The Indiana Jones films

Steven Spielberg, 1981- , rated 12 and below by bbfc

The Indiana Jones franchise contains 5 films with the first one coming out in 1981 and the last one being released in 2008, another film is planned for 2020. The franchise’s main protagonist is Dr. Henry Walton “Indiana” Jones Jr. who is portrayed by Harrison Ford. Indiana Jones or Indy is a professor of archaeology who investigates ancient civilizations mostly involving the supernatural. He often finds himself in dangerous situations, ranging from rolling boulders to fighting the Nazis, a lot of the films do test the laws of physics. Let’s see if they go too far or remain in the realms of possibility.

In the first film, The Raiders of the Lost Ark, there is a scene in which Indy is attempting to gain entry into an ancient temple. He comes across a chasm, with a log hanging right in the middle of the gap. He wraps his trusty bullwhip around the log by whipping it and then uses it to swing himself across the gap to the other side.

Indy’s whip wraps around the rough log

Whether or not this is possible comes down to friction, can the frictional force between the whip and the log, as well as the friction between the whip and other parts of the whip create a large enough force to hold the weight of Indy. Friction depends on the normal force and the friction coefficient, μ.

Here the normal force is acting out from the circle at all points, and Indy is at the bottom of his swing

Now we need the frictional force, F = μR to be equal to the weight of Indy. μ is dependent on the roughness of both the whip and the log, the rougher the two surfaces the higher the value of μ. Both the whip and the log look rough in the film so it does have the highest possible chance of success. The whip and log are also in contact at lots of point as well as the whip being in contact with itself at lots of points meaning that the frictional force could actually be high enough for this to be possible. Once we accept that the force from the friction can be high enough to hold Indy’s weight it’s a simple case of Indy swinging himself across, to do this he has to become a simple pendulum, the two ledges are at the same height so it should be relatively easy for Indy to reach the other side. Them being at the same height means that the gravitational potential energy is the same at both ledges so Indy doesn’t even need to add any kinetic energy. These findings are supported in Mythbuster’s episode 224 where they attempt to do it for real and find it to be plausible in the real world.

Indy swings across the chasm following the laws of physics

Another scene in the franchise that really pushes the boundaries of physics occurs in Indiana Jones and the Kingdom of the Crystal Skull, the most recent film. In the scene Indy finds himself in a Nevada dessert about to experience a test of an atomic bomb he is in a fake city ready to be destroyed, complete with dummies. He thinks quickly and gets into a Lead lined fridge in an attempt to survive the explosion, he survives this unscathed but how would this hold up to the physics of the real world. 

Indy climbing into the fridge

Once he gets in and shuts the door we see the effects of the explosion, including the dummies burning, then the shockwave arrives destroying all the buildings and sending the fridge flying past a speeding car trying to escape, the car subsequently gets destroyed in the blast.

The fridge with Indy in it catches up to and goes flying past the speeding car

Now since this film was set in 1957 and it appears to be Nevada we can actually estimate the power of the bomb and even the actual operation the test bomb was a part of, there were a set of test nuclear operations called Operation Plumbbob that occurred in the Nevada desert in the same time period as the film in the same location. We can also see that it’s a ‘tower drop’ this means that the yield will have to be somewhere between 10 Kilotons and 44 kilotons. Now we can look at how fast the fridge is moving, the best way to do this is to look at how long it takes the fridge to go past the car, this will obviously include some more assumptions; we are gonna estimate the car as 2 metres long, and assume that it’s travelling at least 70 miles per hour (approximately 30 metres per second), one more thing we need to work out from the clip is how long it takes for the fridge to go past the car, we measured it to be 0.5 seconds. So what velocity does this give for the fridge, 

where the subscripts f and c indicate the velocities of the fridge and the car and l is the length of the car and t the time taken for the fridge to overtake the car. When using all our values from above we actually get a value of 34 metres per second for the velocity of the fridge.

