Att WJM: Using the brakes to get traction
Michael
Most of us have heard of the trick of dabbing at the brakes to get going when one wheel ( in the case of a 4x2) is spinning. Opinion is divided over
how effective it is, and under what circumstances it works best/at all.
I can fully understand how this technique works in the case of a torque-biased (or torque sensing) LSD. And I fully understand how ABS-based
electronic traction control (or manual "fiddle brakes") can give additional traction by being applied to one wheel only.
BUT, I can't understand the mechanism by which this works in the case of an axle with an open diff.
Firstly, let's take as a given that the technique *does* work. So I am not looking for anecdotal evidence as to *whether* it works, or how effective
it is. I am trying to understand the way the forces are manipulated/transferred/increased to provide additional traction, and hence motive force.
My basic problem with the theory is this:
1) An open diff always has the same torque on both shafts (barring some internal friction effects, of course).
2) Any additional torque put on a shaft by means of applying a brake to the shaft will, by definition, not be "useful" torque at the wheel.
3) Thus if we apply a braking force equivalent to, say, 500Nm to *both* sides, it is all absorbed by the very act of creating it.
4) So where does the extra torque come from?
Compare this to the case of a 500Nm braking force being applied *to the spinning side only*. Providing the engine and transmission can generate an
extra 1000Nm, the torque on the spinning side will rise by 500Nm (but all used to overcome the brake friction). The open diff will ensure that the
torque on the other side rises by 500Nm as well, and since there is no braking force on that side, it will be "real", gripping torque (traction
conditions permitting, of course).
I only mention the one-sided braking as a comparison, to illustrate why equal-braking (by that logic) would *not* work. As we are accepting that there
is some evidence that it *does* work, I am seeking a descriptive explantion of the mechanics/physics involved.
One theory is that an open diff has a built-in torque bias. While I would concede that internal friction would generate *some* torque imbalace, I
would need some convincing that it is significant enough (some hundreds of Nm?) to have this effect. (Also, why would it be enhanced by braking? A
torque-biased LSD is *designed* to provide ever more friction under higher torque differences. I can't see any components of a normal open diff that
would *multiply* the torque).
I have another theory, but I confess that I can see some snags with it too. Here it is, by all means shoot it down! :-)
The faster spinning wheel has a higher effective gearing than the slower (or stationary) one - it rotates twice as fast as it would if both wheels
were being turned. It seems to me that this would bias the braking - a given pressure on the pads squeezing the brake disc would retard the highly
geared wheel more than the same pressure would retard the other wheel.
If so, there would be a nett gain in torque momentarily, until both wheels were rotating at the same speed. To put some numbers to it, if the initial
braking force on the spinning wheel puts an additional 1000Nm of torque onto that shaft, but only puts an an additional 500Nm onto the stationary
right-hand wheel, then I can see a nett gain of 500Nm of tractive force at the stationary wheel.
Of course, this effect would be short-lived - as soon as the wheels reach the same speed as each other, the gearing becomes the same, the braking
effect becomes the same, and once more perfectly offsets the gain in torque. That would account for the technique being most effective applied by
jabbing at the brakes, rather than applying them progressively (or continuously).
The one snag is that I am not certain that the braking force on each side *would* in fact differ because of the effective difference in gearing.
Certainly there would be a *counter*-effect from the fact that stiction is greater than friction, and this would actually mean a *greater* braking
effect on the stationary wheel than the spinning one.
Willem-Jan, if you see this, your comments? I've looked through your TAD FAQ, but only one brief mention of this issue.
Obviously if anyone else has something to add, I'd appreciate it too.
Regards,
Michael...
Douglas A. Shrader
"Michael" <no...@none.com> wrote in message
news:3df6eb49....@news4.cit-news.com...
> I asked this question as an aside on another thread, and got nowhere:
>
> Most of us have heard of the trick of dabbing at the brakes to get going
when one wheel ( in the case of a 4x2) is spinning. Opinion is divided over
> how effective it is, and under what circumstances it works best/at all.
>
> I can fully understand how this technique works in the case of a
torque-biased (or torque sensing) LSD. And I fully understand how ABS-based
> electronic traction control (or manual "fiddle brakes") can give
additional traction by being applied to one wheel only.
>
> BUT, I can't understand the mechanism by which this works in the case of
an axle with an open diff.
And you never will. It has been well explained to you numerous times, if you
don't get it now you are hopelessly confused, so why not just drop the
subject.
AZGuy
correct or not but I think it's a different theory then what I've seen
in the parts of this thread I've read. Here it is.... and it's only
for rear drum brakes....The reason using the brakes sometimes works
even with an open diff is because the brakes develop their braking
force partly on the basis of how hard you are pushing the pedal AND
ALSO partly on their self-energizing action. If the wheel wasn't
spinning when you put your brakes on all that happens is that both
shoes expand outward as the brake wheel cylinder pushes them out at
the top. However, if the wheel and brake drum is spinning you not
only get the outward expansion which forces the shoes against the
drums but you also have the additional effect created by the torque on
the brake shoe assembly. As the drum tries to turn the entire brake
shoe assembly the rear shoe tries to rotate and in doing so pushes
against the bottom of the front shoe. The overall effect is that the
whole brake shoe assembly is effectively trying to wedge itself into
the front portion of the shoe/drum area. So for any given pedal
pressure you get more brake force in total when the drum is rotating
then you do if the drum is not rotating. Since when you are stuck and
one wheel is not spinning and the other is, applying the brake "just
right" can result in the spinning side having more total braking force
then the none spinning side making it stop spinning, or at least slow
down. Since it has MORE brake force on it then the previously
non-spinning side you can sometimes wind up getting sufficient driving
torque to the non-spinning side to get the vehicle moving.
Erik-Jan Geniets
"Douglas A. Shrader" wrote:
> And you never will. It has been well explained to you numerous times, if you
> don't get it now you are hopelessly confused, so why not just drop the
> subject.
As long as he doesn't understand he is free to ask I think.
That's where newsgroups are for.
Someone might eventually give an understandable explanation.
Kind regards,
Erik-Jan.
Douglas A. Shrader
"Erik-Jan Geniets" <e...@fotograaf.com> wrote in message
news:3DF2002E...@fotograaf.com...
>
>
> "Douglas A. Shrader" wrote:
>
> > And you never will. It has been well explained to you numerous times, if
you
> > don't get it now you are hopelessly confused, so why not just drop the
> > subject.
>
>
> As long as he doesn't understand he is free to ask I think.
> That's where newsgroups are for.
Yes he has the right to ask, but he is never satisfied with any explanation,
and it gets kind of old hearing someone ask "Why?" everytime you explain it
to him. When a kid does that you eventually give up and move on don't you?
> Someone might eventually give an understandable explanation.
He has received several understandable explanations, he simply lacks the
ability to understand. Beating a dead horse never accomplishes anything.
Michael
Douglas:
I am not stupid. The "explanations" I have seen to date to not explain the phenomenon. Most of them have simply been attempts to provide anecdotal
evidence of the technique. Some have obviously failed to grasp the fundamental workings of a diff.
The closest that anyone has come to actually refuting my objections on a logical basis has been one on this thread suggesting that the extra torque
comes from the fact that the torque is retained at the instant when the brakes are released.
If you have no interest in this thread, please go play somewhere else. I am looking for sensible discussion.
Michael...
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Michael
>
>
>For the love of god Michael, READ. If each brake is applying 500Nm of
>force AND the wheels are still turning then there must be over 500
>coming from the drive shaft, right? Now release the brake. For a
>moment there is over 500Nm turning *both* sides. Guess what? Forward
>motion. Gears do not operate at the speed of light Mr. Hawking.
>
There is no problem with my reading. You are about the only person who has at least come up with a rational argument - certainly the first person to
suggest how some kind of "retained" torque might be responsible.
So, as I understand what you are suggesting, the extra traction is brought about at the moment the brakes are released, rather than at the moment they
are applied?
>> Compare this to the case of a 500Nm braking force being applied *to the spinning side only*. Providing the engine and transmission can generate an
>> extra 1000Nm, the torque on the spinning side will rise by 500Nm (but all used to overcome the brake friction). The open diff will ensure that the
>> torque on the other side rises by 500Nm as well, and since there is no braking force on that side, it will be "real", gripping torque (traction
>> conditions permitting, of course).
>
>How is this any different from above?
Err, Chris, if I have to explain to you why braking on one side only is fundamentally different from braking on *both* sides, then I don't think you
understand the entire gist of what I've been saying.
Essentially, braking on the spinning side only means that the torque increase on the *other* side isn't offset by the brakes. Braking on one side
results in 50% of your total additional torque being potentially available as added "grip" torque. Equal braking (i.e. on both sides) results in 100%
of your total additional torque being "braking" torque, and thus not adding to the "grip" torque at all.
Regards,
Michael.
Michael
Thanks for the reply :)
As you can see, I have been getting some flak for asking this question.
Your answer makes sense to me. Basically, you're suggesting that a self-acting drum-brake system will automatically bias the braking towards the
faster-moving wheel? If that happens, it would certainly answer my question about where the extra grip torque comes from. Does this mean it wouldn't
work in reverse? :)
Regards,
Michael...
Michael
Douglas:
I wait with baited breath for your "explanation". Only kidding, of course. I don't expect an explanation from you at all.
I have been amused, if not amazed at your at *your* failure to grasp the nature of my question. When a child asks why the sky is blue, it *is*
possible to give a rational, fact-based explantion, rather than fob him off with the equivalent of, "It's blue because that's the colour of sky." If
he asks again, then obviously reply, "Don't be stupid, you can SEE it's blue! And I've already explained why."
I am really not asking you to contribute to this thread. Could I at least ask you not to make *any* kind of comment, even if you don't understand why
I don't accept certain "explanations"?
Douglas A. Shrader
I've explained it to you three times already. Why should I waste my time
further when you are incapable of understanding anything but numbers on
paper?
> I have been amused, if not amazed at your at *your* failure to grasp the
nature of my question. When a child asks why the sky is blue, it *is*
> possible to give a rational, fact-based explantion, rather than fob him
off with the equivalent of, "It's blue because that's the colour of sky." If
> he asks again, then obviously reply, "Don't be stupid, you can SEE it's
blue! And I've already explained why."
Which is basically how we have answered you, but you persisit in asking
"How, I don't understand. Hell, get off your but and test it, just jack up
one side of your truck, or any two wheel drive with an open dif and put it
in gear. When the wheel in the air is spinning give it some gas and hit the
brakes, your truck will promptly fall off the jack, although you still won't
understand why.
>
> I am really not asking you to contribute to this thread. Could I at least
ask you not to make *any* kind of comment, even if you don't understand why
> I don't accept certain "explanations"?
Sure, if I can ask you to stop asking questions you are unable to comprehend
the answers to.
I not trying to pick on you, simply to make you realize that you will never
accept any explanation you are given so you may as well stop asking it over
and over and over.
Douglas A. Shrader
> On Fri, 6 Dec 2002 15:39:19 -0500, "Douglas A. Shrader"
<dshr...@nospam.com> wrote:
>
> >
> >"Michael" <no...@none.com> wrote in message
> >news:3df6eb49....@news4.cit-news.com...
> >> I asked this question as an aside on another thread, and got nowhere:
> >>
> >> Most of us have heard of the trick of dabbing at the brakes to get
going
> >when one wheel ( in the case of a 4x2) is spinning. Opinion is divided
over
> >> how effective it is, and under what circumstances it works best/at all.
> >>
> >> I can fully understand how this technique works in the case of a
> >torque-biased (or torque sensing) LSD. And I fully understand how
ABS-based
> >> electronic traction control (or manual "fiddle brakes") can give
> >additional traction by being applied to one wheel only.
> >>
> >> BUT, I can't understand the mechanism by which this works in the case
of
> >an axle with an open diff.
> >
> >And you never will. It has been well explained to you numerous times, if
you
> >don't get it now you are hopelessly confused, so why not just drop the
> >subject.
> >
> Douglas:
>
> I am not stupid.
I'm not saying you are, just that you can't understand any answers unless
they are chock full of calculations and numbers. It really is that that
compilcated as I have told you before, once the brake is applied the turning
force on both wheels is EQUAL, you keep trying to throw in figures that
aren't even in the equation.
The "explanations" I have seen to date to not explain the phenomenon. Most
of them have simply been attempts to provide anecdotal
> evidence of the technique. Some have obviously failed to grasp the
fundamental workings of a diff.
Read them all again, we are talking grade school stuff here.
>
> The closest that anyone has come to actually refuting my objections on a
logical basis has been one on this thread suggesting that the extra torque
> comes from the fact that the torque is retained at the instant when the
brakes are released.
You do not need to release the brakes at all to get both sides to pull, the
reason for releasing the brake is simply to take advantage of the power
surge when the brake friction is released. The brake friction equalizing the
turning force between both wheels is what makes the stopped wheel turn, and
as I told you before, once the brake friction gets high enough it is the
only resistance to turning there is, equal resistance means both wheels
turn. Release the brake and the power surge prevents the differetial from
disengaging, thereby maintaining the torque to both wheels. Very basic stuff
here.
>
> If you have no interest in this thread, please go play somewhere else. I
am looking for sensible discussion.
No, you are looking to argue, or you really are to stupid to understand a
very basic fact. Which is it?
Michael
>>
>> I wait with baited breath for your "explanation". Only kidding, of course.
>I don't expect an explanation from you at all.
>>
>
>I've explained it to you three times already. Why should I waste my time
>further when you are incapable of understanding anything but numbers on
>paper?
>
>
>> I have been amused, if not amazed at your at *your* failure to grasp the
>nature of my question. When a child asks why the sky is blue, it *is*
>> possible to give a rational, fact-based explantion, rather than fob him
>off with the equivalent of, "It's blue because that's the colour of sky." If
>> he asks again, then obviously reply, "Don't be stupid, you can SEE it's
>blue! And I've already explained why."
>
>
>Which is basically how we have answered you, but you persisit in asking
>"How, I don't understand.
Yes, that is precisely the nature of your answers. See below.
>Hell, get off your but and test it, just jack up
>one side of your truck, or any two wheel drive with an open dif and put it
>in gear. When the wheel in the air is spinning give it some gas and hit the
>brakes, your truck will promptly fall off the jack, although you still won't
>understand why.
>
I'm not refuting *that* it works. I am asking *why* it works. And I have described, in detail, the reasons that it is mystery to me, in the context of
my understanding of how the various mechnical parts behave, and the forces involved. Only one person has directly challenged my understanding of an
open diff, but since his argument hinged on the concept of torque necessarily involving movement, I can't accept it.
(I am not discounting the two or three possible rational explanations proposed - I will happily invite more comment on them, and confirmation that I
have interpreted the ideas properly).
>
>>
>> I am really not asking you to contribute to this thread. Could I at least
>ask you not to make *any* kind of comment, even if you don't understand why
>> I don't accept certain "explanations"?
