Sandro

iTrader: (1)

Jgrewe,

We cannot be both right as our views are opposite.
And there is no confusion between cause and effect. We are not discussing what causes weight transfer.
The questions under debate are very simple: If you add a weight on top of a spring, is the resulting additional load transfered to the floor instantly, even before the spring starts compressing [DamonB's view], or gradually [my view - according to a harmonic oscillator model]? And, would a stiffer spring cause the additional load to be transfered faster to the floor [my position] or not ? [DamonB's position - by definition, as DamonB's view is that load is transfered instantly whether you have a 30 lb/in or a 1,000 lb/in spring]

- Sandro

jgrewe

iTrader: (1)

OK, no problem. It seems like his posts are dealing with weight transfer when it comes to a chassis and yours are going deeper into the ,uh, "rest of the story".

How about this example for your spring. What if you lay that spring on its side? When you push on it with say 100lbs of force wouldn't you need at least 100lbs of force on the other side to keep it from sliding sideways as soon as you touch it? Because if you only had a 10lb force on the other side THAT would be how much you would be able to push on it.

Does your force that starts an oscillation always get to bounce on the spring? What if the force is from a hydraulic ram? Because there has to be some time frame of your load being supported by something else and the spring holding all the weight. And the oscillation is coming from the movement of the mass when the spring compresses so inertia has to be taken into account.

I promise I won't just breeze over your answers for these questions. I'll probably have to brush up on some of Newton's Laws for this discussion

Sandro

iTrader: (1)

Great questions Jgrewe!

Let me start by pointing to a peculiar characteristic of a “vertical” oscillator first (e.g. the example I used earlier or one of those scale to weight a fish, where the mass is hanged to the spring). Unique of these configurations is that the force resulting from mass x 1G is constantly applied to the end of the spring which moves. No matter how the spring moves, that force is always there and always the same. When you apply forces to an oscillator, different than mx1G, things become quite complicated. Just google for “forced harmonic oscillator” and look at the math. You will also notice the equation is generally limited to a sinusoidal “forcing” function. In reality, forces can be applied in a variety of shapes. Mathematically, is always possible to describe any shape though as a sum of elementary “forcing function” with multiple frequencies and amplitudes. If you want to calculate what goes on mathematically, you have to consider all these various harmonics (or at least the most relevant ones) simultaneously. You can conceptually think of these force-waves actually “traveling” through the spring and interfere with the spring-mass own frequency (like when you move water with a hand in a tub and cause complex wave patterns). As a result of these interactions, the oscillation characteristics – and therefore the forces generated by the spring as it displaces – can be reinforced or weakened. The link below shows an animation of a forced (and damped) oscillator with three different driving frequencies. Note that the resulting motion can be very complex, like move up, reverse a bit, then move up further. http://www.kettering.edu/~drussell/D...***-force.html

Back to your question. let’s consider a vertical oscillator first and see how a horizontal spring can be related to it.
Let's have 20” lb/in spring, 30” long, 100 lb mass on top. At equilibrium, spring top is at 25” (5” compression @ 100 lbf – note this is a force – divided by 20”/lb). Let’s remove the forcing weight (force) by lifting the 100 lb mass 5”. Spring is unloaded, no force generated by the spring, no force on the floor (other than the small spring weight). Drop the weight, spring starts oscillating, overshoots the equilibrium position (25”) because of the inertia and so one. In the meantime, force on the floor matches the force generated by the spring as it moves up and down, smaller when the springs is extended, more than 100 lbf when spring compresses beyond 25”.

Let’s look at a horizontal spring now and assume there is also a 100 lb mass attached to it. This time though the mass does not generate any force on the spring. The equilibrium position is therefore 30”. Let’s compress the spring 5” (assume the other side is held still by a wall) by applying (slowly) up to 100 lbf and stop. Note, force balances are same as in the vertical spring final steady-state case. The spring is compressed 25”, 100 lbf are generated by the spring and applied to the wall. Now, let’s remove the 100 lbf force instantly. The spring will expand (remember that the final steady state position is 30” in this case) and overshoot the 30” because of the inertia associated to the 100 lb mass. If the spring is not anchored to the wall, mass and spring will get lost in space. But this is not your example yet, right?

