.).
The wikia says his jump has been clocked at 473 mph, but also that he's outrun an airplane, jumped 1000 miles, jumped to Mars, jumped from the base to the top of Mt Olympus, yada yada yada.
I'm just using the original Marvel Handbook, which was about 3 miles.
And yeah, if his velocity has to be higher then the speed of sound, when he jumped he'd make a sonic boom, right enough. Note that escape velocity to reach earth orbit is about 7 miles/second...
==Aelryinth
Drahliana Moonrunner |
As soon as he has exerted enough force to lift the center of mass of his 1000 pounds more than 1 meter off the ground, he is no longer in contact with the ground and cannot apply more force.
That's how Nova defeats KLUH, The Hulk's Hulk. wrapping him up in bridge cables, accelerating him to escape velocity and snapping free before KLUH can snap him in half. (I know the one place where his strength is useless..)
Grey Lensman |
sunbeam wrote:That's how Nova defeats KLUH, The Hulk's Hulk. wrapping him up in bridge cables, accelerating him to escape velocity and snapping free before KLUH can snap him in half. (I know the one place where his strength is useless..)As soon as he has exerted enough force to lift the center of mass of his 1000 pounds more than 1 meter off the ground, he is no longer in contact with the ground and cannot apply more force.
So we have the very first thing in AXis that wasn't stupid. Proper use of physics to resolve a fight.
Drahliana Moonrunner |
Drahliana Moonrunner wrote:So we have the very first thing in AXis that wasn't stupid. Proper use of physics to resolve a fight.sunbeam wrote:That's how Nova defeats KLUH, The Hulk's Hulk. wrapping him up in bridge cables, accelerating him to escape velocity and snapping free before KLUH can snap him in half. (I know the one place where his strength is useless..)As soon as he has exerted enough force to lift the center of mass of his 1000 pounds more than 1 meter off the ground, he is no longer in contact with the ground and cannot apply more force.
On a physicist, no less.