Shock Valving

Bham_Mike

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Can someone explain the terminology used for compression/rebound valving. For example, a Bilstein 6100 8" comes with either 275/78 or 180/75. How can I relate that to a given application.
thanks
Mike
 

ntsqd

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According to the Bilstein catalog the numbers are Newtons (N) (of force) needed to move the shaft at a velocity of .52 m/s. Just so that you can get a feel for what a Newton is, one hundred Newtons roughly equals 22 lbs-force. Because of the valving design, if you use a different velocity you will get different numbers so appearently Bilstein has standardized on using .52 m/s as the test velocity for quoted damper values. 0.52 m/s is about 1.7 foot/sec or 1.16 mph.

TS

"It only seems kinky the first time"
-- Bumpersticker seen in Lost Wages
 

ntsqd

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Haven't a clue. Somebody ?

TS

"It only seems kinky the first time"
-- Bumpersticker seen in Lost Wages
 

Kritter

Krittro Campbell
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I posted an equation on the ideal spring rate with explanation of the variables on another page but it could be the ideal damping rate also with a change of variables...it is very common DE used in control systems and a suspension system is very much a control system. As for a quick and dsirty way...I dont know of one

Kris
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Bham_Mike

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Sounds like I'm getting in over my head here. Just trying to gain a little knowledge for a future project.

Mike
 

ntsqd

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I'm not convinced that a DE can be successfully applied because the input freq is too varied and too random. I think you could use it to get you close by using generalities for the input freq(s), but from there it's cut-and-try.

I would start by asking the manufacturer of the damper for a recomendation. that will, worst case, get you in the ball park.

TS

"It only seems kinky the first time"
-- Bumpersticker seen in Lost Wages
 

Kritter

Krittro Campbell
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I agree to the DE but its somewhere to start without tearing down a shock. If you are going to be jumping the truck you use one F(t), if you are going to be running hard in the whoops you use another F(t) and then you bridge them together and get a happy medium.

I actual did contact a company regarding valving and did not get any useful information from them besides the stock shim stack dimensions

Here is what I got from a shock manufacturer-sales engineer, verbatim from the email
"GENERAL RULE OF THUMB FOR VALVING IS IF THE SHOCK BOTTOMS OUT REAL EASILY THEN YOU NEED TO STIFFEN UP THE COMPRESSION. IF THE SHOCK GIVES A BRONCOING EFFECT OR MAKES THE TRUCK BOUNCE ONE MORE TIME AFTER A JUMP THEN YOU NEED TO STIFFEN UP THE REBOUND."









Kris
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ntsqd

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"GENERAL RULE OF THUMB .."

Wow, what a revelation........

I suppose a 39th order DE (Differential Equation for those lost by the abreviation, commonly called a 'Difficult Equation') might put you really on top of things. I sure as heck don't want to develop it.

TS

"It only seems kinky the first time"
-- Bumpersticker seen in Lost Wages
 

Kritter

Krittro Campbell
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Only 2nd order so its not so bad...could be a lot worse. Here is an example using a math package could do same in excel with some effort..scroll down to example 8.3
<A target="_blank" HREF=http://www.awlonline.com/nagle/downloads/pdfpc/Lab08.pdf>http://www.awlonline.com/nagle/downloads/pdfpc/Lab08.pdf</A>

Not saying it is the way to go but it could be used to get a ball park number...experience also works too, but a lot of us dont have valving experience.

Kris
"I was thinking the exact same thing about you..."
 

Kritter

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I agree 100% with the last statement. Great product, great service!

Kris
"I was thinking the exact same thing about you..."
 

