Three simple questions - pls point me

[Rex Schimmer]

If the tractive force is defined as Coefficent of friction x weight then the number of wheels is completely inconsequential. Now where in Hot Rod's formula is there a factor (let's call it the "fudge" factor) that is used to characterize the traction capabilities of the tire therefore the number of tires means nothing. Also his formula for "peak applied force" should be using the drive radius of the tire not the diameter, if you happen to use his formula and have two tires then the divide by number of tires (2) will make this formula work but it is still wrong.

Rex

[lsrengineer]

If the tractive force is defined as Coefficent of friction x weight then the number of wheels is completely inconsequential. Now where in Hot Rod's formula is there a factor (let's call it the "fudge" factor) that is used to characterize the traction capabilities of the tire therefore the number of tires means nothing. Also his formula for "peak applied force" should be using the drive radius of the tire not the diameter, if you happen to use his formula and have two tires then the divide by number of tires (2) will make this formula work but it is still wrong.

Rex

On pavement tires don't completely follow  tractive force available = COF x weight, this is why wider tires give more traction.  I need to dig up the references on this if needed.  If I remember correctly on other types of surfaces this "rule" follows closer.
I was going to use this as a first pass at getting the tractive force available in the model.

[SPARKY]

I think the most important use of the TE formula is to decide the "COMBO" at the hp VS TOTAL DRAG wall 

[lsrengineer]

If the tractive force is defined as Coefficent of friction x weight then the number of wheels is completely inconsequential. Now where in Hot Rod's formula is there a factor (let's call it the "fudge" factor) that is used to characterize the traction capabilities of the tire therefore the number of tires means nothing. Also his formula for "peak applied force" should be using the drive radius of the tire not the diameter, if you happen to use his formula and have two tires then the divide by number of tires (2) will make this formula work but it is still wrong.

Rex

http://insideracingtechnology.com/tirebkexerpt1.htm

some good tire traction discussion in this short exerpt

[Interested Observer]

Wobbly:
Just because the maximum power curves are on the graph doesn’t necessarily mean they were used as such to produce the velocity curve.  At this stage that is probably so, but depending on how lsrengineer chooses to use or display that information, the next refinement taking into account the friction available may be another, different story.

Rex:
Hotrod’s calculation is to determine the tractive effort available from the drivetrain and has nothing to do with the tire’s capabilites.  That comes next, when the surface friction coefficient (your fudge factor) is considered to determine how much of that effort can be utilized.

While you and I are in the “radius” school, Hotrod’s approach is to use the diameter.  This requires the number to be divided by 2 to produce the TE available.  It would probably be safer for him to have simply said  divide by 2 rather than the number of drive wheels.  That would keep people running motorcycles or dualies from being misinformed.

It may not be the way you or I would do it, but that doesn’t make it “wrong”, just fraught with potential pitfalls.


lsrengineer:
"On pavement tires...."     Beware, the salt is not pavement.

[hotrod]

On pavement tires don't completely follow  tractive force available = COF x weight, this is why wider tires give more traction.  I need to dig up the references on this if needed.  If I remember correctly on other types of surfaces this "rule" follows closer.
I was going to use this as a first pass at getting the tractive force available in the model.

Yes that is why engineers in the 1960's said dragsters would never exceed 1 G accelerations. Tire traction on pavement is a combination of 3 different types of behavior.
Classic friction
gearing = the tire conforms to the irregularities in the surface creating a geared effect.
adhesion = brief adhesive bonding between the sticky rubber and the pavement at the molecular level.

If you take all those into account some of the super cars built for ultimate performance primarily because of the type of tires they come with, can achieve acceleration in excess of 1 G on pavement but their total acceleration is still limited by the coeff of friction even if it is greater than 1. High performance tires can achieve coef of friction near 1.2 under the proper conditions.

In the case of bonneville, tire traction is in my view mostly limited by the shear strength of salt on salt, and like you mentioned is probably very similar to other loose surfaces like packed dirt roads.

While you and I are in the “radius” school, Hotrod’s approach is to use the diameter.  This requires the number to be divided by 2 to produce the TE available.  It would probably be safer for him to have simply said  divide by 2 rather than the number of drive wheels.  That would keep people running motorcycles or dualies from being misinformed.

