Tomorrow is the big day ;o)
12-29-06, 08:38 PM
#51
From your explanation is sounds like you are assuming that the coefficient of friction is constant, no matter how much weight is applied. The coefficient of friction is not contant and changes slightly due to how much force is being applied. You will get more friction with a larger tire, because thee coefficient of friction is greater with less weight per unit rubber.
Kevin
Kevin
Originally Posted by wanklin
It's pretty intuitive, as long as you have a basic understanding of physics. The 315s have an additional 1.3" of contact patch over 285s on average. By spreading the weight of the car (mass x acceleration due to gravity), and therefore the normal force (can be called pressure x area), accross a wider contact patch you effectively reduce the force of friction (normal force x coefficient of friction) per unit of rubber.
The net friction force is the same for all tire, regardless of tread width, but wear is more rapid on a narrow tire. The main advantage of big meats is treadlife and the ability to use softer compounds. With that being said, I stand behind my decision for the same reason that Jim put 19" blingers on his FD ;o) Because I can....
If you're wondering why dragsters use such wide tires... It's because the swell-to-deswell change in tire diameter acts as a final gear reduction. There's also the issue of a different type of friction on drag strips which results from the stickiness of the track tires and road surface which creates frictional force which is to a large extent independent of Normal Force.
The net friction force is the same for all tire, regardless of tread width, but wear is more rapid on a narrow tire. The main advantage of big meats is treadlife and the ability to use softer compounds. With that being said, I stand behind my decision for the same reason that Jim put 19" blingers on his FD ;o) Because I can....
If you're wondering why dragsters use such wide tires... It's because the swell-to-deswell change in tire diameter acts as a final gear reduction. There's also the issue of a different type of friction on drag strips which results from the stickiness of the track tires and road surface which creates frictional force which is to a large extent independent of Normal Force.
12-29-06, 09:01 PM
#52
Originally Posted by wanklin
It's pretty intuitive, as long as you have a basic understanding of physics. The 315s have an additional 1.3" of contact patch...
... independent of Normal Force.
... independent of Normal Force.
Originally Posted by CarbonR1
From your explanation is sounds like you are assuming that the coefficient of friction is constant, no matter how much weight is applied. The coefficient of friction is not contant and changes slightly due to how much force is being applied. You will get more friction with a larger tire, because thee coefficient of friction is greater with less weight per unit rubber.
Kevin
Kevin
Someone has a mathematical model out there... But they probably work for a race team and won't share it. I should start another thread for this... Maybe this could be another 'Ask Howard C' thread?
Originally Posted by Fritz Flynn
You of all people don't need any help with this concept
...
...On the street I don't like running more than 255s because you have to go too fast to have any fun.
...
...On the street I don't like running more than 255s because you have to go too fast to have any fun.
12-29-06, 09:15 PM
#53
The whole idea behind it is that the same as why given everything equal, lighter cars will handle better in turns. The lighter car has more relative friction compared to a heavier car, because coefficient of friction has diminishing returns. It decreases as you increase weight and it gets to a point where it levels off, so you could add weight, but not get much more friction out of it.
Widening the tires has a similar effect of decreasing weight of the car, as far as friction is concerned, because you are putting less weight per unit area of rubber, so your coefficient of friction is maximized, giving you the most grip per unit weight.
But the thing is, that decreasing weight per unit rubber on a tire has diminishing returns as well. You can reach a point where you are already maximizing the coefficient of friction with your weight, so reducing the weight per area more wouldn't give you more grip.
Unfortunately I don't know the equation for it, but it would definitely have to do with:
1. tire material/road surface
2. weight acting over the tire
3. type of tire, (how evenly the weight was distributed over the area)
4. contact patch
5. what side of the bed you woke up on, etc, etc
Kevin
Widening the tires has a similar effect of decreasing weight of the car, as far as friction is concerned, because you are putting less weight per unit area of rubber, so your coefficient of friction is maximized, giving you the most grip per unit weight.
But the thing is, that decreasing weight per unit rubber on a tire has diminishing returns as well. You can reach a point where you are already maximizing the coefficient of friction with your weight, so reducing the weight per area more wouldn't give you more grip.