Let’s move a step further then. Let’s assume the 100 lb mass is still there, spring at rest at 30” and now we apply the external 100 lbf force (your ram). And now the catch. How is this 100 lbf applied as the spring moves? Let’s assume your ram is so smart and quick to follow the spring as it moves harmonically and continuously apply 100 lbf on it. Note that the final equilibrium is 25” (100 lbf need to be generated by the spring), exactly as in the vertical case. In this configuration, we have now replaced the 100 lb x 1G force with an external “smart” force. The 100 lb mass is still responsible for the inertia force and cause the spring to overshoot. The load on the wall will match the sinusoidal pattern already described in the vertical oscillator. So, even in this case, if you want to keep the spring from sliding away, you have to apply a sinusoidal force that follows and matches the force generated by the spring as it compresses/decompresses. Quite small at the beginning, but over 100 lb when the spring overshoots. Note, in this example, with the numbers we used, the oscillation frequency is 1.4 Hz, and the load would be close to 200 lbf in less than ½ sec.

Lastly, let’s consider the case where there is no mass attached, or a very small one, 1 lb, just to make the calculation still mathematically possible. The frequency is 14 Hz now, very high. On the other hand, since the mass is a mere 1 lb, the inertia force is very small, and the overshoot will be minimal Assuming you have a super-smart, super-fast ram capable of following such a fast moving spring and constantly apply 100 lbf, the oscillator will rash to its 25” equilibrium position in a mere 0.02 sec, the overshoot a little bit (due to the small inertia force associated to the small mass). Since friction forces increase with frequency (molecules inside the spring rub each other very fast), the spring will vibrate a bit then come to rest at 25”, and so will the ram. Therefore, in this case, in order to avoid the spring to slide, yes you better off anchoring the spring to a good wall to absorb the fast approaching 100 lbf. However, mathematically, the front force wave is still a sinusoid.

- Sandro

KevinK2

CAN SOMEONE SAY AMEN?

But reality is both springs/bars and shocks control transfer rate, as several of your links mention.

The increase in normal force at either outboard tire, as a long corner (say .3g) is initiated from going strait at constant speed, can be simulated by standing on a spring with damper, no preload, and you temporarily support your weight (150 lbs) by an overhead rope. Letting the rope go is like initiating the turn. the damper your standing on may provide 40 lbs of support (and load at the ground) as spring compression starts. Your body initially accelerates down but then decelerates as spring load ( ie tire contact load) increases. As you approach max compression, velocity down aproaches zero, and damper force drops to zip as now the spring fully supports your 150 lbs wt. Assuming the shock rebound is critically damped, no significant rebound occurs.

Full body wt transfer (150lbs) to the ground occurs when you stop compressing the spring. If the spring rate is tripled, the transfer to the ground from 40 to 150 lbs is quicker. If spring is not changed, but the damper was set tight to say 150 lbs viscous damping, then full wt transfer to ground occurs instantly thru shock, and you the are slowly lowered down as weight is transfered from the shock to the spring.

The sprung wt of the car does the same thing during ROLL. Initiall outward acceleration of the CG, relative to roll axis, is resisted by shocks and springs/bars, and final equilibrium deflection is based on springs and bars, as is the weight distribution front to rear. The total wt trans is based on the simple equation using cg, track, and g's.


agree except wheelbase doesn't matter.

Please don't repeat this, 10 times is enough. ALL AGREED on this from the start.

-----------------

This is one of your best links, and it is consistent with all my points.

http://www.smithees-racetech.com.au/...shocktune.html

It also should be read before reading some other links, as a warning:

"... But shock tuning for balance is a different story. Many of the popular Authors on race car set up and handling have got errors in their descriptions of what is happening, or wrong suggestions for changing shock settings. The same applies, amazingly, to some shock absorber company manuals, and tuning advice issued by them."

KevinK2

DamonB, find any place on this site where I said you needed that info for Total Wt Transfer. I bet 1 cup of coffee you can't.

jgrewe

iTrader: (1)

Now I'm going to have to read some other stuff to wrap my brain around all this.

I think I'll go stay at a Holiday Inn Express tonight too...

jgrewe

iTrader: (1)

Ok, now I may have forgotten "The Question" after digging around the net looking up Hooke's Law and other stuff. But doesn't this say what DamonB has been saying? The load gets passed through the spring instantly and the amount of oscillation is more about how the chassis responds.

