what did you guys do to this helpful thread it should be in the llama lounge almost...

no wider than 11.50 MAX ON IFS!!!

you dont need to lift!
lock and armor .

 

The point is that you cannot apply a coefficient of friction to a tire in the dirt, because dirt is not a solid material. The classical euqation Ff = mu * Fn is not applicable to any material that is not both solid and resistant to shear deformation. This means both dirt and tires are out.

That is 100% incorrect, and you have misinterpreted the data I presented. A linear coefficient of friction can only apply to a solid block of material that does not deform. So for example, two pieces of aluminum sliding against each other. It's useless to try to apply such a simplistic approximation to a tire's traction in dirt, because the dirt MOVES. If there really was a constant COF for this interaction, 2" wide tires would perform the same as 20" wide tires in acceleration; it's obvious this isn't the case.

http://buildafastercar.com/node/10

The first graph you are referring to isn't measuring just lateral force, it is showing available lateral grip from a tire that is having an increased amount of normal force put on it. If the coeffcient of friction were constant, the line would be straight and its slope would be the COF.

Instead the graph is a curve. What we see is that the more normal force put on the tire, the less the coefficient of friction is (the second graph visually shows this as COF vs. normal force). Graph 2 shows that a 475% increase in normal load on the tire resulted in a 46% reduction in the coefficient of friction. If the COF were constant, this graph would be a perfect horizontal line.

The point is that if a coefficient of friction could be used to describe the interaction of the tire and the surface it was on, surface area should have no effect! This is of course not the case.

 

you guys are bunch of nerds. i got headaches just by look at these pages, not actually from reading them.

ok, dude was asking why one wants a lift, just tell the truth, a lifted truck is a simple statement: "I HAVE A BIGGER PENIS", or Vagina in rare cases .

i'd suggest the Mod's close this thread forever and the OP go ahead have your truck lifted and be happy with your new improved penis.

 

So much for free speach and free flow of ideas. and your kinda valgar. I'm going to recommend you be banned.... Not my style. If you don't like the thread stop veiwing it. You obviously don't offroad in the same places others do or use your vehicle the sameway. So relax, breath, and listen to the Beetles or something.

 

Haha Tony! Do big and extra aux. lights give me a bigger sack? Hehe

Have you guys thawed out yet back there?

 

then why has 3 LF's hanging up front, huh.... if i didn't spend $1500 on fixing the gdmn blown tranny i'd be buying more.


you took it way too seriously, btw, this is what i'm listening


http://www.youtube.com/watch?v=R7yfISlGLNU

ok, let me be serious. you can see from my profile, i used to have a 94 which i just sold 9 months ago, very similar to yours 95. it's a 15 years old vehicle, it's old, specially the suspensions are pretty much done. if you wanna do some off roading, or even carry some weight on the pavement, the ride won't be good. so, gotta put some new suspension stuff in. you can buy all Toyota parts from dealers, or with the same or less money, get some quality after market shocks, coils, even t-bars, bj spacers. you are doing it any way, why not. yours many be have 120k or 200k miles on the clock, gotta get some tires, well, you can buy some factory spec tires, or just a little bit bigger. why not.

lift or no lift, offroading or mall crawling, it's not a matter of WHY, it's why NOT.

on a side note, i don't care how hi the ground clearance, how fast it can run, how high it can climb, you won't pay me to ride a dune buggy again, i just hate it.

just my opinion, you're at it any way, it's your money your truck, do where ever your happy with, and post some pictures for us.

 

In mathematics, a coefficient is a constant multiplicative factor of a certain object. For example, in the expression 9x2, the coefficient of x2 is 9. It is a constant value that does not vary. The only thing that will change the coefficient of friction is a change in the two interacting surfaces.


Apparently you can. Now consider a bulldozer with tracks. The weight of a dozer gives it a large normal force on any surface. On soft surfaces, such as dirt, the spines on the tracks go down into the surface. This makes the coefficient of friction between the ground and the dozer much greater than one. This larger coefficient of friction allows the dozer to push objects that are multiple times its own weight. The deformation increases the coefficient of friction.

