I did not think anyone argued the basic rule on total steady state wt transfer, side to side.

All of your quotes are about shock behaviour, independent of springs/bars. The example I gave used real data to look at the combined effect. If you disagree with my specific analysis, fine, but explain why.

Also, several of those quotes are not true. What shocks do is allow you to tune the fr/rr wt balance different vs steady state for transient conditions.

"The damper velocities and travel directions resulting from each
cornering phase affect the distribution of load among the four tires."

http://www.turnfast.com/tech_handli...g_springs.shtml

BINGO!

ex: take a properly set up car to Summit Point, then crank a left rear Koni to max, and hit wagon bend (turn 3 left with negative banking) at 80 and see what happens ... gross oversteer because you temporarily have excessive rear weight transfer since the left rear can not properly droop. What you did, in a bad way, is delay wt at the front, and speed it up at the rear too much.

 

by Damon:
"If you're not changing cornering G, CG height, or track width you're not having any effect at all on how much weight your car transfers across the chassis! There is no hidden meaning in any of this. Springs, bars, shocks or tires do not change the total amount of weight transfer! End of story."

correct.

published quote:
"The left side rebound settings should be used to control weight transfer to the right side of the car. Shocks do not change the amount of weight transfer, only the time it takes to transfer the weight."

not correct, violates Damon's quote.

 

???? What do you mean? Shocks can control weight transfer because speed sensitive while springs are not? I miss the meaning of it.

As this horse is still willing to drink, I went through all the links you provided. All of them seem to be dealing with shocks mainly, not springs.

No one has been disputing the notion that shocks are an effective tuning tool to control weight transfer rate. The question under discussion is whether or not springs do affect weight/load transfer rate.

You have been very vocal in positively affirming they don’t and I highly value your experience and opinion. On the other hand, I made an attempt to support the opposite argument that indeed they do, based upon my understanding of basic physical laws. It would be useful – if you still care doing so - if you could please develop some alternate convincing arguments (or refer to some reputable relevant reference/link) to help me and understand why I am wrong.

Thanks.

- Sandro

P.S. one of your link http://www.turnfast.com/tech_handlin...eightxfr.shtml states

“We said that springs, shocks, etc. cannot change the amount of weight transfer. What they can change is the rate of the weight transfer”

A few lines after it adds though

“It turns out that shocks have the largest impact on rate of weight transfer”

This is still different than saying that springs or other suspension components that resist body roll have nothing to do with it. It’s a matter of relate stiffness I guess, as more comprehensively discussed by Kevin in his previous excellent posts.

 

You get oversteer because the rear acts stiffer during the transition and is trying to lift the rear wheel(in your example).

Listen to DamonB, he has up to four people working for him(depending on the shift) to find info to back up what he types

 

There's not a violation there at all. The first paragraph speaks of how MUCH weight transfers and the second speaks of how FAST that weight transfers. No violation. There's nothing mysterious going on here.

 

Impossible! Crazy! Weight is instantly transferred!

For every action there is an equal and opposite reaction. When you push on the spring the spring pushes back against you. In order for the spring to push back against you something must be pushing back against the spring at the opposite end. In this case that is the floor. If the spring is pushing against you, then the spring is also pushing against the floor. The spring can't somehow manipulate the space time continuum and push against you without it itself instantly transferring the load through it and pushing against something else.

In order for the spring not to instantly transfer the weight you apply to the floor then the spring has special super powers that allow it to levitate and travel through time.

 

Hooke's Law shows us how to calculate the force of a spring:

F= kx


where

F is the force exerted by the spring

x is the displacement of the spring

k is the force constant for the spring (a derived constant that tells us how stiff this particular spring is)

Springs that require a large force to stretch or compress them have large k values.


No where in Hooke's Law is speed mentioned because speed doesn't matter. A spring doesn't comprehend speed. All it knows is it displaces a certain amount in response to a certain force applied. Whether that force is applied over 2 seconds or 2 years the spring doesn't know the difference.

