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# The Physics of Fred Flintstone’s Flaming Feet

The views expressed are those of the author and are not necessarily those of Scientific American.

I hope that the father of the “modern Stone Age family” has thick skin, or else he is going to lose his legs.

Let’s put aside the fact that Fred Flintstone basically runs to work and therefore doesn’t really need his wheels (or that he would need the quads of a god to get them moving). What is much more interesting is the way he stops his caveman car. With heels screeching and smoking, Fred famously uses his own feet to stop his forward momentum. Much like how your car’s brake pads work, Fred’s feet absorb all the frictional forces until the stones stop rolling.

But just how much force would his feet need to apply, and what would happen to them?

Friction’s Victory Over DaFeet

The coefficient of friction—a ratio of the force friction provides and the weight pressing down on a surface—is the first thing we would have to find. It has been studied for human skin [PDF], but what we are really looking for is the roughness of the road; how much resistance Fred’s feet would meet. I doubt that The Flintstones had the paved roads of today, so a first guess at the roughness of the road Fred is braking on would be a bit higher than the values we have on highways: around 0.7-0.9.

Heavy loads encounter more friction, so next we have to estimate the weight of Fred’s famous rockmobile (replicas are for sale, by the way).

Most of the car’s mass is going to be bound up in the huge rock rollers at the front and back. Let’s assume that they are granite. If they were one and a half meters long and 80 centimeters in diameter—like huge stone rolling pins—they would be about 360 kilograms (~795 pounds) each. But that’s not all pressing down on the road Fred has to stop on. Fred himself is a hefty fellow, maybe 95 kilograms (210 pounds). For the sake of estimation, we could assume the rest of the car, made of tarp and wood, weighs and additional 50 kilograms (110 pounds). So all in all Fred must use his heels to stop an 865-kilogram (1910 pound) rockmobile. (Note that this is much lighter than cars today due to their metal frames, engines, and other components.)

So we have the weight, now we need the speed. Fred moves his wheels under his own power, but I wouldn’t guess he could go too fast. The fastest human powered bicycle has been clocked at an astonishing 82 miles per hour. There is no way a chubby caveman could get a hunk of granite going that fast with only his legs, and even average highway speeds seem out of the question. The rockmobile is neither a picture of aerodynamics or engineering genius, but it still has to be faster than simply running to make any sense at all as a vehicle. I’ll give Fred the benefit of the doubt and say his top speed is 25 miles per hour.

Now that we have all the numbers, all we have to do is plug them in. Assuming for a moment that Fred’s feet are so callused as to be indestructible, slamming his feet into the rocky road at 25 miles per hour would bring the rockmobile to a stop in about 26 feet (8 meters). That’s not too bad considering his bunions are brake pads. (If you want to see the full calculation for stopping distance, I did the same for Batman and the Joker sliding on the floor here.)

But what if Fred’s feet weren’t indestructible? The force of friction on his feet would be twice the bite force of a large American alligator, so I don’t think his toes would take it too well. To put it another way, it would be like standing on a belt sander, with an adolescent African elephant on your back.

The best-case scenario for Fred’s feet is that they incur a mild “road rash”—the kind of abrasions and burns that motorcyclists and bikers get when they skid along the ground at speed.

The worst-case is much more likely. If Fred pressed and held his feet to the ground, he would most likely lose them. Sadly, we know this would happen because of the unfortunate cases where people’s feet are run over by cars (study link includes some graphic images).

When a material can’t handle the friction, the result is catastrophic heating and disintegration. In all likelihood, Fred Flintstone’s feet after braking would look like airplane landing gear when it can’t handle the friction:

If Fred decides to take his whole family out for a brontosaurus burger and a drive-in movie, the problem of feet-destroying friction only gets worse because the rockmobile would be significantly heavier. All this is assuming of course that the car itself would hold together for more than a few feet. As an engineer, I have no idea how a forward-moving car keeps a rear wheel on that has no backstop.

Engineers today have developed brakes that can take huge amounts of heat and stress precisely because other materials would fail under such strenuous conditions. On a list of those unusable materials is certainly human skin.

Using your bare feet as brakes is a yabba dabba don’t.

Images: Screenshot from Flintstones opening credits; Airplane landing gear from Airlinesafety.com

About the Author: Kyle Hill is a freelance science writer and communicator who specializes in finding the secret science in your favorite fandom. Follow on Twitter @Sci_Phile.

The views expressed are those of the author and are not necessarily those of Scientific American.

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1. 1. ChrisSciAm 1:46 pm 04/22/2013

The weight of the car isn’t on top of Fred when he’s breaking. He only has his own weight, regardless of who’s in the car with him.

2. 2. ChrisSciAm 1:50 pm 04/22/2013

To clarify, I’m curious what the ‘Twice the bite force of an alligator’ is referring to… Is that the force of friction for Fred, alone, pressing his heels down while going 26 mph?

