Swaybar discussion
Rotary Enthusiast
Joined: Dec 2001
Posts: 1,209
Likes: 6
From: Delaware
Originally Posted by jgrewe
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?
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?
Not exactly same with a car. In roll, the extra 250 lbs is a reaction force to the roll of the sprung weight. It would have some effective added mass at the spring, but i'd guess it to be less than the equivilant of 250 lbs of mass on the coil. So i'd expect only moderate increase in the 1/4 period for rise and fall on a hilly road.
Originally Posted by DamonB
Unfortunately my textbooks are too old to agree....
You have to take the inertia force into account, and everything will fall into place.
RCVD Section 6.2 The Spring-Mass-Damper System - basically the same description as in my previous posts.
Page 234 - equation of motion 6.1 F=Inertial Force (reaction) + Spring Force + Damper Force
re. Fig 6.1 or Fig. 6.2
This is what I have been trying to explain. In a dynamic scenario - i.e. when acceleration is present - you need to consider the inertial reaction. The Force acting on the wall (or the floor) are the spring force (K*x) and the damper force (C*v). The balance to the applied force (or the weight in Fig. 6.2) is the inertial reaction. Note, equation 6.1 can be resolved also when F=0 (no external force applied) and no damper. This is the simple "horizontal" oscillator I described earlier.
Page 236 - Undamped Natural Frequency - equation (6.2) - as you can see frequency is a function of the spring rate (K)
Section 22.3 Damping Fundamentals - Page 787 - Basic Ideas - In a damped spring and mass system such as shown in Fig 22.1, there are three forces which govern the dynamic behavior:
1. The inertial force is due to accelerating the mass which is a function of the mass and acceleration
2. The damping force...
3. The spring force....
Nobody is re-inventing the wheel here...
- Sandro
I finally found the time to address your comments in detail. Please see below.
- Sandro
WRONG: It’s not a new theory. But in order to understand it you need to understand Newton’s Law of Motion and be familiar with the concept of inertia first.
Newton’s First Law: Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
Note, it says “external”. “Remain in that state of motion” means either standing still or traveling with the same velocity (i.e. constant speed with no change of direction). Therefore, in order for an object to move, an external force or combination of unbalanced external forces needs to be applied.
With reference to my undamped oscillator example, there are three unique positions where, for an instant, mass is in a uniform state of motion (v=0 or v=constant):
(a) mass has stopped his upward motion and is just about to change direction and “drop” toward the floor,
(b) mass is traveling (downward or upward) through the static equilibrium position (e.g. 15” in a vertical oscillator with a spring 20” long, spring rate (K) = 20 lb/in, and mass = 100 lb), and
(c) spring is fully compressed and mass is about to be “pushed” upward again.
Velocity is zero in (a) and (c),
Velocity is constant in (b)
In any other position, Newton’s First Law suggests there has to be an unbalance of the external forces applied to the mass. This unbalance is the difference between the mass weight and the force K*x exerted by the spring.
Newton’s Second Law: The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma.
This is the most powerful equation in dynamics: a=F/m, where F is the results of any combination of all forces (whether external or reaction – read ahead about inertial force). Once acceleration is determined, velocity and distance over time can be calculated from cinematic equations (a=dv/dt and v=dx/dt).
Newton’s Third Law: For every action there is an equal and opposite reaction.
What you have to realize is that this law does apply all the time, whether the mass is in a state of uniform motion (where Newton’s 1st Law rules) or in dynamic scenario (where Newton’s 2nd Law rules), mass gets accelerated as a result of the application of unbalanced external forces (Newton’s 1st Law).
Example: a body in free fall accelerates as a result of the gravitational field; The external force applied to the body is F=mg, or its weight W.
According to Newton 3rd Law, there has to be an equal and opposite “reaction” force applied to the body, or ma=W.
“ma” is the inertial force, which “balances” the body weight, according to Newton’s 3rd Law. Where “balances” has a dynamic meaning, outside of Newton’s 1st Law, and is the result of the application of Newton’s 2nd and 3rd Law.
As the body’s speed builds up, air friction starts exerting a force to the body, opposite to W, which we call F(v) [as it is a function of speed]. The balance of the external forces is now less than W, and equal to F=W-F(v). The inertial force has still to be equal but opposite of it. We can write the equation as W-F(v)=ma, or W=ma+F(v). This is an important equation, as it relates to the oscillator equation W=ma+Kx.
Although acceleration is now less than “g”, it is still >0, the body keeps accelerating and its velocity increasing, and so does F(v). Eventually, the velocity is high enough and both external forces become equal, W=F(v), and the body stops accelerating (but keeps moving at constant velocity). We are now in Newton’s 1sr Law “regime” and there is no inertial force any more.
Back to the vertical oscillator example, the dynamically balanced forces are W=Kx+ma (where “x” is the compression measured from the spring free length, “20” in our example).