I do not use the diameter of the tire, (see my original computer code) I am using the rolling radius (lever arm) of the axle to determine the actual force applied at the tire contact patch, on the drive wheels. In the case of all wheel drive you have 4 drive wheels each with independent loading (traction) so the total accelerative force available if the the engine can deliver it is 4 times what a single driven wheel with the same weight.

If you have a motorcycle all your traction comes from a single wheel, and it can only apply the force that the traction of a single tire can deliver. Put two driven wheels on it by converting it to a 3 wheeled motor cycle and you roughly double the possible traction (weight unchanged), put the same engine in an all wheel drive streamliner of the same weight and you have approximately 4 x the possible traction for the engine to use.

If a single tire (1 ft loaded radius) loaded with 1000 lbs on a 0.6 coef of friction salt surface, can only deliver a max of 600 ft-lb of axle torque to the salt as a thrust of 600 pounds. Load two of those same tires with 500 lbs weight and you still have the same traction limit because you not only doubled the number of tires, but you cut the individual loading in half. Increase the load on both tires to 1000 lbs with ballast and now you can deliver 600# of thrust with each tire or 1200 lbs of thrust.

I typo'd in my earlier comment I meant loaded radius but said loaded diameter, we are all on the same page, the only thing that counts is the distance (lever arm length) from the axle to the ground contact patch.

Larry

[Interested Observer]

Put two driven wheels on it by converting it to a 3 wheeled motor cycle and you roughly double the possible traction (weight unchanged),

I hope we’re all on the same page, but the way I read it, this statement is in conflict with my “page” and the quote’s own preceeding and succeeding paragraphs.  If F = COF*N and COF and N are unchanged, how does F get doubled?

I think part of the confusion is that Rex and I tend to calculate how much TE is available and then compare it to what traction is usable or supportable by the particular vehicle layout.  Hotrod comes from another direction, discussing the amount of traction supportable by various vehicle layouts.

[hotrod]

I hope we’re all on the same page, but the way I read it, this statement is in conflict with my “page” and the quote’s own preceeding and succeeding paragraphs.  If F = COF*N and COF and N are unchanged, how does F get doubled?

Not sure I understand the question, to me it is obvious. You may be ignoring my qualification that wheel weight stays the same on one wheel vs 2 wheels, ie they both have the same loading.

If the tire contact patch at a certain wheel weight and coeff of friction will support x lbs of thrust before the wheel will spin, as long as you keep the same weight on the wheels in each case, the more driven wheels you have, the more power you can deliver to the ground. If one wheel  can deliver 150 hp to the ground, 2 driven wheels can deliver 300 (150 each) and 4 driven wheels can deliver 600 hp.

Or you can think of it in the reverse, you have a total engine horse power available of 1200 hp, and the wheel salt contact patch of one wheel loaded at 1000 pounds can only deliver 250 hp, how many driven wheels do you need to get all your engines power to the ground? Obvious 4 wheel drive is the solution that comes closest to getting the job done, but even in that case at low speeds you will be power limited to 1000 hp to the ground.


Given the rules package that most "cars" run under they are limited to 2 wheel drive, so the only solution is to maximize the power limit of the 2 wheels you can drive and that is most easily done by increasing the wheel weight on the wheels and setting up the chassis so the wheel weights are as close to equal as possible on your two driven wheels.

Larry

[lsrengineer]

Put two driven wheels on it by converting it to a 3 wheeled motor cycle and you roughly double the possible traction (weight unchanged),

I hope we’re all on the same page, but the way I read it, this statement is in conflict with my “page” and the quote’s own preceeding and succeeding paragraphs.  If F = COF*N and COF and N are unchanged, how does F get doubled?

I think part of the confusion is that Rex and I tend to calculate how much TE is available and then compare it to what traction is usable or supportable by the particular vehicle layout.  Hotrod comes from another direction, discussing the amount of traction supportable by various vehicle layouts.

Maybe another way to say this is to distinguish between TE available or applied from the engine, drive train and tire size and TE supportable from the tire/surface interface.

Adding tires on the ground may provide more supportable TE but probably will not be linear with area (number of tires) - I think this is what Interested Observer is trying to explain.  A car using 4 wheel drive might have some additional supportable TE but won't be 2x of a 2 wheel drive of the same weight and tire type. Right?