3. 3. K.Hill 1:57 pm 04/22/2013

Chris,

If Fred must bring the entire car to a halt, we have to consider both Fred and the entire car to calculate the force of friction. This is because friction depends on the normal force–the weight of the breaking object on the surface. Though the weight is not on top of Fred, he is basically a part of the whole object trying to stop. If you only had to stop your own weight when stopping a car, your brake pads would never wear out.

For stopping distance you are correct, this is independent of weight.

4. 4. nicholasjh1 3:34 pm 04/22/2013

I propose that the ludicrous material of granite is not used, or that they are hollow granite cylinders…. now redo the calculations correctly .

5. 5. greylander 5:11 pm 04/22/2013

Kyle,

You missed Chris’s point. The *normal* force applied by Fred’s feet does not depend on the weight of the car. That weight is born by the rock rollers. So while the M in F=MA is the combined mass of Fred and the rockmobile, we get F from the normal force times coefficient of friction, F=Nc. N will be Fred’s weight plus any additional force he can generate pushing down with his legs.

In the animations, we typically see Fred pulling back and *up* on the steering wheel. The upward component of his force on the steering wheel will be matched by additional downward force on the ground via his feet. Given the awkwardness of the angles involved, this is unlikely to be more than about 100 lbs. If the normal force is just his weight, that is about 10% of the total (rockmobile plus fred) weight. If we are generous and allow Fred can push his feet down with an additional 200lbs or so, then we get a normal force up to about 25% the total weight.

This means the acceleration is between 10% and 25% of that which you used to calculate stopping distance, and a more accurate estimate is between 260 and 104 feet.

6. 6. sangandongo 5:48 pm 04/22/2013

Absolutely a waste of time and resources to have studied this. And people wonder why the right wing is up in arms about wasted science funding.

Go do something useful, this is fecking dumb.

7. 7. K.Hill 6:34 pm 04/22/2013

Greylander,

Using only Fred’s weight would work if we wanted to find out how much friction he alone would have going 25mph, but the calculations I did here assume that the entire vehicle (with him in it) must stop. It’s really not that different than solving the problem for a modern car. It’s just that in this case the brake pads are Fred’s feet. I assume that Fred’s feet are like a component of the car.

8. 8. Shmick 6:55 pm 04/22/2013

Guys, Kyle is correct.

Don’t think about the individual components (Fred or car or rollers…). Think of an object with certain Kinetic Energy (1/2mv^2) with only the force of friction to dissipate that KE.

The only friction in this example is that of Fred’s feet, and it has to turn the KE of the whole object (Fred + Car) into Heat to bring the object to rest (KE=0).

Sanganddongo, I assume you’re being sarcastic. I don’t see where the NSF or other body is cited as a funding source… ;-P

9. 9. K.Hill 7:40 pm 04/22/2013

Shmick,

Yes, you made my point quite well for me!

Here’s another way to think of it (with equations!): http://hyperphysics.phy-astr.gsu.edu/hbase/crstp.html

10. 10. greylander 8:42 pm 04/22/2013

Kyle,

It does not matter if his feet are a component of the car or not. What matters for the force of friction is the downward force of his feet against the road surface — the normal force.

In the link you provided, the force of friction comes from contact between the tires and the road, since the tires bear all the weight of the car, the normal force is the weight of the car.

In the rockmobile, the stone rollers bear the weight, not Fred’s feet. There are no brakes, so the rollers just keep rolling freely, providing no braking force of friction of their own. There are only Fred’s feet to provide a force of friction.

You have naively applied the equations from the link provided. Using those equations, umg (using ‘u’ for greek mu) is the force of friction. mg is just the weight of the car — which in that example is the normal force needed to determine the force of friction. To get the correct result, you must replace the mg in that equation with the downward force of Fred’s feet due to his weight and strength.

If there is any point in talking about the stopping distance for Fred using his feet, then we have to talk about the normal force applied by his feet, which is clearly *not* going to be the weight of the car+driver. You seem to want to calculate the stopping distance as if Fred’s feet would work in the same way as brakes and tires on a modern car, in which case, all you have done is estimate the mass of the rockmobile and plugged it into that equation, which is only applicable when the surface(s) generating the frictional force is(are) also bearing the entire weight of the car.

11. 11. greylander 8:51 pm 04/22/2013

There is actually another problem with naively applying your linked equation, on the kinetic energy side. In a real car, the kinetic energy due to rotation of the wheels can be neglected, as the wheels have little mass as compared the car. But in the rockmobile, the stone rollers account for almost the entire mass. Their moment of inertia is significant. In fact (doing quick calculation) rolling cylinders will have angular kinetic energy equal to half their translational kinetic energy.