>
>Sure, if I can ask you to stop asking questions you are unable to comprehend
>the answers to.
>I not trying to pick on you, simply to make you realize that you will never
>accept any explanation you are given so you may as well stop asking it over
>and over and over.
>
Ok, here's the deal. This is my thread. You think I am the stupid one for not understanding your "explanation". I think you are the stupid one, for
not understanding my question. I have no desire to argue with you, and I hope vice versa. But I still wish to pursue this with people capable and
willing to do so. So please go and play in the mud in your 4x4 for a bit, and leave this thread to me and all the stupid guys.
Bye now,
M...
Douglas A. Shrader
No it doesn't. Once the brakes are applied enough, the force required to
turn the wheel against the braking resistance is greater than the turning
resistance of the wheel against the ground and both wheels turn as long as
sufficient power is provided to overcome braking resistance. Releasing the
brakes has nothing to do with the second wheel pulling, that comes from
applying the brakes. Releasing the brake simply allows you to drive out
without the additional drag of the brakes since, as I have explained before,
once both wheels are pulling they will be locked together by the
differential as long as the power applied doesn't drop. If you were climbing
a hill and didn't make it from one wheel spinning you could back down the
hill part way and then gun the truck forward while it was rolling backward
to accomplish the same thing, differential lock up due to the power surge
while both axles are turning. I have used these methods with great success
for years, and I do understand how it works. What I don't understand is why
you insist on making it more complicated than it is.
Douglas A. Shrader
He does have a good theory, however applying the brakes would work just as
well with disc brakes, which do not see the braking increase from a turning
wheel that drum brakes do. It works because it causes the differential to
lock up, and yes it would work in reverse, read my post in your last thread
where I describe doing just that.
Douglas A. Shrader
But you have, in one of your replys to Jerry you state "if it even works at
all". To me that implies you doubt that it works. Once you accept that it
works you will find it easier to understand why it works.
I am asking *why* it works. And I have described, in detail, the reasons
that it is mystery to me, in the context of
> my understanding of how the various mechnical parts behave, and the forces
involved. Only one person has directly challenged my understanding of an
> open diff, but since his argument hinged on the concept of torque
necessarily involving movement, I can't accept it.
>
> (I am not discounting the two or three possible rational explanations
proposed - I will happily invite more comment on them, and confirmation that
I
> have interpreted the ideas properly).
> >
> Ok, here's the deal. This is my thread.
No, it is not "your thread" Once you posted it you invite comments from
anyone.
You think I am the stupid one for not understanding your "explanation". I
think you are the stupid one, for
> not understanding my question. I have no desire to argue with you, and I
hope vice versa. But I still wish to pursue this with people capable and
> willing to do so. So please go and play in the mud in your 4x4 for a bit,
and leave this thread to me and all the stupid guys.
I don't understand why you can't understand the whole concept. I really am
not trying to belittle you or argue with you, it's just that, to me, it is
very simple and easy to understand. I never play in the mud, I see plenty of
mud in my work and I really hate the stuff, but having driven in it for 32
years I am pretty dam good at avoiding being stuck, and at getting myself
out when I do get stuck. I would gladly tell you anything you wanted to
know, but I am very frustratedat the moment that you just don't get it. So
if after reading my other posts in this thread you still don't get it< I
will leave you alone. But please, stop asking after this thread if you still
can't grasp the concept.
Will Honea
I got irritated enough by this thread to set up a model of a simple
rear axle to see what the numbers crunched out as. What a complex
mother that thing is, from a math standpoint! Anyway, I never got a
complete answer to the original question but the results did open my
eyes to one issue. Do you have any idea how much kinetic energy is
involved with a 30 inch tire and wheel spinning at the equivilant of
30 mph? Wow! Now I can appreciate why axles break when you bounce a
wheel hard enough to get it spinning really fast before it hits the
ground again. Reminds me to get off the skinny pedal when it starts
to hop.
Now back to the differential model... I know what the next question I
submit for the ME portion of the PE exam is going to be.
--
Will Honea <who...@codenet.net>
Douglas A. Shrader
"Will Honea" <who...@codenet.net> wrote in message
news:JxX2tWiP5BNp-p...@anon.none.net...
Yes, and the math standpoint seems to be what Michael wants, which only
makes understanding the whole question needlessly difficult. I can not
provide him with three pages of figures to prove why it works, and it is not
needed, the answers to all his questions have already been posted if he will
simply open his mind to more than the engineering side of it.
Anyway, I never got a
> complete answer to the original question but the results did open my
> eyes to one issue. Do you have any idea how much kinetic energy is
> involved with a 30 inch tire and wheel spinning at the equivilant of
> 30 mph? Wow! Now I can appreciate why axles break when you bounce a
> wheel hard enough to get it spinning really fast before it hits the
> ground again. Reminds me to get off the skinny pedal when it starts
> to hop.
Oh yea. Had the tractor start to wheel hop pulling the spray truck out once,
snapped the chain before I could hit the clutch. Lotta strese there.
Michael
On Sat, 7 Dec 2002 16:45:56 -0500, "Douglas A. Shrader" <dshr...@nospam.com> wrote:
>
>"Michael" <no...@none.com> wrote in message
>news:3df24e8e....@news.cit-news.com...
>> Douglas:
>>
>> I am not stupid.
>
>I'm not saying you are, just that you can't understand any answers unless
>they are chock full of calculations and numbers.
Yup.. that's what scientists do - they quantify statements. As a case in point, it's only when you start trying to account for the torque
*quantitatively* that you realize that the explantions don't actually stack up. If you think there is something wrong with the numbers I have chosen,
by all means refute them, and suggest your own. But don't try to suggest that there is something inherently wrong with a quantitative approach.
> It really is that that
>compilcated as I have told you before, once the brake is applied the turning
>force on both wheels is EQUAL, you keep trying to throw in figures that
>aren't even in the equation.
>
What numbers have I included which aren't in the equation? If I've made a mistake, please point to it. You'll see my comments below indicate precisely
which of your statements I disagree with, and what I believe is correct in their stead.
>The "explanations" I have seen to date to not explain the phenomenon. Most
>of them have simply been attempts to provide anecdotal
>> evidence of the technique. Some have obviously failed to grasp the
>fundamental workings of a diff.
>
>Read them all again, we are talking grade school stuff here.
>
>>
>> The closest that anyone has come to actually refuting my objections on a
>logical basis has been one on this thread suggesting that the extra torque
>> comes from the fact that the torque is retained at the instant when the
>brakes are released.
>
>You do not need to release the brakes at all to get both sides to pull, the
>reason for releasing the brake is simply to take advantage of the power
>surge when the brake friction is released. The brake friction equalizing the
>turning force between both wheels is what makes the stopped wheel turn,
The brake friction doesn't "equalize the turning force on the wheels". The turning force on the wheels is *always* the same (unless you have a reason
for saying it isn't). The braking friction *increases* the turning force on the shafts, equally. As you say, very basic stuff... but you got it
wrong.
Just to recap, the crux of my argument is that the braking friction doesn't increase the turning force on the wheels, it only increases the turning
force on the shafts. There will be a very different torque actually measurable at the wheel versus the shaft. Shall I postulate some numbers to
clarify that?
>and as I told you before, once the brake friction gets high enough it is the
>only resistance to turning there is, equal resistance means both wheels
>turn.
Huh? How does it get to be the only resistance there is? Surely friction between ground and tyre still counts? In fact ground/tyre contact provides
the only *useful* resistance. So.. there is NOT equal resistance, so both wheels DON'T turn. (At least, certainly not for any reasons you've
suggested). I think you've demonstrated the value of a quantitative approach - you somehow forgot the original ground/tyre resistance in your
post-braking equation. And don't tell me you think it's negligible. All the *other* resistance is created by the brakes, and so provides abolutely
*no* motive force.
>Release the brake and the power surge prevents the differetial from
>disengaging, thereby maintaining the torque to both wheels. Very basic stuff
>here.
See, here you lose me. You get a power surge when you release the brakes? Oh, maybe you mean the stored energy held in the spinning wheel? Or the
drive-train? So, we release the brakes, and then where do *you* think the energy is going to go? I think it's going to go down the path of least
resistance, just like it would if you hit the accelerator. It's going to *continue* to spin the low-traction wheel (if we're talking about energy held
in the wheel), or it is going to spin that low-traction wheel even *faster* (if there is any stored energy in the drive-train). Either way, I can see
no mechanism by which that enery is directed to the wheel with good traction, which has been stationary all along. (Hell, if it wasn't stationary,
you'd not be stuck).
Charitably, I'm going to ignore your saying "the power surge prevents the differetial from disengaging" - I am sure that even you didn't actually mean
that.
>>
>> If you have no interest in this thread, please go play somewhere else. I
>am looking for sensible discussion.
>
>No, you are looking to argue, or you really are to stupid to understand a
>very basic fact. Which is it?
>
There is a third possibility which you have overlooked, Dougie-boy... I won't bother explaining it to you.
Bye now.
M...
Michael
>
>"Michael" <no...@none.com> wrote in message
>> Essentially, braking on the spinning side only means that the torque
>>increase on the *other* side isn't offset by the brakes. Braking on one side
>> results in 50% of your total additional torque being potentially available
>>as added "grip" torque. Equal braking (i.e. on both sides) results in 100%
>> of your total additional torque being "braking" torque, and thus not
>>adding to the "grip" torque at all.
>
>No it doesn't. Once the brakes are applied enough, the force required to
>turn the wheel against the braking resistance is greater than the turning
>resistance of the wheel against the ground and both wheels turn as long as
>sufficient power is provided to overcome braking resistance. Releasing the
>brakes has nothing to do with the second wheel pulling, that comes from
>applying the brakes. Releasing the brake simply allows you to drive out
>without the additional drag of the brakes since, as I have explained before,
>once both wheels are pulling they will be locked together by the
>differential as long as the power applied doesn't drop. If you were climbing
>a hill and didn't make it from one wheel spinning you could back down the
>hill part way and then gun the truck forward while it was rolling backward
>to accomplish the same thing, differential lock up due to the power surge
>while both axles are turning. I have used these methods with great success
>for years, and I do understand how it works. What I don't understand is why
>you insist on making it more complicated than it is.
>
>
>
Whoa, there! We're talking about OPEN differentials here. They don't lock up - not under torque, nor under speed. If you're talking about an LSD or a
locker, then I can understand your point. But in my opening of the thread I made it very clear that my question only applied to open diffs. What you
have described here is a torque-activated locker, and makes perfect sense.
Now... back to the question: Why does the technique work for OPEN diffs?
M...
Mike Romain
In one of my posts on this subject I mentioned that the engine can
overcome the rear brake holding power on almost any normal vehicle,
especially if the engine has a running (or tire spinning) head start.
That means that when you put the brakes on hard enough to equal the side
to side torque enough to break free and spin both wheels, the engine
still has enough power to easily overcome the braking force while
continuing to add it's own torque to the driveshaft.
If it didn't, the engine would stall.
Don't forget, we are flooring the gas pedal at the same time we are
hitting the brakes!
That might be the 'added torque' you are looking for, eh?
I think it is and you seem to have been talking a static side to side
torque.
The engine has things to say about that.
Mike
86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail
88 Cherokee 235 BFG AT's
Mike Romain
Open diffs certainly DO 'lock up!
If you have equal torque on and are goosing it up a sand pit wall or on
ice or in mud with spinning tires with open diffs, you get 2 rooster
tails, not 1.
Michael
>Michael wrote:
>
>
>Open diffs certainly DO 'lock up!
>
>If you have equal torque on and are goosing it up a sand pit wall or on
>ice or in mud with spinning tires with open diffs, you get 2 rooster
>tails, not 1.
>
>Mike
>86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail
>88 Cherokee 235 BFG AT's
Erm... that's not "locking up". That's just both wheels turning at similar speeds due to similar traction conditions under both tyres.
M...
P.S. You *always* "have equal torque on" :-)
Michael
>
>I got irritated enough by this thread to set up a model of a simple
>rear axle to see what the numbers crunched out as. What a complex
>mother that thing is, from a math standpoint! Anyway, I never got a
>complete answer to the original question but the results did open my
>eyes to one issue. Do you have any idea how much kinetic energy is
>involved with a 30 inch tire and wheel spinning at the equivilant of
>30 mph? Wow! Now I can appreciate why axles break when you bounce a
>wheel hard enough to get it spinning really fast before it hits the
>ground again. Reminds me to get off the skinny pedal when it starts
>to hop.
>
>Now back to the differential model... I know what the next question I
>submit for the ME portion of the PE exam is going to be.
LOL, yes, I can imagine it must be enough kinetic energy that you don't want want to be in the way if it comes loose. A 40kg flywheel spinning at
700RPM, or thereabouts... :-)
Seriously, though, I like the IDEA of the energy being in the wheel, but I can't see what would force it over onto the opposite side. The same
fundamental "spare" torque considerations exist (it seems to me) whether the motive force comes from the momentum of the wheel, or from a blip of the
accelerator. Namely, that *any* extra torque experienced by the stationary shaft is as a result of the *braking* on the opposite wheel, and if there
is equal braking, it is numerically cancelled out.
What do you think of the gearing concept? Higher effective gearing in the spinning wheel means that for any given brake-pad *pressure*, the
*retardation* will be greater than on the stationary wheel at the same pressure. And once we have unequal braking resistance, we can explain a surplus
"grip torque". Sorta like the self-actuating drum-brake theory.
I have to go to bed now, but I'll look forward to taking up the cudgels in the morning. :-)
G'night all.
M...
P.S. Let me know how the students do on the exam! Publish the model answers here please :-)
Douglas A. Shrader
If one wheel is on ice and the other wheel is on dry pavement, the force
required to turn the wheel on ice is ALL that will be sent to the wheels,
thus the wheel on pavement does not turn. When you apply the brakes the
force required to turn the wheel on ice increases, the other wheel is not
turning at this point so it does not change. NOW, when the turning
resisitance of the wheel on ice is increased by the brakes until it
approximatly equals the turning resistance of the wheel on dry pavement,
BOTH wheels turn The extra number you keep throwing in is that you seem to
think the turning resistance of the wheel on pavement will always be higher
than the wheel on ice, it isn't, the brake equalizes it. Equal resistance
means there is NO wheel which turns easier so BOTH wheels turn. Now isn't
that simple? Do you get it yet, I really don't see how it can be easier to
understand.
The braking friction *increases* the turning force on the shafts, equally.
As you say, very basic stuff... but you got it
> wrong.
>
> Just to recap, the crux of my argument is that the braking friction
doesn't increase the turning force on the wheels, it only increases the
turning
> force on the shafts.
The shafts turn the wheels. Now you are confusing yourself again. The
brakes, as I have stated before, do NOT increase the traction between an
individual wheel and the ground, they increase traction by locking up the
differential and forcing BOTH wheels to turn instead of just one. If both
wheels are already turning then the brakes do nothing to increase traction.
