Hi, i'm sorta confused about something and hoping to get some clarification...this is for a fwd car

as far as I know, a tire's contact area is determined by the weight on the tire, and psi in the tire

so, a heavier car will give the tire a higher contact patch...but since there's more weight on the tire, the tire will be "overworked" and will exceed it's grip threshold at a lower speed than the same tire on a lighter car..the the lighter car can corner at a faster speed since there is less load on the tire correct?

on a fwd car, the rear tires do nothing...and the front tires do all the work...so if we apply the above theory to a fwd car, would it be more beneficial for these cars to run a rake with the rear of the car lower than the front in order to take some load off the front tires? reason i ask is that most cars i see have a rake with the front of the car lower than the rear...so there's even more static weight being applied to an already nose heavy car which i would think would overwhelm the tires even more?

<---confused and needs help

 

Rake will not take weight off any end of the car, unless rake is extreme. Only moving elements that have mass will do that, such as moving the driver's seat back,or thebattery back, or just plain removing a masive element suchas power steering. Rake will affect aerodynamics.

 

Your worrying about something really isn't a concern with streetable cars. Reduced weight in any circumstance is advantagious and I would worry more about where the contact patch is in a corner...I.e. Camber/toe.

 

tire grip does not increase relative to weight and by trying to shift more weight to the front you will only be adding to an already high 60f 40r weight bias and cause understeer

people try to rake there car for various reasons but ultimately a suspension designed to minimize weight transfer is the best way to go

 

also, for the best overall handling, corner weighting your vehicle is very important.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">Rake will not take weight off any end of the car, unless rake is extreme. Only moving elements that have mass will do that, such as moving the driver's seat back,or thebattery back, or just plain removing a masive element suchas power steering. Rake will affect aerodynamics.</TD></TR></TABLE>

I don't think azian is looking to remove weight or relocate mass by means of rake adjustments. That's imposible anyways. He is refering rake and the effect it has on tire LOAD.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by azian21485 &raquo;</TD></TR><TR><TD CLASS="quote">
on a fwd car, the rear tires do nothing...and the front tires do all the work...so if we apply the above theory to a fwd car, would it be more beneficial for these cars to run a rake with the rear of the car lower than the front in order to take some load off the front tires? reason i ask is that most cars i see have a rake with the front of the car lower than the rear...so there's even more static weight being applied to an already nose heavy car which i would think would overwhelm the tires even more?
</TD></TR></TABLE>

my guess is the loss exceeds the gain in performance by putting neg rake on a front biased fwd car. A fwd car is already at a disadvantage. front load is removed during acceleration. purposely lessening it even more will only result in a car with terrible lateral acceleration. Not to mention, raising the front and lowering the rear roll center will compound to this phenomenon we call understeer.

 

It is actually quite common to see the VW's run a rearward rake, especially the Golf's and Rabbits. They tend to have an awkward pitch and roll during corner entry that improves with the reverse raking. Never found it to be beneficial on the Civic/CRX.

Rick

 

maybe i'm nuts or dumb or both, but by raising the rear higher than the front, aren't you moving more weight to the rear?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mike C &raquo;</TD></TR><TR><TD CLASS="quote">maybe i'm nuts or dumb or both, but by raising the rear higher than the front, aren't you moving more weight to the rear?</TD></TR></TABLE>

No.

Think of an example. Have a person sit on the end of a table. Now raise the table legs at that end by sticking say 1" spacers under the two legs under the person end. Do you think the weight on those legs is reduced by having "raked" the table up by 1" (assuming the person did not move of course!).

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by YeahRight! &raquo;</TD></TR><TR><TD CLASS="quote">
I don't think azian is looking to remove weight or relocate mass by means of rake adjustments. That's imposible anyways. He is refering rake and the effect it has on tire LOAD.</TD></TR></TABLE>

Rake has no virtually zero effect on LOAD.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by stormy &raquo;</TD></TR><TR><TD CLASS="quote">tire grip does not increase relative to weight and by trying to shift more weight to the front you will only be adding to an already high 60f 40r weight bias and cause understeer

people try to rake there car for various reasons but ultimately a suspension designed to minimize weight transfer is the best way to go</TD></TR></TABLE>

Tire grip certainly does increase/decrease with a change in weight or load. This is one of the most fundamental aspects of all car handling and the basis for all suspension tuning. As you increase vertical load by a given % on a tire, its lateral grip capacity increases, but by less % than the vertical load % increase. Thus stiffening one end of the car compared to the other end makes more weight transfer occur during cornering at that stiff end. This means at the stiffened end that one tire has more load and the other less load than before stiffening (due to increased weight transfer), and thus the total lateral grip from both tires at that end decreases due to the tire's grip sensitivity to load. Thus if you stiffen the front suspension with springs or sway bars, or add weight to the front, those tires will have less grip than before and the car will tend towards understeer. Rake is just not going to do anything for you in the handling department compared to spring or sway bar changes, or changing the static load on the tries by removing or moving weight around.