Moment of Intertia: I = m r^2 / 2
angular velocity of wheels: w = v / r
Angular kinetic energy = I w^2 / 2

This increases the kinetic energy on the left side of your linked equation by (almost) 50%, and likewise the stopping distance. So now I must revise my estimate: somewhere between about 150 and 400 feet (again, depends on how hard Fred can push down with his feet).

Overlooking the angular kinetic energy of a rolling object is a common physics homework mistake (I just overlooked it myself in my first comment).

12. 12. greylander 9:14 pm 04/22/2013

Kyle, Schmick,

Upon re-reading your responses, I can better point out your mistake. You seem to think that the amount of kinetic energy to be dissipated dictates the force of friction and therefore the stopping distance. What you fail to realize is that the linked equation is only applicable *after* you correctly determine the force of friction (alternatively you could find the force of friction if the stopping distance is known ahead of time).

As I have already pointed out, umg is the force of friction in the equation — works fine for tires and brakes. Does not work for feet, with the wheels still free to roll.

As I’ve also pointed out, the angular kinetic energy cannot be ignored with those massive cylindrical stone rollers, which means the right hand side of the equation also does not work for the rockmobile.

This is, perhaps, a good example of why Engineers should not do physics.

13. 13. Mike Smith 9:34 pm 04/22/2013

This article is silly and the author is clearly losing his grip on reality. He assumes Fred is 210 pounds!? Fred weighs 240 pounds if he weighs an ounce. And he’s probably closer to 260 after a heavy night of snacking on brontoburgers at the drive-in. Which means that every calculation that follows is pure junk. This guy now has no cred whatsoever.

14. 14. jovoc 8:24 am 04/23/2013

When in doubt always go to the raw data. If Fred is 1.8m tall then the car is about 4m long. Examination of the YouTube sequence shows that the car is travelling at about 3 car lengths per second, 12m/s ~ 27mph. The rollers at 80cm diameter and 1.5m length have a volume of ~750 litres and will be 2000kg each in weight. Fred himself has a waist diameter of 75cm and probably weighs as much as a decent sized bull say 600kg+. The car with Fred in it probably weighs about 5000kg. Starting and stopping takes less than a second, about 2 car lengths.
So Fred’s feet are delivering and dissipating about 100kW for a second at the start and end of each journey. They bred them tough in Bedrock….

15. 15. dabble53 8:50 am 04/23/2013

I love all the comments about the normal force being a combination of Fred’s weight and how much force he can apply with his legs, as if they are additive.
Fred cannot apply more force than his weight, unless (as was alluded to in the story) he’s “adding” other weight. In this case, trying to partially lift the rockmoble by the steering wheel. Except for supplying an upper limit on the normal force possible, the weight of the rockmobile is not really a factor in the normal force re: Fred’s feet.

16. 16. jovoc 10:10 am 04/23/2013

Of course Fred could be from Flores in which case he would be 90cm tall and weigh about 70kg. If the wheels were hollow and were made of pumice the their weight would drop to about 40kg each taking the car mass (with Fred in it) down to well under 200kg. Top speed would only be 6m/s(13mph), so the total kinetic energy would be about 100x less. That means that his peak accelerating and braking power would be less than 1kW, well within the capabilities of current athletes.
So it’s all a matter of scale….

17. 17. sangandongo 1:06 pm 04/23/2013

@Shmick – I don’t care how this got funded. It’s a waste of time.

18. 18. Shmick 7:14 pm 04/23/2013

Greylander,

I’ve thought about what you wrote, and I can admit when I’ve made a mistake. Thinking about his “push down” as analogous to brake callipers is instructive, and forgetting angular momentum is short-sighted.

However, “This is, perhaps, a good example of why Engineers should not do physics.” is a wee bit harsh and more than a little rude. Hell, I’m a lawyer by trade so I guess I shouldn’t even be reading the SciAm blogs ;-p
/joke

19. 19. peterjf1 3:58 am 04/24/2013

210 pounds does seem a bit low. Fred Flintstone is considered by many to be based on comedian Jackie Gleason and he was about 250 pounds when he was playing the part of Ralph Kramden in The Honeymooners.

20. 20. jack.123 8:57 am 04/27/2013

The fact that Fred and dinosaurs are living at the same time suggest that this is not Earth.The acceleration of the vehicle with no effect of drag further suggest that it is not even the same universe,with different physical laws than ours.This also means that an extraterrestrial intervention that prevented mass extinctions that happened here,and probably genetic engineering of Freds race’s DNA to incorporate nanotechnology’s that gave him super human strength and friction protection.

21. 21. staciejung 9:10 am 04/29/2013

It wasn’t yabba dabba do-able.

22. 22. staciejung 9:13 am 04/29/2013

@Jack, of course it’s earth. The little martian said
so.

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