I understand the frequency issues with stiffer springs etc but I'm not sure how this discussion got this far away from swaybars(yea they're springs but this is pretty deep). If you need to have the exact load on both sides of the spring to keep it from wanting to move, not just compress, isn't that what is happening? Its just that the ground happens to be an infinite variable on being able to take care of the "push back" part. So the load is passing through in simple terms?

Am I totally lost now or what? Either way this has been fun, I'm always looking to learn something here. If I didn't I wouldn't waste my time!

KevinK2

This is a good example. Springs do have a response rate, or period, that is a function of the spring rate and the applied mass load.

But to clarify, the highest point on the illustrated motion curve would represent where the spring in your examples has no load on it ... just before weight is applied.

And, when your weight (or mass) is 1st applied to the floor spring, ithe inital contact force at your feet is small, so you accelerate down. As the spring compresses, force at your feet increases (as does floor force), and your value of acceleration down decreases. In theory, no damping,when you have compressed the spring enough so that the force at your feet equals your weight, acceleration hits zero and reverses direction (position zero on graph), so you have hit peak velocity down (no damping) and now you are slowing down and overshooting just matching your weight. Dampers on cars usually prevent this overshoot.

-------------

Back to bars .... bars and springs control final steady state FRONT to REAR weight transfer balance, and shocks can alter that front rear balance as the car enters ( or exits) the corner and the sprung weight goes through body roll.

Sandro

iTrader: (1)

[QUOTE=KevinK2]This is a good example. Springs do have a response rate, or period, that is a function of the spring rate and the applied mass load.

But to clarify, the highest point on the illustrated motion curve would represent where the spring in your examples has no load on it ... just before weight is applied.

And, when your weight (or mass) is 1st applied to the floor spring, ithe inital contact force at your feet is small, so you accelerate down. As the spring compresses, force at your feet increases (as does floor force), and your value of acceleration down decreases. In theory, no damping,when you have compressed the spring enough so that the force at your feet equals your weight, acceleration hits zero and reverses direction (position zero on graph), so you have hit peak velocity down (no damping) and now you are slowing down and overshooting just matching your weight. Dampers on cars usually prevent this overshoot.
QUOTE]

100% correct and 100% in agreement. Mass is accelerated by the difference between weight and the force exerted by the spring divided by the mass. a = (weight - k*x)/mass. Force exerted to the floor is always equal to k*x. When spring is at the top and weight is applied, k*x is zero [spring fully extended], force on floor is zero, and acceleartion is max a = (weight - 0)/mass. As spring gets compressed acceleartion decreases bur velocity still increases since still is a>0. When the spring is compressed at the equilibrium point, weight = k*x, acceleration is zero and the mass stops being accelerated. Velocity is at its max at this point and force on floor is equal to k*x=weight. As the spring gets more compressed, mass decelerates since now k*x is > weight and speed is reduced. Force on floor is always equal to k*x, which is now > weight.

- Sandro

Sandro

iTrader: (1)

No, this is exactly the opposite of what DiamonB has been saying. The load does not pass through the spring instantly, unless the spring is infinitely stiff and does not compress. In all the other cases, the load to the floor varies with time and is equal to F=k*x all the time.

Swaybar contributes to the overall spring rate. In one of his previous post, Kevin referenced to how to calculate that.

An infinitely stiff bar will make the spring rate of any spring (even a soft one) infinitely stiff. This would be the only case were spring does not compress further on weight transfer and load is transfered to the floor instantly.

- Sandro

DamonB

There seems to be an emerging theory here that will change history. I haven't seen it named yet so I'll call it the Theory of Spring Levitation. The Theory of Spring Levitation says that springs do not instantly transfer force through them. The spring is instead able to support some mass or force for a short period of time without the spring transferring any of the force it sees at one end to the opposite end of the spring.

This means that springs can levitate mass and/or resist force for some period of time without a counteracting force occuring at the opposite end of the spring. Unfortunately the length of that period of time has not been defined as of yet.

The Theory of Spring Levitation goes on to tell us that the lower the rate of the spring, the longer the spring can support the mass and/or resist force without that force being realized at the opposite end of the spring. As the rate of the spring increases the period of time the spring can levitate mass and/or resist force becomes less and less until the rate of the spring reaches infinity. At this point we no longer have a spring but a rigid body and the theory says the weight will now instantly transfer. According to the Theory of Spring Levitation only rigid bodies can instantly transfer weight. Non-rigid bodies transfer weight at some speed which is inversely proportional to the flexibilty of the body: A highly flexible object transfers weight more slowly than a highly rigid one.