Dylan

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A basic 2nd order DE would work for a sinusoidal input like a really evenly whoop section but I’m guessing that it would result in a damping factor way lower than what we find in the field. This is just my speculation based on the fact that the correct spring rate that would be resonant to the input would yield very little body motion with very little damping needed. But change the frequency a little and watch out… all hell would break loose without the damping there.
I’ve been meaning to look into this more but haven’t yet, but my guess for an input would be a step input or a ramp input for starters.
Also if you look at these inputs at the wheel and what is transferred to the chassis you could equate it like this (for compression only)
F=k deltaX+CV=ma (supper simplified)
In other words what spring force plus what velocity force will cause a given acceleration in the bump direction. With a step input amplitude of less than your bump travel you don’t need much spring or shock… but if the bump is taller than your compression travel you better create enough force to pick up the vehicle before you bottom out. the question is how much bump acceleration do you want???

This is all very interesting but I have to get back to work lunch is over.

Kritter is that thing you posted from a program like” Maple” or something?? Its been a couple of years since I’ve messed with that program but you could probably get some good results from it.

???then what do we do about rebound????
 

JrSyko

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Are we still talking about stuff that goes up/down in the desert? However I'm afraid that you didn't look into the fact of the suspension possibility reaching one of its harmonic oscillations due to under-dampening. Please re-due your report to take that into account. Thank you. :)

See ya in the dirt!
 

Kritter

Krittro Campbell
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Thats taken into account by looking at the graph. Pictures are worth 1000 words...especially graphs.

Kris
"I was thinking the exact same thing about you..."
 

FABRICATOR

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You guys are playing with the DNA of off-road racing! If you can figure this out by purely scientific means, please let me know. Actually Sway A Way has a fairly close spring rate formula in the back of their catalog. The biggest factor affecting spring rates is wheel travel. It's amazing how soft a long travel setup is. Long travel basically needs about zero springing when topped out. Enough to hold up that corner of the car at static ride height. And somewhere around 2-1/4 times the corner weight in full compression. All this does in a dynamic (bumpy and rough) situation is help hold the car up and push the wheel down any time it leaves the ground. The damping does all the heavy duty work. At higher wheel positions (compression) there is an increased need to absorb rather than recycle bump energy. In other words when traversing small bumps it's more like a pavement car and the suspension needs to recover quickly. Over large bumps damping needs to be greatly increased during the last bit of up travel as a tradeoff between recovery of the suspension, and upsetting (raising) of the chassis. Some of the shock makers will tell you that a proper set up does not bottom out, no matter how bad the bump is, period. This is basically true because of the tremendous damping at work when approaching the bottomed out condition. By the time the piston squeezes by that last bit of oil, the vehicle has already cleared what ever it was encountering. This kind of damping can be accomplished by at least 3 ways. The most common is a properly setup bypass shock, the other is to rely on a well tuned bump stop (iffy at best), or a remotely operated shock (proper linkage setup). Rebound damping from the fully compressed position needs to be very stiff to control the fully compressed spring. Then fairly light the rest of the way. A good bypass or remote setup will also provide increased rebound damping toward full extension to prevent harsh topping out of the suspension. A light car can get away with a fairly straight rebound rate all the way through because the spring is not so stiff and unsprung weight is low. A limit strap can suffice for top out. Dynamic ride height (bumpy and rough) is almost entirely determined by the shock absorbers and weight transfer. Springs only play a major role on the side of the car being compressed by weight transfer. This would be the rear under acceleration, left and right when turning and the front when braking. And just to keep it from being too simple, everything here is significantly affected by sprung to unsprung weight ratios, weight distribution, and wheel travel. Another item that should be considered if you are working on a formula is that coil springs have a fairly constant rate throughout only about 85% of their available travel.

<font color=orange>The best ideas are the ones that look obvious to the casual observer.</font color=orange>
 

JrSyko

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Kris - My friend who is an engineering student at UCSB posted that without really looking at all the info. Meant to be funny, guess it wasn't! I have no clue as to what you guys are talking about. But keep it up as it sounds good!

See ya in the dirt!
 

whistler

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Did a search when I was looking for the unit of measure that Fox/King uses.Came across this old post.Does anybody know if there is a way to cross reference the Fox to the Bilstein system.For example, my Fox shocks are rated 90 rebound/40 comp.What would that be in the Bilstein system.Just curious. Thanks.
 
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