In mathematics, we say that quantities are proportional if there exists some number, a multiplicative coefficient, which will produce equality in the relationship. This means that there exists a constant of proportionality that makes the force of friction equal to the product of this constant and F N. This constant is called the coefficient of friction. It is constant.

The amount of area in contact between two discrete surfaces does not affect the force of friction; whether the tire has treads or not is irrelevant. Some tires are completely smooth; tires like this are called 'slicks'. An increase in friction is not provided by the increase in surface area contacting the road; treads are just not used because they are not needed on a dry road. But friction can be increased in a different way. If the tires are spun very fast, the rubber will melt; as it hardens, it becomes rougher, with a higher coefficient of friction. This increases the friction force, and hence the traction. (Continued use wears them smooth again).

Tires also have treads so you can have traction in snow and mud. This works by allowing the snow or mud to get between the treads. You could think of this as increasing the coefficient of friction of the tire as a whole, because it is no longer as smooth. However, what is really happening is that the tires are now grabbing and pushing snow or mud backwards; this requires more power to be transferred to the tires from the engine. A greater force is applied, causing an greater 'equal-but-opposite' reaction force, which is what pushes the vehicle ahead. Better traction comes from more force being applied to the ground, not an increase in friction.

Back to the graph of normal force vs grip. It shows lateral grip due to a lateral force. The reason that a larger vehicle decreases grip while cornering is because of the horozontal force vector perpendicular to the normal force due to the movement of the vehicle. Angular acceleration. F=MA with angular acceleration is larger than Ff. COF does not change here. More normal increases the grip but the amount of increase decreases as the weight of the vehicle gets larger. I dont disagree that what the graph shows is correct but it is a dynamic measurement as a result of cornering. Draw a free body diagram and run through some simple calculations. take a look here http://www.physicsclassroom.com/Class/circles/U6L2a.cfm
Of course in this problem there is no slip. Say we knew ahead of time that the friction force was 3600 newtons. If after calculating we got a force of 4000 newtons then the car would be slipping. This would be the affect of too heavy a car. This is why lighter cars are better for high speed cornering as long as the weight allows enough grip in general.

 

Do you also have a degree in "sctuff?"

 

WOW most of this stuff is way over my head. I'mm liking it. My mind doesn't do math/algebra well (more of the biology type)just barely made it through algebra and physics but allways found it fascinating, just over my head. So now I have a couple more questions... how should I set my torsion bars when I install them? And how do the front and rear swaybars affect this...will disconnecting effect droop only? I don't know quite how they work.

 

muddpigg, pm me about those tires you offered up.

 

I completely removed my swaybar. You can do a quick remove, check Roger Brown's "(4crawler on here) as he has some awesome 411 on it though for a straight axle. Your guess is as good as mine if it'll translate to IFS. The swaybar helps with body roll but limits independant wheel movement for offroading. Use caution when cranking torsion bars as it increases the regular operating angle of the cv axles thus can speed wear. I simply measured mine at the same place on the fender to the rim. Bounced on bumper while adjusting and measuring. Once I got it set to where I thought I wanted it I drove it around to settle the suspension and remeasured.

Have you taken you truck to Tahuya? Take it out before you add these things so you have a baseline to evaluate changes from. http://nwjeepn.com/Tahuya.html should be close to you. Be careful wheelin is addictive.

 

<sigh>

Sure, all that is well and good, but it is useless in describing non-linear friction. I don't think I can explain it any more cleraly than I already did, you have a graph right in front of your face that shows a tire's coefficient of friction is not constant. Sorry.

Also not applicable to a linear relationship. The dozer's traction is not dependent on a linear coefficient of friction, because dirt cannot and does not react linearly. "Digging into" the dirt means you're depending on the dirt's resistance to shear force, not friction.

...and yet, there are many relationships in nature where friction cannot be described as a linear relationship with a constant, but instead it must be described as function. Tires are one of those cases, and there is empirical data all around us that proves it.

Then why do tires even come in different widths? If the surface area of the tire has no effect, all tires should be able to be the same width and perform exactly the same. It's obvious this isn't the case, why do you think drag cars have tires that are 24" wide slicks? To maximize surface contact with the track!