 

there is a key word in my statement

"relative"

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">No.

Think of an example. Have a person sit on the end of a table. Now raise the table legs at that end by sticking say 1" spacers under the two legs under the person end. Do you think the weight on those legs is reduced by having "raked" the table up by 1" (assuming the person did not move of course!).</TD></TR></TABLE>

Reduced? No, wouldn't it be increased? Would the weight distribution remain static? I'm not in a thinking mood today. I was told when I corner-weight my car to raise the corners that are lightest; meaning higher = more weight.

 

think of it like a teeter toter

if you press up on something it's pressing down on the opposite end

 

Ok, so if you raise the rear it gets lighter? That's what always made sense to me, but someone told me differently recently. They have more "car knowledge" than me so I didn't question them

 

perhaps he has a different type of suspension

 

If you have the rear higher than the front, what about braking down force?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mike C &raquo;</TD></TR><TR><TD CLASS="quote">Ok, so if you raise the rear it gets lighter? That's what always made sense to me, but someone told me differently recently. They have more "car knowledge" than me so I didn't question them </TD></TR></TABLE>

Theoretically, yes, raking the car with the rear higher reduces the weight on the rear wheels. But, essentially, you would have to raise the car a lot for any appreciable amount of change. The higher the center of gravity the more the change will be. Most cars have a relatively low center of gravity. Therefore, for all intents and purposes the static weight acting on each wheel stays the same for small changes in rake.

 

Suspension design is not a factor at all. It is purely statics and not kinematics (suspension falls in this catagory) that determines the amount of weight on any given tire.

 

Braking load transfer is a function of tire grip, center of gravity height, wheelbase, and the original static weight on each tire.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Johnny Mac &raquo;</TD></TR><TR><TD CLASS="quote">Braking load transfer is a function of tire grip, center of gravity height, wheelbase, and the original static weight on each tire.</TD></TR></TABLE>

Agghhhtt-tut-tut-tut-tut...

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Johnny Mac &raquo;</TD></TR><TR><TD CLASS="quote">Braking load transfer ... function of ... original static weight on each tire.</TD></TR></TABLE>

WTF Does That Have To Do With Anything?

Scott, who must one of these days find some way of viewing the pilot for "The Paper Chase" - the series - with John Houseman....Hero Role Model Par Excellance...for the pedantic set.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RR98ITR &raquo;</TD></TR><TR><TD CLASS="quote">

WTF Does That Have To Do With Anything?

Scott, who must one of these days find some way of viewing the pilot for "The Paper Chase" - the series - with John Houseman....Hero Role Model Par Excellance...for the pedantic set.</TD></TR></TABLE>


Scott, don't let you're skivvies get into such a bind. My response was to ward off the possible attacks from the pedantic set. BTW, for what it's worth, It has to do with the fact the tire load vs. grip is nonlinear.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Johnny Mac &raquo;</TD></TR><TR><TD CLASS="quote">Braking load transfer is a function of tire grip, center of gravity height, wheelbase, and the original static weight on each tire.</TD></TR></TABLE>

I say braking/accelerating load transfer is a function of longitudinal acceleration, center of gravity height, wheelbase, and the total weight.

Longitudinal Load transfer = Total weight * Long. Accel * CG Height / wheelbase

Cornering load transfer is a function of lateral acceleration, center of gravity height, average track, and total weight.

Lateral Load transfer = Total weight * Lat. Accel * CG Height / wheelbase

Once you know load transfer, you combine that with the original static weight (and thus centre of gravity location) on each tire to get the final weight on each tire, and then you can calculate each tire's grip (force). You will of course need a curve of the tire's vertical load vs grip.

To improve total grip, you want final load on each tire to be as equal as possible. Lowering CG height decreases load transfer and thus makes final load on each tire more equal thus improving grip. Increasing track or wheelbase also decreases load transfer. Moving CG towards rear equalizes load on all 4 tires and thus improves braking (if rear tires come off the ground, they aren't helping you much).
That is why Porsche's are so good under braking. And increasing rake by say lowering one end 1" on a 100" wheelbase car makes a very small change in CG location in plan, and a slight improvement in CG vertical height, and thus may improve things, but only because CG height has been reduced and not because static load on the tires has changed appreciably.