Coil springs are nothing but a rod wound into a helix. The Theory of Spring Levitation shows us that a rod stood on end will instantly transfer weight since the rod is rigid. However if we bend that same rod into a different shape the rod now attains new properties, the most important being that it can resist forces at one end without instantly transferring those forces to the opposite end like it did when it was straight.

What does this ground breaking new theory mean? Here are some examples:

It means that if I were to place a spring between my fingers and hold one finger still while moving the other finger to compress the spring I'd instantly feel resistance against the finger which is compressing the spring, but the finger opposite the spring which I'm holding still would not feel any resistance for some period of time (length of time as of yet undefined by the theory). Since the spring does not instantly transfer the force of my squeaze through itself, my stationery finger doesn't instantly feel the increase in force that my moving finger does. Therefore as I squeaze the spring between my fingers one finger instantly feels the force increase and the other doesn't. Unfortunately the theory doesn't define the length of time over which this phenomenon occurs, but the theory does claim that the forces against the fingers do eventually balance out and both fingers will feel the same resistance. This balance does not occur instantly though, it takes some as of yet undefined period of time. If I were to squeaze a stiffer spring the period of time over which this phenomenon occurs would be shorter. Anyone can disassemble their ballpoint pen and try this experiment.

The Theory of Spring Levitation also insists that springs can levitate masses. The only levitation experts I have heard of are magicians and I know very little of magic so I setup an experiment:

Figure 1 shows a balance that is in equilibrium. When you were little you would call it a "see-saw". At the left of the balance is a spring and at the right is some mass which opposes the weight of the spring and maintains equilibrium of the balance.

Figure 2 shows the same balance at the instant an additional 1 pound mass is added to the left side. The added mass means the balance is no longer in equilibrium and the balance will tilt to the left. According to the Theory of Spring Levitation the balance will instantly begin to tip as soon as the 1 pound weight is hung from the balance since the bodies are rigid. In our example the 1 pound weight disrupts the system so much that the balance instantly tips to the left and strikes the tabletop harshly.

Figure 3 shows the same balance at the instant a 1 pound mass is placed on top of the spring. The Theory of Spring Leviation tells us that the balance will not instantly tip over to the left and strike the table harshly as it did in Figure 2. Instead some undefined period of time passes during which the spring compresses, supporting the mass and achieving equilibrium within the spring itself against the mass. As this happens the balance sees the weight on its left side slowly increase from zero to 1 pound. This happens because the Theory of Spring Leviation says that weight transfers through the spring at some speed which is directly proportional to the rate of the spring. The balance will only slowly tip to the left because the spring is somehow manipulating the additional mass such that the balance doesn't realize the complete 1 pound mass exists the moment the weight is placed. Instead the balance sees a weight that increases from zero to 1 pound at a rate propotional to the rate of the spring. Therefore at first the balance will slowly tip to the left but begin tipping faster and faster as the spring transfers more and more weight onto the balance. This proves that the spring somehow levitates the additional mass in mid-air. The spring must levitate. If the spring pushed against the balance because of the 1 pound weight now on top the balance would slam against the table as it did in Figure 2. Instead the Theory of Spring Leviation shows us that the balance will tilt more slowly to the left and the rate of tilt of the balance is directly proportional to the rate of the spring. Even though the weight is resting on the spring which is resting on the balance the spring on its own can resist some of the weight from the added mass without transfering that weight to the balance. The rate of the spring will determine what period of time it will take before the balance feels the entire 1 pound increase in mass from the added weight above the spring.




Everyone feel free to replicate these experiments. While you're doing so you may begin to wonder what would happen if you maintained the masses at each end of the system in such a way that after compressing the spring you could cause the spring and the 1 pound weight above it as well as the balance to oscillate. Oops. Getting ahead of myself again. The balance wouldn't oscillate. The Theory of Spring Leviation shows us that the weight the balance sees is determined by the spring. Even though the 1 pound weight rests on the spring which rests on the balance, the balance only sees the weight the spring decides it will until the spring itself has achieved equilibrium. Only after the spring finishes compressing enough to achieve equilibrium in itself against the 1 pound weight will the spring allow the balance to feel the entire force of the 1 pound mass. Until that time the balance will "feel" less than 1 pound of additional weight.