Again, the brakes DO NOT INCREASE TRACTION/FRICTION BETWEEN THE TIRES AND
THE GROUND. They simply force BOTH wheels to pull instead of only one, which
has the effect of increasing traction by virtue of the fact you now have TWO
wheels turning instead of only one, and the second wheel has higher traction
ability,. we know this because if it was equal to or greater than the first
wheel it would already be turning.
There will be a very different torque actually measurable at the wheel
versus the shaft. Shall I postulate some numbers to
> clarify that?
That has NOTHING to do with anything, the point is to get BOTH wheels
pulling, the torque on the axles will be equal, which is why both wheels
turn. As stated before, you are confusing yourself here, it is very simple.
>
> >and as I told you before, once the brake friction gets high enough it is
the
> >only resistance to turning there is, equal resistance means both wheels
> >turn.
>
> Huh? How does it get to be the only resistance there is? Surely friction
between ground and tyre still counts? In fact ground/tyre contact provides
> the only *useful* resistance.
Resistance between the tire and the ground, even if one tire is on ice and
one is on hard pavement will not be a factor, the resistance between axle
shafts WILL be equal once the brakes are applied and both wheels WILL pull
at that point once you apply enough throttle to overcome the brakes. AS you
pointed out above, the brakes are between the tires and the axle shafts, so
they are out of the loop here, forget all about the resistance to the
ground, this is the useless extra number you keep using to confuse yourself.
So.. there is NOT equal resistance, so both wheels DON'T turn. (At least,
certainly not for any reasons you've
> suggested).
I've proven you wrong here so often it is tiring.
I think you've demonstrated the value of a quantitative approach - you
somehow forgot the original ground/tyre resistance in your
> post-braking equation. And don't tell me you think it's negligible. All
the *other* resistance is created by the brakes, and so provides abolutely
> *no* motive force.
It is not forgotten, it doesn't count. It has nothing to do with anything.
Once the brakes equalize the turning resistance on the axles the
differential locks up and will remain locked until power to the shafts is
eased.
>
> >Release the brake and the power surge prevents the differetial from
> >disengaging, thereby maintaining the torque to both wheels. Very basic
stuff
> >here.
>
> See, here you lose me. You get a power surge when you release the brakes?
Oh, maybe you mean the stored energy held in the spinning wheel? Or the
> drive-train?
While the brakes are held you will need very high engine RPM's to turn the
wheel. When you release the brakes you DO NOT let up on the throttle. Same
power applied but greatly reduced turning resistance equals a net power
surge, same priciple planes use to take off from short runways, hold the
brakes until full RPMs are reached, then release the brakes and go.
So, we release the brakes, and then where do *you* think the energy is going
to go? I think it's going to go down the path of least
> resistance, just like it would if you hit the accelerator.
It won't, because by the time you release the brakes BOTH wheels will
already be turning, objects in motion tend to stay in motion, and
differentials like turning both wheels anyway, thus it stays locked up until
something occurs tostop it, which is normally when you finally let off the
throttle after getting out.
It's going to *continue* to spin the low-traction wheel (if we're talking
about energy held
> in the wheel), or it is going to spin that low-traction wheel even
*faster* (if there is any stored energy in the drive-train). Either way, I
can see
> no mechanism by which that enery is directed to the wheel with good
traction, which has been stationary all along. (Hell, if it wasn't
stationary,
> you'd not be stuck).
It isn't stationary by the time you release the brakes, if done correctly
both wheels will be turning by this point and will continue to spin for
reasons already stated.
>
> Charitably, I'm going to ignore your saying "the power surge prevents the
differetial from disengaging" - I am sure that even you didn't actually mean
> that.
Hard to untie a knot while pulling on the rope, hard to disengage an axle
shaft while it's already pulling hard to. As long as you keep the power on
both wheels will keep turning.
>
> >>
> >> If you have no interest in this thread, please go play somewhere else.
I
> >am looking for sensible discussion.
> >
> >No, you are looking to argue, or you really are to stupid to understand a
> >very basic fact. Which is it?
> >
> There is a third possibility which you have overlooked, Dougie-boy... I
won't bother explaining it to you.
>
Dougie-boy now? Cool.
Douglas A. Shrader
Yes they do, as explained repeatedly. Once again you reveal you have no idea
what we are talking about even though it has been explained in every
possible way. Think what you want, the problem here is not the explanations
but your inabilty to understand what is being said.
Douglas A. Shrader
"Mike Romain" <rom...@sympatico.ca> wrote in message
news:3DF28442...@sympatico.ca...
> Michael,
>
> In one of my posts on this subject I mentioned that the engine can
> overcome the rear brake holding power on almost any normal vehicle,
> especially if the engine has a running (or tire spinning) head start.
It certainly can. Although on the big 6x6's I drive at work you have to be
very carefull with the brake trick, hit the brakes to hard in low range and
you twist the driveshafts into, as has happened to a couple of drivers.
...
Jerry Bransford
> Yes he has the right to ask, but he is never satisfied with any explanation,
> and it gets kind of old hearing someone ask "Why?" everytime you explain it
> to him.
Exactly, and is why I moved on myself.
Jerry
--
Jerry Bransford
PP-ASEL KC6TAY
The Zen Hotdog... make me one with everything!
Geezer Jeep: http://www.jjournal.net/jeep/gallery/JBransfordsTJ/
Michael
Michael was in the habit of gazing at thse sea, and it didn't take him many weeks to pick up on the fact that the sea level seemed to rise and fall
every 12 hours or so.
"Grandma, why does that happen? " he asked.
"Don't ask stupid questions, boy!" she snapped. "It's been doing that since I were a wee lass, so there's no point debating it."
Michael, being in his forties (early forties, I hasten to add), was not easily deterred. So he wandered over to Grandad.
"Grandad, tell me about the rise and fall in sea level... What causes it?"
"Arrr, lad. That's the tides."
"Mmmm... but what makes the tides occur?".
"You should have learnt that in grade school, boy. It's the wind, blowing out in the middle of the ocean, creating currents."
Well, on the face of it, it seemed reasonable. But on thinking about it a bit more, the explanation didn't quite fit the facts. For one thing why
would it be such a regular cycle? (Yet not a perfect cycle). Very strange, but no point arguing with Grandad, who just got irritated, and accused him
of arguing for the sake of it. "How many times to do I have to explain it, lad?"
Then Dad had a quiet word in Michael's ear. "It's the moon. As the earth rotates, the moon's gravity pulls on the water and sucks it all up towards
that side of the earth. So the part facing the moon has high tide, the part facing away from the moon has low tide."
Grandad and Grandma overheard, and nodded wisely. "It's what I keep telling you", said Grandad, "The wind and the moon cause the tides. I've far more
experience in life than you, me boy, so next time listen to me in the first place".
Ignoring the old folks, but pleased that he understood at last, Michael went about his business.
But then he started to think. "How odd. The earth turns a full circle in 24 hours. But the tide rises here, on my side of the earth, every 12 hours.
So I have an observable phenomenon on the one hand, and a superficially plausible theory on the other hand, but the arithmetric doesn't stack up...
Something's not quite right. Well, maybe I should just let it go. I mean, it's only a small detail."
But Michael doesn't like an unscratched itch, so to speak, so he broached the subject again.
Well, as you can imagine, Dad and co were not pleased.
"What? That old chestnut again? How many times do we have to explain it?"
"Didn't you understand the first time? Why don't you go on out right now, and WATCH the ocean. You will see, we were right, it rises and falls as
regular as clockwork. I don't see why you want to make it so complicated. You may think you're being clever by trying to put numbers on the problem,
but the numbers just confuse things. You're in the REAL world, not some silly world of theory and numbers."
Well, Michael was a bit disheartened, especially when Dad went off reminding him that when HE was Michael's age HE had already sailed around the world
single handed - blindfolded, in fact - and in the light of THAT kind of extensive experience, who was Michael to question his explanations?
Still, the problem bugged Michael. It wasn't that he didn't know about gravity (in fact he studied up lots about gravity), or that he doubted the
tides themselves. He just couldn't explain, from his own knowledge of gravity, or from the explanations he got from Grandma, Grandad, or Dad, WHY
there were two high tides a day, when the Earth spun around on its axis in front of the Moon only once a day...
To find out how this thrilling tale ends, read the next episode, entitled "Uncle Bob's Red Herring - The Sun"
Mike Romain
>
> Since it does work in reverse and with disc brakes I think it may be of
> partial benefit but is not the answer.
> --
This 'book' guy still doesn't know that the definition of torque
involves movement.
I hope his parents didn't have to spend too much money on his
education....
Mike
AZGuy
<Xcphi...@ns.sympatico.ca> wrote:
>In article <3df454ad....@news.cit-news.com>, no...@none.com says...
>Since it does work in reverse and with disc brakes I think it may be of
>partial benefit but is not the answer.
Will it work with disk brakes and an open diff or only if you have a
LS? None of my 4x4s has ever been fancy enough to have rear disks.
AZGuy
wrote:
>
>>
Before you pat yourself too much on the back keep in mind that torque
does NOT involve movement. You can have as much torque as you want in
a non-moving system. POWER requires movement.
Michael
<dshr...@nospam.com> wrote:
>
>> >
>> >You do not need to release the brakes at all to get both sides to pull, the
>> >reason for releasing the brake is simply to take advantage of the power
>> >surge when the brake friction is released. The brake friction equalizing the
>> >turning force between both wheels is what makes the stopped wheel turn,
>>
>> The brake friction doesn't "equalize the turning force on the wheels". The
>turning force on the wheels is *always* the same (unless you have a reason
>> for saying it isn't).
>
>If one wheel is on ice and the other wheel is on dry pavement, the force
>required to turn the wheel on ice is ALL that will be sent to the wheels,
>thus the wheel on pavement does not turn.
Agreed.
>When you apply the brakes the
>force required to turn the wheel on ice increases, the other wheel is not
>turning at this point so it does not change.
Wrong - it does change. The torque on BOTH shafts increases
simultaneously, even though one is not moving.
>NOW, when the turning
>resisitance of the wheel on ice is increased by the brakes until it
>approximatly equals the turning resistance of the wheel on dry pavement,
>BOTH wheels turn
No, they don't. Because BOTH brakes are on. Yes, if the ice-wheel's
brake alone were applied, you would be correct. But because the other
brake is also on, all that "extra" force is absorbed by the brake.
>The extra number you keep throwing in is that you seem to
>think the turning resistance of the wheel on pavement will always be higher
>than the wheel on ice, it isn't, the brake equalizes it.
If you're adding the same braking force to EACH side, then how does that
equalize the resistance? The one side started out with more resistance,
then we added the *same* extra braking resistance to both sides... It's
an arithmetic fact that the net result will still be unequal. (Unless
you come up with a plausible reason why the braking itself might be
unequal).
>Equal resistance
>means there is NO wheel which turns easier so BOTH wheels turn.
That would indeed be the case, IF there were equal resistance.
>Now isn't
>that simple? Do you get it yet, I really don't see how it can be easier to
>understand.
>
>>
>> Just to recap, the crux of my argument is that the braking friction
>doesn't increase the turning force on the wheels, it only increases the turning
>> force on the shafts.
>The shafts turn the wheels. Now you are confusing yourself again.
I know that the shaft turns the wheels. That doesn't mean the wheel and
the shaft have the same torque. See a bit further down, where I
elucidate..
<snip>
>
>>There will be a very different torque actually measurable at the wheel
>>versus the shaft. Shall I postulate some numbers to
>> clarify that?
>
>That has NOTHING to do with anything, the point is to get BOTH wheels
>pulling, the torque on the axles will be equal, which is why both wheels
>turn. As stated before, you are confusing yourself here, it is very simple.
Sorry, but that is just plain wrong. The same torque on the *axles*
doesn't mean the same torque on the *wheels*. Why? Because there are
two reasons for the axle torque - two sources of resistance. The first
is the friction between the ground and the tyre. (We're agreed we have
no control over that). The second is the friction between the brake pads
and discs. An increase in braking resistance will increase torque on the
shaft, but will not increase the tractive force on the tyre.
If I turn a shaft as hard as I can from one end, while you hold it as
hard as you can at the other end, there will be torque on the shaft, but
no force on a wheel mounted in the centre of the shaft. Yes? And that's
what the brakes on the pavement-wheel are doing - they are taking all
the additional torque generated by the opposite brake, and holding it.
Generating a torque, but no movement.
That's what I mean about the shaft-torque increasing, but the
grip-torque, or wheel-torque NOT increasing.
>
>>
>> >and as I told you before, once the brake friction gets high enough it is
>the
>> >only resistance to turning there is, equal resistance means both wheels
>> >turn.
>>
>> Huh? How does it get to be the only resistance there is? Surely friction
>between ground and tyre still counts? In fact ground/tyre contact provides
>> the only *useful* resistance.
>
>Resistance between the tire and the ground, even if one tire is on ice and
>one is on hard pavement will not be a factor, the resistance between axle
>shafts WILL be equal once the brakes are applied and both wheels WILL pull
>at that point once you apply enough throttle to overcome the brakes. AS you
>pointed out above, the brakes are between the tires and the axle shafts, so
>they are out of the loop here, forget all about the resistance to the
>ground, this is the useless extra number you keep using to confuse yourself.
>
This doesn't address my point. The only resistance that provides any
MOTIVE force is that between tyre and ground. And it WILL be a huge
factor. The total resistance on each side will be the braking
resistanceon that side, plus the traction resistance on that side. Since
the braking resistance is the same each side, it follows that, once you
have braked, the pavement-side will have more total resistance than the
ice-side. Yes?
If it has more resistance, and the same torque, then it won't turn.
<snip>
>Once the brakes equalize the turning resistance on the axles the
>differential locks up and will remain locked until power to the shafts is
>eased.
>
Well, here we have a serious fallacy on your part. I address this a bit
further down...
<snip>
>
>It won't, because by the time you release the brakes BOTH wheels will
>already be turning,
No, they won't - see my explantion above.
>objects in motion tend to stay in motion, and
>differentials like turning both wheels anyway, thus it stays locked up until
>something occurs tostop it, which is normally when you finally let off the
>throttle after getting out.
>
No argument with this part, IF we somehow had conjured up the motion in
the first place. But we've not.
>It isn't stationary by the time you release the brakes, if done correctly
>both wheels will be turning by this point and will continue to spin for
>reasons already stated.
See above.
>> Charitably, I'm going to ignore your saying "the power surge prevents the
>>differetial from disengaging" - I am sure that even you didn't actually mean
>> that.
>
>Hard to untie a knot while pulling on the rope, hard to disengage an axle
>shaft while it's already pulling hard to. As long as you keep the power on
>both wheels will keep turning.
>
Ok, here, the part about an open diff "locking".
