Unusual Physics: Understanding a Strange Truck Phenomenon in Reality

[physicsisphirst]

There is no 'friction energy' part of the system. It is a dissipative power. Once energ is converted in this 'friction energy' (ie heat, sound etc.) it is gone. So having less friction doesn't speed you up, absence of friction doesn't add energy to the motion. As said before, this only works for an applied force, even when the breaks are on.

ok thanks for the explanation - it makes sense.
but wouldn't the absence of friction decrease the decelerating force and therefore create an acceleration?
what about a curling stone that you slide over a carpet that is on an ice rink. what will happen when the rock hits the edge of the carpet and slides onto the ice? will it speed up or not?
is this situation similar?

 

[marrsc]

The speedometer takes its reading from the transmission. Most automotive braking braking is done by the front wheels. These trucks have a high center of gravity. In the initial moments of braking the front wheels react first and unload the rear wheels breaking their traction and allowing them to spin faster than their true speed. Then the rear brakes must overcome the considerable rotating mass of the tires wheels axles and drive shaft transmission and engine before everything settles down and the speedometer again shows a true reading.
I hope I have stated my theory clearly enough. I am more of a mechanic than a physicist.

 

[Integral]

Marrsc.
We may be getting closer. Did you look at the pics of these rigs? The front wheels are jacked up, not in contact with anything but air! Looks like the only braking is from the rear.

 

[Gokul43201]

ok thanks for the explanation - it makes sense.
but wouldn't the absence of friction decrease the decelerating force and therefore create an acceleration?

No it does not. To have a forward acceleration you need a net forward force. This is Newton's Second Law.

By reducing the net backward force, you do not create a net forward force.

what about a curling stone that you slide over a carpet that is on an ice rink. what will happen when the rock hits the edge of the carpet and slides onto the ice? will it speed up or not?
is this situation similar?

This situation is similar, though not identical. But here too, there will not be any speeding up... only a reduction in the rate of slowing down.

 

[Gokul43201]

Marrsc.
We may be getting closer. Did you look at the pics of these rigs? The front wheels are jacked up, not in contact with anything but air! Looks like the only braking is from the rear.

I think the front wheels get jacked up only during maintenance, when the truck is stationary. Look at the pics of the moving trucks. It does look like they have their front wheels on the ground. If not, you'd need a rear wheel drive.

 

[marrsc]

I assumed that the trucks in the pictures with the front wheels in the air were in the process of being set up for the tracks. In some of the other photos they appear to be in contact. When I have seen these in action they always have their front wheels in contact with the rails. Probably so their front brakes will work. I am assuming that the cylinders that lower the guide wheels allow some travel, like a shock absorber. If they are rigid like a hydraulic cylinder then the shift of the center of gravity would be a lot less but still possible.

 

[munky99999]

there is 2 different Highrail systems, it just depends on if its front wheel drive or rear wheel. But CPR only uses all wheel drive which you can change to RWD, FWD, AWD

Also I believe it doesn't matter which you want.

Anyway. I still go along with the changing friction acceleration. Monday-tuesday I should be getting the thing.

 

[Integral]

Ahh... Yep, Now that you mention it, the ones that I have seen tooling down the tracks had all four wheels on the tracks. I think you have hit on the root cause.

 

[Gokul43201]

It certainly looks like a good plausibility argument.

I'm not entirely convinced about this part though.

...the front wheels react first and unload the rear wheels breaking their traction and allowing them to spin faster than their true speed.

Unless you have an increased torque on the rear wheels, they won't spin faster. Where does this torque come from ? Are the rear wheels coupled to something else ?

 

[physicsisphirst]

No it does not. To have a forward acceleration you need a net forward force. This is Newton's Second Law.

By reducing the net backward force, you do not create a net forward force.

This situation is similar, though not identical. But here too, there will not be any speeding up... only a reduction in the rate of slowing down.

oh dear
i think my physics needs more help than i realized if i have forgotten this sort of thing.
thank you for explaining all this, Gokul.