The problem is that we are not dealing with discrete surfaces, and rubber is a tricky material to describe it terms of how it will stick in a given situation.

That makes no logical sense.

A slick has excellent traction on dry racing surfaces because it A) has a softer rubber compound than road-tires that provides increased adhesion to the track, and B) it maximizes surface area in contact with the ground by getting rid of the tread.

The rubber comound in slicks is also specially chosen because as it gets hotter it gets stickier! Drag racers do a burnout with slicks before the race because the hotter rubber sticks even better to the track. Your whole "roughness" argument is complete nonsense gibberish.

It says right on the graph that it is Lateral Grip vs Vertical Force. It says it right on the graph! If the friction between the tire and the road was constant, the graph would be a straight line. But it's not, so the friction isn't constant. The second graph, which is the derivative of the first graph, quantifies that.

I don't know what else I can say, the evidence is right in front of you.

 

PM sent with follow up email w/pics.

 

The easy question first - the swaybar:
The swaybar acts as an extra torsion spring across the truck, so as one side tries to compress, it has to act not only against the spring rate of the "normal" suspension on that side, but also against the spring rate of the sway bar. In reality, this means it takes more force to make the suspension work independently side-to-side. It rides much rougher and travel CAN be sacrificed offroad because of the higher spring rate.

As for removing it ... I hardly noticed when I broke the front swaybar. Now, when I broke the rear swaybar (and had neither hooked up), THAT was noticeable. I highly encourage people who use their trucks offroad to at least experiment with no front swaybar.

The next simplest question regards droop travel and spring rates:
Springs compress a certain distance by the weight on them - for example 300 lb/in. So, let's use the rear suspension of an empty 1st gen 4Runner (this is actual data from AxleIke's new suspension) which weighs about 1200 lbs. So that's 600 lbs per side, so the weight of the truck compresses said spring 2". So, let's say you have no compression travel to make it easier for now. When the spring extends 2", you are applying no force to that wheel, so there is no friction. As we allow compression travel, the axle can apply leverage so there will be some traction on the drooped wheel. The idea of 4" up travel and 8" downtravel is folly because once you pass 6" down, the wheel is not applying any force on the ground and therefore generating no traction.

Lastly, the idea of front/rear travel "balance":
The front and rear should work similarly. IMHO, THIS is the true benefit to a SAS. Let's say you have up/down 3" in the front, for a total travel of 6", and let's go crazy here, up/down 10" for a total of 20" in the rear. To eliminate a variable, let's say the spring rates are the same front/rear. You are on the trail and come upon a rock on the driver's side that is 12" tall. So, the front suspension compresses it's 3" and hits the bumpstop, then the driver front of the truck comes up. The back still has PLENTY of compression travel, so it allows the truck to lean over and get more and more tippy until the compression of the spring generates enough force to offset the weight of the leaning - and that generally takes quite a bit.

NOW, imagine that the truck has the same front suspension, but only 3" up/down in the rear to exactly match the front. Same situation, the truck essentially stops leaning and picks up the passsenger front tire when you hit the rear compression bumpstop, resulting in a MUCH more stable situation than the previous guy and his ton of rear travel.

 

Okay, first off, completely agree on the experimenting with the front swaybar. With the droop I agree somewhat and that is why my bj spacers have not had the tbars crank the lift out of them, I can pull up to a rock and have X amount of inches before the LCA hits the bumpstop and starts lifting the front of the truck.

As for the spring rates at droop and everything, I kind of agree, but it isn't as simple as that. YOu also have the shock force pushing down as well as a tire in positive contact with another surface (I'll leave the tire talk to you guys ). I'll use a ramp as an example (I know vert ramps aren't necessarily the tell all for suspensions but it makes this more simple). Open/open vehicle, driverside front wheel on the ramp, obviously with IFS, we are going to start drooping that driver's rear pretty quick as we move up the ramp. If it was as simple as the spring rate, an open/open vehicle is going to start spinning that driverside tire pretty quick even though it is still on the ground because it has no traction. Alas, it doesn't start to spin till right before it lifts off the ground or lifts off the ground (depending on the type of surface it is on). I have noticed this due to me having to wheel so much with open diffs. So while it does not have nearly the force applied to that wheel due to the spring rates that you talked about, it can still have enough traction, even though due to spring rates it shouldn't have much downforce on it. Does that make sense? I'm having a hard time figuring out how to write it exactly how I'm picturing it.

