As a long-time aficionado of the original Star Trek series, it's always exciting for me when I hear that Captain Kirk and Mr. Spock are going to make a reappearance on the big screen. Although it'll be a bit strange without William Shatner and Leonard Nimoy running the show, what recourse is there? We've got the next generation playing the previous one.
Anyway, in the trailer we get a glimpse of the juvenile origins of the future Captain Kirk's daredevil thrill-seeking persona, not to mention his incredible physical prowess. In the scene in question we see young James T. leap out of his classic convertible sports coupe moments before it projects itself off of a several-thousand-foot precipice. James saves himself by gripping the sandy ground and pulling himself to a stop just as he reaches the edge of the cliff.
Just for fun, we are going to apply a little physics and do a couple of quick back-of-the-envelope calculations to demonstrate how truly formidable the young Kirk really is. Our program will be to estimate a) the speed at which the car goes off the edge b) the speed of Kirk as he starts his slide towards the cliff and c) how much force he is going to have to apply to the road to bring himself to a stop at the brink. Now this one is going to be a little bit quantitative, so for all of you physics aficionados out there... here we go.
We see the car's speedometer approaching 80 miles per hour (36 meters per second) in the trailer, so let's estimate that the car is moving at that speed when it goes into its slide. The car appears to be about 30 meters from the edge when it starts skidding through soft dirt and sand. According to Newton's Second Law we get:
Fnet = Ffriction = µmg = ma
where the acceleration of the car is completely due to the friction force. M is the mass of the car, g is equal to the acceleration due to gravity (9.8m/s2) , µ is the coefficient of sliding friction between sand and tires (0.5 at most), and a is the acceleration of the car. Solving for a we get:
a = µg = (0.5)(9.8m/s2) = 4.9 m/s2
If we assume a relatively constant acceleration then
a = (v2 - v02)/2x
where v0 = 36m/s, a = - 4.9 m/s2 and x = 30 m.
Solving for v we find that the car goes over the cliff at a speed v = 32 m/s.
Now the scene does a nice job of showing relative motion between the car, Jim, and the ground. When Kirk jumps off the car, although he is moving backwards relative to the car, he's also still moving forward relative to the ground. A professional basketball player can jump with an initial liftoff velocity of few meters per second, so let's generously give Kirk an initial takeoff speed of 4 m/s. But even so he's still moving at a speed of 32m/s - 4m/s = 28 m/s towards the edge! And he's only at most about 5 meters away when he jumps. In order to come to a stop in that distance Kirk will have to accelerate at a rate of
a = (v2 - v02)/2x = (0 - 28m/s)2/2(5.0m) = 78 m/s2
requiring a force of
Fnet = Ffingers = ma
and assuming that the young Kirk has a mass of around 50 kg we get
Ffingers = ma = (50kg)(78m/s2) = 3900 Newtons (or 877 pounds)
So what does all of this mean? It means that since the only thing(s) slowing him down are his fingers digging into the ground, James T. Kirk will have to exert a force of almost 900 pounds with those fingers to stop from being flung over the precipice. Although Dr. McCoy could never pull that one off, it is Captain Kirk after all. Or at least he will be Captain!
Adam Weiner is the author of Don't Try This at Home! The Physics of Hollywood Movies.
One important thing I'm not seeing is taking into account the friction of his body on the ground. It isn't all just his fingers doing the stopping, after all, since his whole body is dragging, you have all that friction applied, plus whatever force he seems to be imparting by trying to dig in with his feet. Taking all of that into account would decrease the required strength significantly.
I totally agree that his body must have applied a large amount of friction on the ground to cause him to stop. Also, I want to point out that little James pounds on the break pedal and down shifts (or shifted) when he is reaching the precipice. Car people all know older model cars (in this case context way older) did not have anti-lock breaks. So again, Jimmy boy panic stops, the cars breaks lock up, it goes into a slide, at which point he would have bled off a lot of speed. I don’t think the Vette was doing more than 40 mph when it went over the edge.