 

[da_willem]

The speedometer takes its reading from the transmission. Most automotive braking braking is done by the front wheels. These trucks have a high center of gravity. In the initial moments of braking the front wheels react first and unload the rear wheels breaking their traction and allowing them to spin faster than their true speed. Then the rear brakes must overcome the considerable rotating mass of the tires wheels axles and drive shaft transmission and engine before everything settles down and the speedometer again shows a true reading.
I hope I have stated my theory clearly enough. I am more of a mechanic than a physicist.

This only works if the breaking is on the front wheels, and the engine and speedometer are attached to the rear wheels. And then only when the engine doesn't automatically stops driving the wheels when breaking, allowing for the extra torque Gokul whas referring to.

Anyway. I still go along with the changing friction acceleration. Monday-tuesday I should be getting the thing.

Lowering the friction doesn't cause any net acceleration only a smaller deceleration, so this is no explanation! And what thing...?

 

[PBRMEASAP]

It certainly looks like a good plausibility argument.

I'm not entirely convinced about this part though.
Unless you have an increased torque on the rear wheels, they won't spin faster. Where does this torque come from ?

Are these trucks either rear-wheel or all-wheel drive? If not, then that is a good question.

 

[marrsc]

The trucks are rear wheel drive. There would be no reason to keep the four wheel drive engaged while set up for running on the rails. There is some small lag between letting off the accelerator pedal and the reponse of the engine and drive train. When you blip the throttle the engine does not instantly return to the idle setting. When you begin braking the front wheels with less stored rotating energy react first. This is because the mass is less and the braking force is biased to the front wheels. The throttle lag and the much greater rotating energy of the rear drive wheels require more time to react to the initial braking. During this period the rear wheels become unloaded and start to spin in a forward direction giving you the speedometer reading indicating that you have increased your speed. If you had a radar gun reading the true speed of the vehicle there would be NO acceleration.

 

[PBRMEASAP]

The trucks are rear wheel drive. There would be no reason to keep the four wheel drive engaged while set up for running on the rails. There is some small lag between letting off the accelerator pedal and the reponse of the engine and drive train. When you blip the throttle the engine does not instantly return to the idle setting. When you begin braking the front wheels with less stored rotating energy react first. This is because the mass is less and the braking force is biased to the front wheels. The throttle lag and the much greater rotating energy of the rear drive wheels require more time to react to the initial braking. During this period the rear wheels become unloaded and start to spin in a forward direction giving you the speedometer reading indicating that you have increased your speed. If you had a radar gun reading the true speed of the vehicle there would be NO acceleration.

I like that explanation, marrsc.

Now where exactly does the wetness of the track come in? I can see how the reduced friction would allow the back wheels to spin up more. However, it seems that the reduced friction would also reduce the effect of unloading the back wheels. The friction between the front tires and the ground produces the torque that causes the truck to rotate about its center of mass, and this is what unloads the back wheels, right? If the friction is less because the track is wet, the truck doesn't rotate much, and not much weight is shifted from the back. I wonder this "speeding up" also happens on a dry track, and maybe nobody bothered to look at the speedometer?

 

[T.Roc]

ok

the coefficient of friction(cof) in rolling friction is 100 to 1000 times greater than that of sliding (kinetic) friction, I will use the more conservative 100 to show a minimum effect possible.

let's say the weight of the truck is 1000 Newtons. let's assume the cof (dry) between track steel and wheel metal is .80 and the cof (wet) is .20. This means that the rolling cof is .008 (dry) and .002 (wet).

This rolling truck then, has a frictional force of 80 Newtons on dry track. The engine force would be 80 newt. in the opposite direction in a state of equilibrium. (actually a little more because of dissipation of energy in the deformation of the wheel as it rolls) When the truck moves onto wet track and locks up the brakes, the frictional force is just 20 newt. , while the engines 80 newt. force is conserved (for a short declining time period) by momentum until the speed increases (also by the addition of the energy previously being dissipated through deformation) to what it would have been if the resistance were 20 newt. all along, then the force = 0 and the normal braking curve will apply until the vehicle stops.