Maybe you should have another look inside an open diff. Nothing engages
or disengages. There are no dog clutches, everything is in constant
mesh. There is nothing to lock in place, or parts to get jammed
together, even under the most horrendous torques. If something in there
"locks" it's going to be a permanent condition. (I have heard that if
you spin a wheel fast, for a long time, the spiders can actually weld
themselves to their shafts, but that's not likely to get you OUT of a
mud-hole...)
So, do you mind telling me which parts you think "lock" together?
Here is a summary of your opinion, as I understand it.
1) If one wheel is on ice and the other on pavement, the one on ice will
have a very low resistive force, and the pavement wheel will have a much
higher resistive force. Both shafts will have the same torque as each
other.
2) If you apply an equal braking force to each side, this will equalize
the resistance on each shaft.
3)Since there is an equal resistance on each side, there will be the
same torque, so both wheels will turn, and then, since one wheel is on
pavement, the vehicle will move.
4) Since everything is under high torque, the open diff will lock up,
and not "disengage" as long as you keep the power on.
5)You can help thing along a bit by releasing the brakes. This provides
another surge of power to the wheels.
Have I got all that about right?
Ok, humour me now: let's try some numbers in there. You can tell me
which numbers you don't like.
1) Left wheel (ice), say 100Nm resistive force. Right wheel 1000Nm.
Total requirement to move the vehicle, say 500Nm. But torque on shafts
is only 2x100Nm. Left wheel spins.
2) Apply braking force of 300Nm per side. Resistive force on both sides
now 300+100=400Nm per side (?). Total is 800Nm, so the vehicle moves.
If you're happy so far, I'll tell you what I don't agree with...
Michael...
Michael
wrote:
>Michael,
>
>In one of my posts on this subject I mentioned that the engine can
>overcome the rear brake holding power on almost any normal vehicle,
>especially if the engine has a running (or tire spinning) head start.
>
Agreed.
>That means that when you put the brakes on hard enough to equal the side
>to side torque enough to break free and spin both wheels, the engine
>still has enough power to easily overcome the braking force while
>continuing to add it's own torque to the driveshaft.
>
Well, it's the "put the brakes on hard enough to equal the side to side
torque enough to break free and spin both wheels" part that we
fundamentally disagree on. Arithmetically, that doesn't happen.
>If it didn't, the engine would stall.
Not necessarily. Under my reasoning the brakes are on and the engine is
running. I just maintain that the resulting movement will be still on
the ice-wheel only, not both wheels.
>
>Don't forget, we are flooring the gas pedal at the same time we are
>hitting the brakes!
>
>That might be the 'added torque' you are looking for, eh?
>
>I think it is and you seem to have been talking a static side to side
>torque.
>
>The engine has things to say about that.
Well, I am looking for more torque, yes, to explain how the
pavement-wheel starts to turn. But the problem has never been (in this
discussion) that the engine can't produce enough. It's accounting for
the *consumption* of that power that we differ on.
Prior to braking, the torque on the prop-shaft might be, say, 200Nm,
distributed evenly between the wheels (as it always is). It's only 200Nm
because one wheel is on ice and only supports 100Nm.
What happens to that 200Nm of torque? Well, on the ice-wheel, it is
translated into motion, because the total resistance on that side is not
greater than the torque. On the pavement wheel it remains static torque
because the total resistance on that side is greater than the torque
When we apply the brakes to both rear wheels, the prop-shaft torque
might rise to, say, 2000Nm, still distributed evenly between the two
wheels. So now each wheel has 1000Nm of torque.
What happens to the 2000Nm of torque? Well, on the ice side, the total
resistance is now 1000Nm, made up of 100Nm of grip torque (that caused
by tyre/ground friction), and the remaining 900Nm is made up of brake
torque (overcoming brake pad/disc friction). That wheel will continue
to move, since it still has 100Nm of torque that isn't being consumed by
the brakes, and that amount of torque is sufficient to overcome the low
ground/tyre resistance..
On the pavement-side, we know that the braking force will be the same as
on the other side, namely 900Nm (equal braking force). This allows us to
*calculate* that we are still only getting 100Nm on that side too. That
wheel will continue to remain stationary, since it still has only 100Nm
of torque that isn't being consumed by the brakes, and that amount is
insufficient to overcome the high ground/tyre resistance.
Regards,
Michael...
Thus of the 2000Nm of propshaft torque, a total of 1800Nm is being taken
by the brakes, and 200Nm is
>
>Mike
>86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail
>88 Cherokee 235 BFG AT's
-----------== Posted via Newsfeed.Com - Uncensored Usenet News ==----------
Douglas A. Shrader
the technique wouldn't work, but it does work, which proves you are wrong.
You are simply to hard headed and dense to understand what everyone else
here knows. As I said in my very first post in this thread you will never
understand it. Now give it up, your hopeless.
Douglas A. Shrader
"AZGuy" <jimnaz...@cox.net> wrote in message
news:g906vukg6s31sda4b...@4ax.com...
Yes, it will work in reverse with disc brakes. There is no need to have
greater braking force on the spinning wheel, the brakes apply equal
resistance to both sides which is what forces both wheels to turn.
Douglas A. Shrader
> On Sat, 07 Dec 2002 18:29:06 -0500, Mike Romain <rom...@sympatico.ca>
> wrote:
>
> >Michael,
> >
> >In one of my posts on this subject I mentioned that the engine can
> >overcome the rear brake holding power on almost any normal vehicle,
> >especially if the engine has a running (or tire spinning) head start.
> >
> Agreed.
>
> >That means that when you put the brakes on hard enough to equal the side
> >to side torque enough to break free and spin both wheels, the engine
> >still has enough power to easily overcome the braking force while
> >continuing to add it's own torque to the driveshaft.
> >
> Well, it's the "put the brakes on hard enough to equal the side to side
> torque enough to break free and spin both wheels" part that we
> fundamentally disagree on. Arithmetically, that doesn't happen.
As I have told you before you are adding the friction between the tires and
the ground into your figures and it doesn't belong there. The only force at
this point that will affect the turning of the axles is the braking, and it
will be equal. God, I have tried to be polite but I can't for the life of me
see why this is so hard for you to understand.
>
> >If it didn't, the engine would stall.
> Not necessarily. Under my reasoning the brakes are on and the engine is
> running. I just maintain that the resulting movement will be still on
> the ice-wheel only, not both wheels.
You have been givenm several experiments to try which would prove you wrong.
You can maintain whatever you want, you have been proven wrong by actual use
and experimentation.
> >
> >Don't forget, we are flooring the gas pedal at the same time we are
> >hitting the brakes!
> >
> >That might be the 'added torque' you are looking for, eh?
> >
> >I think it is and you seem to have been talking a static side to side
> >torque.
> >
> >The engine has things to say about that.
>
> Well, I am looking for more torque, yes, to explain how the
> pavement-wheel starts to turn. But the problem has never been (in this
> discussion) that the engine can't produce enough. It's accounting for
> the *consumption* of that power that we differ on.
>
> Prior to braking, the torque on the prop-shaft might be, say, 200Nm,
> distributed evenly between the wheels (as it always is). It's only 200Nm
> because one wheel is on ice and only supports 100Nm
>
> What happens to that 200Nm of torque? Well, on the ice-wheel, it is
> translated into motion, because the total resistance on that side is not
> greater than the torque. On the pavement wheel it remains static torque
> because the total resistance on that side is greater than the torque
>
> When we apply the brakes to both rear wheels, the prop-shaft torque
> might rise to, say, 2000Nm, still distributed evenly between the two
> wheels. So now each wheel has 1000Nm of torque.
>
> What happens to the 2000Nm of torque? Well, on the ice side, the total
> resistance is now 1000Nm, made up of 100Nm of grip torque (that caused
> by tyre/ground friction), and the remaining 900Nm is made up of brake
> torque (overcoming brake pad/disc friction). That wheel will continue
> to move, since it still has 100Nm of torque that isn't being consumed by
> the brakes, and that amount of torque is sufficient to overcome the low
> ground/tyre resistance..
>
> On the pavement-side, we know that the braking force will be the same as
> on the other side, namely 900Nm (equal braking force). This allows us to
> *calculate* that we are still only getting 100Nm on that side too. That
> wheel will continue to remain stationary, since it still has only 100Nm
> of torque that isn't being consumed by the brakes, and that amount is
> insufficient to overcome the high ground/tyre resistance.
THE GROUND/TIRE RESISTANCE IS NOT PART OF THE FORMULA!! Once you apply the
brakes the ONLY resistance the differetial will feel is FROM THE BRAKES!!
The brakes are BETWEEN the tires and the Axle. Sheesh, your entire failure
to understand how it works is because you refusee to understand this basic
point. THINK about it. Two locomotives are pulling a line of cars, you
disconnect the first locomotive, now do you include it in your numbers when
you figure out how much pulling force is on the cars? NO, it's out of the
loop, and so is you ground/tire resistance. A 6 year old could understand
that. Give it up man, you have a mental block you don't even know about.
Douglas A. Shrader
"Jerry Bransford" <jer...@cox.net> wrote in message
news:3DF2965D...@cox.net...
> "Douglas A. Shrader" wrote:
> > Yes he has the right to ask, but he is never satisfied with any
explanation,
> > and it gets kind of old hearing someone ask "Why?" everytime you explain
it
> > to him.
>
> Exactly, and is why I moved on myself.
>
> Jerry
Your right, but hope springs eternal. I just kept hoping he would get it
eventually but as long as he refuses to believe the truth when he hears it
he will never learn.
Michael
wrote:
>
>This 'book' guy still doesn't know that the definition of torque
>involves movement.
>
>I hope his parents didn't have to spend too much money on his
>education....
>
>Mike
No, torque doesn't involve movement. Your parents very obviously didn't
spend much on your education... :-)
However, with this one statement, you demonstrate perfectly why your
arguments don't hold water.
TORQUE DOES NOT INVOLVE MOVEMENT.
Douglas A. Shrader
> The Tide
>
> Michael was in the habit of gazing at thse sea, and it didn't take him
many weeks to pick up on the fact that the sea level seemed to rise and fall
> every 12 hours or so.
>
> "Grandma, why does that happen? " he asked.
>
Do we have to explain tides to Micheal now? I know how this story goes,
One day Michael takes his question to a group of wise men. At first the Wise
Men welcomed Michael, and explained to him how the Tides really work. But
Michael was not satisfgied, and kept asking them, Why? So the Wise men gave
Michael some experiments to try which would prove to Michael that his
assumptions were wrong, but Michael refused to try them, prefering instead
to keep asking Why? The Wise explained it to Michael in every possible way,
keeping the explanations so simple a child could understand them, but
Michael refused to listen to them, and just kept asking Why? Finally the
Wise men concluded correctly that Michael was incapable of learning and left
him alone with his willful ignorance, and asa they left he asked them Why? I
just want to learn, I am not stupid, you just havn't explained it to me yet.
Michael
<dshr...@nospam.com> wrote:
>
>"Michael" <no...@none.com> wrote in message
>news:3df50c72....@news.cit-news.com...
>> On Sat, 07 Dec 2002 18:29:06 -0500, Mike Romain <rom...@sympatico.ca>
>> wrote:
>>
>> >Michael,
>> >
>> >In one of my posts on this subject I mentioned that the engine can
>> >overcome the rear brake holding power on almost any normal vehicle,
>> >especially if the engine has a running (or tire spinning) head start.
>> >
>> Agreed.
>>
>> >That means that when you put the brakes on hard enough to equal the side
>> >to side torque enough to break free and spin both wheels, the engine
>> >still has enough power to easily overcome the braking force while
>> >continuing to add it's own torque to the driveshaft.
>> >
>> Well, it's the "put the brakes on hard enough to equal the side to side
>> torque enough to break free and spin both wheels" part that we
>> fundamentally disagree on. Arithmetically, that doesn't happen.
>
>As I have told you before you are adding the friction between the tires and
>the ground into your figures and it doesn't belong there. The only force at
>this point that will affect the turning of the axles is the braking, and it
>will be equal. God, I have tried to be polite but I can't for the life of me
>see why this is so hard for you to understand.
>
It's hard for me to understand, because it isn't true... See below.
>
>>
>> >If it didn't, the engine would stall.
>> Not necessarily. Under my reasoning the brakes are on and the engine is
>> running. I just maintain that the resulting movement will be still on
>> the ice-wheel only, not both wheels.
>
>You have been givenm several experiments to try which would prove you wrong.
>You can maintain whatever you want, you have been proven wrong by actual use
>and experimentation.
>
<snip>
>> What happens to the 2000Nm of torque? Well, on the ice side, the total
>> resistance is now 1000Nm, made up of 100Nm of grip torque (that caused
>> by tyre/ground friction), and the remaining 900Nm is made up of brake
>> torque (overcoming brake pad/disc friction). That wheel will continue
>> to move, since it still has 100Nm of torque that isn't being consumed by
>> the brakes, and that amount of torque is sufficient to overcome the low
>> ground/tyre resistance..
>>
>> On the pavement-side, we know that the braking force will be the same as
>> on the other side, namely 900Nm (equal braking force). This allows us to
>> *calculate* that we are still only getting 100Nm on that side too. That
>> wheel will continue to remain stationary, since it still has only 100Nm
>> of torque that isn't being consumed by the brakes, and that amount is
>> insufficient to overcome the high ground/tyre resistance.
>
>THE GROUND/TIRE RESISTANCE IS NOT PART OF THE FORMULA!! Once you apply the
>brakes the ONLY resistance the differetial will feel is FROM THE BRAKES!!
>The brakes are BETWEEN the tires and the Axle. Sheesh, your entire failure
>to understand how it works is because you refusee to understand this basic
>point. THINK about it. Two locomotives are pulling a line of cars, you
>disconnect the first locomotive, now do you include it in your numbers when
>you figure out how much pulling force is on the cars? NO, it's out of the
>loop, and so is you ground/tire resistance. A 6 year old could understand
>that. Give it up man, you have a mental block you don't even know about.
>
The ground/tyre resistance IS part of the equation.
Your analogy of locomotives isn't appropriate. There are two forces
resisting the transmission from turning the side-shaft. One is the
brake. The other is the ground. You can't pretend that one doesn't
exist.
Here's a thought experiment to show this:
Imagine we have an electric motor driving a single shaft, with a wheel
at the end of it. The wheel is rigged up on a roller-bed in such a way
that, by applying more or less friction to the rollers, we can vary (and
measure) the amount of torque at the wheel.
Just inboard of the wheel, we mount a brake disc onto the shaft.
We also put suitable strain guages on the shaft, so that we can measure
the twisting force (torque) being experience by the shaft at any time.
One strain guage measures the torque on the shaft between the motor and
the disc, and the other, between the disc and the wheel.
The motor is powerful enough that it can overcome any kind of retarding
force we can put on it.
