So I just recently picked up 4, 2" spacers to run 4thgen five spoke wheels. The question is not directly with the spacers but more of a performace difference after putting them on. I have noticed that my car rides a little bit different then it used to, turing is much more touchy and the front suspesion seems to "bounce" a bit more. I am thining that this all has to do with the fact that the wheels are mounted two inches further out then original. If anyone has some similar happening after doing spacers, please let me know. I am going to continue to run my spacers so this is pure curiosity.

Thanks
Chris

 

Spacers change the leverage geometry, of how the road/ride is transferred to the suspension, so I'm really not surprised if you feel a difference.

For the best results, you don't want the mounting surface so far from what supports the rim ring. Centerline mounting is the best.

 

You did put the 4th gen wheels/tires on, right? How do the wheels/tires compare with what you had before? Lower profile? Wider? Spacers don't change the suspension geometry or how the load is transferred to the suspension, provided that you are using the right spacer and wheel combination. However, there are a lot of factors that go into making spacers that are right for the combination. Especially in the front, and especially if you go with a wider tire.
You may need to play with the alignment a bit to get it to drive better. Is it possible that the car is angled 'up' more in the front than before?

 

did you go from stock 15" tires/wheels to 16" wheels?
tell us what size and brand/model tires you had and what you are running now.

I notice a difference with my 17" 275/40/17 street tires and my 16" 265/45/16 road race tires.

I am running 2" spacers as well'

 

It's called 'leverage". Where the wheel mounts to the suspension effect point, most certainly effects transfer.

If you push down on a 4' long board do you have more force than pushing directly down on the same abject? Of course you do. Same thing for the different offset/backspacing/spacers of a wheel.

 

If the center of the wheel is in the same place, it doesnt make a bit of difference.

 

Do you REALLY think that the forces being mechanically transferred through a spacer/long offset to the center of a wheel, are the SAME as if the mechanically connection is directly in the center line of a wheel?

 

Alright, draw up a free body diagram to prove your point. Better be careful with that shovel you're holding though.

 

I will....Wouldn't wanna damage my wheels, getting through what your shoveling.


I'm not a teacher. Do some research on your own.

 

In other words you dont know how to draw one. Thanks for showing your expertise.

 

Feel free to show YOUR expertise then & draw something up.

 

Sorry not going to educate you, just everyone else here that you don't know what you are talking about. You say the forces are different, draw up the diagram. You only have one transmission point and 4 reaction points, isnt that hard. Without having a free body diagram, you cannot do an analysis, and therefore have no conclusion to draw.

 

I didn't ask for the education. Your not drawing anything because you lack the ability to, or you would be so eager to disprove me. After all, you took to time to contradict me, so now prove it. You can't, or you would.

 

Sure, thats your only argument left. I'm not here to show you I know more than you, I already know that. I'm here to show that what you said is wrong and you cant even back up your statement with ANYTHING so it just proves it right there.

2 things wrong with what you said:
1. A properly bolted connection doesnt move, its static.
2. Properly designed and/or sized parts dont deflect excessively. Seeing as how the aluminum wheels arent breaking left and right, I think they're ok to transmit the forces.

So given that the contact patch and center are in the same location (talking about not moving the entire wheel in or out relative to the spindle), the force transmitted to the tire is the same, and given that the wheel and spacer didn't break off the car the force transmitted at the hub surface will be the same as well. If its different either the wheel or the spacer or the bolts are deflecting excessively, and with them being fatigue prone parts they would not last long at all.

 

THINK? I thought you were gonna PROVE something. We already know what you THINK. Thoughts prove nothing.

Bust out your magic crayon & draw. You were SOOO eager for me to draw. No you can't.....

 

I asked you to prove your point, and you are still refusing to do so. Thats because your opinion is wrong, as usual.

Draw up that free body diagram. Show where the force is different.

 

I see.....You have no way to prove I'm wrong (because I'm not), so you just wanna spout & spout. Wow. I'd think you'd rather prove someone wrong, than type useless dribble.

YOU said *I* am wrong. PROVE I'm wrong. You can't, so STFU!

I'm done littering the OPs thread (sorry Chris), until you have something CONCRETE worth responding to.

 

I dont have a way? Thats a laugh. You will never learn. Go to school, take some classes. You are obviously not even to the smart enough to be dangerous level.