COF is a proportionate ratio of forces (friction/load); it is dimensionless and area and speed are not used as determiners. A similar "phenomenon" is a regular landslide vs. a "great" landslide where the point of friction becomes "liquified" and cof drops to near zero, making the slide travel faster and farther that your "normal physics" will allow based on mass and incline.

live long and prosper.

TRoc

 

[Cliff_J]

So if you are driving along in this vehicle, and a slight grade comes along, you needn't use extra fuel to make it up the grade? You can just hit the brakes hard enough to lock the wheels and so gain 8-10km/h, then let go the brakes, and sail right up the grade? I don't think so!

I think it is true that if you have parallel tracks, 2 trucks at 50km/h side-by-side, and one brakes normally, while the other locks the wheels, the latter truck can at some point be traveling 8-10km/h faster than the former.

Thank you, a first plausible explanation of the data by changing the problem. Instead of an acceleration in the truck from t=0, its the net velocity of a brake lockup versus ABS over the same distance. Now its a simple sliding friction/static friction problem.

To add to Gokul43201's posts here on the second page, if there is an acceleration, where is the force from?

Cliff

 

[Bystander]

Whence the force on the system truck plus wheels when decoupled from the track? Or, from 747 plus wheels after liftoff? Upon braking, the rotational energy of the wheels can be dissipated as heat, OR transferred to the rest of the system by a "lock-up." "How?" you say. The force necessary to accelerate the stationary point on a wheel to axle velocity is less than that required to decelarate the opposite point from twice axle velocity to axle velocity, hence, a net force on the axle and on the non-rotating body attached to it. The same argument applies to all pairs of points above and below the axle.

Kapiche?

 

[Cliff_J]

What? A truck of over 2500kg is going to be accelerated by some 30kg discs spinning at a few hundred RPM? Even with the mass concentrated soley around the circumference to maximize the moment of inertia of the wheel/tire, and using a sharp impulse to minimize the losses to friction, and applying the entire force as a straight vector and not a torque, it still seems extremely low. Just not enough momentum to justify a 20% increase in velocity much less a 2%. And this is rotation, the force is a vector torque and not a straight vector and would be applied to the vehicle as such.

Care to show the math for a 20% increase in velocity using what, a plastic collision as the model to transfer the tire's rotational momentum to the truck?

 

[Bystander]

What? A truck of over 2500kg is going to be accelerated by some 30kg discs spinning at a few hundred RPM?

Yes. At no point in the thread have I specified a magnitude for that acceleration.

Even with the mass concentrated soley around the circumference to maximize the moment of inertia of the wheel/tire, and using a sharp impulse to minimize the losses to friction, and applying the entire force as a straight vector and not a torque, it still seems extremely low.

Intuition? Intuitive solutions do NOT work in this world. da willem laid out as good a quantitative recipe as is necessary. Plug in what ever numbers you wish.

Just not enough momentum to justify a 20% increase in velocity much less a 2%.

Yes, Mrs. Malaprops? (You meant to state the inverse?)

And this is rotation, the force is a vector torque and not a straight vector and would be applied to the vehicle as such.

You (and Gokul) wanted a description of the couple yielding a net force on the axle (vehicle), you got one.

Care to show the math for a 20% increase in velocity using what, a plastic collision as the model to transfer the tire's rotational momentum to the truck?

Angular momentum is conserved for "plastic" or elastic collisions of wheel with brake shoe (vehicle frame --- how many more hints do you need?) --- that was an earlier challenge to da willem for "bringing his grade up to B" following his conservation of kinetic energy for the system.

Get cute and get cut, Cliff. This thread is the most beautiful demonstration of what's right and what's wrong with this forum --- what's right is some great questions show up; what's wrong is the threads develop like a game of Hollywood Squares with bullsh!t arguments based on friction, failures to realize that ω for the vehicle propshaft is the differential ratio times the sum of the ωs for the wheels, intuition (if "I" can't see a mechanism for it, it can't happen) --- and, worst of all, the inability to back off from mistakes --- munky asked if it can happen (yes), magnitude he's heard is 20 % at 50 km/hr (da willem posted the mass ratio necessary), and I've postulated a separate "measuring" wheel driving an old fashioned hysteresis type speedometer (plenty of mass in the needle to overshoot) for the indicated 8-10km/hr "step" increase in speed.

