I finally saw the movie Iron Man. It was good. I feel that I am qualified to evaluate the movie. When I was in high school, I was totally into comic books. Mostly Spider-man, but I still have a significant collection of Iron Man comics. Ok, now you know I am not an Iron Man attacker. I will now attack the movie. Sorry, it's what I do (remember, I already said I liked it). There are several things I could comment on, in fact I recall some other blog talking about the physics of Iron Man.
My attack will center on the scene where Tony Stark (Iron Man) escapes from captivity with his home made iron man suit. He uses some type of rocket boots to fly away. But alas, the rockets fail leaving Tony Stark to go plummeting towards the ground. The problem I would like to look at is when Tony Stark crashes into the sand. I question whether he could have survived the landing even with the suit on. What does the suit do? Maybe it would prevent broken bones and provide an evenly distributed force. However, Tony would still have a large acceleration. This large acceleration could cause internal damage. So, the goal is to estimate his acceleration when he crashes into the sand. I am going to start with him at his highest point.
How high did he go?
This is a tough question. I am just going to estimate this one. From the above image, it seems like he went pretty high. I am going to use 500 meter. There is a good possibility that in real life in the movie he went much higher. Hopefully this will be a low estimate. Note that after reaching the highest point, Tony was essentially in free fall. This brings me to the next question.
Was air resistance significant in this fall?
The best way to answer this question is to run a numerical calculation (in vpython - my favorite numerical calculation tool). I need to make some estimates:
- Mass: 300 kg. I am not sure what metal he was using, but I will guess it was steel (or something close to that). Steel has a density of around 7800 kg/m3. If he had an average of 2 cm thick suit on him (that includes all the spaces for stuff) and I estimate a human to have a surface area of 2 x 2 x 0.2 m2 (the first two was for front and back) = 0.8 m2. This would give a mass of the suit of m = (0.8 m2)(0.02 m)(7800 kg/m3) = 124.8 kg. Ok, so with the man, I will call the total mass 185 kg.
- Surface area - I already did this - I will use 0.4 m2.
- Coefficient of drag. Wikipedia lists the drag coefficient for a man in the upright position as 1.0 - 1.3. I will use 1.5 because Iron Man is bigger than a man.
Putting these parameters into my numerical calculation, I get a final speed of 84 m/s. This is similar to the speed he would have without air resistance (99 m/s) - so, it really doesn't too much which one I use. Anyway, he hits the sand. I want to calculate the average acceleration he would have. One way to do this would be to use the work-energy principle. As a general rule, if you are looking at motion and you know the time that a force is applied for a certain time, then use the momentum principle:
If you know the distance a force is applied, you can use the work-energy principle:
In this case, I can estimate the distance that the force from the sand is acting on iron man, so I will use work-energy. (Note, you could also do this purely from a kinematics stand point, but I just like work energy, it's cool) So, how far did he move while being stopped? Here is a shot after he landed.
From that image, it kind of looks like he landed feet first and his feet are around 1 to 1.5 meters deep. I will call it 1.5 meters (to be conservative). Now, I am ready to do the calculation (I will do this in a general sense so that if you complain about one of my assumptions - like the starting height, you can easily recalculate). Here are my starting values:
- y1 = 500 meters. This is the initial height he starts his fall from.
- d = 1.5 meters. This is the depth he falls into the sand, or the distance the force of the sand exerts on him.
- m = 185 kg
I am going to approach this in as few steps as possible. In the work-energy principle, you need to pick a starting and an ending point. I am going to pick iron man at the top of his fall for the starting point, and iron man in the sand for the ending point. I also need to pick my system. I am going to take JUST iron man as my system. This means that there will not be any gravitational potential energy, but there WILL be work done by the gravitational force. I will have to break the work done into two parts since the sand does not always exert a force on him. I like diagrams, so here is one:
The reason I choose these two positions is first - this will include the force of the sand (needed). Second, the change in energy for that case is zero. He both starts and ends at rest and there is no potential energy. And here is the work-energy principle for this situation:
Note the dropping of vector notation due to ultimate laziness. The work done by gravity is in the same direction as the displacement. Also notice that I used y1 + d as the distance for the work done by gravity (just to be complete). Even as iron man is slowing down in the sand, gravity is still acting on him. For the work done by the sand, it is a negative value since the force is in the opposite direction as the displacement. I want the acceleration of iron man during this time, so I can use Newton's second law:
The net force is only in the y-direction, so solving for the acceleration in the y-direction:
I need the net force during that time, so that is the force from the sand plus gravity (plus in the vector-sense). So, plugging in:
Yes, that looks confusing. First, it can look a little better by canceling the masses - which do in fact cancel.