Motor Inner shaft Disc Wheel
[======]----------------------------I-------------------------------{}
First, we tighten up the rollers to the point where the strain guages
both read, say 1000Nm. (The brake has not been applied).
The we apply the brake progressively, and watch what the strain guages
read.
I predict that the inner guage will immediately start rising from
1000Nm, showing the increased torque resulting from the brakes. The
outer strain guage will remain at 1000Nm.
Or, if we did the experiment the other way around, and started with zero
resistance at the wheel, and 1000Nm at the brakes, we could observe what
the strain guages read as we progressively increase firction at the
rollers: At first, the inner guage would register 1000Nm, and the outer
would read zero. Then as we apply "ground" resistance at the rollers,
BOTH guages will increase at the same rate. (i.e. the inner gauge will
always be 1000Nm MORE than the outer guage.
Either way, when the brakes AND the rollers are EACH applying 1000Nm,
the outer guage will register 1000Nm, and the inner guage will register
2000Nm.
For the purposes of a differential, the *inner* shaft torque will be the
one that is "carried" over to the other side. The *outer* shaft torque
will be that which provides motive force.
This doesn't seem hard to grasp.
Michael...
Imagine a rear axle rigged up the rear wheel of a motorcycle (one wheel,
for simplicity). It is on a suface with good traction. We apply a
certain torque to the drivetrain, say 100Nm. (It doesn't matter if it
results in movement, but for aguments' sake, let's assume that there is
in fact movement).
Now we apply the brake
Douglas A. Shrader
It is true, you are simply to ignorant to understand and to stupid to
realize it. I'm through with you, there is no way to educate a mindless fool
who can't understand grade school problems. Go away, with you to dumb for
words.
Wallow in your ignorance and declare yourself the winner, there are perhaps
two people on earth who agree with you, the evidence does not, as you could
easily find out for yourself if you truly wanted an explanation instead of a
round-in-circles argument.
Mike Romain
>
> On Sat, 07 Dec 2002 22:10:17 -0500, Mike Romain <rom...@sympatico.ca>
> wrote:
>
>
> TORQUE DOES NOT INVOLVE MOVEMENT.
>
>
Do you want to show me a formula for torque that has no motion involved?
Take an engine:
(horsepower X 5250) / RPM = Foot pounds.
Note RPM in the equation. Add zero for rpm. Divide any numbers you
want to use for hp by 0 and you have 0 = foot pounds.
Erik-Jan Geniets
Mike Romain wrote:
> Take an engine:
>
> (horsepower X 5250) / RPM = Foot pounds.
>
> Note RPM in the equation. Add zero for rpm. Divide any numbers you
> want to use for hp by 0 and you have 0 = foot pounds.
Your mathematics are wrong. Dividing through zero (0) is impossible.
Try you electronic calculater with 1 HPx5250=ERROR.
So I agree: Torque does not involve movement, as I did already explain
in an earlier thread.
Kind regards,
Erik-Jan.
Mike Romain
Unless there is an RPM in the equation, the equation won't work.
RPM is movement.
What am I missing?
Michael
wrote:
>Unless there is an RPM in the equation, the equation won't work.
>
>RPM is movement.
>
>What am I missing?
>
Since you were casting aspersions on MY education, dare I suggest that
elementary knowledge of physics is what you are missing?
For simplicity, think of torque on a shaft being the force that "wants"
to twist the shaft into a pretzel. If you apply a twisting force against
a resistance, the shaft will "feel" that force, even if it is unable to
turn.
When one wheel is on a muddy slope, and the other on a slope with good
grip, BOTH shafts will experience the same torque, even though one is
stationary, and the other is turning. (Open diff, obviously).
It is also true that BOTH wheels are contributing equally and positively
to the forward force acting on the vehicle, even though that force may
be insufficient to be successful. Hence the oft-heard "a 4x4 is really
only 2 wheel drive" simply isn't true.
Michael...
>Mike
>86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail
>88 Cherokee 235 BFG AT's
Michael
Funnily enough, no matter how much you proclaim your argument is right,
it doesn't *make* it right. I've detailed the factual inconsistencies in
your arguments, and quantified my own. I have yet to see someone point
out a simple factual error or false assumption of mine.
I can point out several of yours - one-liners that stand out as errors
on their own. You have said all of these things at one time or another.
1) An open diff will lock up.
2) An open diff will not disengage as long as it remains under torque.
3) Equal braking on both rear wheels will equalizes the total resistance
of the two wheels, regardless of the fact that they had different
traction conditions prior to braking.
4) Torque is only present when there is motion. (OK, I can't be bothered
to check if that particular error is shared by you, but you and Mike nod
knowingly to each other as you cast your pearls, so you'll forgive me if
I add this one to your list).
If all of the above WERE true, it would probably prove your point. Mabe
even if some of them were true. But I know them to be false, and so does
anyone else with an elementary knowledge physics, and of how a diff
works.
Michael...
Douglas A. Shrader
> On Sun, 8 Dec 2002 13:19:28 -0500, "Douglas A. Shrader"
> <dshr...@nospam.com> wrote:
>
>
> >It is true, you are simply to ignorant to understand and to stupid to
> >realize it. I'm through with you, there is no way to educate a mindless
fool
> >who can't understand grade school problems. Go away, with you to dumb for
> >words.
> >Wallow in your ignorance and declare yourself the winner, there are
perhaps
> >two people on earth who agree with you, the evidence does not, as you
could
> >easily find out for yourself if you truly wanted an explanation instead
of a
> >round-in-circles argument.
> >
>
> Funnily enough, no matter how much you proclaim your argument is right,
> it doesn't *make* it right. I've detailed the factual inconsistencies in
> your arguments, and quantified my own.
You state that the wheel on the hard surface CANNOT turn when the brakes are
applied because the ground resistance is always greater than the wheel on
ice, or in the air. This is easily proven false by simply jacking up on side
of your car, putting it in gear, giving it enough throttle to keep the wheel
spinning and slowly applying the brakes until the wheel on the ground starts
pulling. Your assumption has been proven wrong but you cling to it. You
claim you are using scientific methods, but scientist dicard theories when
they are proven wrong. YOUR theory has been debunked, which proves YOUR
assumptions are wrong. Since you are incapable of understanding this, and
unwilling to perform the experiment yourself, you are not only wrong but
about the dumbest person to post here since Lloyd Parker. You have not
proven anything but your inability to understand basic facts, and no matter
how many times you declare me wrong I am not, you are. It has been proven
for 80 years and remains true today.
Erik-Jan Geniets
Mike Romain wrote:
>
> Erik-Jan Geniets wrote:
> >
> > Mike Romain wrote:
> >
> > > Take an engine:
> > >
> > > (horsepower X 5250) / RPM = Foot pounds.
> > >
> > > Note RPM in the equation. Add zero for rpm. Divide any numbers you
> > > want to use for hp by 0 and you have 0 = foot pounds.
> >
> > Your mathematics are wrong. Dividing through zero (0) is impossible.
> > Try you electronic calculater with 1 HPx5250=ERROR.
> > So I agree: Torque does not involve movement, as I did already explain
> > in an earlier thread.
>
> Unless there is an RPM in the equation, the equation won't work.
>
> RPM is movement.
>
> What am I missing?
Your not missing anything. I made a typing error:
I meant this: (1 x 5250) / 0 = ERROR
Kind regards,
Erik-Jan.
Erik-Jan Geniets
Michael wrote:
> Hence the oft-heard "a 4x4 is really
> only 2 wheel drive" simply isn't true.
Worst case scenario (stuck) with center diff locked or T-case this is
very true. With an open center diff you even could end-up with a 1 wheel
drive vehicle.
Kind regards,
Erik-Jan.
http://www.fotograaf.com/trooper
Michael
wrote:
Well, of course, it's really zero-wheel drive, then, isn't it? :)
My point was just that in a 3-open-diff situation, all four wheels will
be contributing, or NONE of them (if one has lifted). But never one,
two, or three wheels providing force.
M...
.
Michael
If you read my opening to this thread, I am looking for explanations for
the phenomenon, not evidence of it. I HAVE no theroy of how it works,
so I can't discard it. But it's perfectly valid for a scientist to
reject an explanation of a phenomenon, if the *explanation* voids other
principles he knows or believes to be true. So I am looking for
alternative theories, two of which are under consideration. Sorr, but
yours are not really under consideration, as you show a dismal lack of
understanding of basic physics and mechanics.
I am also willing to re-examine my assumptions and principles, such as
if someone with credibility (Erik?) tells me that an open diff locks up.
In that instance, they simply need to say which components stop working
the way they used to, and why that happens. At least then that would
isolate the debate to a single point of disagreement, which is what
generally happens when scientists disagree.
Michael.
Michael
<Xcphi...@ns.sympatico.ca> wrote:
>
>So now it's a phenomenon? Mystical powers are at work here I suppose?
>
Phenomenon: (noun) Any fact or experience that is apparent to the
senses, and can be scientifically described. e.g. an eclipse.
You watch too much X-Files.
Erik-Jan Geniets
Michael wrote:
>
> Well, of course, it's really zero-wheel drive, then, isn't it? :)
>
> My point was just that in a 3-open-diff situation, all four wheels will
> be contributing, or NONE of them (if one has lifted). But never one,
> two, or three wheels providing force.
>
I think it is still a one wheel drive in your example... But it won't
move the car...;-)
Erik-Jan.
Erik-Jan Geniets
Michael wrote:
> I am also willing to re-examine my assumptions and principles, such as
> if someone with credibility (Erik?) tells me that an open diff locks up.
Haven't seen any other Erik in this thread, but it wasn't me.
No offense.
Kind regards,
Erik-Jan.
P.S. I think this is a really interesting thread after all. I don't
understand all the ins an outs about diffs either. I'm following this
one closely just to learn and understand more about the subject. Thanks
to you all so far.
Douglas A. Shrader
> On Sun, 8 Dec 2002 17:47:44 -0500, "Douglas A. Shrader"
> <dshr...@nospam.com> wrote:
>
>
> >You state that the wheel on the hard surface CANNOT turn when the brakes
are
> >applied because the ground resistance is always greater than the wheel on
> >ice, or in the air. This is easily proven false by simply jacking up on
side
> >of your car, putting it in gear, giving it enough throttle to keep the
wheel
> >spinning and slowly applying the brakes until the wheel on the ground
starts
> >pulling. Your assumption has been proven wrong but you cling to it. You
> >claim you are using scientific methods, but scientist dicard theories
when
> >they are proven wrong. YOUR theory has been debunked, which proves YOUR
> >assumptions are wrong. Since you are incapable of understanding this, and
> >unwilling to perform the experiment yourself, you are not only wrong but
> >about the dumbest person to post here since Lloyd Parker. You have not
> >proven anything but your inability to understand basic facts, and no
matter
> >how many times you declare me wrong I am not, you are. It has been proven
> >for 80 years and remains true today.
> >
>
> If you read my opening to this thread, I am looking for explanations for
> the phenomenon, not evidence of it. I HAVE no theroy
That much is obvios. Neither do you want to hear any theories, much easier
to whine that no one has told you what you want to hear.
of how it works,
> so I can't discard it. But it's perfectly valid for a scientist to
> reject an explanation of a phenomenon, if the *explanation* voids other
> principles he knows or believes to be true. So I am looking for
> alternative theories, two of which are under consideration.
Yes, I saw the two you like, both of which are easily disproven by anyone
who desires to try. A man searching for the truth would be a little more
open than you are.
Sorr, but
> yours are not really under consideration, as you show a dismal lack of
> understanding of basic physics and mechanics.
Coming from an idiot of your caliber this is high praise indeed.
>
> I am also willing to re-examine my assumptions and principles, such as
> if someone with credibility (Erik?) tells me that an open diff locks up.
Contact GM, Ford, or any automobile manufacturer, but tell me this, what the
hell do you call it when both shafts are spinning even though only one
should be under your assumptions? No, there is no mechanical coupling per
say, but you would be hard pressed to say the least to stop one wheel under
those conditions. I call that "locked up".
> In that instance, they simply need to say which components stop working
> the way they used to, and why that happens. At least then that would
> isolate the debate to a single point of disagreement, which is what
> generally happens when scientists disagree.
What debate? You won't listen to anything that isn't exactly what you
already think, therefore no debate is possible. Pull your fingers out of
your ears and open your mind, if, by some miracle, you ever do learn the
truth, you will see then that I am right and you are a horses behind.
Until then you are simply a clueless idiot with delusions of grander, and NO
understanding of differentials or anything mechanical. I do find it kind of
amusing that you accuse me of having no mechanical understanding,
considering the fact that I have been repairing cars, trucks and farm
equipment from the time I was first able to hold a wrench, and aced every
class on automechanics I took from vocational school to technical school. My
whole life has been dedicated to understanding WHY something works as it
does, only by truly understanding how something works can you diagnose and
repair it when it doesn't work. What the hell did you ever learn about it,
how to turn the key? No wonder your engineering students give you blank
looks all the times, they are no doubt wondering how anyone can believe the
shit you promote as truth.
Douglas A. Shrader
"Erik-Jan Geniets" <e...@fotograaf.com> wrote in message
news:3DF3EE18...@fotograaf.com...
Hope all this helps someone. Sorry to be so rude to Michael but it is more
than a little irritating trying to explain light to the blind.
Bombsite
Mike,
An example for you, electric motors deliver Maximum torque at zero rpm.
Edd
--
This space for rent.
http://www.lrovernut.btinternet.co.uk
http://www.londoncrawling.com (drink and dash not drink and drive)
Mike Romain
Bull shit.
I asked for a formula for torque that does not involve movement mr book
man!
You conveniently didn't answer THAT post, just every other one I have
made.
Show me a formula that will work for torque that has no movement vector
involved or that the movement vector or moment arm is 0 and the formula
still works.
Like this one:
(horsepower X 5250) / RPM = Foot pounds.
Put 0 in for the rpm and the formula won't work.
Mike
86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail
88 Cherokee 235 BFG AT's
Mike Romain
Mike
Jerry Bransford
rpm... that's because it is working into the most possible resistance
which has reduced its rpm to zero.
--
Jerry Bransford
PP-ASEL KC6TAY
The Zen Hotdog... make me one with everything!
Geezer Jeep: http://www.jjournal.net/jeep/gallery/JBransfordsTJ/
Mike Romain
(horsepower X 5250) / RPM = Foot pounds.
Mike
Rob Munach
>
> Then it all messes up if you use 0 for rpm to get the motor torque...
>
> (horsepower X 5250) / RPM = Foot pounds.
>
> Mike
>
> Jerry Bransford wrote:
> >
> > That should be true of any engine if it were capable of running at zero
> > rpm... that's because it is working into the most possible resistance
> > which has reduced its rpm to zero.
> >
> > Bombsite wrote:
> >
Actually, it doesnt. At zero RPM, you cannot have any HP becuase no work
it being done. So you end up with 0 = torque x 0 which balances.
