I rolled my tractor

[wushaw]

Soundguy, the rear brakes only accout for 30% braking for the vehicle no matter how good they work or what vehicle it is.

 

[Soundguy]

Um.. logically thinking out what you said.. it seems incorrect.

If you have a vehicle with ONLY rear brakes.. and you apply them.. those brakes account for 100% of that vehicles breaking power.... even if they are ineficient.. they are accounting for 100% of the braking occouring on that vehicle.. less wind resistance, rolling resistance and mechanical drag, and engine compression braking.. etc...

I think you might revise that statement to preface it with: for vehicles with front disc and rear drum.. and possibly 4 wheel drum.. the rear brakes only account for X % of the total braking of the vehicle.. no matter how good they work.

The "no matter what vehicle it is in" is the falacy... there are plenty of rear brake only vehicles, with those brakes being drum brakes....

For instance.. my bycicle.. has only rear brakes.. yet they are the 100% sum total of the braking force on that vehicle.. I.e. no front brakes.. etc.. same with a rear brake only truck.. like an oldie/antique... PLENTY of those still in service..

Soundguy

 

[ultrarunner]

I would have to agree...Both my 1912 Model T and 1905 Oldsmobile only brake the rear wheels. The T has external contracting bands and the Olds has a single band that slows the flywheel (You need to be in neutral to stop!!)

 

[tjkadar]

On the flip side of that, some vehicles have 100% of their braking force on the front wheel. A modern sportbike can generate enough stopping power to lift the rear tire off of the ground. So, 100% of braking power is on the front wheel. It all depends on the vehicle and its use at the time of braking.

 

[Soundguy]

Yep.. good point. The previous blanket statement about rear brakes accounting for 30% of breaking on a vechicle no matter the make/model/efficiency was just too all-encompassing to be very correct. Plenty of exceptions exist..

Heck.. my new CUV has rear brakes only.. some sort of small cable actuated 'canister' style breakes situated on either side of the axle ends.. just inboard of the tires.... I'd wager they are 100% of the vehicles brakeing.. since no other brakes exist /forums/images/graemlins/smirk.gif

Soundguy

 

[RedRocker]

It may be more accurate to say that 70% of your stopping power comes from your front brakes. This would assume you have brakes on both ends. If you lift your rear wheel off the ground then you've eleminated 30% of your stopping power. If you only have rear brakes, you only have 30% of the stopping power you'd have with fronts added.

 

[Soundguy]

I'd say that is a more acurate way to present it than the statement that was mentioned earlier.

Soundguy

 

[tjkadar]

That would totally negate the physics at hand. Stopping a vehicle is the act of taking the energy of forward motion and translating it into heat energy. How fast you can do that depends on a variety of factors, including weight shift, available traction, the ability of your brakes to dissipate heat, etc.

Any vehicle equipped with brakes has a maximum braking capacity, which we shall call 100%. It doesn’t matter if the brakes are disc, drum, or Fred Flintstone’s feet. That 100% of braking capacity is not a static number. It is very dynamic. In this case, we shall use a motorcycle with disc brakes at the front and rear.

Braking is directly related to the amount of traction available at a given wheel. That traction is dependent on tire condition, road condition, and the forces being applied to the tire. To make things more interesting, we’ll have to discuss traction too. A tire has a certain level of grip before it will slide or skid. This is 100% of available traction.

If you are accelerating, part of this available traction will be given to forward motion. If you exceed the available traction limits of the tires, your wheel will start to spin and can reach the point where you have no forward motion at all. You have exceeded the limits of available traction.

Now, in this example, a motorcycle is approaching a corner at a high rate of speed. The rider wants to bleed off excessive speed at the last moment possible in the least amount of time. He begins his braking efforts while the machine is still upright.

For the first few milliseconds his available traction and braking force is distributed between the front and rear tires. However, as weight shifts forward more and more of his available traction and braking force is on the front wheel. In the space of a very short span of time, the rear wheel is off of the ground and the front wheel is at impending lock-up. One hundred percent of the available braking force is on the front wheel and the front wheel only. The rear tire at this point has 0% braking force and 0% traction.

As the rider enters the corner, he will slowly release the pressure on the front brakes. Now the tire is being asked to help slow the bike and turn at the same time. The available traction of the tires is now being tasked to do two different things, so the available traction is now divided between those two things. The rider is still at 100% braking, but that 100% is now dependent on only 50% of the traction.

If the rider exceeds his available braking force, he will also exceed his available traction and quickly wind up on his head. Braking and the forces at work during the act of braking are not static. You cannot simply say, 70% of your braking force is on the front and 30% is on the rear because those relationships are very dynamic. They change the instant you start applying brakes and change the forces at work on the machine.

 

[RedRocker]

You'll have to argue all that with the really smart folks that came up with that ratio, it wasn't me. The way I understand it, 70% of your stopping power is on the front brakes, if you choose to not use them, use them a little bit, lock up the rear, lock up the front, whatever, that would be your choice. But, if you stop something as fast as possible with front and rear brakes, the majority of the power is up front. Maybe I'll get out my Keith Code book and we can talk cycle racing. /forums/images/graemlins/grin.gif

 

[Tim_in_IA]

Comparing motorcycles to cars to tractors might not be apples to apples to apples here. I think all soundguy was saying was that a machine with ONLY rear brakes would use them for 100% of it's stopping power, since it only has rear brakes to work with. Vehicles equipped with front brakes also might be a different matter altogether.