This is only the horizontal velocity of the fridge it has also been accelerated in the vertical direction but we will ignore this, so all estimates for forces from now on will actually be more like lower estimates. We now need to look at the mass of the fridge with Indy in it, we will say that Indy weighs 90 kilograms and the fridge ≈70 kilograms from looking at similar fridges from the same time period. We now need to know how long the shockwave acted on the fridge for and this will then gives us an estimate for the amount of acceleration the fridge and Indy experienced. The shockwaves for atomic bombs can produce speeds in excess of ∼300 metres per second, however to again provide a lower estimate we have gone with 250 metres per second as a rough estimate. We now only need the length of the fridge which looks to be around 0.7 metres. The time that the shockwave acted upon the fridge for is 

this gives us a time of  2.8 milliseconds for the shockwave to act upon the fridge (note that we are assuming the shockwave is moving parallel to the fridge). Using this we can finally calculate the acceleration of the fridge and therefore Indy,

using 34 metres per second for the velocity change and 2.8 milliseconds for the time taken, we get the acceleration to be ≈ 12100 metres per second squared or 1200 g’s comfortably enough to kill Indy, in a car crash anything more than 80 g’s is taken to be certain death so 1200 g’s is beyond an unlikely survival especially given the most g’s ever recorded and survived by a human is only 214 g’s. This doesn’t even include the heat of the blast effecting the shape of the fridge and the deceleration of the landing which would certainly kill Indy. So this scene really does ignore basic laws of physics in exchange for a good scene visually.

In a scene in Indiana Jones and the Temple of Doom, Indy, his 11 year old sidekick Short Round and nightclub singer Willie Scott find themselves in a plane that is going down without any parachutes. Indy gets all three of them to climb into an inflatable raft, pulls the chord so it begins to inflate before jumping out of the plane with the raft. The raft inflates mid air and they all land safely on the snowy mountain before sliding their way down the mountain. Let’s look at this from a physics standpoint.

The raft inflates mid-air and floats down safely

Let’s see if the fall would have been survivable, it comes down to air resistance and what air resistance the raft experiences on the way down from the plane. We will assume that the raft and it’s passengers reach terminal velocity before they hit the ground, there are also a couple of other assumptions we have to make. We are going to assume that the raft is a flat plate so it has a drag coefficient of ≈1.3, we also estimate the area of the bottom of the raft to be 3 metres squared after looking at the clip and other similar rafts (length of 3 metres and width of 1 metre). Finally we have to estimate the weight of the raft and it’s passengers, we are going to take the weight of the raft to be 45 kilograms as that is very similar to other rafts of similar sizes and we will say that Indy weighs 90 kilograms, Willie 50 kilograms and Short round 40 kilograms. Now for this we are going to use the air resistance equation where ρ=1.2 kilograms per metre cubed, the drag coefficient and area are as we said earlier. We now need to set the force from the air resistance equal to the force from gravity, F = mg, where m is 224 kilograms. Rearranging we get this gives us a velocity of  ≈31 metres per second. If we look at experimentation done by NASA into impacts on snow, they actually found that humans have tolerated impacts at velocities of 53.6 metres per second without serious injuries when landing on snow. The fact that they are in a raft would lower the deceleration and therefore the forces experienced by Indy and his friends. This all leads us to believe that this scene is incredibly reasonable as long as the raft doesn’t flip in the air while it is trying to inflate and that it has definitely inflated in time for it too slow itself down.

Throughout the films there are obviously some very large liberties taken with the laws of physics, however there is an attempt to get basic scientific principles correct throughout the films even if the execution of these principles does often leave things to be desired.

The Indiana Jones films therefore get a rating of 7/10 for accuracy:

So we have looked in depth at a couple of key scenes in the Indiana Jones films and the physics behind them as well as giving them our rating on accuracy, but we would also like your input on how accurate you think the Indiana Jones films are in their use of physics, so it would be great if you could vote in our poll.