I kind of get that, especially when you are talking front driver's and rear passenger are both moving up on something. Now think about what you said with just the front drivers climbing up something while the others are on flat ground. When that rear hits the compression stop on the passenger side, it does not lift the passenger front like you said. Instead now that rear bumpstop has created a "fulcrum" of sorts that connects in a line to that front drivers wheel. If what you said was true, then you could drive right up a ramp with your driverside tire because you have magically lifted your front passenger tire once you hit your rear bumpstop. The only way you are going to lift that front passenger tire is if that rear reaches its bumpstop and it is also traveling up something.

 

Yep - kinda like this
http://www.youtube.com/watch?v=f2cZXAie4lU

(NOTE about the end - he was WAY up that ramp, you can see as it goes over that the lean angle was outside the tire - even if it had all the front flex in the world, and had that front tire down, it would go over)

 

Okay, now show me one of ours going up a vert ramp with the other tire up in the air.

That was pretty funny how it dumped after he got out. Actually, I missed it the first time, I didn't see that he lifted up on that back end. Haha Glad that kid on the passenger side was fine.

 

You just dont get it. Its not my argument. Pulled most of that stuff of of other reputable websites. Let me just ask this question. Have you ever taken a physics course in your life. That graph refers to traction not COF though COF is the term used it is incorrect. The reason TRACTION is less is because of the angular acceleration times the weight of the vehicle gives a larger F in the F=ma equation than the F in the F=N(cof). See the difference there. The graph shows correct information but you are misunderstanding the reason for the results. The reason drag tires are larger is because they have adhesive properties. This is the only case where surface area matters, but fundamentally it does not. So yes in a way contact patch on a tire is a factor but not more than the weight of the vehicle. Also it allows a larger amount of force from the tire to be transfered to the ground without exceeding the Force of friction. The kicker is in that you are applying a force (torque) not just simply sliding. the muK between the two does not change( I should say once the tires have reached operation temp say after a burn out). Yes the temperature of a tire will change the muK I am not debating that. I am debating that normal force does not change muk but will change the force of friction. Non linear friction only really applies in damped oscillation in which the higher the frequency the more friction is present even with the same fluid.

I suggest you check this study out
http://www.itk.ntnu.no/ansatte/Grip_...ip_CDC2006.pdf

 

That paper is a very interesting study into vehicle dynamics; thanks for posting it. I especially like section II.B which says:

And at the end of the paper they say this:

Quote:

Experimental results support this and show that the lateral velocity estimate remains accurate even when the initial estimate of the friction parameter is completely wrong.

Basically, the control system is able to compensate for poor guesses at the coefficient of friction as long as the real coefficient of friction is always less than their upper bound maximum value. Luckily, they don't have to go into extreme detail trying to calculate an accurate frictional coefficient, because the system doesn't require it.

Look, I've taken lots of physics and dynamics. The fact is that the coefficient of friction is not constant between tires and the road; even your posted paper admits this is the case. The purpose of the paper you posted is not to accurately determine friction on the road anyway, they are estimating lateral velocities and admit they don't need an accurate coefficient of friction to do it.

This isn't the topic of this particular thread, and I think mods ought to cut all of these posts into a new thread altogether.

 

The reason the cof is not uniform is due to varying road conditions. Like mud Ice etc. But they assume then that the cof is then what it would be if the contact were ideal. Again it has nothing to do with the normal force varying the coefficient of friction which is my point again. Normal force wont affect COF(assuming no deformation, however tires do deform so that makes things interesting but one can still measure the cof between tires and road experimentally). It will however affect the force of friction and traction or lateral grip due to the affect of the angular acceleration on the weight of the car being greater than the force of friction which is what that graph is referring to. I have enjoyed very much this friendly argument of sorts. Cheers for making things interesting.