Coil Over Comfort Determination

[PMF]

Hi All,

I have an insurance group coilover set on my car and wanted to get the optimum ride comfort as well as lowering of the vehicle.

I found this site and it that has tons of information and need some help to understand this....

Site:http://www.rqriley.com/suspensn.htm

"The static deflection rate of the suspension is not the same as the spring rate. Springs are located inboard of the wheels where they are normally subjected to the mechanical advantage of the suspension linkages. Static deflection is related to the distance the sprung mass (essentially the body) moves downward in response to weight. A static deflection of 10 inches in response to a weight equal to that of the sprung mass will produce a natural frequency of 1 Hz. A 5 inch deflection produces a 1.4 Hz frequency, and a 1 inch deflection results in a 3.13 Hz frequency. The natural frequency of a suspension can be determined by a simple formula expressed as follows:

NF=186/√SD

NF = Natural Frequency in Cycles Per Minute (divided by 60=Hz).

SD = Static Deflection in Inches"

I am not exactly sure what they mean by Static Deflection... can someone please help explain this to me.

Thanks!

[maxnix]

This would apply more to something like the V35 suspension design rather than the FGY33 design which is MacPherson strut up front and coil over shock multi link in the rear, both of which tend to be more linear. The implication of the V35 design or say modern F1 cars with the levers and cams operating inboard springs is that the spring rate is typically higher and as a consequence of the geometries involved, non-linear. In short, time to crack the old trigonometry texts if you want to calculate true spring rate.

There is a good previous post by Q45tech with some links.

Let us know how the Insurance set-up works as most others have employed the less well regarded JIC system.
Modified by maxnix at 7:35 AM 11/19/2008

[Q45tech]

The static deflection is the amount of compression the spring suffers when the body weight is placed atop it,

A typical 90-96 Q will have springs whose free [unmounted] length is ~~ 16" so when mounted the spring compresses to ~ 8" so 8 x 146 [inch stiffness] = ~ 1168 pounds of body weight.

Sqrt of 8 = 2.8284 so 186/2.8284 = 65.76 cycles per minute or 65.76/60 =1.096 cyles per second or Hertz.

A spring that only compressed 7" ~ =186/2.64575 =..................1.172 Hertz and would have a per inch stiffness of 167 pounds per inch with same body weight.

See how important knowing exact 4 corner body weight is.

See how having a passenger or not can change ride height 1/2" or more on a side.

Shocks usually have a 7" stroke [limit to limit] so necessary to make sure the springs cannot rip them apart under extremes.

Adjustable shocks are necessary to optimize non oem springs, but after market shocks are usually 10% stiffer than oem to compensate for the 60k of wear in suspemsion that caused the replacement in the first place. Beyond ~ 120k the rubber isolators are done and need total replacement if as new ride is desired.

Back in 1953 they were simulating springs and shocks in lab to optimize suspensions:http://deepblue.lib.umich.edu/...1.pdf