--
________________
Rob Munach, PE
Excel Engineering
Carrboro, NC
Erik-Jan Geniets
Mike Romain wrote:
>
> Show me a formula that will work for torque that has no movement vector
> involved or that the movement vector or moment arm is 0 and the formula
> still works.
Just too make things clear for me.
I would translate 'moment arm' and 'torque' as being the same thing in
my language (Dutch). Do I go wrong here ???
Kind regards,
Erik-Jan.
Erik-Jan Geniets
Rob Munach wrote:
> > >
>
> Actually, it doesnt. At zero RPM, you cannot have any HP becuase no work
> it being done. So you end up with 0 = torque x 0 which balances.
Where does that leave the force in your example.
I assume there is still a force working here which does not result in a
movement.
Kind regards,
Erik-Jan.
J5
pressure to each wheel (assuming the brakes are in order) and should be
"loading" them up equally. Since your theory that both wheels will receive
equal torque and the brakes acting to give each wheel the same resistance,
there will come a point when the resistance is equal that both wheels will
spin. Thus the wheel that was not spinning previously will now be spinning
at the same rate as the other thus freeing the stuck vehicle. I've seen it
work but I like my locker much better :) And after digging into this thread
out of curiosity as to what could possibly keep it going for so long, that
is all I'm going to say. You guys have fun.
j5
--
J5's jeep page at http://www.users.qwest.net/~j5/
"Michael" <no...@none.com> wrote in message
news:3df6eb49....@news4.cit-news.com...
> I asked this question as an aside on another thread, and got nowhere:
>
> Most of us have heard of the trick of dabbing at the brakes to get going
when one wheel ( in the case of a 4x2) is spinning. Opinion is divided over
> how effective it is, and under what circumstances it works best/at all.
>
> I can fully understand how this technique works in the case of a
torque-biased (or torque sensing) LSD. And I fully understand how ABS-based
> electronic traction control (or manual "fiddle brakes") can give
additional traction by being applied to one wheel only.
>
> BUT, I can't understand the mechanism by which this works in the case of
an axle with an open diff.
>
> Firstly, let's take as a given that the technique *does* work. So I am not
looking for anecdotal evidence as to *whether* it works, or how effective
> it is. I am trying to understand the way the forces are
manipulated/transferred/increased to provide additional traction, and hence
motive force.
>
> My basic problem with the theory is this:
>
> 1) An open diff always has the same torque on both shafts (barring some
internal friction effects, of course).
> 2) Any additional torque put on a shaft by means of applying a brake to
the shaft will, by definition, not be "useful" torque at the wheel.
> 3) Thus if we apply a braking force equivalent to, say, 500Nm to *both*
sides, it is all absorbed by the very act of creating it.
> 4) So where does the extra torque come from?
>
> Compare this to the case of a 500Nm braking force being applied *to the
spinning side only*. Providing the engine and transmission can generate an
> extra 1000Nm, the torque on the spinning side will rise by 500Nm (but all
used to overcome the brake friction). The open diff will ensure that the
> torque on the other side rises by 500Nm as well, and since there is no
braking force on that side, it will be "real", gripping torque (traction
> conditions permitting, of course).
>
> I only mention the one-sided braking as a comparison, to illustrate why
equal-braking (by that logic) would *not* work. As we are accepting that
there
> is some evidence that it *does* work, I am seeking a descriptive
explantion of the mechanics/physics involved.
>
> One theory is that an open diff has a built-in torque bias. While I would
concede that internal friction would generate *some* torque imbalace, I
> would need some convincing that it is significant enough (some hundreds of
Nm?) to have this effect. (Also, why would it be enhanced by braking? A
> torque-biased LSD is *designed* to provide ever more friction under higher
torque differences. I can't see any components of a normal open diff that
> would *multiply* the torque).
>
> I have another theory, but I confess that I can see some snags with it
too. Here it is, by all means shoot it down! :-)
>
> The faster spinning wheel has a higher effective gearing than the slower
(or stationary) one - it rotates twice as fast as it would if both wheels
> were being turned. It seems to me that this would bias the braking - a
given pressure on the pads squeezing the brake disc would retard the highly
> geared wheel more than the same pressure would retard the other wheel.
>
> If so, there would be a nett gain in torque momentarily, until both wheels
were rotating at the same speed. To put some numbers to it, if the initial
> braking force on the spinning wheel puts an additional 1000Nm of torque
onto that shaft, but only puts an an additional 500Nm onto the stationary
> right-hand wheel, then I can see a nett gain of 500Nm of tractive force at
the stationary wheel.
>
> Of course, this effect would be short-lived - as soon as the wheels reach
the same speed as each other, the gearing becomes the same, the braking
> effect becomes the same, and once more perfectly offsets the gain in
torque. That would account for the technique being most effective applied by
> jabbing at the brakes, rather than applying them progressively (or
continuously).
>
> The one snag is that I am not certain that the braking force on each side
*would* in fact differ because of the effective difference in gearing.
> Certainly there would be a *counter*-effect from the fact that stiction is
greater than friction, and this would actually mean a *greater* braking
> effect on the stationary wheel than the spinning one.
>
> Willem-Jan, if you see this, your comments? I've looked through your TAD
FAQ, but only one brief mention of this issue.
>
> Obviously if anyone else has something to add, I'd appreciate it too.
>
> Regards,
>
> Michael...
>
>
Bombsite
Michael
You state the solution very concisely.
For the record though, the essential point that I don't agree with is
that there will be equal resistance at each wheel.
If you start with unequal resistance (the grip torque is unequal), and
then add the same amount of braking resistance to each side, you end up
with precisely the same "gap" in resistance at each side. (My attempt
to quantify the problem with numbers is to highlight this).
The objection would fall away if somehow the braking force on the
spinning side were HIGHER than on the other side. Then the resistance
would indeed be equalized, and both wheels would spin.
Two theories have been proposed that would suggest that there may be a
naturally higher braking force on the spinning side.
One of them argues that a drum brake will always work more effectively
on a moving wheel. (That one seems very plausible to me, though some
people say the technique works with disc-brakes too, which makes me want
an alternative explanation).
The other is that the gearing effect of a diff means that the faster
moving wheel is always going to experience greater braking, from a given
brake pressure. This is my own theory, and I like it when I say it
quickly, but when I think about it some more I am not so sure...
Regards,
Michael
though there has been little debate over those, as not everyone (ahem)
On Mon, 9 Dec 2002 22:44:20 -0700, "J5" <j...@qwest.net> wrote:
>Think of the brakes as being an equalizer. They're applying the same
>pressure to each wheel (assuming the brakes are in order) and should be
>"loading" them up equally. Since your theory that both wheels will receive
>equal torque and the brakes acting to give each wheel the same resistance,
>there will come a point when the resistance is equal that both wheels will
>spin. Thus the wheel that was not spinning previously will now be spinning
>at the same rate as the other thus freeing the stuck vehicle. I've seen it
>work but I like my locker much better :) And after digging into this thread
>out of curiosity as to what could possibly keep it going for so long, that
>is all I'm going to say. You guys have fun.
>j5
-----------== Posted via Newsfeed.Com - Uncensored Usenet News ==----------
Michael
mainly arithmetic. The answers should be quite interesting :)
Seriously, though, I am not looking for flames here. I would be
interested to see how others would do these calculations. FWIW, I have
added *my* answers in my next message.
*************************************
A rear-wheel drive truck is attempting to accelerate from rest up a
steep slope. The left-hand (rear) wheel is on some slushy ice, which
will support a maximum torque at the wheel of 200Nm. The right-hand
wheel is on pavement, which will support a maximum wheel torque of
2000Nm. If the torque at the surface of either tyre exceeds these
maximum figures, the wheel will start to spin.
The vehicle requires 1500Nm of torque in order to start moving up the
slope. Assume that the engine and drivetrain can deliver unlimited
torque.
Q1: If the vehicle has an open diff, will the vehicle be able to move?
If not, will the left, right, or both wheels spin? What will be the
torques on the left- and right-hand wheels respectively when the vehicle
starts to move, or the wheel(s) starts to spin? (3)
A1:
--------------------------------------------------------------------
Q2: As for Q1, but assume the vehicle's diff is fully locked (using an
air-locker, for example). (3)
A2:
--------------------------------------------------------------------
Q3: If the vehicle has rear brakes that can be applied to the left- and
right-hand sides individually, and the left-hand brake is applied
progressively until the vehicle starts to move: a) What is the total
torque on the left-hand shaft? b) What is the total torque on the
right-hand shaft? (4)
A3:
--------------------------------------------------------------------
Q4: In Q3 above, why did the vehicle not start to move when the combined
torques on the two shafts reached 1500Nm? (2)
A4:
--------------------------------------------------------------------
Q5: In Q3 above, at the instant the vehicle begins to move: a) What is
the torque at surface of the left-hand tyre? b) And the right-hand one?
(4)
A5:
--------------------------------------------------------------------
Q6: In Q3 above, account fully for the torque being produced by the
engine and drive-train. I.e. explain how that torque has been
distributed, and what the torque is on each component at the instant the
vehicle starts to move. (5)
A6:
--------------------------------------------------------------------
Q7: If the two rear brakes cannot be operated independently, and always
apply the same retarding force, and they are applied hard enough to add
3000Nm of resistance to each side, what will the the torque be: a) On
the left- and right-hand shafts respectively? b) On the diff ring gear?
c) At the respective points of contact between the ground and the left-
and right-hand tyres. d) How much torque is being used to overcome the
brakes? (4)
A7:
--------------------------------------------------------------------
Notes: Assume simple systems, i.e. disregard static and dynamic friction
differences, internal resistance within the diff, inertia and momentum,
brakes getting hot, etc.
As is always good exam technique, answer the questions asked, not the
ones you think should have been asked :-) But by all means pose your
own questions in a different part of the thread.
Michael
>*************************************
>
>A rear-wheel drive truck is attempting to accelerate from rest up a
>steep slope. The left-hand (rear) wheel is on some slushy ice, which
>will support a maximum torque at the wheel of 200Nm. The right-hand
>wheel is on pavement, which will support a maximum wheel torque of
>2000Nm. If the torque at the surface of either tyre exceeds these
>maximum figures, the wheel will start to spin.
>
>The vehicle requires 1500Nm of torque in order to start moving up the
>slope. Assume that the engine and drivetrain can deliver unlimited
>torque.
>
>Q1: If the vehicle has an open diff, will the vehicle be able to move?
>If not, will the left, right, or both wheels spin? What will be the
>torques on the left- and right-hand wheels respectively when the vehicle
>starts to move, or the wheel(s) starts to spin? (3)
>
>
>
>--------------------------------------------------------------------
>Q2: As for Q1, but assume the vehicle's diff is fully locked (using an
>air-locker, for example). (3)
>
>
>--------------------------------------------------------------------
>Q3: If the vehicle has rear brakes that can be applied to the left- and
>right-hand sides individually, and the left-hand brake is applied
>progressively until the vehicle starts to move: a) What is the total
>torque on the left-hand shaft? b) What is the total torque on the
>right-hand shaft? (4)
>
>
>>--------------------------------------------------------------------
>Q4: In Q3 above, why did the vehicle not start to move when the combined
>torques on the two shafts reached 1500Nm? (2)
>
therefore not adding motive force.
>
>
>--------------------------------------------------------------------
>Q5: In Q3 above, at the instant the vehicle begins to move: a) What is
>the torque at surface of the left-hand tyre? b) And the right-hand one?
>(4)
>
>
>
>--------------------------------------------------------------------
>Q6: In Q3 above, account fully for the torque being produced by the
>engine and drive-train. I.e. explain how that torque has been
>distributed, and what the torque is on each component at the instant the
>vehicle starts to move. (5)
>
resulting from friction between tyre and ground) is 200Nm. Therefore te
right-hand wheel needs a minimum of 1300Nm to make the vehicle move.
Therefore the left brake must be applied hard enough to provide 1100Nm
of additional resistance at that wheel, bringing the left-hand total to
1300Nm.
(The drive-train will then be applying a torque of 2600Nm on the centre
of the diff (the ring-gear)).
>
>--------------------------------------------------------------------
>Q7: If the two rear brakes cannot be operated independently, and always
>apply the same retarding force, and they are applied hard enough to add
>3000Nm of resistance to each side, what will the the torque be: a) On
>the left- and right-hand shafts respectively? b) On the diff ring gear?
>c) At the respective points of contact between the ground and the left-
>and right-hand tyres. d) How much torque is being used to overcome the
>brakes? (4)
>
>
wheel=200Nm, Right wheel=200Nm, Brakes absorbing 6000Nm.
Mike Romain
I know DC motors have the most torque at low rpm and the best (lowest)
power drain at high rpm. (I have used a lot of electric fishing motors
in the past, the ones with a gearbox to allow the DC motor to spin up
fast got the best battery life)
That site states that the highest effective torque is just before the
stall torque that occurs at 0 rpm.
Once you hit 0 rpm, there is no more torque, just burned brushes and
windings.
Mike
Michael
Power is a measure of how fast you can rotate an object at a given
torque. If you stop rotating it, there will be zero power. However, that
doesn't mean the torque value is zero.
The torque on a shaft is the force which attempts to twist it. If the
other end is held firmly, there will be no movement, but the shaft will
still be under torque, "trying" to turn. Remove or reduce the resistance
sufficently, and there will be movement (along with a possible reduction
in torque).
Regards,
Michael...
On Mon, 09 Dec 2002 11:29:24 -0500, Mike Romain <rom...@sympatico.ca>
wrote:
>Then it all messes up if you use 0 for rpm to get the motor torque...
>
>(horsepower X 5250) / RPM = Foot pounds.
>
>Mike
>
-----------== Posted via Newsfeed.Com - Uncensored Usenet News ==----------
Douglas A. Shrader
> Hi J5,
>
> You state the solution very concisely.
>
> For the record though, the essential point that I don't agree with is
> that there will be equal resistance at each wheel.
>
> If you start with unequal resistance (the grip torque is unequal), and
> then add the same amount of braking resistance to each side, you end up
> with precisely the same "gap" in resistance at each side. (My attempt
> to quantify the problem with numbers is to highlight this).
You are still wrong Michael, as I will attempt to explain once more.
Lets start with an analogy. First your theory, we will say there is 50 lbs
of ground resistance, and use 50 gallons of water to represent this. Now
there is one hundred pounds of braking resistance represented by 100 gallons
of water. Applied to the differential this would indeed equal 150 gallons
(pounds) of resistance. However a far more accurate analogy is to represent
the 50 pounds of ground resistance with 50 PSI of air in a tank, and the
braking resistance with 100PSI of air in a second tank. Combine the tanks
and apply it to the differential and you will never have over 100 PSI of
air, you cannot add them together to get 150 PSI, which is what you are
trying to do.