Thus for mechanical grip:
Lower car = better
Wider track = better
Longer wheelbase (within reason) = better
Lower CG height = better
More even static wheel weights = better
Lighter car = better

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mike C &raquo;</TD></TR><TR><TD CLASS="quote">Ok, so if you raise the rear it gets lighter? That's what always made sense to me, but someone told me differently recently. They have more "car knowledge" than me so I didn't question them </TD></TR></TABLE>

When you raise the ride height on a vehicle that corner and the diagonal corner will increase in weight, while the other adjacent corners decreqase by an equal amount. This is cornerweighting. If you raise ride heights of both rear wheels, then the effect cancels itself out. You cannot change the rear static distribution or the right or left side dist. You can only change cornerweights.

So, unless there is a CG migration (in any direction) the static weight doesn't change.

Adding weight to a tire will increase it's grip but it is only that simple in the static conditions. Adding mass (weight) increases force, which is good, and adding mass also reduces the effective wheel vertical frequency. This is particularly important at low frequencies (handling) which is "mass-driven". Basically, more mass slows down the tire vibrations when you move it up, which is good for tire grip (bad for response and damping). Unfortuantely, more mass means more load transfer = bad for grip, and this additional mass requires more force to accelerate it: a = F/m. More mass means more force generated, but that mass also reduces acceleration, so it's a compromise. The ideal situation is to reduce mass everywhere as much as possible and use tires that generate nearly linear grip under your loadings. So don't use Azenis on a kart, or use 10" Hoosiers on your Integra.

Descartesfool: you mean "track" in equation 2.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">[good stuff]</TD></TR></TABLE>

Thanks, I'm glad I didn't have to figure that out and/or post it.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">So, unless there is a CG migration (in any direction) the static weight doesn't change. </TD></TR></TABLE>
Very true, assuming you overlook the <U>small</U> shift in cg towards the front of the car when you raise the rear. Like a kid rocking back on a 4 legged chair, the center of gravity can move "over" a pair of tires, but at the angles we're talking about it's negligible.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">Adding weight to a tire will increase it's grip but it is only that simple in the static conditions. Adding mass (weight) increases force, which is good, and adding mass also reduces the effective wheel vertical frequency. This is particularly important at low frequencies (handling) which is "mass-driven". Basically, more mass slows down the tire vibrations when you move it up, which is good for tire grip (bad for response and damping). Unfortuantely, more mass means more load transfer = bad for grip, and this additional mass requires more force to accelerate it: a = F/m. More mass means more force generated, but that mass also reduces acceleration, so it's a compromise. The ideal situation is to reduce mass everywhere as much as possible and use tires that generate nearly linear grip under your loadings. So don't use Azenis on a kart, or use 10" Hoosiers on your Integra. </TD></TR></TABLE>

Ok, I'm lost. When you say "adding weight to a tire", do you mean a heavier tire or shifting the vehicle's weight to that tire (via spring perch or moving mass)?

"effective wheel vertical frequency" : Huh??

"mass-driven" : Eh?

"more mass slows down the tire vibrations when you move it up, which is good for tire grip" : Que?

"a = F/m" : Well, at least I understand this part


What frequency are you talking about, the rotational velocity? The natural frequency of the spring with the mass of the car on it? The squealing sound they make when you **** them off?

 

Alright, I admit I use complicated language cuz I suck at explaining things otherwise. The stuff I posted is only interesting for people interested in vehicle dynamics.

On a 1/4 car model (one corner of the vehicle), you have the sprung mass at that corner, the unsprung mass at that corner, the spring, the damper, and the tire spring. As you're driving around, the tire is moving vertically because of 2 main things: vehicle motions (roll, pitch) and bumps. These motions are often described as sinusoidal, mostly because people like to talk about sinusoids when springs are involved. The equations of motion are solved in the vertical direction (or you can include roll and pitch to make it worse) and you can back out frequency and magnitude when you hit a bump or when you dive under braking, for example. The magnitude is important, because a large magnitude means the tire is lifting off the ground. If your frequencies are really high, then your tire normal force is changing rapidly (may lead to wheel hop).

The bottom line is that tires like a lot of normal force and they dislike it when you change that force quickly. Adhesion takes a finite amount of time to occur. All that **** about mass-driven doens't make sense to anyone unless you have an intermediate dynamics textbook in front of you, so I'm a bastard for talking about it.


Modified by GSpeedR at 10:34 AM 6/27/2005

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">

Descartesfool: you mean "track" in equation 2. </TD></TR></TABLE>

Track it is! Damn that cut and paste feature.