If one were to investigate how to calculate the period of oscillation of an undamped or even a damped system we find that the idea which makes this calculation possible is Hooke's Law. In fact the reason we are able to calculate the oscillation frequency of a system in the first place is because the rate of the spring is constant at all times as Hooke's Law tells us. As opposed to a theory which is a model which proves true under carefully defined circumstances, a law is true at all times. That's why we call it a Law. If we derive the calculations which compute oscillation we find they come from Hooke's Law. If we derive Hooke's Law we find it comes from Newton's F=ma, which is the basis for the entire works of classical physics. The Theory of Spring Leviation flies in the face of all of this and will define a new moment in science, one even greater than Newton himself.

Unfortunately I've just learned that not just anyone can put forth a nomination for the Nobel Prize:

http://nobelprize.org/nomination/physics/index.html

Basically it says that if the Nobel Foundation hasn't contacted me and asked for my nominations I can't nominate anybody. Maybe someday the news will spread from our little forum here to someone who is asked to submit Nobel nominations.

jgrewe

iTrader: (1)

Now I understand it very clearly. Thanks for taking the time DamonB.


If you've been following along on this thread you will now know that it is about to get very noisey or very quiet in here. I'm heading for cover for a while and will lurk to see how this turns out.

KevinK2

Yikes ... that cold weather must be getting to you!

There is a shock wave that travels through a supported vertical rod subject to a hammer load. As I recall speed of sound in the material is involved, and we are talkin nanoseconds. Same with wacking a vertical spring with hammer. Instead a compression wave as occured in the rod, you get a twist wave along the spiral wire ... again near instant. Not levitation, basic impact load analysis.

Levitation? Copperfield stuff.

Case 3 is a simple dynamics problem. Limits:

very stiff spring and light weights: spring will hardly compress, balance hits table with same speed as case 2, but instead of instant impact load, contact force will have less peak as spring now compresses some and absorbs energy of the impact.

spring with minimal stiffnees, large weight: Spring offers negligible force and goes into coil bind as weight now impacts balance before significant tipping.

Properly selected (light) weight and spring and balance length:

Just like "step on spring" case sandro and then I discussed, except spring not grounded. Your basic f=ma equilibrium equation is more complex as you need to relate motion at top of spring to bottom of spring and different acceleration rates of ball vs spring cg vs sq weight on right side. No magic levitation, just physics.

Mahjik

iTrader: (1)

I don't believe Damon is referring at all to how much or how less the spring compresses. To me, it seems like he's only talking about weight transfer.

i.e. if you have a scale and you set a spring on the scale. Now, the scale displays the weight of the spring. If you then put a 2 lb object on top of the spring, does the scale show the spring + 2lbs regardless of how much or little the spring compresses?

Sandro

iTrader: (1)

DamonB,

First of all thank you for hinting we may have invented a new theory. But, actually, we have not.

Secondly, I apologize if this post will be short. I am not able to address each of your considerations as I will be traveling for the rest of the week and the time I will be able to spent on the Internet will be limited. I may come back on commenting each of your observations at a later date.

More or less the description you are giving are quite correct as far of the load being transferred as a function of the spring rate, although there are certain inaccuracies here and there, like what would you feel on your finger (which indeed is the force exerted by the spring).

In the meantime, please consider the following. As I read over your posts, what seem to be bothering you the most, is that you cannot reconcile the 3rd Newton’s Law (action and reaction) with the descriptions I had given. How is it possible, you are asking, that the force of the bottom of the spring is not all the time in equilibrium with the force that pushes on top of the spring? Indeed, it is, if you take into consideration the inertial forces that are also in play.

I admit it is not obvious at first, but let me attempt to explain how all this works and how all the 3 Newton’s Laws come into play.

Let’s consider a real simple oscillator first (with no external forces applied as it oscillates). See how that it works first, then over impose a constant force with the same frequency (the example with a weight on a spring placed vertically), and see how does it work this time.