Unfortunately I don't know the equation for it, but it would definitely have to do with:
1. tire material/road surface
2. weight acting over the tire
3. type of tire, (how evenly the weight was distributed over the area)
4. contact patch
5. what side of the bed you woke up on, etc, etc
Kevin
12-29-06, 10:53 PM
#54
Originally Posted by CarbonR1
From your explanation is sounds like you are assuming that the coefficient of friction is constant, no matter how much weight is applied. The coefficient of friction is not contant and changes slightly due to how much force is being applied. You will get more friction with a larger tire, because thee coefficient of friction is greater with less weight per unit rubber.
Kevin
Kevin
3.2.2 Friction of polymers.
When either or both of the surfaces are polymers, Equation 3.1 no longer applies. Polymers deform viscoelastically: the deformation depends not only on the normal load N but also on the geometry and time of loading. With fixed geometry and duration of loading, the area of true contact is proportional to
where 
. For a truly elastic solid (for example rubber), 
[Lincoln, 1952]. From Figure 3.2 for
(typical for a hard sphere on a Plexiglas surface), a 16-fold increase in load from 1 to 16 pounds produces an 8-fold increase in frictional force, hence the coefficient of friction is halved. This is a general feature of polymers; the effective coefficient of friction reduces at higher loads. 
- Figure 3.2: Friction of a hard sphere sliding over a clean flat surface of Plexiglas. As the contact force is increased the ratio of frictional force to contact force is reduced, this is a general feature of polymeric materials. [Bowden and Tabor, 1974]
Informational reading from an engineering forum:
Is the force required to slide apart two surfaces related only to the reaction force and coefficient of friction or does the amount of contact surface area play a part perhaps by 'ploughing' effects or by elastic deformation of the surfaces? corus
I was always taught that the frictional force was proportional to the normal load, the coefficient of friction and was independant of surface area. desertfox (Mechanical) 6 Aug 06 7:21
Hi corus
Assuming were talking dry friction here then yes the force required for one surface to slide over the other surface is
related only by the coefficient of friction and the reaction force.However according to "Applied Mechanics by
Hannah & Hillier this is only true for average loads because the area of true contact is proportional to the load applied and is almost independent of the apparent area of contact. Sadly they don't define what they class as an average load, however in all my experience of design, friction as always been treated as defined earlier that it is independent of apparent contact area.
desertfox GregLocock (Automotive) 6 Aug 06 8:12
In general, for engineering type materials, if you increase the surface area you will increase the apparent coefficient of /static/ friction.
On the other hand I can think of one obvious case where reducing the surface area reduces the apparent coefficient of /dynamic/ friction.
The problem is that there are several different mechanisms at play, trying to capture them all in one single number is ridiculous.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
MintJulep (Mechanical) 6 Aug 06 10:39
Normal force and contact area define an average contact pressure. However in real world materials, the contact pressure is almost always nonuniform.
If you break the contact surface into infinitesimal elements you can cause mu to be just about anything you like depending upon the assumptions you use in your calculation.
This is why you should never rely on handbook values for mu when designing things that rely on friction. You need to test your application. corus (Mechanical) 6 Aug 06 11:45
In this particular case the area in contact is not the same as the 'average' area as perceived, and within that contact area the contact pressure varies considerably. Primarily it's the effect on the static coefficient I'd be interested in, but the effect on the dynamic value would be of interest too. The materials involved are ceramics with a relatively smooth surface.
At the macro level of the surfaces does localised yielding of peaks and troughs in the surface cause a non-linear effect to the value of the apparent coefficient of friction as the load increases?
Greg doesn't give any examples or references but I presume that the obvious case of reducing the apparent coefficient of dynamic friction with area is in the case of ice skating, ie. through melting at the ice surface, perhaps an example of macro changes at the surface referred to above.
Any references would be useful. corus
GregLocock (Automotive) 6 Aug 06 20:13
How about a flat knife in butter? flat side to the butter is very draggy, compared with sharp side down.
The references I have are concerned with tires and road surfaces, probably not very helpful, here's the best: John Dixon : Tires, Suspension, and Handling Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
BigInch (Petroleum) 7 Aug 06 9:43
Greglocock,
The butter effect is viscosity related, which is not to say it doesn't involve friction, but it is a friction of a different... "flavor" shall we say? Going the Big Inch!
http://virtualpipeline.spaces.msn.com
GregLocock (Automotive) 7 Aug 06 19:47
Sure, and that illustrates my point. There are at least 4 mechanisms I can think of that generate "friction", hence the difficulty with assigning a one size fits all factor to them.