Note that acceleration “a” can be calculated directly from this equation as two of the three terms of the equation (W and Kx) are independent from it. Since W and K are constant, “a” and “x” result related directly. a=(W-Kx)/m=(mg – Kx)/m. For instance, in our example, if x=0” (spring fully extended) and the mass is about to start moving downward, then a=g. But if a=g, then the inertial force “ma” is also equal to “mg”, and the equation of motion simply becomes W=mg, i.e. the inertial force is equal to the mass weight. Note, this is exactly the same equation of the “free fall” motion. Therefore, when the spring is fully extended and W is applied, for an instant the mass “levitates”, like at the starting of a free falling.
So, how does all this related to the force on the floor?
Assuming the floor doesn’t move, the only “reaction” force from the floor is the opposite of the force exerted from the spring to it, or Kx. Note Kx=W only when x=5” and there are no inertial forces at play. In all other cases, Kx is either <W (when x<5”) or is Kx>W (when x>5”). The difference between W and Kx is “absorbed” by the inertial force.
x [in] 5 2.5 0 7.5 10
Kx [lbf] 100 50 0 150 200
a [g] 0 0.5 1 -0.5 -1
ma [lbf] 0 50 100 -50 -100
WRONG: Springs transfer force instantly and across themselves, but the amount of the force is Kx=W-ma, not W.
ALMOST CORRECT BUT INACCURATELY STATED: If spring is fully extended, by definition it doesn’t “see” any force, and otherwise it would be compressed already. As such, there is no force to be transmitted across it. Spring itself does not levitate mass. Mass is instantly suspended if initially applied when spring is fully extended, before gravity starts accelerating it.
WRONG: Mass is instantly suspended if initially applied when spring is fully extended before gravity starts accelerating it. The time it takes to get Kx=W for the first time is T/4, where T is the cycle period or the inverse of the oscillation frequency, which can be calculated from K and m.
ALMOST CORRECT BUT INACCURATELY STATED: As the spring gets compressed, the force exerted by the spring is Kx. This is realized at both end of the spring. Depending on the amount of compression, this force can be less or more of the mass weight.
ALMOST CORRECT BUT INACCURATELY STATED: Mass “levitation” only occurs when the spring is fully extended, independently from the spring rate. The period of time it takes for the spring to transfer the weight to the floor is indeed a function of the spring rate. The natural frequency of a mass-spring oscillator is higher with a stiffer spring. As such, the time it takes for the spring to get compressed to the static equilibrium point is shorter.
WRONG: The properties of the material (spring steel) are the same. However, the material compression modulus (which determines the rod elasticity on compression) is several orders of magnitude higher than its shear modulus). If you use a rod, the force is applied in the direction of compression. If you use a coil (the very same rod wound into a helix) the force is applied “across” the wire and the shear module determines the elastic properties. If you would apply a torque to the rod, rather than an axial force, the elastic properties would be the same. This is how torsion springs work. Re: RCVD 21.2 Coil Springs page 762 – “”Coil springs utilize the elastic properties of a wire in torsion to produce a rectilinear spring rate”
ALMOST CORRECT BUT INACCURATELY STATED: The force that the spring “sees” is weights less inertial force. That is equal to Kx, which is indeed transferred instantly to the floor.
ALMOST CORRECT BUT INACCURATELY STATED: Assuming that you would be able to move your finger in sync with the spring natural frequency but apply the same force all the time, that would indeed be the case. Not an easy experiment to perform with a ballpoint pen spring though. The experiment you described resembles more a force being applied gradually. In such conditions, the spring does not oscillate. If the spring is not allowed to oscillate, the force will just compress gradually such that Kx=the force you apply with your finger. Your other finger will feel Kx as usual. Another more accurate experiment you could do with a ballpoint pen spring is the following. Squeeze the spring between your two fingers until fully compressed, and then remove one of your fingers very quickly. This approximates a sudden de-application of force. If your theory were right, that an external force is transmitted instantly, that should work also when a force is suddenly removed, and the spring should experience zero force at both ends. You will notice instead that the spring will actually jump away from your remaining finger, as it still would exert a force Kx, even though the external force at its other end is zero.
CORRECT: Assuming you would have the right accurate measurement instruments and the experiment is done correctly, that would indeed be the case. I would suggest using a combination mass/spring with a very low natural frequency, to overcome differences otherwise attributable to inevitable measurement errors. I am reading Kevin has actually modeled this.
CORRECT: Again, given the right experimental rig you would be able to measure the dynamic unbalance.
CORRECT. You have also to take the mass in consideration, as the frequency is equal to (k/m)^0.5 Just wondering though, Why in your post # 73 did you state that “OSCILLATION HAS EXXACTLY ZERO TO DO WITH THE RATE OF ANY SPRING”?
WRONG: Nobody has invented the wheel here, only applied the Newton’s Laws accurately. On the other hand, I have seen no demonstration on how would it be possible to have a force transferred to the floor, different from F=Kx. You have maintained that the spring resist the mass weight all the time as it oscillates and that is what is transmitted to the floor because you have ignored the effect inertial forces that come into play in dynamic systems. Your theory that the weight is transferred instantly independently for oscillations carries conclusions difficult to explain. For instance, when the spring overshoots beyond the equilibrium point and the force exerted by the spring is higher than the weight applied to, where would the excess load go?