This is simple, fundamental and unarguable. Just as the weight transfer equations don't have any elements representing spring or shock forces, Hooke's Law doesn't involve speed because speed doesn't effect the calculation!

 

"This is simple, fundamental and unarguable. Just as the weight transfer equations don't have any elements representing spring or shock forces,..."

Again, true for total steady state 1g wt trans from one side to the other. But to determine fr vs rr wt transfer balance as a 1g corner is initiated, the REAL equations include bar, shock, and spring rates at all 4 corners.

at t=0 in the 1g corner, total wt transfer is basic, but the front/rear transfer bias will transition after t=0 based on the springs bars and shocks, and the rate of transition to basic steady state fr/rr transfer bias will be based on these same factors.

again, my example illustrates this.

 

the 2nd is not correct. If you are at 1G, total wt tranfer is done ... it's called force/moment equilibrium. The inside shocks may be temprorarily providng some of the unload that eventually will be done by the springs/bars. and, if shocks are set stiffer in rear, wough will have some temporary oversteer.

It's best to make a clear distinction between basic total wt transfer, and it's steady state bias, and transient wt transfer and it's variable bias due to shocks.

 

DamonD,

I believe your rationale is faulted. You are assuming religiously that your body weight HAS TO BE transferred instantly to the floor. In order to defend this assumption, you are driving conclusions that negate elementary Newton’s laws.

If forces were in equilibrium all the time on top and bottom of the spring – as you assume uncritically – the spring CG would not move at all upon you stepping onto the spring (some force needs to be applied for the spring’s CG to move).

In the example, there are only two conditions when forces are in equilibrium and the spring CG is not moving:

- before you step onto the spring (when the normal force from the floor reacts – and is in equilibrium) with the spring weight and whatever other weight the spring is supporting already
- after the spring has finished compressing enough to exert a force (Hooke’s Law) equal to your weight

Between these two conditions, the top and CG of the spring have moved. Therefore, there must be a force of some sort which is applied initially and make the spring move, which force then fades out as the spring compresses. Such force is the difference between your constant weight and the variable force generated by spring as it compresses (according to the Hooke’s Law). Its amplitude is the largest at the instant you step onto the spring – as the spring has not started moving yet – and becomes zero when the spring is compressed to the point to be able to hold your weight without further deflection.

Let’s go to the floor now. Since the floor doesn’t move while you are on the spring and the spring compresses, the forces must be in equilibrium all the time there. The normal (reaction) force exerted by the floor must therefore be equal all the time with the force exerted by the spring. But such force is the very same one derived from the Hooke’s Law, which would only include the entirety of your weight only when and after the spring has stopped compressing, but not before such time.

I hope this helps more this time.

- Sandro

 

I've already listed the REAL equations.

You're not discussing the amount of weight transfer here, you're discussing how that amount is shared by the front and rear axles. The total amount of weight transfer will be the same. What you're confusing above is weight transfer versus how that weight transfer is shared between the front and rear axles.

You give me a vehicle's weight, its CG height, its track width and the cornering g it's experiencing and I can tell you exactly how much weight is transferring off the inside tires and onto the outside ones. That's all I need to know. If instead you want to know how much weight transfers onto the outside front tire versus the outside rear tire then now I need to know the anti-roll rate of the vehicle which means knowing the spring and bar rates (I don't need to know anything about the shocks. The shocks don't effect how much, they effect how fast). Now I could tell you what the load increases of the front and rear outside tires are, but no matter what you do with the springs, shocks and bars the sum of the front and rear weight transfers will always equal the total that the weight transfer equation said it would!

When I put a stiffer front swaybar on my car I do not increase the total weight transfer of the vehicle, I merely force the front end to accept a greater percentage of the total weight transfer the chassis experiences. The total MUST stay the same so whatever percentage increase the front receives, the rear must decrease by the same amount.

 

They need to look up the definition of the word; "cause" and the word "effect".

 

For steady state, 1g corner, big sweeper, all this is true, I AGREE. Total wt transfer per basic cg/track eq'n, and ususal fr/rr bias calc'n.

