Ford F-150 vs. a Plane. Really?

Big trucks are not as popular as they used to be, but gas prices are going down so maybe this commercial for the ford F-150 will start showing up again:

They say its a real demonstration, but it has seemed odd to me. (they also say this is a professional driver on a closed course and don't try this at home - damn! I was SO going to do that this weekend). Here is my analysis:

Here is the important data I have gathered from the Internets.

  • The truck (a Ford F-150) stops a plane C-123 Provider
  • The curb weight of the F-150 is around 5,000 lbs - depending on options and stuff.
  • The empty weight of the C-123 is around 35,000 lbs.
  • The landing speed of the C-123 is 105 mph - http://airheritage.org/c123.html.
  • The static coefficient of friction between rubber and concrete is 1.0

So, how fast does the plane slow down (on the commercial)? The reason this bothers me is that there are forces on the plane that would make it stop anyway. I could ride behind the C-123 on a skate board and the thing would still stop.

Acceleration

Look at this frame in the video around time 0:25 seconds.

i-1c734ff0ea6f7bad81c88d7b861cbf4f-page-0-blog-entry-2-1.jpg

This shot shows the upper right gauge decreasing. It must be the speedometer (or whatever that is called in an airplane). I assume this is measured in knots per hour. At the first frame of this shot, it appears the gauge is on 40 kph (46 mph). WHAT???? The plane has already slowed down from 105 mph to 46 mph or 40 kph (if the gauge is in kph).

Anyway, that first frame is at 0:00:25,13. The last frame of this shot is at 0:00:26,18 and the final speed appears to be 17 mph or knots per hour (19.6 mph). I am not what the comma means. My best estimate is that this shot is 1 second and 5/30 or 1.167 seconds. This would give an acceleration of

i-dac050248541549351433736770ee610-page-0-blog-entry-2-2.jpg

Now, to convert this to m/s2

i-eaf63e3f32d8205b9f54fac0a9b55913-page-0-blog-entry-2-3.jpg

So, is this reasonable? Let me first apply a simple model (that may not quite work in real life). I will make the following assumptions:

  • No rolling friction.
  • A simple model for friction (that doesn't always work) that says Ffriction is less than or equal to a constant times the normal force.
  • No air resistance
  • The plane has no friction
  • The truck can apply the brakes hard enough to make the wheels lock up, but it has anti-lock brakes so it uses the maximum friction force possible.

Here is a diagram with the forces on both the truck and the air plane (the blue thing is a truck, in case you can't tell)

i-87714adc838bc405ad936bffa433c10a-page-0-blog-entry-2-4.jpg

In this diagram (called a free-body diagram), the tension on the truck is the same magnitude as the tension on the plane (because its the same rope). In this analysis, I will use Newton's second law, which states:

i-25392edb9e6bb37cbaa37f290b577b1a-page-0-blog-entry-2-5.jpg

Where this is a vector equation. We could write this as two scalar equations:

i-cab9c44f8e3a0c357419c4254544b2b2-page-0-blog-entry-2-6.jpg

So, here is the plan. I will use these equations (with the acceleration from above) on the plane to determine the tension. Then I will use that tension and these equations on the truck to determine the coefficient of friction and I will compare this to acceptable values.

The plane

I only need to look at the x direction here. In the y-direction, the acceleration is zero so the weight (gravity) is equal in magnitude to the ground pushing up (normal force). In the x-direction: (note: I am calling to the right the positive x-direction)

i-3df059191aef796a1a3a9541cd1bad26-page-0-blog-entry-2-7.jpg

(in case you couldn't guess, mp is the mass of the plane)

The truck

For the y-direction the same is true (that the normal force is equal in magnitude to the weight). For the x-direction:

i-6c7f8894b616f15db2f1f2ef943b6788-page-0-blog-entry-2-8.jpg

mt is the mass of the truck. The acceleration for the truck is the same as the plane (they are tied together). Now, I can put in the model for friction as well as an expression for the tension (from the plane)

i-8390e5941f00e4016195411cd4b5d737-page-0-blog-entry-2-9.jpg

First some notes: This answer is a unitless quantity (as it should be). The numerator has units of kg m/s2. The denominator has units of Newtons. In fact these two units are the same. This is always a good sign - that the units work out the way the should. It doesn't meant that the answer has to be correct, but if the units didn't work out it would mean it is wrong. Finally, remember this calculation starts with some assumptions. That being said: WOW.

The truck has to have a coefficient of static friction of 6.33??? Rubber on concrete has a coefficient of 1.0 (approximately). So, why is this value so large? (impossibly large) Here are a couple of reasons:

  1. My model for friction simply doesn't apply for this case. I understand that friction is very complicated, but this model is usually pretty good. Note: I used a model for friction (as does everyone). This model is based on empirical data, not fundamental forces. But, it usually works. I do not have a lot of experience with the forces on cars, so I really could be wrong here - but probably not 6.8 times wrong.
  2. There are other forces that are important - like rolling friction and air resistance. Rolling friction is not that great of a factor for the car, but maybe this will make a big deal with the plane. Also perhaps air resistance of the plane plays a large roll.
  3. Perhaps my acceleration is really wrong. I assumed the dial in the video clip was in knots per hour. Why? I guess I assumed that planes used knots per hour. Also, maybe this clip is not in real time but faster than real time for effect (and to make the commercial only 30 seconds total).
  4. Finally, maybe the plane had its reverse thrusters on or its brakes or was in some way contributing to the braking process.
  5. One more - maybe they actually do have super duper high-sticky tires that have a coefficient of friction that high.