References:

Featured image – https://www.google.co.uk/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=2ahUKEwitwu-DpMnZAhVlD8AKHccuBhYQjRx6BAgAEAY&url=http%3A%2F%2Fwww.nme.com%2Fnews%2Ffilm%2Findiana-jones-costume-mansplaining-2149437&psig=AOvVaw28_4bIVl-lWC19n2QG1MXg&ust=1519930266640953

Image of Indy’s whip wrapped around the log, Image of Indy closing the fridge, Image of the fridge going flying past the car, Image of the raft mid-air – all screen grabs off youtube movie clips channel.

Power of atomic bomb tests – https://en.wikipedia.org/wiki/Operation_Plumbbob

Blast speed of shock waves for nuclear explosions – https://en.wikipedia.org/wiki/Effects_of_nuclear_explosions

g-force survival – https://en.wikipedia.org/wiki/G-force

weight of inflatable raft – http://www.saturnrafts.com/technical-specs-boat.html

diagram describing friction – http://www.schoolphysics.co.uk/age16-19/Mechanics/Statics/text/Friction_/index.html

Survival of impact with snow – https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930020462.pdf

The Fast and Furious films

multiple directors , 2001- , rated 12 and below by bbfc

The Fast and Furious franchise currently contains 8 films with the first one coming out in 2001. The films mainly centre on street racers performing more and more extreme stunts to evade capture by various law enforcement agencies all involving different luxury sports or muscle cars. The main characters include Paul Walker’s ex FBI agent Brian O’Connor and Vin Diesel’s street racer Dominic Toretto. Throughout the films a lot of possibly physics defying stunts are performed and some real liberties are taken with the laws of physics, so we will have a look at some of these instances now.

Throughout the films we see a lot of people jumping between cars and trucks that are moving at high speeds, would this be possible in the real world or is it just Hollywood trying to make the most interesting scenes possible? Here we will take a look at the physics behind it.

Roman Pearce jumps between two vehicles in Fast and Furious 6

This being possible is due to relative motion, when two vehicles are moving at the same speed no matter how high that speed is, it’s the same as if they are two completely still vehicles.
The only problem with this is that in the films the cars are moving at very high speeds, this means that there would be a large force from air resistance and this could slow you down considerably in the time you are in the air between the two cars, causing you to miss your target. This force from air resistance is proportional to the velocity squared so as the speeds get higher it very quickly gets a lot harder to jump from vehicle to vehicle. The majority of the jumps in the films are relatively small though, so this shouldn’t have too much of an effect. In a lot of these scenes the drivers of the vehicles can always decelerate or accelerate to compensate for this air resistance and even then it’s never shown to be easy to jump from vehicle to vehicle. If two vehicles were travelling at largely different speeds then the characters would however have difficulty in jumping between vehicles. For example if there was a 40 kilometre per hour difference between the speeds of the two cars and someone tried to jump from the slower vehicle to the faster vehicle it would be equivalent to getting hit by a car moving at 40 kilometres per hour. Relative motion also works the other way though, if two cars were travelling towards each other at 40 kilometres per hour and you jumped from one car to the other it would be the same as getting hit by a car moving at 80 kilometres per hour.

In the seventh instalment of the franchise, Fast and Furious 7, we see the main characters all sitting in their vehicles on a plane before they all reverse out and free fall in their vehicles before they pull their parachutes out of the back of their cars and land in the most part softly, but would this be possible?

Dominic Toretto’s muscle car pulls it’s parachute before landing safely

First of all the cars are shown free falling nose down in the most part, this would be the case as when objects free fall the heaviest ends do tend to point downwards and since the engines are in the front, these cars will be heaviest at the front.

For this next part we are going to try and work out how large these parachutes would have to be to bring the cars down safely leaving them driveable. We will need to make a couple of assumptions for this though, we will assume that the cars weigh 1.5 tonnes as the one that Dominic Toretto is driving is a large muscle car and even has it’s chassis reinforced. Now the equation for the drag force from a parachute is given by

Where ρ = the density of air, which for this we will take to be 1.2 kilograms per metre cubed. A is the area of the parachute and v the velocity of the object. C with subscript D is the drag coefficient and is ≈ 1.5 for round parachutes which appear to be the parachutes we see in the clip.

