In the movie Pacific Rim, humans pilot giant, 250 foot tall mechs called jaegers. Of course these aren't real, but how do you make them look real in a movie? I love when things like this have to do with changing scale. I'm not sure why.
But how do you make something that isn't real look really really big? One trick is to make it move slower - well, not actually slower but slower apparent motion. Instead of moving an arm from up to down in half a second, maybe it takes 1 second. Why does this work? Let's consider something else. Suppose I take a ball on top of my head and let it fall off. How long will it take to hit the ground?
Once the ball falls off my head, there is only one force acting on it - the gravitational force. Since both the acceleration and the gravitational force depend on the mass, all objects will fall with the same vertical acceleration of around 9.8 m/s2. Now, if the ball starts at a position of h above the ground with an initial velocity of 0 m/s, then I can write the following kinematic equation:
With a height of about 1.7 meters, I get a drop time of about 0.6 seconds. Now, what if I dropped a ball off the top of the Jaeger Gypsy Danger with a height of about 250 feet (76 meters)? Using the same equation above, I get a fall time (ignoring air resistance) of almost 4 seconds.
Ok, this seems easy enough to recreate in a movie, right? Well, let's see. I found two scenes from the trailer (here it is on youtube) that show a jaeger in free fall. Here is the first one showing a jaeger dropped into water.
Now I can use my favorite video analysis tool - Tracker Video to plot the motion of this jaeger as it falls.
If I scale the video based on the length of the jaeger (76 meters), I get the following plot for the position.
The first thing we can see from this plot is that the jaeger fell a distance of 27.3 meters in just 0.6 seconds. If I use a distance of 27.2 meters in the same expression above for falling time, I get an expected free fall time of 2.4 seconds. So, it should take longer to fall. A good bit longer.
What about the acceleration? I can fit a quadratic equation to this falling data (even though there isn't much data). The coefficient in front of the t2 term is 1/2 the acceleration. That would put the vertical acceleration at 43.5 m/s2. That's clearly not correct.
How about another example? Here is a scene where a jaeger gets knocked back a flies over a bridge.
This motion is a bit different. The jaeger obviously starts at the ground with some initial velocity at an angle. Also, the motion in the frame is not perpendicular to the camera so I can't really get a good plot of the position. Let me just look at the time the jaeger is in the air.
From the clip, I find an air time of 3 seconds. Assuming the effects of air resistance are small, this is just plain old projectile motion. The key to projectile motion is that I can treat the horizontal motion and the vertical motion as two separate one dimensional motions. The vertical motion has a constant acceleration (of -9.8 m/s2) and the vertical motion has a constant speed. Here is a diagram.
Let me first look at the vertical motion. What do I know? I just know the time and the acceleration. I can also assume the initial and final vertical velocities have the same magnitude (but different directions). From this, I can solve for this initial velocity based on the definition of the acceleration.
With a time of 3 seconds, this gives an initial speed of 14.7 m/s (32 mph). If you don't think that is very fast for a giant jaeger, I might agree - but let's carry on. Next, let me look at how high this jaeger would go after being walloped by a monster. Without air resistance, the trip from the highest point back to the ground would be half of the total time. Since this trip starts with a zero y-velocity, I can use the same equation above for the dropping ball off the head and solve for the height.
Using a time of 1.5 seconds, I get a height of 11 meters (36 feet). That's not very high compared to the size of a 250 foot tall jaeger. I'm not sure he could even clear that bridge shown in the clip. Ok, what about the horizontal distance traveled? If I know the vertical velocity and the launch angle, I can find the horizontal velocity.
Using a launch angle of 20° I get a horizontal velocity of 40.4 m/s. Since the motion in the horizontal direction has no acceleration, I can use the time to find out how far this jaeger was thrown.
With a time of 3 seconds, I get a horizontal distance of 121.2 meters or just under 400 feet. That's not too bad, but I would suspect he would go farther.
The Problem With Time
Here is the real problem. If you have these giant things, they are going to move giant distances and this takes some time. Say you want to show a jaeger falling over, if you did the physics it might take several seconds for this fall (tipping over is different than dropping a ball). However, people watching the movie don't want to wait 4 seconds or so watching a jaeger fall - that's a pretty long time in an action sequence.
So here is where the special effects people have to make choices. Should they use completely realistic physics or not? They have to decide how long these motion should take. If they are too short, the jaegers won't "feel" very big. If the motions are too slow, people get bored. In the end, you have the scenes in the movie with the times above. I suspect that there won't be too many complaints about the time.