Recalculate

One more - maybe they actually do have super duper high-sticky tires that have a coefficient of friction that high.

  • Speed. I will assume the gauge on the plane is in mph. This would give an acceleration of -8.8 m/s2 maybe even call it 8 m/s2
  • The mass of the truck could be a little larger and the plane a little smaller. I will say the mass of the truck is 6000 lbs and the plane 30,000 lbs.

Redoing with this calculations gives:

i-e3eca5a97f0adf5439c7843d324700e9-page-0-blog-entry-2-10.jpg

This is still unbelievably huge. Are there any other clues in the video? One assumption I am making is on the acceleration. I am assuming the decrease in speed on the air plane gauge is the acceleration due to the truck. Why wouldn't I assume that? They show that clip right after they show the truck "apparently" stopping the plane. Is this enough to claim a foul? Not sure.

Conclusion

This commercial has always bothered me, perhaps because I don't really understand cars. What is the big deal with braking? Is it really that hard? It seems (with my limited understanding) that the main factors for braking while pulling something are the vehicle's mass and coefficient of friction between the tires and the road. I would think that any car of the same mass and same tires could stop the same. Perhaps this assumption is based on my limited experience with cars. I have always been able to "lock up" the brakes of a car indicating the limiting factor in braking is the tires. Nonetheless, I claim that truck did not stop that plane all by itself. That commercial sits on a throne of lies and smells like beef and cheese.

i-11d9b50549630c6c1ff8adfec2e65ae1-page-0-blog-entry-2-11.jpg

The other point that annoys me is that they (Ford) claims that truck is so awesome because it can stop a plane (at least that is what it appears they are claiming). When actually, any car could stop that plane, just not as quickly.

UPDATES

Since this is a repost from earlier, I have some corrections:

First, my mistakes. I am an idiot. I referred to the speed of the air plane in knots per hour. Duh - knots is short for nautical miles per hour, so the speed would just be knots. Also, feedback from experienced pilots suggest that the airspeed is likely in miles per hour anyway.

Next, a comment: The most common comment I received was that the point of the demonstration was to show that the F-150's brakes didn't overheat. Ok, I can accept that. If that was the case, it wouldn't matter if the plane was helping it stop - would it? Regardless of the intent, I think the commercial seems to imply the truck did the stopping.

The magic of commercials: Another experienced pilot claims that the speed gauge from the video is fake and should not be in that location in the instrument panel anyway. I can believe this is possible. Perhaps, the speed gauge did not slow down dramatically enough so they made it dramatic. Perhaps they had to speed up the shot of it slowing down so the commercial would not go over 30 seconds. So, this might NOT be the actual acceleration - although the commercial explicitly says "An ACTUAL BRAKING DEMONSTRATION" - so .....

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Here are a couple of things to think about. Firstly, is the chain from the truck to the plane exactly horizontal? If it was vertical, how would that change the calculations. For that matter, is the runway horizontal, or could it be sloping upwards? (Question everything!)

Next, it might be interesting to calculate the amount of energy dissipated in the brakes. Since you know (guessed) the masses of the plane and truck, and the approximate speed, it might be interesting to see how much energy (and, given a suitable time interval, power) is dissipated by the brakes. If you can find/guess the mass and material (steel?), so you could figure the specific heat, you could then calculate an approximate temperature rise.

One of the tricks for braking is that an appropriate force needs to be directed to each wheel. Since the mass is not necessarily evenly distributed, the braking force is not the same for each wheel. Note that static friction is generally greater than dynamic (sliding) friction, so you don't really want a wheel to lock. Thus, the popularity of ABS systems.

Also, note that there are materials which can increase the friction of rubber on asphalt. These are typically used by drag racers to increase their traction. It might be interesting to do some research to see what coefficient of friction these guys can achieve in their rails.

Dave

Don't worry, you are spot on with the idea the commercial is trying to convey: "Look at our big-ass truck!! It can stop a huge plane!!! Go buy it now!!!". Have advertisers EVER been accused of honesty and transparency?

I'm glad you performed this little exercise.

youre so stupid...

in this comercial plane stopped with own brakes..you can see it on its wings and at the end, when using ITS OWN brakes gets a little "down down down" on its nose...

almost any car can stop a plane landing, becose if a track is 500miles it is clearly sure that it has to stop some time

btw, nice idea from ford to make a advertise...

It's kind of interesting when you think about it. A pick up bringing a plane to a halt. What does Ford know and when did Ford know it? The plane that was shot down over Nicaragua creating the Iran-Contra scandal (a C-123) was also the plane that Barry Seal used to import cocaine into this country back in the 1980's. The government has failed to stop drugs from entering the country. Maybe Ford could do a better job of stopping the flow of drugs into the country. Just a thought.

By Bruce Allen (not verified) on 17 Jan 2010 #permalink