The only exception to this with tractors might be a unit that has 4wd engaged, which would still only use the rear brakes to stop, but distribute the braking force through the drivetrain to the front tires as well.

 

[tjkadar]

Even a 2WD tractor with only rear brakes has some front wheel assistance when stopping. The friction (traction, if you will) of the front tires on the ground will aid in stopping the tractor. Granted, this might be a miniscule amount, but it is still there.

It doesn't really matter what vehicle you are stopping. The basic physics of braking doesn't change simply because you are stopping a car, truck, tractor, or motorcycle. The dynamics might be larger or smaller, but the same rules of physics still apply.

I'll admit it, Keith Code did a much, much better job explaining the traction theory then I can. He can take a complex idea or theory, break it down into its basic parts, and then make someone feel as if they really do understand it. I wish my Physics professors had the same talents. /forums/images/graemlins/grin.gif

Now how about aerodynamic braking? Would a tractor with a FEL stop quicker then a tractor without - all other things being equal? /forums/images/graemlins/confused.gif

 

[Tim_in_IA]

I think we need a 3pt mounted emergency parachute! This of course should be connected to the draft control in some anti-opposing force. /forums/images/graemlins/smile.gif /forums/images/graemlins/smile.gif

 

[tjkadar]

Awesome idea! With a little work, maybe the parachute could be used in some manner to prevent tractors from rolling over too! But would it be an option on CUTs with no draft control?

Caveat: The above was a very lame attempt to try and bring the subject back to the original topic. /forums/images/graemlins/crazy.gif

 

[KentT]

</font><font color="blue" class="small">( Awesome idea! With a little work, maybe the parachute could be used in some manner to prevent tractors from rolling over too! But would it be an option on CUTs with no draft control?

Caveat: The above was a very lame attempt to try and bring the subject back to the original topic. /forums/images/graemlins/crazy.gif )</font>

How about a giant, inflatable airbag instead? That could work in rollovers, and the "balloon effect" could serve as brakes -- of course when it deployed, it'd be like a rocket booster... /forums/images/graemlins/shocked.gif /forums/images/graemlins/shocked.gif

 

[RedRocker]

Ejection seat, just watch out for the sunshade.

 

[trlong]

Where this thread has gone has been interesting, thoughtful, and informative. However, the "physics" of stopping vs. braking has pretty much hijacked the thread. I believe (and some one correct me if I'm wrong), the original problem had nothing to do with tire condition, "road" condition, coefficient of friction between tires and surface (sorry, I expanded on that), or much of anything besides...

</font><font color="blue" class="small">( <font color="blue"> I stomped on the breaks the tractor did not slow down. </font> )</font>

My interpretation of that, and subsequent statements was that the problem had nothing to do with traction but, that the wheels' d(rotation)/dt /forums/images/graemlins/grin.gifwas not a negative number /forums/images/graemlins/shocked.gif.

My experience with that function,

f(stomp on the brake) = 0

when releasing the "parking brake" on an incline without 1st placing the HST in gear, has brought me to the same conclusion as SoundGuy. The "parking brake" is pretty much just that, a Parking Brake. Never trust it to do much for you, regardless of traction. Just make sure you're in gear before you release it and don't rely on it on an incline.

As said, I kinda like the RB dropped (or BH with stabilizers down), the FEL dug in, and the wheels turned hard when stopped on an incline. Not great on the lawn but provides a pretty decent coefficient of friction. QED

'Nuff pontificating on physics? /forums/images/graemlins/smirk.gif

Tom

 

[Soundguy]

</font><font color="blue" class="small">( a machine with ONLY rear brakes would use them for 100% of it's stopping power, since it only has rear brakes to work with )</font>

Yep..

Soundguy

 

[Soundguy]

</font><font color="blue" class="small">( Even a 2WD tractor with only rear brakes has some front wheel assistance when stopping. The friction (traction, if you will) of the front tires on the ground will aid in stopping the tractor. Granted, this might be a miniscule amount, but it is still there.
)</font>

That's a red herring argument.. lets stick to the tangibles. and leave the 'epsilon' constant forces of bearing drag and wind resistance out, and focus on dynamic braking.

A 2wd vehicle with rear brakes only.. 100% of the dynamic braking comes from the only brakes present.. the rear in this case.

Lets forget about tail winds, gradient, and what grade of axle grease you are using... ( if your bearings drag so much that your tractor will stop rolling on a grade.. then the preload is too high or the lube is insuficient.. or the bearing is already shelled! /forums/images/graemlins/shocked.gif )

Soundguy

 

[Junkman]

I just finished doing the brakes on the wifes car... all four wheels... pads and rotors. Just like last summer when I did them last, the rears were down to the rivets and the fronts still had lining material on them. I replace all rotors and pads that time also. If the rears do less braking than the fronts, then why are the rears wearing out so quickly? I have my ideas as to why, but will hold onto them for a while and want to see what your opinions are....
The car is a 1997 Ford Explorer with 4 wheel drive, automatic with 6 cylinder engine... and a female driver 100% of the time. /forums/images/graemlins/frown.gif

 

[ronjhall]

I'll take a guess at this one. When pushing on the brake pedal the rear brakes always energize first. If someone is riding the brake pedal, rear brakes drag and wear out. /forums/images/graemlins/tongue.gif