And this is were you go off track. Yes TOTAL W.T. stays the same, but tuning front and rear shocks will alter that normal front-rear bias during the TRANSIENT, as you start and finish the 1g corner. High damping means this temporary change in bias will last longer until steady state bias occurs (steady state = no shock influence). If normal bias is 60% roll stiffness up front, stiff rear shock settings will temporarily give mabe 60% REAR bias as the corner is entered. At the extreme, with shocks off a 100 ton earth mover installed in the rear, you would essentially lock up the rear suspenion and have 100% roll stiffness in the rear as you enter the corner. If you had turned onto a constant radius test track, eventually the shocks would move and spring and bars would again allow the normal 60% front roll stiffness.

I'm not confused. Before responding, a few deep breaths, think about my specific examples, and consider the possibility that this consultant that has published and lectured in the field of Mechanical Engineering, just might be right. And mabe, we are kind of saying the same thing. I'm sure you know exactly how shock changes will alter how the car enters corners.

 

wow, that's great technical input. thanks

 

No, it's because you won't allow yourself to see it's really this simple.

During the transient g is changing; by definition that's what a transient IS. G is not constant throughout the turn. G doesn't instantly peak as the car corners, g builds until it peaks, and then it declines again. If you want to know what the weight transfer is at any given instant around the corner just plug in the instantaneous g for that moment in time. You still don't need to know the shocks settings because monkeying with those settings will merely create various g numbers at different instances around the corner. The equations I've posted will still be valid. Look at any of my data from my datalogging thread and you'll see exactly what cornering g does during a turn.

The shock settings allow you to tune the speed of the total weight transfer as well as front versus rear when the shocks are moving (during the time the car is transitioning). While the car is transitioning g is constantly changing. For instance if you set the rear shocks in such a manner that the car instantly goes loose during turn in the cornering g of the vehicle plummets. If I plug the instant cornering g of the oversteering vehicle into the equation I can tell you what the total weight transfer is at that instant. I don't need to know how the shocks are set, their asjustments will show as changes in g. The weight transfer eqautions I listed are 100% valid. You will not ever find a case or instant where they are not.

It's very fundamental. Vehicles exhibit weight transfer because the tires' reaction forces occur at ground level and the CG of the vehicle is some distance above ground level. You can see from the equations that if CG height is zero weight transfer is zero. This is why racecars are designed and run so low to the ground. If you don't change the CG height, the track width or the wheelbase of the vehicle, you will not do anything that changes the amount of weight transfer at a given g!!!! Springs, bars and shocks don't change the CG location or tire locations, therefore it's impossible for them to change the total amount of weight transfer of the vehicle!

 

This argument can be ended very quickly.

KevinK2, find an equation that calculates the total amount of weight transfer which insists you need to know anything about the spring, shock or bar rates.

Open Milliken and Milliken to section 18.4 and explain how they have it all wrong.

Sandro, find an equation for the rate of a spring which insists you need to know anything about time.

Open Milliken and Milliken to section 21.4 and explain how they have it all wrong.

If you can't do that, you can't prove me incorrect.

 

DamonB,

In my previous posts I compared initial and final states to demonstrate that weight cannot be transferred instantly to the floor. Below is a more realistic example, which takes into account elasticity (the spring) and inertia (related to mass and velocity, which causes the spring to overshoot). While the conclusion is the same (it takes time to transfer your full weight to the floor), this example is maybe easier to grasp.

The system I had described previously – a mass (your body) on op of a spring – is a simple harmonic oscillator. Elastic (spring) and inertia (mass and velocity) properties cause oscillatory motion to occur. The frequency of the oscillation is proportional to the square of the k/m ratio, where k is the constant of the Hooke’s Law (i.e. spring stiffness) and m is the system mass. Once put in motion, a perfect oscillator (no friction or dampening) will continue oscillating for ever, by keeping swapping potential energy and kinetic energy. The amplitude of the oscillation is determined by the initial condition (initial displacement and velocity). In real life, because of the spring’s internal friction, the oscillation amplitude will keep deceasing (but not the frequency) until the oscillation fades out.