The best approximation of a simple harmonic oscillator is a spring placed horizontally on a frictionless table with a mass attached to an end and the other end solidly linked to a wall., Since the gravitational field is at 90 degrees to the spring, there is no force on the spring (caused by mass x 1G) to consider. We take our usual spring, 20” long, k=20 lb/in, and a 100 lb mass. At rest, since no forces are applied to the spring, the spring is fully extended, force applied to the mass in the direction of the spring is zero, and force at the wall is zero. Now, we apply “slowly” up to 100 lbf force horizontally toward the wall. “Slowly” means that the force is applied gradually in order not to induce oscillation, so start with one lb and keep adding up force up until it gets to 100 lb. As you apply force, the spring compresses in order to balance the force that is gradually being applied. This is basically equivalent to your ball pen spring – finger example. At all the time, the same force is felt by the wall. The force amount is equal to k time x (were x is the amount of compression). With 100 lb applied, the spring has compressed 5”, its compressed length is 15” now. Before the external force was applied, remember the equilibrium status was 20”. So, no force applied equilibrium 20”, with force applied new equilibrium at 15”. CG (basically the mass CG) has moved 5”. Why? Because forces have been applied (1st Newton Law) in the meantime, action/reaction principle (3rd Newton Law) has been verified all the time while the spring was moving.

Next, let’s remove the 100 lbf, but this time “suddenly” no “slowly”. Now, all the fun starts. First of all, guess what, at the very instant I remove the 100 lbf external force, at time 0.000000000… sec, the force at the wall is still 100 lbf! I know I am losing at this point whatever credibility I had hardly gained with you so far. But please, bear with me, I will show you later on why this is indeed the case. So, let’s forget the wall at the moment and see what happen at the mass. With the external force removed, the only force exerted on the mass is the 100 lbf from the spring (5” x 20 lb/in). Let’s consider Newton 2nd law, a=F/m, where F=100 lbf (5” x 20 lb/in) and m=100 lb. Therefore, a=1G. What happen next? Well mass starts accelerating away from the wall. As the spring starts decompressing, the force on the mass from the spring decreases, although still a>1. Provided that the mass is subject to a force pushing the mass away from the wall, the mass keep accelerating, and its velocity keeps increasing. We eventually get to the point where the spring has extended 20” and the F exerted from the spring to the mass is zero. The mass has reaches its max speed. What is the force at the wall now? I guess we can safely say it’s zero. No force from the spring, no other external forces around to balance out. So, zero lb for sure. But, as the mass transit through the 20” sign, it is doing is while moving at max speed. Now, 1st Newton Law, we need some force to stop this mass, otherwise it will just keep moving at constant speed. As the mass moves on beyond the 20” sign, the spring starts stretching. The stretched spring starts pulling the mass with a force F=k*x (x is the spring extension this time). a=x*k/m is now decelerating the mass velocity slows down. Let’s go to the wall now. The same force “pulling” the mass is also “pulling the wall, and with the same intensity x*k/m. Once the spring has stretched to 25”, a=0 and v=0 as well, the mass has stopped. The spring “pulls” both the mass and the wall with a max force F=k*5”. Next, the mass, which is still pulled with a 100 lbf start moving toward the wall and accelerates at a=k*x/m. Velocity (toward the wall) increases. When the mass pass through the 20” sign, F=0 and v=max again. The spring has stopped pulling the wall and force at the wall is 0 lb again. Next, the mass with no forces applied to it at this instant, overshoots the 20” equilibrium point and start getting compressed. As it compresses, the spring builds up a force F=k*x, which decelerates the mass again and “pushes” on the wall at the same time and with the same force F=k*x. When the spring is finally compressed 15”, the force exerted by the spring on both the mass and the wall is 100 lbf again, which is equal to the force applied at the start of the process, before removing the 100 lbf external force. Needless to say, since the spring is still pushing on the mass but the mass has came to a stop, the mass gets accelerated again away from the wall. I hope you found this example intuitive enough and have no doubt that, as the mass oscillates, the wall gets “pushed” and “pulled” with a force equal to k*x all the time. Note, there is no external force applied any more. We “used” the external force just to move the mass from its 20” equilibrium position, the removed it “suddenly”. As a result, the mass started oscillating around its 20” equilibrium point, with a max extension of 25’ and a max compression of 15”. Note, there is no longer any external force applied, yet the force at wall keeps oscillating between +100 lb and -100 lb. On the average, it’s still zero, and eventually gets to zero once the oscillations are faded out. Again, I hope you find all this still enough intuitive. But, how to reconcile this intuitive behavior of this oscillator with a lack of force balance at the mass? At the wall things seem OK with the Newton’s 3rd Law, as there is an action/reaction between the spring pushing or pulling and the reaction force at the wall, but, at the mass, or, at the end of the spring attached to the mass? There are no external balanced forces applied there, yet, the mass gets pushed (or pulled) up to 100 lbs, but there is no an equal and opposite force to react there. Well, actually there is one, and has to do with inertia. Newton 2nd Law: F=ma (where mass is 100 lb and acceleration is the result of the force exerted by the spring a=k*x/m. So any time the mass gets accelerated by the spring the inertia “pushes” on that end of the spring as well, although there is no “real” external force. If you compare the two equations, you can deduct that the inertia force is equal all the time to F=k*x. So in conclusion, At each interface point, wall/spring and spring/mass forces are in equilibrium all the time and equal all the time to F=k*x.