The worst case I remember was polyurethane tubing against itself. Depending on the lubrication ( a mixture of seawater and kerosene in arbitrary proportions) and other grot that was hanging around on the deck, I got a factor of about 3 in stiction, maybe 1.5 in sliding contact. All figures from memory. Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
vonlueke (Structural) 8 Aug 06 1:59 corus: For dry, unlubricated surfaces, the following applies. For low surface pressure, the coefficient of friction (COF) is approximately independent of contact area, over wide limits, regardless of whether the contact pressure is uniform or nonuniform. There is always elastic and plastic deformation of asperities (peaks), at least to some extent; but at low surface pressure, friction is mainly a shearing of the surface film. For very high surface pressure, the surface oxide apparently becomes ineffective, or is pushed aside, and the parent materials begin to cold weld (gall), such that the COF quickly increases. If this continues, the surfaces seize.
Surface roughness surprisingly has relatively little effect on COF (provided the surfaces are not so extremely rough that you enter the realm of what could be termed visible mechanical interlocking, instead of friction). COF is highly sensitive to what the parent materials are made of (intermolecular electromagnetic attraction), and highly sensitive to surface films and oxides. 25362 (Chemical) 8 Aug 06 3:39
Engineering Tribology by Stachowiak and Batchelor, Chapter 16: Wear and Friction of Ceramics. Tribology Series, 24. Elsevier. FACS (Mechanical) 8 Aug 06 11:56
"Is the force required to slide apart two surfaces related only to the reaction force and coefficient of friction or does the amount of contact surface area play a part perhaps by 'ploughing' effects or by elastic deformation of the surfaces?
corus"
Everyone seems to be using "Pressure" and "force" as if they are the same thing.
Under constant load, the friction does not increase or decrease as the area increases or decrease.
The frictional coefficient will always remain the same.
The required force to move the load will only change if the load on the surface area changes.
Deforming the surface, as a result of the load/friction damage, will change the coefficient of friction and thus change the required force to move. Charlie
www.facsco.com
electricpete (Electrical) 8 Aug 06 13:13
FACS - scroll down to the bottom of page 1 for the Stribeck curve:
http://www.me.utexas.edu/~bryant/courses/me383s/DownloadFiles/LectureNotes/BoundaryLubrication.pdf
Wouldn't you agree this represents friction coefficient changing as a function of pressure (among other things). =====================================
Eng-tips forums: The best place on the web for engineering discussions.
corus (Mechanical) 9 Aug 06 3:44
Just to add to previous comments, this particular application involves ceramic type materials at very high temperatures so the comments regarding lubrication/surface films/oxides/welding in relation to pressure are very relevant with regard to chemical releases at the surfaces. I'll refer to the book recommended by 25362 to see if there is more information there.
corus
EnglishMuffin (Mechanical) 10 Aug 06 14:54 For rigid materials, friction force is often said to be largely independent of the apparent contact surface area. Friction force is mainly dependent on the actual contact area, which is usually much less, and is heavily influenced by the normal force, having little relationship with the apparent area of contact. But for other materials such as elastomers, for example, this is not so true, perhaps because the actual and apparent contact areas are becoming similar (they also change greatly with time and temperature). Everyone knows that large tires have more grip than small ones. And Teflon behaves strangely too - I recently made some special hydraulic cylinders with all Teflon-to-metal contact and sealing surfaces, and the friction force was considerable, but totally independent of hydraulic pressure up to 2000 psi. blfarrar (Automotive) 16 Aug 06 14:41
The coefficiaent is independant of surface area, the actual efficiency of device is dependant on the surface area - hence bigger brakes on faster cars. Bruce L Farrar.
Works Engineering Manager
Marshalls Mono PLC.Brookfoot Works.
Halifax W.Yorks UK
corus (Mechanical) 17 Aug 06 3:50
I would describe efficiency as the ratio of the force required to overcome friction in relation to the applied force, ie. the coefficient of friction. Whether this is due to other physical phenomena occuring is unsure, but it does seem that area does play a part whether it's actual or apparent surfaces in contact.