- Sandro
Originally Posted by 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.
Newton’s First Law: Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
Note, it says “external”. “Remain in that state of motion” means either standing still or traveling with the same velocity (i.e. constant speed with no change of direction). Therefore, in order for an object to move, an external force or combination of unbalanced external forces needs to be applied.
With reference to my undamped oscillator example, there are three unique positions where, for an instant, mass is in a uniform state of motion (v=0 or v=constant):
(a) mass has stopped his upward motion and is just about to change direction and “drop” toward the floor,
(b) mass is traveling (downward or upward) through the static equilibrium position (e.g. 15” in a vertical oscillator with a spring 20” long, spring rate (K) = 20 lb/in, and mass = 100 lb), and
(c) spring is fully compressed and mass is about to be “pushed” upward again.
Velocity is zero in (a) and (c),
Velocity is constant in (b)
In any other position, Newton’s First Law suggests there has to be an unbalance of the external forces applied to the mass. This unbalance is the difference between the mass weight and the force K*x exerted by the spring.
Newton’s Second Law: The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma.
This is the most powerful equation in dynamics: a=F/m, where F is the results of any combination of all forces (whether external or reaction – read ahead about inertial force). Once acceleration is determined, velocity and distance over time can be calculated from cinematic equations (a=dv/dt and v=dx/dt).
Newton’s Third Law: For every action there is an equal and opposite reaction.
What you have to realize is that this law does apply all the time, whether the mass is in a state of uniform motion (where Newton’s 1st Law rules) or in dynamic scenario (where Newton’s 2nd Law rules), mass gets accelerated as a result of the application of unbalanced external forces (Newton’s 1st Law).
Example: a body in free fall accelerates as a result of the gravitational field; The external force applied to the body is F=mg, or its weight W.
According to Newton 3rd Law, there has to be an equal and opposite “reaction” force applied to the body, or ma=W.
“ma” is the inertial force, which “balances” the body weight, according to Newton’s 3rd Law. Where “balances” has a dynamic meaning, outside of Newton’s 1st Law, and is the result of the application of Newton’s 2nd and 3rd Law.
As the body’s speed builds up, air friction starts exerting a force to the body, opposite to W, which we call F(v) [as it is a function of speed]. The balance of the external forces is now less than W, and equal to F=W-F(v). The inertial force has still to be equal but opposite of it. We can write the equation as W-F(v)=ma, or W=ma+F(v). This is an important equation, as it relates to the oscillator equation W=ma+Kx.
Although acceleration is now less than “g”, it is still >0, the body keeps accelerating and its velocity increasing, and so does F(v). Eventually, the velocity is high enough and both external forces become equal, W=F(v), and the body stops accelerating (but keeps moving at constant velocity). We are now in Newton’s 1sr Law “regime” and there is no inertial force any more.
Back to the vertical oscillator example, the dynamically balanced forces are W=Kx+ma (where “x” is the compression measured from the spring free length, “20” in our example).
Note that acceleration “a” can be calculated directly from this equation as two of the three terms of the equation (W and Kx) are independent from it. Since W and K are constant, “a” and “x” result related directly. a=(W-Kx)/m=(mg – Kx)/m. For instance, in our example, if x=0” (spring fully extended) and the mass is about to start moving downward, then a=g. But if a=g, then the inertial force “ma” is also equal to “mg”, and the equation of motion simply becomes W=mg, i.e. the inertial force is equal to the mass weight. Note, this is exactly the same equation of the “free fall” motion. Therefore, when the spring is fully extended and W is applied, for an instant the mass “levitates”, like at the starting of a free falling.
So, how does all this related to the force on the floor?
Assuming the floor doesn’t move, the only “reaction” force from the floor is the opposite of the force exerted from the spring to it, or Kx. Note Kx=W only when x=5” and there are no inertial forces at play. In all other cases, Kx is either <W (when x<5”) or is Kx>W (when x>5”). The difference between W and Kx is “absorbed” by the inertial force.
x [in] 5 2.5 0 7.5 10
Kx [lbf] 100 50 0 150 200
a [g] 0 0.5 1 -0.5 -1
ma [lbf] 0 50 100 -50 -100
Originally Posted by DamonB
The Theory of Spring Levitation says that springs do not instantly transfer force through them.
Originally Posted by DamonB
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.
Originally Posted by DamonB
Unfortunately the length of that period of time has not been defined as of yet.
Originally Posted by DamonB
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.
Originally Posted by DamonB
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.
Originally Posted by DamonB
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
Originally Posted by DamonB
, 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.
Originally Posted by DamonB
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.
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.
Originally Posted by DamonB
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.
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.
Originally Posted by DamonB
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.
Originally Posted by DamonB
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.
Originally Posted by DamonB
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.
I was gone for a couple days on a road trip to NJ to pick up an Elva Courier project car. I would like to thank Sandro and KevinK2 for taking the time to enlighten us on the finer points of springs. I wold also like to thank DamonB for keeping things interesting. I learned a lot. I wonder if the original poster had any idea where this would end up?
