Unusual Physics: Understanding a Strange Truck Phenomenon in Reality

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In summary, when driving a truck with high wheels on dry railroad tracks at a speed of 50km/h, it stops easily and quickly. However, if it is raining and the truck tries to stop at the same speed, it gains an average of 8-10km/h. This is because when the wheels are rolling, the friction is static friction, but when they start to slide, the friction becomes kinetic friction. This reduces the force due to friction and results in a net increase in applied force, causing the train to accelerate. This phenomenon has been observed by the University of Toronto, but the exact explanation is still not fully understood. Factors such as speedometer measurement and the engine's continued driving force may also
  • #36
Rogerio said:
Well, as I don't believe in the real truck behavior, I won't explain the toy till the end of the thread. :smile:

Oh, I get it. You were just pulling our legs. Sorry, I'm a little slow when it comes to message board sarcasm.
 
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  • #37
munky99999 said:
I'm pretty sure i understand the friction change to increase speed. I do believe it would explain very well as to why.

But it does not ! It can only explain why the deceleration is not as great.

Unless your truck is being towed/pushed by something, the only way to gain speed is from the rotational KE of the wheels. And even then, stopping them will result in heating of the brake pads, so it doesn't seem like there's anything to be gained there either.


Moreover, you claim that the speedometer shows this increase in speed. The only way a speedometer will show a greater speed is if the wheel (shaft) is spinning faster. There's no way that this can happen when you hit the brakes...unless of course, your speedometer works differently than most.

I'd like to see a source or reference that makes this claim. Or is this personal experience ?
 
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  • #38
Some vehicles gather the speed from one tire. If there was a traction difference between the tires and its a tire connected to the other by an open differential than it could result in one tire spinning faster. The traction difference could even be a braking difference where one drum brake's leading shoe is grabby because the drum surface is irregular. Remote possibility, but it doesn't sound like multiple controls and data points were gathered here.

If the test data couldn't be gathered using a radar gun or fifth wheel to eliminate system interactions then the data may not reflect reality at all but simply a measurement error.

Otherwise did the people in the truck get set back in their seat like someone had accelerated instead of braked? Probably not. A rock hanging from a string tied to the rear-view mirror would be an indication of how to draw the vector...

Cliff
 
  • #39
  • #40
let the speed of the train after considering the retardation due to friction be X
let the speed of train withought considering the friction be x
let retardation due to friction be y

X = x - y

now, if the friction decreases by an amount m

X = x - (y - m)
= x - y + m

so the speed increases by m.
 
  • #41
but how do you explain that "the" friction decreases when hiting the brakes?
(and which friction do you mean?)
 
  • #42
WORLD-HEN said:
let the speed of the train after considering the retardation due to friction be X
let the speed of train withought considering the friction be x
let retardation due to friction be y

X = x - y

now, if the friction decreases by an amount m

X = x - (y - m)
= x - y + m

so the speed increases by m.

x, X are velocities; y is an acceleration; m is either a dimensionless number, or a force but you also claim it is a velocity.

You are now in a dimensional mess ! :biggrin:
 
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  • #43
It's more of a logical one...Adding accelerations and velocities...

Daniel.
 
  • #44
I envisioned a train, not a truck, I have seen these rigs, looks like a lot of fun( as long as nothing else BIG is using the tracks!) . With a full train behind you a lot of strange things can happen, but this is not a train.

I retract my overly quick guess.

More data is needed. External speed measurement or an accelerometer would be necessary to determine what is happening. It is not clear how a trucks speedometer could read anything with the rear wheels locked up.
 
  • #45
maybe munky99999 wanted a formula to explain, but i thought a common sense explanation was what he sought. ??

(keep it simple, stupid)

now i realize no one here has experienced this, so you are viewing it differently than me. to me this is like someone wondering why it is harder to walk uphill (who has never walked up a hill).

on my feet (running from ground to ice, "locking" up legs), on skis (as mentioned), in full size pickup (regardless of 4WD & ABS), and in 13 ton tracked armored personnel carriers, i have experienced momentary increase of speed due to my momentum crossing over to a medium with much less friction (ice).

a geared wheel has both engine resistance and road friction (also resistance). when you lock up a wheel, you are bypassing the engines braking effect, as well as decreasing the surface area of the wheel used to resist. this lower resistance, small surface area will travel across the next distance of "slick" road/track at a faster rate than would have been allowed by the circumference of the wheel on "normal" conditions.

also, the odometer does not need to be connected to a drive wheel, allowing for correct speed indication even when drive wheels are locked up.

TRoc
 
  • #46
If the speed really increases, there is a forward acceleration, implying a net forward force.

Where does this force come from ?
 
  • #47
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.
 
  • #48
the biggest ego...

dextercioby said:
It's more of a logical one...Adding accelerations and velocities...

Daniel.
You said nothing. This kind of contribuition is just ridiculous!


dextercioby said:
Then University of Toronto should start reading...phyics... :rolleyes:

Daniel.


Certainly ur going to explain the phenomenon before someone does it, or...

should you start reading physics, too?

:smile:
 
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  • #49
my dad is still out of town in the safety class, so the books and stuff aren't available to me till like wednesday i think. I'll see what i can do.
 
  • #50
also, the odometer does not need to be connected to a drive wheel, allowing for correct speed indication even when drive wheels are locked up.

I believe that speedometers and odometers make the basic assumption that the wheels are in rolling contact with the road surface. The actual sensor is at some point in the drive train but if the wheels are not rolling, any indication of speed is based on a incorrect assumption, and is therefore suspect. The only way to get correct speed data in this situation is to have either an external device (radar gun) or an on board accelerometer.

A big difference between one of these rigs an a normal car or truck is that the drivers hands and eyes are much freer to roam. Notice that the steering wheel is completely useless, there is no possibility of losing control and smashing into the ditch. All a driver has to do is control the speed, I would bet that it is very easy to lock the wheels up when braking, it is also a lot more likely that the driver will be watching the speedometer when braking, perhaps if someone where to do the same sort of thing in a car on the road you may see similar effects, but who, in a skidding car, pays a lot of attention to the speedometer?

The real question here is where is the speedometer sensor and how does it interact with the drive train. This is a more of question for a mechanic then a physicist.
 
  • #51
da_willem said:
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?
 
  • #52
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. :smile:
 
  • #53
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.
 
  • #54
physicsisphirst said:
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.
 
  • #55
Integral said:
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.
 
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  • #56
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.
 
  • #57
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.
 
  • #58
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.
 
  • #59
It certainly looks like a good plausibility argument.

I'm not entirely convinced about this part though.
marrsc said:
...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 ?
 
  • #60
Gokul43201 said:
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 :frown:
i think my physics needs more help than i realized if i have forgotten this sort of thing. :redface:
thank you for explaining all this, Gokul.
 
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  • #61
marrsc said:
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. :smile:

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.

munky99999 said:
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...?
 
  • #62
Gokul43201 said:
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.
 
  • #63
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.
 
  • #64
marrsc said:
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?
 
  • #65
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
 
  • #66
krab said:
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
 
  • #67
Cliff_J(snip)To add to Gokul43201's posts here on the second page said:
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?
 
  • #68
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?
 
  • #69
Cliff_J said:
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.
 
  • #70
Fliption said:
Then why not just make a train that slides on a wet track instead of rolling wheels? :smile:

It's called a Mag Lev.
 

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