In this application I've found that a change in temperature has a significant affect on friction. It's well known that two surfaces in frction will generate heat. Can an increase in contact area between the two surfaces (or vice versa), for the same applied reaction force, affect the temperature increase between two sliding surfaces? corus
EnglishMuffin (Mechanical) 16 Sep 06 11:29 I did not notice the two posts after my last one until now. Even today, friction is a controversial subject, but the following quotes are taken from a relatively old book I possess entitled : "The Friction and Lubrication of Elastomers". I think they basically support the comments I have made in posts above.
Beginning of quote :
The classic laws of friction as they evolved from the early work of da Vinci, Amonton and Coulomb may be summarized as follows :
1. Friction force is proportional to load
2. Coefficient of friction is independent of apparent contact area
3. Static coefficient is greater than the kinetic coefficient
4. Coefficient of friction is independent of sliding speed
...... the classic laws have survived the years ....... most of them have been found to be incorrect.
The first law is correct except at high pressure when the actual contact area approaches the apparent area in magnitude. (emphasis mine)
The second law appears to be valid only for materials possessing a definite yield point such as metals and it does not apply to elastic and viscoelastic materials (such as rubber). (emphasis not mine)
The third law is not obeyed by any viscoelastic material - indeed, a controversy exists today as to whether viscoelastic materials possess any coefficient of friction at all. (emphasis not mine)
The fourth law is not valid for any material, and it is now well establihed that elastomer friction exhibits distinct viscoelastic properties. (emphasis not mine)
End of quote.
Personally, I think the concept of a "coefficient of friction" is a somewhat misleading and outdated one in many cases, if applied without broader knowlege. But like many other historical yet admittedly insightful concepts, it will probably never die. zifos (Mechanical) 20 Sep 06 17:15
Probably enough technical info so I'm going to not be technical.
I remember when I first heard that firction was independant of contact area, I couldn't believe it. I grew up around cars and I know that bigger tires = more traction.
Then I came to the realization (or I made it up to help justify) that "traction" is different then "friction." Friction, in my definition, is based on two "smooth" surfaces. Tires have tread lugs and asphalt has relatively large (macroscopic) bumps that dig into the rubber and catch on the edges of the lugs, thus a worn tire (no/rounded lugs) has less traction. Tires rely on surfaces interacting with perpendicular components (not unlike pushing against a wall). Hold your hands flat and rub them together, then interlock your fingers and try to rub them together.
Probably not a very good technical definition and the line between friction and traction may be somewhat ethereal but it helps me feel better about friction.
Also I believe brakes are larger on some cars to help deal with cooling and wear. For heavier cars with more energy to dissipate, it heats the brakes less if you spread it out over larger areas. For a given friction force the normal force is the same but if you spread it out over a larger surface area you have a smaller pressure which means less heat and wear. Breaks are a good example of friction because contact surfaces are quite smooth. i.e. machined steel against compressed asbestos with binders and what not. EdDanzer (Mechanical) 1 Oct 06 23:47
MY CONCLUSION:
Larger tires provide more grip, not because of "friction force" per se but because they provide more traction. With that being said, if two tires are of the same rubber compound and one is wider than the other, the wider tire will have a higher coefficient of friction further improving it's ability to grab the road surface. Friction can be seen as a micro-traction. Granted, the importance contact patch surface area goes out the window around the apex of a sharp corner but the surface area of the contact patch becomes more important upon exiting the corner as lateral forces subside and allow traction along the full tread width.
12-30-06, 02:13 AM
#58
tard of teh century
I think part of the advantage of the wider tire is the same as why you want to keep an ideal camber all the time. With a narrow tire the contact patch will be more heavily loaded on the edge and as the tire is a circle, it has to deform a good bit to keep the same contact patch.
/`| narrow tire
|
\_|
/```| wide tire
\___|
The wider tire keeps a more spread out pressure. Tires make the same amount of traction for every pound of weight added up until their peak, after that they still make more traction but at less and less per pound. 3000 lbs car has more traction but less per pound than a 1000 lbs car. Same as we all know with horsepower, it can be better at times to have less HP but also have a lot less weight. With all the weight loaded up so close to the edge on a narrow tire it goes over the peak and wastes what the wider tire doesn't because it was still below the peak.