Now, let’s place the spring on a scale on the floor. Scale should not be spring operated like your bathroom scale but use a rigid thin strain gauge instead. Step onto the spring and wait until any residual oscillation stops. All forces are in equilibrium now and the scale would read your weight [equal to your mass x 1G, as we can assume the spring has negligible mass in comparison].

Let’s assume spring k=20 lb/in, spring length 30” at rest, and your mass m=200 lb. The spring is therefore compressed 10”, top of the [compressed] spring is 20” from the floor/scale. Force on top of the spring is 200 lbf and the scale would read 200 lb.
Now, have a crane (or a few friends) lift you up 10” by applying slowly an increasing force up to 200+ lbf force, and stop when spring top is at 30”. Your mass is levitating in the air now held by a 200 lbf force applied by the crane. Obviously, the scale on the floor would read 0+ lb (just the spring weight). The force exerted by the spring is 0 lbf both at your feet and onto the scale. While the crane was lifting you, the scale reading had kept decreasing from 200 lb to 0 lb. The scale reading was indicative of the force exerted by the spring both at your feet and onto the scale = (30”-X”) x 20 lb/in, where X” is the height of the top of the spring from the floor.
Now, the crane or your friends stop holding you suddenly – note this instant is equivalent to my previous static example of you stepping onto the spring – and you start dropping and compress the spring.
The scale would read 0+
From now your body and the spring will start oscillating like a harmonic oscillator with a frequency = 1 Hz
When X” reaches 20” the scale reads 200”. This happens after 1/3 sec. However, because of the inertia, your mass will keep dropping and the scale reading increasing as the springs continue to compress. Were no friction present, the spring would compress another 10” and the scale would read 400 lb. In reality, it will compress some less and the scale will read something in between 200 and 400 lb; equal to [200 lb + (20” – X”min) x 20 lb/in]. The spring will start decompress now and push up your mass. After another 1/3 sec X” = 20” again and the scale reads 200 lb. Again, if there were no friction losses, the mass would be raised again up to X” = 30” and the scale would read 0 lb. In reality, X”max will be less than 30” and scale will read more than 0 lb. At this point, a new cycle begins and after 1/3 sec X” = 20” and the scale reads 200 lb again.

Let’s change spring now and use a stiffer one, k = 80 lb/in. Frequency is now 0.5 Hz. It takes now only 2.5” compression to balance the mass weight. The first time the scale reads 200 lb is now 1/6 sec instead of 1/3 sec with the previous spring.

Finally, let’s use a 1,000 lb/in spring. The scale would read 200 lb this time after less than 0.05 sec and compression would be a very small 0.2”

The example above demonstrates that applying a weight to a spring, the load is transferred to the floor (or a tire) faster through a stiff spring rather than a soft one, even though the spring factor k does not depend upon velocity.

- Sandro

 

DamonB is doing enough typing for five people to explain this.

You guys are confusing wheel loading with weight transfer it's that simple.

 

And this a link to a nice animation of the example above

http://www.kettering.edu/~drussell/Demos/SHO/damp.html

- Sandro

 

That shows what a shock/damper will do in relation to deflection. Not that the blue spring is stiffer.

 

Yes, of course. Where did you get the idea that the two curves represent different springs?

Read again: the animated gif at right shows two mass-spring systems initially at rest, but displaced from equilibrium by x=xmax . The black mass is undamped and the blue mass is damped (underdamped). After being released from rest the undamped (black) mass exhibits simple harmonic motion while the damped (blue) mass exhibits an oscillatory motion which decays with time.

I posted the link just to show an animation of what I described happening with an "ideal" spring and with frictional losses - oscillation fading away, which is what happens in reality. Stiffer springs cause higher frequency, since frequency is equal to square root of k/m.