If y1 is much greater than d, this simplifies even further, but I will leave it at this. Note that I am assuming no air resistance (which is mostly ok). Now to plug in my values, here is where if you disagree, you can plug in your own numbers.
Granted, this is a big acceleration - but is it too big? Who knows? NASA knows. Yes, they have data - via wikipedia about what kind of accelerations the body can take. I listed this before when I talked about professor splash's jump into 1 foot of water. Here is the whole data table from wikipedia:
Note that the above table is in unit of "g's" where 1 g = 9.8 m/s2. Iron Man's acceleration of 3267 m/s2 is 333 g's. If Iron Man landed the way indicated by his final position, he would accelerate "+Gz - blood towards the feet". NASA lists this direction with a max acceleration of 18 g's for less than 0.01 seconds. The iron man suit doesn't decrease the acceleration of Tony Stark's internal organs even if it does give him super strength and a built in cell phone. (actually, it's probably a built in satellite phone). Even if he only fell from 100 meters, he would have an acceleration of 65 g's.
Ok - stop for a second. Think before you act.
I am not asking for the Science and Entertainment Exchange to intervene and make Tony dead in this scene. That would make for a boring moving.
Yes, there are other ways to solve the problem above to find the acceleration. If you know the speed of Iron Man right before he hits the sand and you know how far it takes him to stop, you could use the following kinematic equation:
But, I like my way better.
Other things I could have complained about:
- This energy source thing that he wears.
- Using an electromagnet to prevent shrapnel from getting to his heart (couldn't they just surgically remove this later? Also, is most shrapnel ferromagnetic?)
- So, say this energy thing has tons of energy stored in it. How does this make thrust for flying? It looks like he has rocket boots. But rocket boots have to shoot something out to make it go.
- Momentum is not conserved when he uses his repulsor hand shooter things. He moves a car back, but he is just staying there.
Notice that I could have complained about these things, but I didn't. Really, I thought it was a good representation of the comic books.
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The energy source - the necessary palladium ring - is cold fusion. Respect the film for its delicate handling of the absurd. A repulsar field's push is anchored against Heisenberg uncertainty vacuum fluctuations, 10^90 g/cm^2 equivalent. It's the stuff that covariantly sources the Casimir effect, Lamb shift, Purcell effect, Rabi vacuum oscillation, electron anomalous g-factor...
If you can believe string theory has 10^(10^5) acceptable vacuum solutions you can hardly criticize a mere 126 minutes of gadget porn. Ask yourself, "how did Iron Man take a whiz?" Now we shall see if the movie is true to itself - "They say the best weapon is one you never have to fire. I prefer the weapon you only need to fire once."
Since his repulsor hand shooter things are powered by the arc reactor thing which would have a gradual power output falloff curve and not some liquid fuel source that would have an abrupt boundary between "on" and "off", it's plausible (within the movie's rule set) that they were able to act as retro-rockets, retarding his fall, even if they couldn't prevent it.
While we watch the film my brother and I wondered why he didn't look for more uses for the anti-momentum device he had also invented. I figure it was in his pocket or something. It is clearly also in use when he is testing his suit and he slams into the ceiling and back again. Tony was just too modest to mention it.
I liked your analysis, but I question the assumptions that went into your value for "d". Given typical impact crater formation, I think it would be safe to assume the local sand height was significantly higher pre-impact than it is post-impact.
I think Professor Jones's trip in the refrigerator (in Indiana Jones 4) is a more egregious example of that kind of bad movie physics. Or maybe I'm just saying that because I liked Iron Man more.
your just ruining the movie so shut your silly mouth
Amusingly enough I was able to handwave all the physics problems. It was only when the tank hit him in mid air that my mind went "Wait a minute...".
like the move ad many years ago said--"it's only a movie-it's only a movie-it's only a movie"..just no guys with chain saws to freak you out in this one..just close your eyes and say "it's only a movie--it's only a movie"---
You killed my Iron Man :) now he is not my super hero...Spider Man rulz
These two would-be nitpicks are actually the same:
# So, say this energy thing has tons of energy stored in it. How does this make thrust for flying? It looks like he has rocket boots. But rocket boots have to shoot something out to make it go.
# Momentum is not conserved when he uses his repulsor hand shooter things. He moves a car back, but he is just staying there.
The repulsors are inertialess (or nearly inertialess) thrusters. The handwavium that lets them work both serves to propel the suit (without reaction mass) and lets him blast things around without getting pushed back himself.
It blatantly violates real-world motion, but it's the major suspension of disbelief in the Iron Man story.
Matt at Built on Facts did a post on the rockets