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#15 | ||
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Hombre
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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!!
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#16 | ||
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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 Quote:
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. |
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#17 | |
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Good luck with the applications. |
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#18 | |
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Engineering grad school he'd get a thump for not elaborating. |
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#19 | ||
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You are correct differential equations and ones ability to classify each type is an important aspect in itself not just with respect to grad school. I'm am mostly using my forum responses to hone in my communication abilities. One of the biggest issues I noticed in my undergrad experience was that a large portion of both students and professors had poor communication skills when it came to conveying ideas and concepts to others in a digestible way toward a lay audience. After presenting my research to an audience twice and having to write a these paper on that research as a graduation requirement I became aware of the importance to practice this sort of thing. So by all means if you want to knit pick and correct anything I place out there please do. Better to get these kinks out of the way now than later down the line when it comes to scrutinizing my dissertation or I'm being paid to conduct and present research with an attached teaching requirement. Quote:
So are all of you guys engineers (OND. Strat61caster, silver supra)? Every job I come across here in the bay area seems to want an engineering degree. Right now my goal is a Ph.D. in cosmology, theoretical gravity or something similar. But I have a couple of friends that just got their masters degree in software engineering a few months ago from san jose state... lets just say their salaries can pay off my student loans, credit card debt and brz loan in less than a year with spare money to play. It honestly has me thinking about a software engineering masters program and just calling it quits. |
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#20 | |
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In my experience Masters an PHD does not mean the person is more competent or capable of doing a job unless they studied and researched a particular field and continue to work in that field. The higher starting pay can be quickly closed in upon by someone who is good at negotiating and job hopping with a "lesser" degree. And there are plenty in software around here who make great money without a degree, reputation can carry you a long way if you can network well. |
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| The Following User Says Thank You to strat61caster For This Useful Post: | TachyonBomb (07-20-2017) |
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#21 | ||
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Read up on the definitions for terms like linear and nonlinear. They sound obvious, but have specific meanings. We tend to think of springs as linear - i.e., a simple spring rate defining the ratio of force in the spring to displacement. However, many springs are not linear, the rate changes with various properties including load rate (dynamics), material properties, or geometric influences. Even the supported mass could be nonlinear - that 300-lb fat guy in your car may leave his seat, no longer be supported by that spring, after you hit a big bump. When he hits the seat again, his mass returns, but with added incremental acceleration (you know, that "theoretical gravity"!). I wrote the simple equation of motion for s single degree of freedom (SDOF) model above. Same equation applies, in general, for a multi-degree-of-freedom (MDOF) model (where the terms are 1- and 2-dimensional arrays). I run a lot of dynamic, nonlinear analyses of MDOF models of assemblages and buildings as well as doing back-of-the-envelope SDOF calcs in a pinch. Yes, I am an engineer with MS and PhD and offer this advice. Get the degree you want for you - not for some near-term job or even occupation. I did that because I wanted to learn more in my field and, incidentally, wanted to meet future requirements to teach in a major University if I decided to. I also incidentally learned to program computers long, long, long ago and have also applied that throughout my career. I have worked in many different areas using my background knowledge - don't get pigeon-holed! Anyway, we have jacked this thread beyond belief , so I need to get off my soapbox!
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Anyway, should not palindrome be spelled palinilap? |
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#22 | ||
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So that is why a more dynamic spring system with more degrees of freedom where a fat guy gets lifted off his seat for a split second didn't cross my mind haha. I pretty much got introduced to non-linear systems when I decided to do my undergrad research on quadrupolar gravitational radiation and detection. So my limited experience with non-linear concepts is in the context of mass densities/gravitons being self interacting and tensor field comparisons to their scalar field Maxwell counterparts. The whole thing was a 6 month crash course in self teaching myself Einstein tensor notation and general relativity... I learned a hell of a lot but lord knows I only scratched the surface on the subject haha. Also thanks for the wise advise as well We should really give this guy his thread back To sum up for the OP -springs in this context function the same installed upside down -spring rate and damper rate should be determined in a dynamic system |
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#23 |
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Only users lose drugs.
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All that brainpower and yet nothing about the obvious increase in unsprung weight at the motion end of the spring when upside down? There are ~twice as many coils on the tighter wound end; that end of the spring is twice as heavy, and springs aren't light to begin with. While it doesn't affect the spring rate or how the spring works in a sanitary, theoretical world, in real life you've doubled the spring's contribution to unsprung weight. The spring itself functions the same, the car's suspension does not. Take the free performance improvement and flip the spring.
Last edited by venturaII; 07-20-2017 at 10:23 AM. |
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#24 | |
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#25 | |
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Only users lose drugs.
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Granted, the location of the coil in relationship to the suspension's pivot point has a different impact than static weight lost further outboard, like lighter brake calipers or control arms or whatever. But considering this reduction is a freebee, and no serious car guy looking for performance improvement would ever question the value of reducing unsprung weight, it'd be kinda stupid not to take advantage of the benefit in simply flipping them. I suppose if you were hard parking it, then there's no benefit to the OP... |
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