This is another link you may find the animation more interesting as it let you change the values of k and m and see the change in frequency.

http://www.lon-capa.org/~mmp/kap13/cd361a.htm

- Sandro

 

I thought you were trying to show the stiff spring/ soft spring difference, my bad

 

All that was demonstrated was that you don't understand the concept. Oscillation has exactly zero to do with the rate of any spring.

You criticized me earlier and claimed you'd read Milliken and Milliken's book. Open Milliken and Milliken to section 21.4 and explain how they're incorrect. If you in fact don't have your own copy of M&M go to any library and check out any elementary physics book. Derive your own math which proves spring rate is a function of time and post it.

I'm through here until someone can post equations which support their theorys and explain those equations in a manner which proves you comprehend them. No more nonsense gobbledy gook.

 

DamonB,

1. I never criticized you. I indicated that I respect your experience and knowledge and just asked to explain and support your statement that springs do not play any role on how fast the load is transferred to the tires as a result of a weight transfer.
2. Your explanation so far has been that “a spring has absolutely no idea how fast something is happening and therefore has no effect on the speed of weight transfer”, and that, as such, the load resulting from a weight transfer must transfer to the tires instantaneously (!?)
3. I attempted to explain why if that were indeed the case, it would contradict 400+ years of universally accepted physics laws.
4. Your “scientific” replies have been of the kind “Impossible! Crazy! Weight is instantly transferred!....” and repeating your mantra that since no velocity term is present in the Hooke’s Law, then springs must have no influence on how fast weight get transferred to the floor.
5. Finally, I invoked the more comprehensive harmonic oscillator model, to show that a load resulting from the instant application of a weight on top of a spring is transferred to the floor according to a sinusoidal pattern.
6. In response, you asked me to derive my own math and prove that spring rate is a function of time. It appears that in your mind this should still be required in order to transfer load gradually. Indeed it is not. The harmonic oscillator model shows that the frequency of the oscillation is a function of the spring rate, even though the spring rate is not a function of time. Therefore, there is no need to prove that it is. I have no reason to doubt that the harmonic oscillator physics works and I am perfectly happy with the conclusions that can be derived from its application. On the contrary, you keep maintaining that a load resulting from the instant application of a weight on top of a spring is also transferred to the floor instantly, which contradicts the harmonic model. So, why don’t we reverse this. If you are not happy with the harmonic model and its conclusions, and still want to support DamonB's Law, why don't you derive your own math, possibly consistent with existing universally accepted physics law, and let your peers understand your theory. And let me repeat once more that the simple harmonic model works with a spring rate that is not dependent from time or velocity, which indeed is not (we are 100% in agreement on this one). Therefore, you cannot invoke this lack of property to demonstrate that the harmonic model does not work.
7. Also, please explain your novel statement that “Oscillation has exactly zero to do with the rate of any spring.” On the contrary, it has to do a lot in my book. The frequency of the oscillation is equal to square of k/m, where k is the spring rate, and m is the mass attached to the spring. So here too, what did you mean?
8. Finally, I can’t refrain any more from commenting on the rudeness of your posts. Why do you do that? I never insulted you – while you seem to be doing it all the time. I enjoy discussion and debate but certainly do not enjoy your rudeness.

- Sandro

 

What this all comes down to is that you both are right, well I know DamonB is, but your talking about different things. From Sandros and KevinK2's posts I sort of breezed over them once I got the idea of what you were talking about, I didn't get out my calculator to check your math or anything because I knew you were getting away from what weight transfer IS and how you figure it.
You both(Sandros& Kevin) have done good jobs describing a few of the many things that have to be controlled when you start dealing with weight transfer and how it loads the tires. Thats where the cause and effect post came from. Your getting ahead of yourself with chassis dynamics and still trying to call it weight transfer.

How the chassis and suspension splits up the weight transfer and sends it to the ground at any given moment is not the same thing. That is where all the magic happens and it involves every part of the suspension from the air in the tires to the bushings and everything in between. So all your posts about springs are probably correct, they just don't mean anything when it comes to figuring weight transfer.