 

[ohwilleke]

Then why not just make a train that slides on a wet track instead of rolling wheels?

It's called a Mag Lev.

 

[ohwilleke]

This is set in reality.

You have a truck, set with high wheels(railroad wheels). Your driving on the railroadtracks going 50km/h. It's completely dry. You stop easily and quickly.

The weird thing. If its raining and you try to stop from 50km/h, you gain 8-10km/h on average.

How is this explained?

How do you observe your speed? Are you using the speedometer (which really is a proxy for wheel RPMs) or are you using a radar gun?

If the former, the reduced friction should increase RPMs without necessarily producing a like increase in radar gun speed.

 

[ohwilleke]

Your going a constant 50km/h. You hit the brakes with constant force. It also doesn't matter what type of brakes(air, anti-lock, and ABS(whatever that means) were tried, all same result) While keeping your foot on the brake. The first thing the truck will do is gain 8-10km/h ontop of the 50km/h so the odometer and inertia feeling, saying 60km/h or so. After that you start to actually brake and you loose speed. all the way to stop.

Sounds to me like you are using the truck's speedometer to measure speed. Problem solved.

 

[Cliff_J]

A meta comment on what wrong with a forum mixed with a personal attack?

I've only gone on ice and spun my rear tires at 160MPH (2160 RPM at the wheels) and grabbed the handbrake and yet experienced no forward acceleration. Must just be some repeatable isolated personal incident where your physics don't apply.

 

[Bystander]

Do you understand the mechanism by which the final velocity of a vehicle with locked wheels can be greater than that of the same vehicle with wheels rolling synchronously along a track with no change of energy, or not?

That was the question posed by Gokul and echoed by you. No friction, no magic in differentials, no intuition --- just old-fashioned, smash-mouth physics.

 

[CronoSpark]

This is set in reality.

You have a truck, set with high wheels(railroad wheels). Your driving on the railroadtracks going 50km/h. It's completely dry. You stop easily and quickly.

The weird thing. If its raining and you try to stop from 50km/h, you gain 8-10km/h on average.

How is this explained?

The rotational energy is transferred to the truck and not to the ground.
Since it is raining, there is a small (and probably negligable) amount of that energy that is loss due to friction.

 

[Kittel Knight]

The rotational energy is transferred to the truck and not to the ground.
Since it is raining, there is a small (and probably negligable) amount of that energy that is loss due to friction.

It would be ok, if the truck was rolling...

 

[Cliff_J]

Nope.

A wheel sliding on a track offers less friction than a roller bearing? If so, as Fliption said on page 1, why even bother with rolling wheels when a McCoy oiler dripping water in front of a steel skid would be superior and cheaper?

Torque applied to the axle would cause angular acceleration - this is rotational and it would then be transferred to the vehicle. That is similar to the classic statics problem of a bridge with a fixed end pivot and the other end supported vertically but not laterally. Even if we deform the structure or apply thermal coefficients of expansion to the bridge, its net lateral movement as a structure is near zero. And even on the free end, only a few inches.

At best the combined torques would serve to rotate the entire vehicle and compress the front springs/expand the rear springs. So if we measured from the top of the vehicle there might be a small lateral displacement of that point. But if the measuring point were as low from the center of rotation as the previous was high, the results would negate themselves to measurement errors.

Where is the lateral acceleration coming from? If this is basic physics, please provide the simple homework problem where the simple structure experiences an acceleration from the translation of angular to lateral.

 

[CronoSpark]

oh... what i was trying to say was that the force of friction acting on the wheels is negligable compared to the force produced by the wheels onto the truck when it is raining, which results in that acceleration when braking.

i guess a catapult design might be useful to illustrate this idea...