So can we agree on about 200 lbs of force all around, underwear full of sand, a few cracked ribs and one pissed off Corvette owner?
One other thing that was not taken into account in the calculation : how heavy were Kirk's balls.
Thorn, did you even take Physics?
The coefficient of friction is different for Static friction and Kinetic friction. When you panic stop and your tires lock up, you change the coefficient of friction from Static to Kinetic between your tires and the ground. It happens that Kinetic friction is LESS than Static. You get LESS braking power when you lock up your brakes.
Why are people trying to even defend this?
Even if the article is off by an order of magnitude, you can't get 90lbs of pressure from your fingers, let alone 900. Your shirt and feet aren't going to fix things.
What a bunch of silly fanboys trying to cling to some belief that the movie director gives a rats ass about the physics of the shot. It's about the drama of the shot.
Adam, if you look carefully at the movie clip, you'll notice that young Kirk's shoes are similar to the artificial gravity boots worn by the rogue federation officers in Star Trek VI.
Thus, there are two additional factors at play. Kirk knew he was never <i>truly</i> in any danger (he doesn't believe in the no-win scenario even at this young age.)
Second, while we do not know the setting of the artificial gravity boots, even at a minimal state the boots would have greatly retarded his progress towards the cliff.
As you can see, Kirk was completely in command of the situation.
I saw the film preview. I remember remarking to myself of its kinematic abuse. Glad someone out there had enough free time to quantify the scene's implausibility. Regarding the curmudgeonly fellow above, while the coefficient of kinetic friction is always less than that of static friction, it's hardly ever an order of magnitude less. I applaud Adam Weiner's estimate of mu = 0.5, and in fact I applaud all of his approximations – all sensible in the absence of other known quantities.
Furthermore, while it's extremely hard to invoke a force greater than your own weight via friction alone, your body’s limbs can indeed exert a much greater force by simply grabbing or pressing against things. The question then remains, can a person using all muscles available exert a force of 900 pounds? Personally, I can squeeze a bathroom scale to 250lbs using both hands. I can curl 125 pounds. I can leg-press 500 lbs. That sums to just the right amount of total force needed. But I weigh twice as much as the prepubescent Kirk. So the only way for him to stop without flying off the cliff-face to his death is if he had my strength compacted into his body weight.
We accept this premise for two reasons – it’s he who grows to become Captain Kirk, and, equally as important, the story is fiction.
Neil deGrasse Tyson
Could that possibly really be Dr. Tyson?
If so, wow. I love your Colbert visits. I feel honored to be called a curmudgeon by you, but I think you misread my intent. I applaud the original article as well, it is one of the commenters (Thorn) to whom I was replying who implied that skidding the car would stop it faster than anti-lock brakes. This is why I went on my Kinetic versus Static friction screed.
I do have a problem with the assertion that the sum of your gym workouts is enough to stop you, however. The implication is that you can do all of them at once and that you have something to push against. If you exherted that much force straight down you would fly off the ground. You need a backboard for your bech press, you need the isometric pressure to squeeze the scale. You need the ground and a back rest to push back on your leg press. A 120lb Kirk only has gravity to work with. Then again, it's Kirk. I'm sure his belt buckle added extra stopping power due to 'testicular mass.'
With all that said, here's another piece... the article calculates a 78 m/s2 decleration. Assuming hands above his head sliding feet first, this would be the same type of additional downward force that fighter pilots feel. The blood would pool in his legs. The translation is about 8Gs. Trained fighter pilots WITH pressurized flight suits blackout at around 7Gs. It would probablty also rip his arms off (900lbs?), but I've overstayed my welocome here already so I won't go into that.
Michael van Huystee
Physics Major and born skeptic.
You can't quite make out the large nautical themed belt buckle on Kirk's belt that has a huge anchor on it. I am sure that explains his stopping ability, as well as the 50mph headwind he was driving into.