Now lets further explain the concept of braking for traction with an
exagerated example. Our poor open differential truck is stuck, the spinning
wheel takes 50 pounds of torque to turn it, the non-spinning wheel needs 100
pounds of torque. Now, lets floor the brakes, just shove that old pedal
clear through the firewall.The brakes are locked so tight no force on earth
can turn that wheel. Now will the differential still feel that little extra
50 pounds on the one wheel? No way, it's out of the equation. Now, while
the brake is locked, put the tranny in gear, forward or reverse makes no
difference, disc brakes or drum brakes make no difference, just put it in
gear. Now, lets floor the throttle, all that lovely torque is sent down the
driveshaft to the differential, which splits it 50/50 between the axles, it
can do nothing else. Now we have the brakes locked and maximum torque to the
axles, so lets release the brakes. WOW, all that stored up energy is turned
loose all at once, which starts both wheels spinning like mad, the
differential is locked up by the torque, and will remain locked all the way
out, twin rooster tails flying, until you slow down. In actual practice
there is no need to go to this extreme, because as soon as the brake
resistance exceeds the 100 pounds of torque required to turn the
non-spinning wheel both will start pulling, we know this because you cannot
add the resistance between ground and brakes any more than you can add the
50 PSI of air to the 100 PSI of air in the second tank to get 150 PSI of
air. Now if you finally understand this Michael than you just might be
smart enough to start pre-school, if it's still over your head I assure you
there won't be a soul in this newsgroup who will credit you with any brains,
and you are a prime candidate for a Darwin award. Given the Dismal lack of
knowledge you have demonstrated concerning physics and mechanics, especially
differentials, I'm betting you still don't understand.
Benjamin Lee
> Here are what I think are the right answers.
> >*************************************
> >
> >A rear-wheel drive truck is attempting to accelerate from rest up a
> >steep slope. The left-hand (rear) wheel is on some slushy ice, which
> >will support a maximum torque at the wheel of 200Nm. The right-hand
> >wheel is on pavement, which will support a maximum wheel torque of
> >2000Nm. If the torque at the surface of either tyre exceeds these
> >maximum figures, the wheel will start to spin.
> >
> >The vehicle requires 1500Nm of torque in order to start moving up the
> >slope. Assume that the engine and drivetrain can deliver unlimited
> >torque.
> >
> >Q1: If the vehicle has an open diff, will the vehicle be able to move?
> >If not, will the left, right, or both wheels spin? What will be the
> >torques on the left- and right-hand wheels respectively when the vehicle
> >starts to move, or the wheel(s) starts to spin? (3)
> >
> A1: No vehicle movement. Left wheel will spin. Left=200Nm, right=200Nm
Assume frictionless differential, yes. Real world, no. Left=200nm, right=
maybe 230nm or so.
> >
> >
> >--------------------------------------------------------------------
> >Q2: As for Q1, but assume the vehicle's diff is fully locked (using an
> >air-locker, for example). (3)
> >
> A2: Vehicle will move. Left=200Nm, Right=1300Nm
> >
Is indeterminate. Anywhere from L=200, R=1300 to L=0,R=1500. Both wheels can
grab, don't know which will carry the load.
> >--------------------------------------------------------------------
> >Q3: If the vehicle has rear brakes that can be applied to the left- and
> >right-hand sides individually, and the left-hand brake is applied
> >progressively until the vehicle starts to move: a) What is the total
> >torque on the left-hand shaft? b) What is the total torque on the
> >right-hand shaft? (4)
> >
> A3: Left=1300Nm, Right=1300Nm
Again, indeterminate. From L=1500, R=1500 to L=1300,R=1300
> >
> >>--------------------------------------------------------------------
> >Q4: In Q3 above, why did the vehicle not start to move when the combined
> >torques on the two shafts reached 1500Nm? (2)
> >
> A4: Of the 2600Nm total torque, 1100Nm is created by the brakes, and
> therefore not adding motive force.
Correct
> >
> >
> >--------------------------------------------------------------------
> >Q5: In Q3 above, at the instant the vehicle begins to move: a) What is
> >the torque at surface of the left-hand tyre? b) And the right-hand one?
> >(4)
> >
> A5: Left=200Nm, Right=1300Nm
> >
> >
> >--------------------------------------------------------------------
> >Q6: In Q3 above, account fully for the torque being produced by the
> >engine and drive-train. I.e. explain how that torque has been
> >distributed, and what the torque is on each component at the instant the
> >vehicle starts to move. (5)
> >
> A6: The maximum possible "grip torque" on the left wheel (torque
> resulting from friction between tyre and ground) is 200Nm. Therefore te
> right-hand wheel needs a minimum of 1300Nm to make the vehicle move.
> Therefore the left brake must be applied hard enough to provide 1100Nm
> of additional resistance at that wheel, bringing the left-hand total to
> 1300Nm.
> (The drive-train will then be applying a torque of 2600Nm on the centre
> of the diff (the ring-gear)).
Correct
> >
> >--------------------------------------------------------------------
> >Q7: If the two rear brakes cannot be operated independently, and always
> >apply the same retarding force, and they are applied hard enough to add
> >3000Nm of resistance to each side, what will the the torque be: a) On
> >the left- and right-hand shafts respectively? b) On the diff ring gear?
> >c) At the respective points of contact between the ground and the left-
> >and right-hand tyres. d) How much torque is being used to overcome the
> >brakes? (4)
> >
> >
> A7: Left shaft=3200Nm, Right shaft=3200Nm, Ring gear=6400Nm, Left
> wheel=200Nm, Right wheel=200Nm, Brakes absorbing 6000Nm.
This is where you are not getting it. You just described a frictionless
differential. Try again to include a normal force at each sliding and
rotational interface inside the differential gear, shaft, and bearing, and a
coefficient of friction in each of those interface. You will complicate the
math by 50 times, but you will get a different answer that reflect the real
world.
Remember, when two gears carry a load, because of the geometry of the
involute gear, there is a force that spread the gears apart. Therefore, when
the side gears in the differential tries to carry the 3200nm of torque in
the diff, there is a tremendous amount of force in the diff that tries to
explode the differential if it was not for the strong housing. The side
gears tries to separate from each other. The back of the side gear then rubs
on the diff housing, and create friction. This friction acts like a brake
that stops the rotation of the side gears. If I put a brake liner between
the side gear, and the differential housing, I have just created a crude
limited slip diff. Only difference is that a limited silp have springs
between the side gears to spread them apart initially. Even without a brake
liner on the back side of the gear, and it is just metal to metal rub, there
is still braking force that stop the spin of the side gears.
For the calculation, assume 10% of torque the side gears carries is
friction.
Now we have:
13000 on left shaft. 200 carried by the left wheel, and 12800 by the left
brake. 13000 is going into the side gears of the differential. 10% of 13000
which is 1300Nm becomes friction between side gear on the differential
housing, so the right shaft has 13000+1300=14300Nm. Take away 13000 absorbed
by the right brake, and you are left with 1300 going to the right ground.
The total input torque to the diff would be 27,300Nm. Of that 26000 is
absorbed by the brake. Chances are something broke at this point already. I
believe this case is what is called 10% lock in limited slip differential
terms.
Ben
Mike Romain
LOL!
My bet is he won't even believe that!
Mike
86/00 CJ7 Laredo, 33x9.5 BFG Muds, 'glass nose to tail
88 Cherokee 235 BFG AT's
<snipped good extra reason for an open diff to lock>
Michael
<dshr...@nospam.com> wrote:
>
>"Michael" <no...@none.com> wrote in message
>news:3df5ca81...@news.cit-news.com...
>> Hi J5,
>>
>> You state the solution very concisely.
>>
>> For the record though, the essential point that I don't agree with is
>> that there will be equal resistance at each wheel.
>>
>> If you start with unequal resistance (the grip torque is unequal), and
>> then add the same amount of braking resistance to each side, you end up
>> with precisely the same "gap" in resistance at each side. (My attempt
>> to quantify the problem with numbers is to highlight this).
>
>You are still wrong Michael, as I will attempt to explain once more.
>Lets start with an analogy. First your theory, we will say there is 50 lbs
>of ground resistance, and use 50 gallons of water to represent this. Now
>there is one hundred pounds of braking resistance represented by 100 gallons
>of water. Applied to the differential this would indeed equal 150 gallons
>(pounds) of resistance. However a far more accurate analogy is to represent
>the 50 pounds of ground resistance with 50 PSI of air in a tank, and the
>braking resistance with 100PSI of air in a second tank. Combine the tanks
>and apply it to the differential and you will never have over 100 PSI of
>air, you cannot add them together to get 150 PSI, which is what you are
>trying to do.
Ahh, now we're getting to one of the roots of our disagreement, perhaps.
Yes, you MUST add the two resistances, unlike your air-pressure
"addition" analogy.
You are not adding the air pressures because you are averaging them.
When you apply two separate resistances to the same shaft, you must add
them. Well, slightly more complicated than that because there will be
two different torques, on different parts of the shaft depending on
where the resistamces are being applied.
If you apply a brake to the chuck of an angle-grinder while you are
cutting, the motor will deliver additional torque (assuming it can), so
that the torque becomes the SUM of the two resistances.
I attach my previous model to the bottom of this message.
Why would you ignore the resistance put up by the gound, just because
the brakes are on? And if you do, then by that reasoning you could move
even if you had insufficient grip under BOTH wheels, simply by creating
torque with the brakes!
>Now lets further explain the concept of braking for traction with an
>exagerated example. Our poor open differential truck is stuck, the spinning
>wheel takes 50 pounds of torque to turn it, the non-spinning wheel needs 100
>pounds of torque. Now, lets floor the brakes, just shove that old pedal
>clear through the firewall.The brakes are locked so tight no force on earth
>can turn that wheel. Now will the differential still feel that little extra
>50 pounds on the one wheel? No way, it's out of the equation.
It's not out of the equation. I'll accept that it's close to zero
though, in some cases. (But in your exampe, it's 25% of the torque
required to move the truck, so it's material for the purposes of the
equation).
>Now, while
>the brake is locked, put the tranny in gear, forward or reverse makes no
>difference, disc brakes or drum brakes make no difference, just put it in
>gear. Now, lets floor the throttle, all that lovely torque is sent down the
>driveshaft to the differential, which splits it 50/50 between the axles, it
>can do nothing else. Now we have the brakes locked and maximum torque to the
>axles, so lets release the brakes. WOW, all that stored up energy is turned
Stored up? Where? This is just the same as hitting the accelerator.
Holding the brakes and then releasing them doesn't allow an engine to
deliver more power.
>loose all at once, which starts both wheels spinning like mad, the
>differential is locked up by the torque, and will remain locked all the way
>out, twin rooster tails flying, until you slow down. In actual practice
>there is no need to go to this extreme, because as soon as the brake
>resistance exceeds the 100 pounds of torque required to turn the
>non-spinning wheel both will start pulling, we know this because you cannot
>add the resistance between ground and brakes any more than you can add the
>50 PSI of air to the 100 PSI of air in the second tank to get 150 PSI of
>air. Now if you finally understand this Michael than you just might be
>smart enough to start pre-school, if it's still over your head I assure you
>there won't be a soul in this newsgroup who will credit you with any brains,
>and you are a prime candidate for a Darwin award. Given the Dismal lack of
>knowledge you have demonstrated concerning physics and mechanics, especially
>differentials, I'm betting you still don't understand.
>
You're still SURE you can't add those resistances?? Think again. If I'm
wrong, and they can't be added, then your argument holds water.
Unfortunately they MUST be added.
Paste from previous post:
**************************
Imagine we have an electric motor driving a single shaft, with a wheel
at the end of it. The wheel is rigged up on a roller-bed in such a way
that, by applying more or less friction to the rollers, we can vary (and
measure) the amount of torque at the wheel.
Just inboard of the wheel, we mount a brake disc onto the shaft.
We also put suitable strain guages on the shaft, so that we can measure
the twisting force (torque) being experience by the shaft at any time.
One strain guage measures the torque on the shaft between the motor and
the disc, and the other, between the disc and the wheel.
The motor is powerful enough that it can overcome any kind of retarding
force we can put on it.
Motor shaft Disc shaft Wheel
[======]----------------------------I-----------------------------{}
First, we tighten up the rollers to the point where the strain guages
both read, say 1000Nm. (The brake has not been applied).
The we apply the brake progressively, and watch what the strain guages
read.
I predict that the inner guage will immediately start rising from
1000Nm, showing the increased torque resulting from the brakes. The
outer strain guage will remain at 1000Nm.
Or, if we did the experiment the other way around, and started with zero
resistance at the wheel, and 1000Nm at the brakes, we could observe what
the strain guages read as we progressively increase firction at the
rollers: At first, the inner guage would register 1000Nm, and the outer
would read zero. Then as we apply "ground" resistance at the rollers,
BOTH guages will increase at the same rate. (i.e. the inner gauge will
always be 1000Nm MORE than the outer guage.
Either way, when the brakes AND the rollers are EACH applying 1000Nm,
the outer guage will register 1000Nm, and the inner guage will register
2000Nm.
For the purposes of a differential, the *inner* shaft torque will be the
one that is "carried" over to the other side. The *outer* shaft torque
will be that which provides motive force.
Michael
Thanks for the scientific approach! It's exhausting replying to several
different people's views, especially when one gets side-tracked into
arguing lots of small but important points of physics, which can be
soooo difficult to prove, and just lead to swapping insults. ("No, I'M
not a moron, YOU'RE a moron!")
No major disagreements with your answers, nor with your explantion of
the braking phenomenon. My comments/questions in the text below.
Regards,
Michael.
On Wed, 11 Dec 2002 05:00:53 GMT, "Benjamin Lee"
<ben...@worldnet.att.net> wrote:
>
>"Michael" <no...@none.com> wrote in message
>news:3dfd0eb0...@news.cit-news.com...
>> Here are what I think are the right answers.
>> >*************************************
>> >
>> >A rear-wheel drive truck is attempting to accelerate from rest up a
>> >steep slope. The left-hand (rear) wheel is on some slushy ice, which
>> >will support a maximum torque at the wheel of 200Nm. The right-hand
>> >wheel is on pavement, which will support a maximum wheel torque of
>> >2000Nm. If the torque at the surface of either tyre exceeds these
>> >maximum figures, the wheel will start to spin.
>> >
>> >The vehicle requires 1500Nm of torque in order to start moving up the
>> >slope. Assume that the engine and drivetrain can deliver unlimited
>> >torque.
>> >
>> >Q1: If the vehicle has an open diff, will the vehicle be able to move?
>> >If not, will the left, right, or both wheels spin? What will be the
>> >torques on the left- and right-hand wheels respectively when the vehicle
>> >starts to move, or the wheel(s) starts to spin? (3)
>> >
>> A1: No vehicle movement. Left wheel will spin. Left=200Nm, right=200Nm
>
>Assume frictionless differential, yes. Real world, no. Left=200nm, right=
>maybe 230nm or so.