Finally, let’s evolve the example and make it similar to the configuration I spoke about in my previous posts. We now have to apply a constant external force, applied in phase and with the same natural frequency of the simple oscillator examined above. Well it could be a smart ram, or more simply, a field able to exert a force on the mass, no matter where it is. A field like this could be a magnetic field (assuming the mass has magnetic properties) or – guess what – a gravitational field. This can easily obtained by rotating our system 90 deg counter clock wise, and make the wall become the floor. Now, the equilibrium point is 15” (vs. the 20” before as the 100 lbf weight needs to be balanced by 5” compression). If you start oscillation by lifting the 100 lb mass by 5” – zero lb on the floor - or pushing it by 5” (200 lbf on the floor), then leave it alone, the 100 lb will oscillate and the force on the floor will be in between 0 (when spring is at 20”) and 200 lbf (at 15”). Basically, same as the example below, with 100 lbf super imposed.

As you can see, once the effect of inertia is taken into consideration to balance the Newton’s 3rd Law as well, there is no basis any more to claim for any new revolutionary theory. Anyhow, thank you for the thought.

On the other hand, I believe that your theories, load transmitted instantly to the floor independently from spring rate, and also frequency of harmonic oscillator independent from spring rate, could be revolutionary indeed. I warmly invite you to develop them further.

- Sandro

jgrewe

iTrader: (1)

I'm learning more with every post here!

I'm wondering how half of the oscillation equation works on a car since we never get to stretch our springs during normal operation(not a rally car)? And that's not to say I don't know how to figure the natural frequency of a suspension. So are we discussing something that only F1 engineers worry about on race cars? Or do they not even care

DamonB

Unfortunately my textbooks are too old to agree with such a cutting edge new theory as Spring Levitation. I can't take any credit for the theory because all I did was put the information from others' post into one place and provide experimental examples. I'll leave it to you guys to further expound on them.

The next hurdle I see is someone needs to derive how to compute the period of time over which Spring Leviation takes place. This would be a formula that tells us what the length of time would be for a particular rate of spring which is countering some mass to completely transfer through the spring 100% of the weight of the mass which the spring is countering.

KevinK2

Simple ... simple harmonic non-levitation motion.

spring on ground, hold mass just touching top of spring, let go.

total period for it to go down and back up is 2 x pi x (m/k)^.5 (no damping)

time for force at each end of spring to simultaneously go from zero to mass's weight is 1/4 of the total period.


----------------

back to balance beam case 3. I derived equations of motion for a similar case, where lever is more like a pulley system, to eliminate rotary inertial effects. If spring point on balance goes down opposit side must go up. Also assumed spring had no mass, but two equal weights were fixed at ends of "beam".

assumed each of the 3 masses weighed 10 lbs, spring rate 1 lb/ inch.

-> at instant load is applied, ball had 1g acc'n, bench had zero acc'n, load in spring zip.

-> if no spring were there, acc'n in theory would be 1/3g for 2M on one side and 1M on other.

I plugged in 1/3G at the ball in the model, and bottom of spring was at same 1/3g, behaving like no spring at that instant. Spring load was 6.6 lbs, and spring compression 6.6 inches.

-> plugged in 0 g's at ball ( ball likely at max drop speed ), and other end of spring was accelerating at 1/2g. Force in spring 10 lbs match ball weight, 10" sprng compression.

With all same except 1000lb/in spring, same g results but spring deflection was 1/1000 values mentioned, suggesting load transfer was much quicker.

jgrewe

iTrader: (1)

So would increasing the load from a static ride weight that compresses a spring already to then adding the weight being transfered have the same time element to be figured.

