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TORSION BAR QUESTION?

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Old 09-04-2010, 04:54 PM
  #76  
ehall
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Originally Posted by Chris White
OK, but the first step in preping your car will be taking the gun rack out of the rear window!
I can make you up a set of number stickers for your car….the background might just look remarkably like a target!

BTW – I am quite sure you will be very good at it. The detailed control under pressure needed is similar to some of the skill sets of a good marksman – being the least bit jerky will make you miss…
STOLEN. Saved as...AVATAR!
Old 09-04-2010, 04:57 PM
  #77  
ehall
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Originally Posted by Chris White
OK, but the first step in preping your car will be taking the gun rack out of the rear window!
I can make you up a set of number stickers for your car….the background might just look remarkably like a target!

BTW – I am quite sure you will be very good at it. The detailed control under pressure needed is similar to some of the skill sets of a good marksman – being the least bit jerky will make you miss…
mmm....jerky!


oh...you meant...nvrmnd.

Gun rack is to disuade tailgaters!


Your point is interesting. My guess is that after figuring out the proper line, the real callenge is in finding out the limits of how late one can break and how hard.
Old 09-06-2010, 11:49 PM
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Oddjob
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Not sure if I am taking a risk of being shot at by posting in this thread...



Originally Posted by mikey_audiogeek
FORCE can be additive or subtractive, however the rate remains constant. F=-kx for a linear spring, whether in compression or tension. The slope remains the same.
Yes, we are talking about the forces acting on the car and suspension, based on effective combined spring rates and suspension travel/deflection.

The absolute spring rate constant (k) of the t-bar by itself or the linear spring alone, is a…constant. But when installed in the 944 suspension system, we are now talking about the effective spring rate at the rear wheel. And depending on the index point of the torsion bar, the t-bar and the coil spring either act in conjunction or opposition. So the combined effective rate is either the summation or the difference of the t-bar and coil spring depending on suspension travel.

Originally Posted by mikey_audiogeek
Put more simply: Hooke's law states that the rate of a linear spring remains constant, independent of the displacement - whether positive or negative. For linear springs in parallel, k(eq) = k(1) + k(2), irrespective of displacement.
This is for a single mass w/ two linear springs attached, basic force balance equation that assumes x1 = x2 (not irrespective of, its dependent on) for a simplistic academic model. It is too simplified for the two spring types, different attachment types, preloads, and range of motion for the 944 rear end.

Originally Posted by mikey_audiogeek
Where I see that torsion bar preload would make a difference, is where there is a non-linear component to the rate. This would be the case for a torsion bar supported in rubber bushes, as the rubber has a non-linear rate which would create a "wobble" in the force/displacement chart about the crossover point. This effect is exacerbated by any play in the system.
In general, rubber has a spring and damping rate which, if significant, would need to be accounted for in-series. But the torsion bar has metal to metal contact on both ends, anchored to the torsion bar carrier and the spring plate. Theoretically, the rubber bushings on some coilover shocks would influence the effective rate of the coilover spring, but it's a relatively small impact. Compare the deflection of the rubber bushings to the spring (a few mm to a few inches, so 10-20 times higher spring rate). Primarily, removing the rubber will change feel, response time, reduce dynamic alignment deflection but does not significantly alter the over all handling balance of the car.



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