To find the required parachute diameter we must first note that the force from gravity is F = mg, this must be equal to the force from the air resistance when the cars reach terminal velocities which they will be at when they land. The area of a circle is equal to

so plugging this into our equation for the force from air resistance and setting it equal to mg, we get the required diameter of the parachute to be

to use this equation we need to make a couple more assumptions we have already taken the density of air to 1.2 kilograms per metre cubed, and the drag coefficient of the parachute to be 1.5, we take the gravity to be 9.81 metres per second squared as always. For the purpose of this calculation we take the weight of the cars to be 1.5 tonnes or 1500 kilograms as we said earlier. We must now estimate the velocity that the cars can land at, the safe speed for humans to land at is 3 metres per second, so we will go with 7 metres per second for the cars. Plugging all of this into the equation for the diameter we get D ≈21 metres, looking at the clips 21 metres doesn’t actually look to be too high for the parachutes we see in the films. It should be taken with caution though as 7 metres per second is still a very high velocity to be landing at vertically and it’s unlikely all the cars would be able to survive the landing and drive off.

In Fast and Furious 6, there is a chase scene involving Dominic Toretto and Letty, Letty gets flung off the top of a tank that is moving at high speed. It appears like she is going to fall off the bridge that they are on but Dom dives off his own vehicle and collides with Letty mid-air catching her before they both land on the windscreen of a car on the other side of the bridge.

Dom and Letty collide mid-air

The first problem with this scene is conservation of momentum. The law of momentum conservation states that for a collision between two objects (people in this case) the total momentum before the collision must be the same as the momentum after the collision. Momentum is a vector so direction does matter, when the two collide and then carry on together the direction should therefore be much closer to parallel to the road whereas in the film Dom doesn’t even change direction. The only thing that could possibly explain this could be the fact that momentum depends on mass and Dom’s mass will be much greater than that of Letty’s, however this difference isn’t large enough to explain the massive change of direction that Letty experiences.

The next thing to look at is whether or not Dom and Letty could have actually survived the landing on the windscreen and how much deceleration they actually experienced. From the clip we are gonna assume that the vehicles and therefore Dom and Letty are travelling at about 100 kilometres per hour or ≈28 m/s. From looking at the windscreen we can say it bends about 20 centimetres or 0.2 metres.

To calculate the acceleration we are going to use the equation of kinematic motion where the subscripts f and i indicate the final and initial velocities, a is the acceleration and d is the distance Dom and Letty were decelerating for. Plugging in the estimates from above and a final velocity of zero, we get an acceleration of  ≈1960 metres per second squared or 200 g’s. To put that into context we make the following graph, 

This graph shows the different g’s experienced in other situations that involve large g-forces, normally for car accidents 80 g’s is taken to be certain death. The highest ever g-force recorded where the driver survived occurred in an Indycar race where the driver was in a vehicle with extensive safety features and even then the driver was incredibly lucky to survive. Dom and Letty had no protection so it would be extremely unlikely that both Dom and Letty would have been able to survive the collision let alone walk away unharmed. So this is one case where the film really does ignore basic physics.

Throughout the films we have seen a lot of pretty incredible stunts some of which are relatively realistic (people being able to jump from moving vehicle to moving vehicle) and some obviously less realistic (Dom and Letty surviving almost certain death with no injuries to speak of).

So because of all this the Fast and Furious franchise gets a 4/10 rating for accuracy:

So we have analysed the physics behind a couple of the more interesting scenes in the Fast and Furious franchise and have given the films our rating but we would also like to get your opinion on how accurate the films were in their use of physics so it would be great if you could vote in our poll.

References:

featured image – http://www.vulture.com/2015/04/could-furious-7s-skyscraper-jump-really-happen.html

all three photos – screen grabs from youtube channel movieclips,

Drag Coefficients for parachutes – http://www.rocketmime.com/rockets/descent.html

g’s used in the graph – https://en.wikipedia.org/wiki/G-force#Human_tolerance_of_g-force