It is somewhat interesting to see the black lines from tires spinning near the starting line.  Agree salt is not pavement but it is not completely marbles either.

Maybe I should post a graph showing TE applied and AR (air resistance),  it is interesting that the TE applied goes way down as the vehicle speeds up and higher gears are used.  Does this mean low powered cars are less likely to slip the tires at the top end?

I need to go look at other or past posts and see if someone will share data,  who has been using a data acquisition system.

[jl222]

  Look at post 1577 [ getting ready for Bonneville] build diarys.

  F.A.S.T Data logger shows different throttle positions [tps] between this years Bville loose track to last years better track.

  Also shows Map or manifold absolute pressure in KPA or kilopascals [wish it was psi] 101 KPA = 14.7 or
atmosphereic pressure at sea level [10 or 15%] less at Bville. 6.9 kpa = 1 psi.

  The boost level shows the power level that could be applied.

    JL222

[Moxnix]

.

[Interested Observer]

Ref. Three-wheeled cycle example

Larry,
We both know what is correct.  My point is simply that the way the statement is written it says that if you add a second driven wheel without changing the weight, that one can magically create twice as much traction (leaving aside any second or third order effects having to do with contact pressure etc.)  The second wheel has no bearing on the traction available unless weight is added along with the wheel.  And, if the added weight were simply added to the single wheel, it would have as much traction as the dual setup.  So, what’s the point of adding the wheel from a traction standpoint?   

Gee, in order to get more traction we need to add weight to the driven wheel(s)!  Has that ever happened at Bonneville?

[hotrod]

Ahhh now I understand how you are misreading the statement!

My comment "without changing the weight" was referring to the weight applied to each wheel, ie the load on each wheel, I was trying to explicitly state that you had to increase the weight of the motor cycle so the driven wheels still had the same load as the single wheel.


Poor wording yes, but I was whipping out the answer while here at work and did not take a lot of time to proof it.

Now that we have an edit time out, you can no longer go back and fix that sort of poor verbage.

So, what’s the point of adding the wheel from a traction standpoint?  

It matters in a theoretical context for example converting a 4 wheel drive car to a 2 wheel drive car to fit the rules package. Wheel weights do not change but the number of driven wheels do. Same applies to streamliners deciding if they should run 2 or 4 wheel drive. Individual wheel weight will matter regarding rolling resistance  due to deformation of the salt surface, two equally loaded tires will always have less rolling resistance than a single similar wheel under twice the load.

On a motorcycle probably not so much unless you wanted to set up a driven front wheel as well, I was just trying to state the general case. It could make a difference on three wheel motorcycles if a decision was to drive only one wheel as in the side cars setups or change to two driven rear wheels in a motorcycle streamliner. Would a side car bike be legal if both rear wheels were driven?

As I said it was a quick note that was knocked out under pressure -- as you stated we both know what was meant, so the real question is why are you beating a dead horse and cluttering up the thread with this?


Larry

[jl222]

  Makes one wonder why rally cars have 4 wheel drive

   JL222

  have to look in my Carroll Smith Tune To Win book were it explains adding weigh detracts from tire performance.

  But later have company.

[hotrod]

Makes one wonder why rally cars have 4 wheel drive

No wonder needed, it is because when Jeep ran the rally circuit in 1972-73 with their new Quadratrac full time all wheel drive they totally ran off from the other competitors. They won the FIA rally champion ship that year. They beat the European cars that normally dominated the sport so resoundingly that the FIA outlawed All wheel drive until Audi and Subaru developed commercial all wheel drive systems.

It was a case of no competition and total dominance. In loose traction conditions like rally (salt) all wheel drive is the drive of choice if you want to get power to the ground reliably.

It also gives you a secondary advantage over the absolute increase in traction. It provides "terrain advantage", by using multiple drive wheels you reduce the chance that all wheels lose significant traction at the same time. The only down side of all wheel drive is drive train losses and complexity/packaging issues. If those can be managed it will out perform 2 wheel drive in almost every case.


Of course that is why it is also outlawed in land speed racing except in special construction and production even though it comes standard on a good number of production performance cars. Adjust the rules to accommodate the lowest common denominator, rather than deal with all wheel drive classes in other than production and special construction.

Larry