>
Fair comment, of course, though I did explicitly specify frictionless.
If we can agree on the simple model, then we won't get, umm, bogged down
by misunderstandings when things get more complex.
>> >
>> >
>> >--------------------------------------------------------------------
>> >Q2: As for Q1, but assume the vehicle's diff is fully locked (using an
>> >air-locker, for example). (3)
>> >
>> A2: Vehicle will move. Left=200Nm, Right=1300Nm
>> >
>
>Is indeterminate. Anywhere from L=200, R=1300 to L=0,R=1500. Both wheels can
>grab, don't know which will carry the load.
I don't think it matters in this case, but why would it be
indeterminate? There is essentially a solid spool, the system will
minimize the maximum torque (a "path of least resistance" principle). So
the higher torque side won't go to 1500, because it only needs to go to
1300. And that will leave the other one at 200 at the moment the vehicle
starts to move.
>
>
>> >--------------------------------------------------------------------
>> >Q3: If the vehicle has rear brakes that can be applied to the left- and
>> >right-hand sides individually, and the left-hand brake is applied
>> >progressively until the vehicle starts to move: a) What is the total
>> >torque on the left-hand shaft? b) What is the total torque on the
>> >right-hand shaft? (4)
>> >
>> A3: Left=1300Nm, Right=1300Nm
>
>Again, indeterminate. From L=1500, R=1500 to L=1300,R=1300
There is no need to go to 1500 per shaft. As soon as either shaft
reaches 1300, the other one will be 1300, and there will be sufficient
torque to start moving.
>
>> >
>> >>--------------------------------------------------------------------
>> >Q4: In Q3 above, why did the vehicle not start to move when the combined
>> >torques on the two shafts reached 1500Nm? (2)
>> >
>> A4: Of the 2600Nm total torque, 1100Nm is created by the brakes, and
>> therefore not adding motive force.
>
>Correct
>
>> >
>> >
>> >--------------------------------------------------------------------
>> >Q5: In Q3 above, at the instant the vehicle begins to move: a) What is
>> >the torque at surface of the left-hand tyre? b) And the right-hand one?
>> >(4)
>> >
>> A5: Left=200Nm, Right=1300Nm
No comment? :-)
That is true, I have assumed a frictionless diff. I assume, though, that
you would agree with my figures if the diff WERE frictionless?
>Try again to include a normal force at each sliding and
>rotational interface inside the differential gear, shaft, and bearing, and a
>coefficient of friction in each of those interface. You will complicate the
>math by 50 times, but you will get a different answer that reflect the real
>world.
>
>Remember, when two gears carry a load, because of the geometry of the
>involute gear, there is a force that spread the gears apart. Therefore, when
>the side gears in the differential tries to carry the 3200nm of torque in
>the diff, there is a tremendous amount of force in the diff that tries to
>explode the differential if it was not for the strong housing. The side
>gears tries to separate from each other. The back of the side gear then rubs
>on the diff housing, and create friction. This friction acts like a brake
>that stops the rotation of the side gears. If I put a brake liner between
>the side gear, and the differential housing, I have just created a crude
>limited slip diff. Only difference is that a limited silp have springs
>between the side gears to spread them apart initially. Even without a brake
>liner on the back side of the gear, and it is just metal to metal rub, there
>is still braking force that stop the spin of the side gears.
>
Your argument makes sense, and it's precisely what I was looking for - a
reasoned explanation of which part of the diff starts to locks.I can
see the side gears being forced back up against the housing... The
success of the method hinges entirely upon the degree of friction within
the diff, and that friction being related to the total torque.
>For the calculation, assume 10% of torque the side gears carries is
>friction.
>Now we have:
>13000 on left shaft. 200 carried by the left wheel, and 12800 by the left
>brake. 13000 is going into the side gears of the differential. 10% of 13000
>which is 1300Nm becomes friction between side gear on the differential
>housing, so the right shaft has 13000+1300=14300Nm. Take away 13000 absorbed
>by the right brake, and you are left with 1300 going to the right ground.
>The total input torque to the diff would be 27,300Nm. Of that 26000 is
>absorbed by the brake. Chances are something broke at this point already. I
>believe this case is what is called 10% lock in limited slip differential
>terms.
>
Ok, I follow that, and I agree with it. (Apart from minor arithmetic -
you need to deduct 12800 from the right, not 13000, since braking force
is 12800. That will mess up the rest of the figures, but I follow the
principle.
I will do a little research (and experimentation) to see what kind of
friction the side gears might generate. Is your 10% figure just an
easy-to-work-with example, or is that the kind of proportional loss one
might actually expect?
Thanks again for your well-reasoned (and quantified) answer.
Michael
wrote:
>Good one!
Ride his coat-tails, Mike.
When you told me that the diff "locks", and won't "disengage", all I
asked you for was the mechanism by which that lockup occurs. The
difference is that Ben could explain what parts of the diff are
affected, and what the nature of the torque bias is. There is no
"engaging" or "disengaging".
So Ben has provided a concise, and in fact the only *explanation* of why
the brakes might might cause a torque bias due to diff friction.
I'm afraid I stopped taking your contributions seriously when I realised
that you don't know what torque is.
Michael
<dshr...@nospam.com> wrote:
>
>"Michael" <no...@none.com> wrote in message
>news:3df5ca81...@news.cit-news.com...
>> If you start with unequal resistance (the grip torque is unequal), and
>> then add the same amount of braking resistance to each side, you end up
>> with precisely the same "gap" in resistance at each side. (My attempt
>> to quantify the problem with numbers is to highlight this).
>
>You are still wrong Michael, as I will attempt to explain once more.
<snip>
Douglas, see Ben's discussion below. Note how he adds the grip torque
and the brake torque?
I am pretty certain that you agree with his conclusion. Does it bother
you that you disagree with his reasoning?
The reasoning is the only important thing in this discussion, NOT the
conclusion, since I was asking WHY the phenomenon occurs, not IF it
occurs.
Ben's explanation of why it works, (uniquely thus far) doesn't
contradict nor alter my understanding of physics or how an open diff
works. (He did, however, draw to my attention a valid source of internal
friction that I had not previously thought of. His description of the
forces involved are eminently plausible).
The meat of the issue is this. Now that I have a rational explanation, I
can understand what factors are likely to make the technique effective,
or under what circumstances the technique is likely to work. I know what
new questions to ask.
For example, how *much* internal friction is there, as a percentage of
total torque (assming roughly proportional)? If it's around 10% (say),
then I know that I need to apply vast amounts of braking effort, to
provide a relatively small gain in useful torque.
I am going to trawl around for some figures. Or does anyone have any
idea of what the internal friction is?
Michael...
P.S. I have just done the practical experiment you suggested - see my
other most recent post. It would suggest that at best the effect is
small, at worst it's virtually zero. That would suggest that the
friction bias in a diff must be pretty small - I would have said less
than 10% (though not *strongly* convinced it's *that* small.)
Michael
<dshr...@nospam.com> wrote:
<SNIP>
>This is easily proven false by simply jacking up on side
>of your car, putting it in gear, giving it enough throttle to keep the wheel
>spinning and slowly applying the brakes until the wheel on the ground starts
>pulling. Your assumption has been proven wrong but you cling to it.
Just as a matter of interest, I never said that that experiment wouldn't
work. I was looking for an explanation of WHY it might work, that
doesn't contradict physics!
However, as I do, in fact, have some doubts about the effectiveness of
braking to increase torque, I tried your experiment this morning.
I was using a 1993 Land Rover Discovery (TDi, permanent 4x4, lockable
centre diff, open axle diffs, disc-brakes front and rear. Transmission
parking brake on the rear transfer box output shaft).
I parked the vehicle on a very slight uphill slope, jacked up one front
and one rear wheel. Chocks behind the other two to prevent it from
rolling back. No chocks in front of the wheels.
Then I locked the centre diff, put the vehicle in gear, and allowed the
two wheels to spin. Then I applied the (foot) brakes, slowly. Guess
what? It didn't come off the jacks.
Same in low range. In low range it was almost impossible to stall the
vehicle with the brakes (I did just manage, at idle), yet even so, it
didn't drive off the jacks.
The truck did move a bit, it tilted to one side, I am pretty sure due to
the torque on the prop-shafts transferring the rotational force to the
chassis. But it never felt like it was going to move forward when I
pressed the brakes steadily. If I *tramped* the brake at slightly higher
revs, there was a jerk, and the vehicle rocked, but I couldn't make out
if it was a fore-and-aft movement, or some other kind of movement.
Either way, I would expect *some* kind of shock when a heavy spinning
wheel is suddenly brought to a stop. And either way it didn't drive or
fall off the jacks.
So... either this force is small, or it is zero. (OK, not zero, becase
*any* internal friction in the diff will make the force non-zero).
Quantifying this isn't easy, but I estimate that in low range, first
gear, with the amount of throttle I was applying, the vehicle was
developing lots more than 5 times the minimum torque it would have
needed to push itself off the jacks. (Low range is about 2.5x lower
than High, and I was braking the engine gradually all the way down from
above max torque to below max torque).
Based on this, it seems that the bias due to internal friction must be
well under 20%, probably lower than 10%, since it didn't seem to be even
close to running off the jacks. It'd be interesting to do this on a test
rig, and measure the wheel- and shaft-torques with a strain guage.
(Just for kicks, I played around with the transmission brake too, with
predictably no effect. With the centre diff unlocked, and only one rear
wheel jacked up, applying the transmission brake to the rear output
shaft would obviously make the vehicle drive off the jack.)
If anyone else intends to try this experiment to confirm my result,
please keep these in mind:
1) If your vehicle has a torque-biased LSD, or a locker, or traction
control, it *will* drive off the jack!
2) Don't spin up the jacked-up wheel(s) too fast for too long. The diff
wasn't designed to have the side-gears turning at high speed, or
continuously.
3) Make a note of whether your vehicle has drums or discs on the axles
concerned. Mine has discs, so I was unable to tell if drum brakes would
work. (I would be *really* interested to find out, since, if so, it
would seem to confirm Jim's (AZGuy) theory about the braking bias on
drums making the technique more effective).
Michael...
Douglas A. Shrader
"Mike Romain" <rom...@sympatico.ca> wrote in message
news:3DF6DAD4...@sympatico.ca...
> Good one!
>
> LOL!
>
> My bet is he won't even believe that!
He won't believe anything he doesn't already "know" to be true. That most of
his "facts" defy the laws of physics doesn't faze him, because he "knows" he
is right. He has discarded everything that every credible, informed,
knowledgable poster on this group has told him and refuses to consider any
theory that isn't word for word what he has already said. He has now joined
my extremely small killfile list, in three years I have killfiled only 4
people, he is now the fifth. His moronic ramblings are just to illogical to
follow.
Benjamin Lee
>
> Thanks for the scientific approach! It's exhausting replying to several
> different people's views, especially when one gets side-tracked into
> arguing lots of small but important points of physics, which can be
> soooo difficult to prove, and just lead to swapping insults. ("No, I'M
> not a moron, YOU'RE a moron!")
>
> No major disagreements with your answers, nor with your explantion of
> the braking phenomenon. My comments/questions in the text below.
>
> Regards,
>
> Michael.
>
>
Your questions were tedious, but I figure if you actually spent the time to
figure it all out and write it down, you must be serious about knowing the
answer. Some people just have to spend all the time to know these thing. I
can understand that :)
> >> >
> >> >--------------------------------------------------------------------
> >> >Q7: If the two rear brakes cannot be operated independently, and
always
> >> >apply the same retarding force, and they are applied hard enough to
add
> >> >3000Nm of resistance to each side, what will the the torque be: a) On
> >> >the left- and right-hand shafts respectively? b) On the diff ring
gear?
> >> >c) At the respective points of contact between the ground and the
left-
> >> >and right-hand tyres. d) How much torque is being used to overcome
the
> >> >brakes? (4)
> >> >
> >> >
> >> A7: Left shaft=3200Nm, Right shaft=3200Nm, Ring gear=6400Nm, Left
> >> wheel=200Nm, Right wheel=200Nm, Brakes absorbing 6000Nm.
> >
> >This is where you are not getting it. You just described a frictionless
> >differential.
> That is true, I have assumed a frictionless diff. I assume, though, that
> you would agree with my figures if the diff WERE frictionless?
>
If diff were frictionless, yes.
...
>
> >For the calculation, assume 10% of torque the side gears carries is
> >friction.
> >Now we have:
> >13000 on left shaft. 200 carried by the left wheel, and 12800 by the left
> >brake. 13000 is going into the side gears of the differential. 10% of
13000
> >which is 1300Nm becomes friction between side gear on the differential
> >housing, so the right shaft has 13000+1300=14300Nm. Take away 13000
absorbed
> >by the right brake, and you are left with 1300 going to the right ground.
> >The total input torque to the diff would be 27,300Nm. Of that 26000 is
> >absorbed by the brake. Chances are something broke at this point already.
I
> >believe this case is what is called 10% lock in limited slip differential
> >terms.
> >
> Ok, I follow that, and I agree with it. (Apart from minor arithmetic -
> you need to deduct 12800 from the right, not 13000, since braking force
> is 12800. That will mess up the rest of the figures, but I follow the
> principle.
That is correct, should be 12800.
>
> I will do a little research (and experimentation) to see what kind of
> friction the side gears might generate. Is your 10% figure just an
> easy-to-work-with example, or is that the kind of proportional loss one
> might actually expect?
If you know the answer, let me know.
WJM's page said it could be up to 1.5:1 bias ratio, or 50%. That seems a
little hight. In real life at least 10% should be expected. In my
experiance, the seat of the pants guess is that 20% is a fair number. Also,
static friction is greater than running friction. So when starting from a
stop, you get better traction when you pull the hand brake. But once one
wheel start to spin, you don't get as much traction gain. It is because once
one wheel spins, it becomes running friction in the carrier, which is lower
than static friction. Is a careful balance between gas, hand brake and
clutch when starting. It takes some time to master it. I have not tried
using foot brake because it is too powerful for a 4 cyl to overcome, and you
don't get as much feel doing a toe-heal on the gas and brake pedal. Plus,
birdfield joint in the front of Toyota is weak, so is better not to put too
much torque on it. Easiest use of hand brake is when truck is still moving,
and you feel you are loosing traction. Of course, once you pull the hand
brake, you slow the truck down, which is counter productive to the forward
momentum that you need to overcome the obstacle. Usually, if the truck is
moving, I don't use the handbrake. I just gas the truck to get as much
momentum as possible. The handbrake trick worked for me once while stuck.
The truck was in mud. I did the rock, then on the last rock, pulled the
brake, and floored the gas. The truck went out of the ditch, but got stuck
in another on 20 feet ahead. Had to be pulled out anyway :( but it proved
it worked.
Ben