Lets say a 250lb/in spring already has 500lbs on it. Is there the same delay when 250 more pounds are added? We know on car the weight isn't going to just appear from nowhere it will build as the car turns. So the ramp of increased weight will be very steep, not instant or = to 1G. With the inertia of the body leaning and the spring compressing passed the equalibrium point(no damper) we end up with spring loaded higher than the 250 lbs for a split second but then going back toward that equalibrium. Now with this increase in load happening at a certain rate is this spring load transfer issue even something to try to factor in?

So, I'm beginning to wonder if we can say, for our use, that a spring transfers the load "instantly enough"? If you are talking about an increased load taking 1/4 of the total period to reach its full ,uh, effect, your dealing with time frames and values that I think only NASA would care about. So go back to your job at the Cape and leave us car guys alone

DamonB, I think, has been refering to the load to the top of the spring not changing speed. You have been talking about the amount of time to get to the wheels with changes in the spring rate affecting how quickly the loaded spring would reach that "1/4 of the total period" point. The stiffer spring cycles faster so the time to the "1/4" value will also be faster. You also say that the load reaches the full value at the same time on both ends of the spring, isn't that like saying the load passes through?

KevinK2

"You also say that the load reaches the full value at the same time on both ends of the spring, isn't that like saying the load passes through?"

No. With your example, 500lb static load on strait, yank wheel for constant say .3g corner with no speed change. The SPRUNG weight of the chassis near instantly picks up that .3g lateral load, and just like the guy suddenly presents his 1g load on an unloaded spring, the spring load increases (at both ends) with body roll, going from 500 to 750 lbs. Shocks should be stiffer with stiffer springs and bars to provide more damping to avoid overshooting the extra 250 lbs.

DamonB, insists the load is different at the spring ends during transients. He calls it levitation, I say it's true if there is a "slinky" (google it) stretched between us 8 feet apart, and I jerk one end. You can see the tension pulse makes it's way down the spring, and eventually you will feel the steady slight force increase. Cars don't ride on toys designed to walk down stairs. Coil surge is when a vibration occurs in a very suddenly loaded spring, more of a vale train issue. It just alters the normal linear response curve by superimposing a high speed ripple on it ... not what damonB dreaming up.

DamonB has done a tremendous job on this suspension forum, with great contibutions. But don't accept as fact what anyone posts on the web ... go with guys with good batting averages.

Sandro

iTrader: (1)

Sorry for the belated reply. I am not sure I understand your question. From your posts, I believe you are familiar with how to calculate natural frequency with different spring rates and so on. So maybe your question deals with shocks? With shocks, the load on the floor will be the sum of both forces (spring and shock). So will the combined forces applied to the the mass. As the mass moves down between 20" and 15", the mass is subject to an acceleration that is now lower than the undamped case as a=(Force spring+Force shock)/m. This causes the mass to travel at a lower speed. The period will be higher (more time to reach 15") and so will be the overshoot as the mass pass through 15" with less speed.

- Sandro

DamonB

No I have not.

A spring is equally stressed along its entire length. The fact that a spring is evenly stressed long its entire length is proof that the loads at each end of the spring are equal. The loads at each end being equal is proof that the spring cannot support a force at one end without experiencing an equal and opposite force at the other end. This is proof that springs instantly transfer weight or force.

Sandro

iTrader: (1)

Kevin has properly answered you questions already.

Whether or not all this matter, this is of course subjective and up to you to decide. You may consider it just an academic exercise, which in the end only shows that tires get loaded "slightly" faster with stiffer springs, or you may want to drive conclusions that may affect your suspension choices. If you just drive your car on the road and non-competitively, you are likely perfectly happy with your stock springs. If you drive it competitively and the rules allow you to change springs, you may prefer (or not) stiffer springs, which settle the car and load the tires faster, but may make the car more difficult to control. If you want to try the feeling of a higher wheel rate, just inflate the tires at 60 lb and go out for a ride.

- Sandro

KevinK2

recent DamonB quote:

" The spring is instead able to support some mass or force for a short period of time without the spring transferring any of the force it sees at one end to the opposite end of the spring."

later said the time period of this spring/mass ststem (case 3) is linked to period of oscillation.... oscillation you initially said had nothing to do with spring response time.

Hope you stay with above statement.

DamonB

I don't personally believe a word I wrote concerning the Theory of Spring Levitation. It's unfortunate that some do.