You wanted a crayon drawing to prove you dont know what youre doing, here it is. Its a free body diagram, you just learned step one in analysis. It took me a minute to draw. Real hard. Notice that the wheel design.. where the face is relative to the tire... is a complete non-factor. The force isnt transmitted there or reacted against there. The places it would be are the tire contact patch and the control arm mounts, spring, and strut and if you want to be picky you could throw the anti-roll bar in there.

Might be time to crack open a book, Newton's third law is not a secret or hard to understand.

 

this were about 2" spacers on stock 3rd gen wheels then Stephens diagnostic would be more valid but since the spacers presumably reset the 4th gen wheels to match the 3rd gen wheels it's much less acurate.

The 4th gen wheels with proper spacers are essentialy larger 3rd gen wheels as far as the forces being applied.

Madmax may be right in assuming that the wheels (with a proven track record of not failing or deforming excessively) and spacers (it would depend on the spacer and the instalation) are sound enough to be considered a single unit and there for the relevant surface is the one between the hub and the spacer.



Looking at what is actualy different though can give some clues. The contact patch may be wider and the side wall may be shorter, both contribute to a sensation of the front end floating in cases where an inertia valve is used in the strut/shock.

Also wider wheels (assuming a wider contact patch) can drasticaly change the steering feel since it will tend to exagerate any camber present in the suspension, lower profile tires will also tend to sharpen turn in and compared to the tall "baloon like" tires seen on most american cars prior to the 90's the difference can be suprising.

 

Wow....Just wow...

You just proved my point for me. Zero is better. And here it sounded like you meant the positive offsets were just as strong & you go & draw a zero offset wheel.

 

 

X2 LOL

 

Maybe you'd like to point out where that shows zero offset? In your mind only huh?

Where is YOUR proof Stephen? You have none, thats why you keep posting here and posting no proof of your ignorant opinions.

Here, I modified your misconceived notions of a zero offset (thats not the mounting flange above but I can see how someone grasping at straws with zero knowledge of physics would think that) and made up a massive 20" offset wheel for you with a 20" spacer with my new crayon drawing. Looky there, the applied force is still in the same spot, MIDDLE OF THE TIRE. What a concept. Newton is still right. Come back and reply when you have some mathematical proof that the forces applied on the chassis by using a 2" offset wheel and 2" spacer are different, otherwise you're just wasting more space with your useless drivel.

 

I like how you changed your pic ( https://www.thirdgen.org/forums/post...60-post18.html ) from showing a zero offset wheel.

Tell me.....

With the load on each wheel I'm showing (the green line showing the load which is stronger? Almost directly over? Or from farther out off of the center??

Would you rather stand on the top of a wall? Or on the end of a 4' long board hangin off the wall?

 

that's all from me.

 

Wow, still no factual evidence to support your opinion. Wonderful. Shows how much proof you have to back up your statement. Zero.

Your picture does not show any change of force transmitted to the suspension itself because you omitted it. Your free body diagram=Fail.

If zero offset is the best way, you had better tell Chevy they're doing it wrong. They're making Corvettes by the dozen with wheel faces at the edge of the tire and huge offsets.

 

i can't see it. unless the wheel's contact patch is out further from the car. but hey i'm no expert.

cale

 

I guess its time for a basic lesson in High School Physics. Newton's Third Law, from nasa.gov http://www.grc.nasa.gov/WWW/K-12/airplane/newton3.html
So here's a diagram. Top wheel is 2.000" offset with a 2.000" spacer bolted to it. Second wheel is a 0.000" offset no spacer. Third wheel is a different face with 0.000" offset no spacer. The applied force on the tire is a summation of the forces across the contact patch, identical in each situation. Notice the reaction force, that would be at the brake rotor flange or drum flange, is the same. 10lb up, the resisting force is 10lb down. Maybe Stephen can explain how that red spacer creates something other than a 10lb force at the brake rotor flange. Love to hear that explanation.

 

I NEVER said squat about suspension. Not ONCE. Never.

You ABSOLUTELY, CANNOT DENY, that weight directly on top of a wall (the wheel center) is stronger than the supported weight being off center.

I wanna see what kinda lie to come up. Otherwise, we wouldn't use ratchet wrenches. We'd just use ratchet screwdrivers. LEVERAGE is exactly what I've been repeating, yet you ignore.

 

And the example of aerodynamics applies to supported weight HOW here?

As for your picture.......It means nothing, except that you can draw.

 

Wow, great way to the refute the facts you cant get around that clearly show you dont even know basic physics. Stop posting and go get your GED.