I don't see any comments about his time in the air. You need an additional calculation for his rate of fall. THEN, you can see how much time he has to decelerate. (or add in a few seconds of air resistance...)
Sorry, Adam. Your calculations fail because you ASSUME he is traveling 80 mph upon exiting the cliff. Anyone with a brain can see he has slowed down considerably by then.
Why don't you re-do the calculations based on the actual footage of the car skidding towards the cliff?
I believe the key assumption made by the author, the speed at which the car goes over the cliff, is flawed.
If you look at 27s into the clip, you will see the car is tilted at an angle of at least 45 degrees, yet half of it still appears to be on solid ground. If it were indeed moving at the assumed speed of 32 m/s it wouldn't 'topple' over the edge, it would fly off, and the degree of downward tilt of the car from the perspective of the cliff edge would not be near that high.
That said, there appears to be a disconnect between the speed of the car in the previous frame as it skids towards the edge, and the speed of the car as it falls off the ledge which the downshifting, braking and friction cannot seemingly account for.
Thanks for your article, Mr. Weiner! I would add only one thing to Kirk's nigh-impossible feat. At the beginning of his leap, Kirk appears to be approximately 1 meter off the ground already. If we give him the benefit of the doubt and assume his leap imparted him with no upward momentum, his rate of downward falling is determined only by gravity:
1/2 x (9.8 m/sec^2) x (y sec)^2 = 1 m
y^2 = ~0.2 sec^2
y = ~.45 seconds
Accounting for his momentum toward the cliff (28 m/s), we see that Kirk should travel 12 meters before hitting the ground -- thus being well over the cliff. So, either Kirk falls faster than gravity, or this stunt did not take place on earth!
at least it looks cool.
IT IS A MOVIE!!!!! Use your minds for something a little more productive please.
It's a magic braking car, not magic fingers. I'm surprised everyone missed it, but when you look down the cliff after the car flies off, it lands only a short distance from the cliff. And if you watch it fly off, it appears to just barely go off the cliff. He's about half a car-length away from the edge when he jumps, so he'd easily be able to stop himself. If anything, he slid too much.
Think of daredevils that jump multiple car lengths. That car only fell a few lengths from the edge when it went down. It was going REALLY slow when he jumped.
michaelgvh, putting on the brake in your car does not change from kinetic to static friction. Static friction occurs when the object is not moving, while kinetic friction occurs when the object is moving. If the car is moving, it's kinetic. It doesn't care whether your wheels are locked or not.
Who cares about the physics of some movie? I sure don't. All I care about is if the movie is entertaining, and to some extent, realistic.
Your reasoning regarding static vs kinetic friction is slightly off. Anti-lock breaks are designed with the purpose of allowing the tires to remain in static friction with the road. If the breaks on a car are not locked, there is always one part of the wheel with a velocity of zero in relation to the road (the part of the tire that is in contact with the road.) If the breaks lock and the tires stop rotating, then it becomes kinetic friction, i.e. the tires are "sliding" on the road. So yes, believe it or not, even a car moving at 120 mph is still utilizing static friction. Ironically, in the absence of anti-lock breaks, your tires are in static friction while the car is moving, but kinetic friction as soon as you jam on the breaks too hard.
Assuming that Kirk has ample time to hit the ground after his leap, it seems logical to assume that the ensuing collision will be inelastic. Therefore, energy will be lost on impact and his kinetic energy will have been considerably reduced from that of the initial leap. Furthermore, the energy of the leap itself will be reduced by the surrounding air through which he is traveling, before he is forced to grasp aggressively at the cliff with all of his available appendages. Before he begins to slide towards the cliff's edge, his motion towards it will already have been decelerated. Certainly, these factors play a role in realizing his dramatic leap of faith as a masterfully executed plan of action.
In a way he seemed to be dumb luck that he got out before the car and him went over the cliff. It would be funny if he stole that cop's bike.