/`| narrow tire
|
\_|
/```| wide tire
\___|
The wider tire keeps a more spread out pressure. Tires make the same amount of traction for every pound of weight added up until their peak, after that they still make more traction but at less and less per pound. 3000 lbs car has more traction but less per pound than a 1000 lbs car. Same as we all know with horsepower, it can be better at times to have less HP but also have a lot less weight. With all the weight loaded up so close to the edge on a narrow tire it goes over the peak and wastes what the wider tire doesn't because it was still below the peak.
Last edited by madbouncy; 12-30-06 at 02:19 AM.
12-30-06, 08:33 AM
#59
Originally Posted by madbouncy
I think part of the advantage of the wider tire is the same as why you want to keep an ideal camber all the time. With a narrow tire the contact patch will be more heavily loaded on the edge and as the tire is a circle, it has to deform a good bit to keep the same contact patch.
/`| narrow tire
|
\_|
/```| wide tire
\___|
The wider tire keeps a more spread out pressure. Tires make the same amount of traction for every pound of weight added up until their peak, after that they still make more traction but at less and less per pound. 3000 lbs car has more traction but less per pound than a 1000 lbs car. Same as we all know with horsepower, it can be better at times to have less HP but also have a lot less weight. With all the weight loaded up so close to the edge on a narrow tire it goes over the peak and wastes what the wider tire doesn't because it was still below the peak.
/`| narrow tire
|
\_|
/```| wide tire
\___|
The wider tire keeps a more spread out pressure. Tires make the same amount of traction for every pound of weight added up until their peak, after that they still make more traction but at less and less per pound. 3000 lbs car has more traction but less per pound than a 1000 lbs car. Same as we all know with horsepower, it can be better at times to have less HP but also have a lot less weight. With all the weight loaded up so close to the edge on a narrow tire it goes over the peak and wastes what the wider tire doesn't because it was still below the peak.
I believe that everything is a trial and error.. Reality is that you can't mathmatically calculate whats best for your car(s).. You have to try it out and see what works for you.. as there are so many variables you have to include.. ie suspension, alignment, wheels, tires, weight, etc.. No race car or race team just use mathmatics or just pure physics to set up a car.. they use it as a base line.
12-30-06, 10:41 PM
#61
Originally Posted by madbouncy
I think part of the advantage of the wider tire is the same as why you want to keep an ideal camber all the time. With a narrow tire the contact patch will be more heavily loaded on the edge and as the tire is a circle, it has to deform a good bit to keep the same contact patch.
/`| narrow tire
|
\_|
/```| wide tire
\___|
The wider tire keeps a more spread out pressure. Tires make the same amount of traction for every pound of weight added up until their peak, after that they still make more traction but at less and less per pound. 3000 lbs car has more traction but less per pound than a 1000 lbs car. Same as we all know with horsepower, it can be better at times to have less HP but also have a lot less weight. With all the weight loaded up so close to the edge on a narrow tire it goes over the peak and wastes what the wider tire doesn't because it was still below the peak.
/`| narrow tire
|
\_|
/```| wide tire
\___|
The wider tire keeps a more spread out pressure. Tires make the same amount of traction for every pound of weight added up until their peak, after that they still make more traction but at less and less per pound. 3000 lbs car has more traction but less per pound than a 1000 lbs car. Same as we all know with horsepower, it can be better at times to have less HP but also have a lot less weight. With all the weight loaded up so close to the edge on a narrow tire it goes over the peak and wastes what the wider tire doesn't because it was still below the peak.
3 more days....
01-02-07, 10:39 PM
#63
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Originally Posted by wanklin
Just got off the phone with the shop. I've decided to pick the car on Tuesday Jan 2nd instead to allow them more time to detail the car. Better to wait a few more days than to create a rush scenario....
01-03-07, 08:17 AM
#68
Please somebody help!!!
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This is just not right, i read through 5 pages. (Thoroughly enjoyed the physics lesson though) getting it tomorrow, picking it up in a couple of days. Buggary! Where do you live? I'll take some pics!
Don't be cruel to animals.
Don't be cruel to animals.