 

Wow, just wow. Reading comprehension = zero.
I clicked a link on the page there and guess what, his third law didnt change! And it doesnt reference aerodynamics! What a concept!

http://www.grc.nasa.gov/WWW/K-12/airplane/newton.html

There you go, same law, no mention of aerodynamics. Now what? Going to call his law false?

 

How nice....You provide a DIFFERENT URL this time & try to play it off.

Here's the one you posted above. Go look, if you think I posted a differnt one.
" Originally Posted by madmax View Post
I guess its time for a basic lesson in High School Physics. Newton's Third Law, from nasa.gov http://www.grc.nasa.gov/WWW/K-12/airplane/newton3.html"

Your pathetic.....You provide useless crap.....Change your story at whim......You have FAILED to provide a single argument against the law of leverage. I love how you keep ignoring that FACT. Leverage = more force. ZERO way around that, yet you keep skirting that issue. I'm done with you.

 

I have 100K miles on my wheel spacers and my bearings are fine, wouldn't my bearings fail sooner using a spacer. It has the same bearing on it when I bought the car 6 years ago.

 

And I have 190,000 miles on my engine, starts within the first revolution when the engine is cold & roasts my 285s at will. But that doesn't mean there isn't any wear. No, of course not. Same thing with your bearings.

Leverage is leverage. Period. That is why a 377 make more torque than an equally cammed/induction/intake etc., 350. Besides the extra cubes, the rod to stroke is whats responsible for that. Remember the word 'leverage'?

 

I think you are using leverage in the wrong context here.
Let's say I was really really strong and grabbed the front wheel and lifted the car up, would the spacers change the leverage required to lift the car up?

 

You mean"Would it change the amount of force needed?' Yes. The farther out you are, the less you'd need.

In reverse of your example, more weight/force would be transferred from the chassis to wheel wheel, with less effort.

 

why would I need more force if the spacer is keeping the wheels that "require" the spacers in the same position as the stock wheels?

 

So what? Its the same law. You're pathetic, the best you can do is tell me the link is different? Ok...
http://www.glenbrook.k12.il.us/GBSSC...aws/u2l4a.html
http://www.usoe.k12.ut.us/CURR/scien...er/newton3.htm
http://www.iki.rssi.ru/mirrors/stern...e/Snewton3.htm
http://csep10.phys.utk.edu/astr161/l...wton3laws.html
http://swift.sonoma.edu/education/ne...l/newton3.html
http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html
http://www.iit.edu/~smile/ph9712.html
http://www.batesville.k12.in.us/phys...ton3Intro.html

That's from a google search of newton's third law. I did not look at a single one of them, but I will bet you your car they all are the same law. Walls and boards, bridges, car wheels, airplane wings, the law does not change. You obviously have no education in physics or reading for that matter if you cant see the text of the law is the same.

Leverage? What leverage? Whats with this wall and board nonsense? Its a wheel. Maybe you would like to explain, like I asked the first time, how the force changes at the suspension or to make it even simpler at the brake rotor face by using a spacer with an equally offset rim? As I said way before you went on a rant about how little you know, if the contact patch of the tire has not moved the spacer does not matter. Its that simple. There is no leverage going on here, the location of the force relative to the car has not moved if you use an offset rim and a spacer to put the wheel and tire right back where they were with a zero offset. Thinking otherwise just shows you are uneducated and from your attitude you will never learn it either.

I'm done with this post, anyone with half a brain will see the error in Stephens logic and non-existent proof of said logic, and Stephen... you're just hopeless and will never learn.

 

Talk about hopeless, Mr. Hide-behind-an-internet-screen-name. Easy to be dumb.

Your obviously not capable of simple physics understand, so I'm done with trying to explain the effects of leverage too you. Use of a standard ratchet vs a breaker bar, will be obvious. Or should be, to the half intelligent.

I can't explain it any clearer, AGAIN, so (this tine) I AM done with you.

