[TachyonBomb]

Originally Posted by Silver Supra

Oh, you will be in academics. Trust me.

I am pretty sure the manufacturers test single units in a dynamic loading apparatus, not on a car.
The equation of motion for axial force and displacement is relatively simple:
Say axial displacement = X
velocity V = dX/dt
acceleration A = dV/dt
relevant mass = M
Damping = C
spring rate = K
equation is
F(t) = M*A + C*V + K*X
but the damping C and spring rate K (and even mass, M) may be nonlinear with X so it gets interesting.

It's all to say it may make some difference in response - or maybe not, but the manufacturer had a reason for the defined configuration.

I suspect the OP will think twice about starting another thread!!

We all physics high jacked this poor thread!!

I was trying to maybe get a simplified static (not dependent on time) spring rate value which would be the maximum spring rate for a set testing downward force. Meaning a junk 86 that will never be touched again will have to be used to keep things constant. Also like I stated to keep things simple the shock can't damp at all.

But you are completely correct. I didn't want to get into the equation of motion but that is exactly what is needed to find the spring constant(spring rate denoted k in both our equations) and damping rate/coefficient (denoted by C in your equation) due to a time dependent (also called dynamic) force.

Since we are in agreement to springy-ness being based on a spring constant/rate and that damping is based on a damping constant, I'm very curious as to how manufacture come up with a spring rate for a shock. It always throws me off in how to handle that definition and how it relates to the damping the shock does. Maybe racecomp or one of the other members more experienced with these numbers can chime in and define it for me.

Also in your given equation of motion:


F(t) = M*A + C*V + K*X

you mentioned that it doesn't need to be linear which is 100% true let me blow your mind!

Set the force to 0. The system is in equilibrum.

0=M*A+C*V+K*X

you stated:
axial displacement = X
velocity V = dX/dt
acceleration A = dV/dt

Now lets change what you said to this:
axial displacement = X
velocity V = dX/dt = X' (X first derivative)
acceleration A = dV/dt = (dX/dt)/dt = X'' (2nd derivative of X)

so 0=X''+X'+X

Doesn't that look alot like a non-linear equation since the leading coefficient is pretty much (X') squared.

So now lets take the damping term to the left side:

-X'= X''+X

X is the initial condition so time (t)= 0. So this is can be treated as a hidden 2nd derivative of X.

-X'=X''+X'' ~ X^2+Y^2 = the equation of a circle that is based on the damping term (X') on the left side.

To further add the spring coefficient K and damping coefficient C can 100% be based on changes in time. So K(t) and C(t) can be found, this can be further noticed when you follow the units.

Force = kg*m/(sec*sec)
K*X = (kg/(sec*sec))*meters
C*V= (kg/sec)(meter/sec)

So to conclude.... I suppose we should include time to be more accurate lol

Originally Posted by strat61caster

Ok, you're back into grad school, nobody wants to hire someone who would go through all that trouble.

I just used a drill press, ruler, and a scale rated to a few hundred or ideally thousand pounds, but you do you. Race teams will get into how each spring acts dynamically but the rest of us can't afford a sample size to actually do anything about it (other than put the slightly stiffer spring on the heavier side of the car), maybe something useful there for measuring lifespan of a spring (I'd presume elasticity degrades over time but who knows) but the deltas are probably too small on quality springs to be measured with amateurs equipment.

You hit the nail on the head if you want to include dynamics (changes as time goes on) well that will cost a pretty penny. This holds true for most fields of physics research.

You also touched on one of my favorite subjects and newest research fields in physics called soft matter physics! When it comes to measuring the lifetime of a spring if you used time according to physics all things will break down to a state of equilibrium as time goes to infinity. The 2nd law of thermodynamics literally can be used to troll and shut down most physics questions lol. So given enough time the springs internal elastic energy will be lost as the molecules of the material break down :-P You were very correct to say that the delta would be small over time.

So even though we are talking about metal/steel/some hard material we can still treat it like it is silly putty or a memory foam bed. The graph you will want to look at will be a stress over strain.

Stess = Force/Area
Strain = The distance you stretch something out to divided by the objects original length.

So as you stretch or compress a spring/silly putty/a foam bed it will tell you the force needed to accommodate this much stretch in a material.

So if you keep stretching a spring you are going to see a lot of force needed when you get started but after a certain point you will notice the force needed to continue the stretching will go down as the spring looses elasticity.

Remember that springs work that same whether stretching or compressing so this will give you a value of how much force your spring can take before it starts to lose elasticity. Since force = mass*acceleration you can calculate how much load or how fast your spring can be compressed and efficiently recover from before it starts to lose performance.

To get these values is were someone needs those expensive machines that can squish and stretch any material. :-( I actually got to play with some of these in college.