 

I think it might be time to shut this post down. I am sorry, I did not mean to start any aguments, just wanted to hear if anyone had any similar happenings when they added 2" spacers. Honestly, I think the leverage point is changed, even if the point of contact from the tire to the ground is in the same location. The point at which the wheel originaly mounted to the rotor/spindal has now been changed by two inches. this is just my two cents and philosiphy aswell. I am not saying that this is the end all be all answer here, just my opinion. Also, other things coming into play, my firebird did originaly come with 16x8 rims. Now, somewhere along its 17 year life, some goofball decided that a set of chrome 15x7(or narrower) wheels would look better and perform better. the tires where 215 60 15. With that said, I did/do now have a set of fourth gen, 16x8 rims that have 245 50 16s. The last thing that might come into play here aswell is the fact that my suspesion probably has not seen much attention is the cars 150000 mile vetur in life. I honestly think I am going to be one of the few people who has a car that will not drop when I go to the Eibach Prokit.

Thanks for the entertaining input and again, I did not want to cause any arguments.

 

stephen, i hate to say it but you are DEAD WRONG on this...
you have the right idea, but are missing a very important part of the equation.

look at it this way.

the stock IROC wheels are zero (0) offset.

4th gen wheels and corvette wheels are 56mm offset.
to get the wheel back to where it should be, you need a 56mm spacer.
(56)/(25.4) = 2.204 inches.

since the OP said he used a 2" spacer, he actually moved the wheel INBOARD of where the stock wheels were.


NOW


if the wheel was moved farther out, then the leverage on the suspension would be greater, causing MORE ROLL, MORE DEFLECTION per unit force, and LESS FEELING...


madmax is completely right that no matter what, as long as the centerline of the wheel is in the same place as it was originally, NOTHING CHANGED.
you could have a 8" spacer on there, but as long as the centerline of the wheel is moved to a final position equal to where it was before, you have no change.




as for your argument about the breaker bar and ratchet thing, that has ABSOLUTELY NOTHING TO DO WITH THIS QUESTION

the only way it relates is back to my previous statement
look at this... it will help
the top is your standard wrench
the bottom one is your extended wrench with a spacer added on...

 

forgot attachment

 

another example

 

I swear...This is my LAST post here...

1 LAST explanation......Get it or not, I don't F-ing care. UNSUBSCRIBING! Trash me if you want...I won't even see it.

Do you want the load riding STRAIGHT down on a 0 offset whell? Or do you want the load riding on a big offset wheel?

Can NOBODY see my logic is load transfer here??????? What is stronger? A load ON an object, or offset from the object?



"since the OP said he used a 2" spacer, he actually moved the wheel INBOARD of where the stock wheels were."<----WTF????? Since when do spacers move thing IN???? Hello??? Spacers move thing OUT!

 

That'd be when they are using a wheel with a 56mm positive offset.

 

max, i think its time to just let him go into the corner and pout.
i think he is under the impression that the reaction from the wheel pushes up at the mounting surface, no matter where it is...
the reaction of the vehicles weight acts at the centerline of the wheel. if the mount is anywhere other than the centerline, then there is a force and a moment (aka, a force couple).
the force is equal to the vehicles weight on the wheel, the moment is equal to the distance from center times perpindicular distance from centerline to mounting surface.
the mounting surface can be changed with spacers.

 

madmax, your terminology is slightly off. Thats just a simple diagram. A free body diagram is a single point and all forces denoted by vectors as they affect the point. Your point is the same thou

Stephens leverage argument was how the weight of the car was transmitted to the ground thru the bearings. His drawing is exaggerated showing the spacer to move the load when they actually moved the wheel to realign the load to the factory location. His references to boards and walls while difficult to relate to were about this leverage, not suspension geometry

pic:


As state before, the ideal distance to space the wheel for equal offset would be 2.2". Since he only spaced it 2" the additional .2" is negative, so going by offsets the wheel actually moved inwards



Going back to the OP, I also felt a big difference in handling when upgrading to Y2K wheels. Refer to ls six's post about the effect. You can also visually see it when you install the wheels and the car is higher off the ground since there is less tire to bulge out. Most of the effect you will just have to get used to, but some will go away with a proper alignment

 

FBD for a 'rigid' body, thats what I drew up in crayon model. I did not bother with magnitudes or dimensions for such a basic explanation. Sure the second cad line drawing I put up could be reduced to a point FBD because the forces act on the same line but I chose not to because they do not act on the same point. From a book:

Free-Body Diagram (1)

• Establish the x, y coordinate axes in any suitable orientation.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on the body.
• Label all the loadings and specify their directions relative to the x, y axes. The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed.
• Indicate the dimensions of the body necessary for computing the moments of forces.

(1) Engineering mechanics. Statics. R.C. Hibbeler.-8th ed. pp. 210 ISBN 0-13-577032-7

Nothing wrong with my terminology.

To add:
A "correct" FBD of the 'rigid body' would be the wheel, spacer (if used) the rotor, bearings, and spindle, with angles, dimensions, and loads... none of which I have. The rotation of the rotor can be ignored because the load from the tire would likely be considered a static load for a basic analysis of spacer and offset vs no offset. Did not even want to get into dynamics. I also did not want to get into trying to explain that the only counteracting forces would be from the strut mounting point and lower ball joint and not the tie rod end because I did not feel like trying to explain degrees of freedom of the joint at the tie rod. I'm not sure, maybe Racecraft would know, but the lower ball joint hole centerline may be perpendicular to the spindle bearing shaft so that would require a 'rigid body' FBD to properly show the forces unless you want to explain moving the force from the ball joint to the intersection of the strut and wheel centerline (if those even intersect) and how that adds a moment and so on... certainly way over the head of most people. You could also add in the location of each center of mass (mentioned above, the only factor that would make any difference where the wheel center is). Best to draw a 'rigid body' FBD for a 3 dimensional object with force vectors that may never intersect. More often used in dynamics but not unused in statics.

I think where someone got really lost and mistaken is there being some sort of added leverage with an offset wheel center vs centered wheel center, irregardless of the offset of the wheel. The 91-92 Camaro wheels have the spokes at the outer face of the wheel vs the GTA rims that are fairly close to centered and both are zero offset, I think where he got lost in his thoughts of how the force travels (and its magnitude, sorry but it doesnt change) is along those lines rather than talking about spacers. The best way to counteract traveling down that wrong path of thought is to realize that irregardless of the location of the wheel center, the tire itself will wear evenly across the tread. There are literally millions of cars on the roads that prove this point every single day. That might clue you in that the normal force is in the center of that tire, and the center of that wheel, irregardless of the wheel center's location. Might have something to do with the air in there distributing the force....

You can stand on a wall or at the end of a board 4' off the wall if you want, does not matter to you or the Earth same as it does not matter to the car or the Earth. You are still 200lbs (or whatever) and the Earth still sees 200lbs on the wall or cantilevered on the end of a board. Go get your bathroom scale and a piece of wood, put a heavy weight on the wood and cantilever the wood off the scale. Place a weight on the scale, read the number. Place the same weight on the end of the wood 1' off, 2' off, 5' off, as far as you can without it all falling apart or tipping the scale plate off its spring plate, and read the number on the scale. I can save you the time and tell you its the same number, but go ahead and try it if you want. You can support this by experiment or by math, results are the same either way. Only thing standing on the end of a board will do is stress the board and the wall, irrelevant to the discussion of force transmitted to the suspension.

 

Thanks for those tire specs, I bet you can get a more desirable ride with a little playing with the front end alignment, and there may be some getting used to the new equipment. I don't remember having trouble with 245's, but with 275's up front, I have found that a little more camber took away most of the pulling the car used to do when driven over lumpy pavement / lane changes on highways (where the tire would ride up or catch grooves in the direction of travel.) That may not be exactly what you are experiencing, but was the first thing that came to mind with 'touchy.' The Milliken book does a better job of describing it.

Quotes are from Race Car Vehicle Dynamics, Milliken & Milliken.
Effects of tire and rim size changes (p. 392):

Cornering: "Cornering stiffness is often a function of tire/rim size, aspect ratio, and width. As has been mentioned earlier, higher cornering stiffness tires require lower slip angles to produce a given amount of lateral force." (p. 394)Steering force & ratio: "Changes in tire cornering stiffness change the effective steering ratio. ... Simply changing the rim width can change the cornering stiffness; changing to larger size or wider wheels will likely raise cornering stiffness. Steering wheel force will be affected if tires are fitted that have different aligning torque characteristics." (p. 400)Steering kickback: "The early wide street tires had a tendency to 'nibble'; that is, follow longitudinal ridges in the road. While this has been improved, the current use of very wide tires has brought it back. If this is a problem, little can be done but play with tire pressures or change to narrower tires." (p. 400)I am sticking with "Spacers don't change the suspension geometry or how the load is transferred to the suspension" because of my own analysis, and because in all the suspension geometry analysis in the Milliken & Milliken book, no mention of the wheel/hub geometry is made. If the force generated by the contact patch is in the same place relative to the suspension points, what connects it to the upright only matters if it deforms significantly (like the tire sidewalls) under the changing loads.