Yeah but you're taking such a long time. You are set to break the record for the longest SAS ever. <img src="/forums/images/graemlins/butwiggle.gif" alt="" />

I have no tech to add to what's already been said, but I have tried different torsion bars, shocks, ride heights etc and I found that the stock torsion bars set 1" higher than stock and Bilstein 5100 gas shocks give an excellent ride and offroad handling with 32" tires.

If you want to stay off the bumpstops, try increasing the ride height on the stock t-bars before you upgrade to larger ones.

I know but that to easy!! <img src="/forums/images/graemlins/smile.gif" alt="" />

Frank.

Perhaps a picture will help, this is a typical motocycle shock design.



Regardless of the theory you guys choose to espouse the reality is that cranking t-bars always has and always will sitffen the suspension. If you take some leaf springs and bring the ends closer together thus resulting in a steeper arc and some lift do you not think they would be stiffer than they were? Have you ever used a bow and arrow? Do you think it is as easy to pull the last 6" as the first? No way unless it is a compound. I don't care what any of you say about the "linear" springs rates, all springs will require increasing force as they compress, it is a fact of life. Yes the steel in springs can deform, that's why they are springs, but they also resist deformation, or they would collapse with a load on them.

My username is robinhood. <img src="/forums/images/graemlins/smile.gif" alt="" />

That is not the same situation because you are changing the geometry of the spring, unlike when a tbar is cranked. Like I said before, a better analogy is lift blocks under the leaf springs. Will adding lift blocks make the ride stiffer? Will the leaf be preloaded? Ask yourself why not?


Yes, nobody is disputing that. A spring rate of 100lbs per inch means if 100lbs were loaded onto the spring, then the spring will compress 1inch. If 300lbs were exerted onto the spring then the spring will compress 3 inches. I think what you are missing is that when a spacer is used or a tbar is cranked and the truck is stationary, the spring is not compressed (or torqued) anymore than normal. This also applies when you are driving down a smooth road.

Now, when you hit a BIG bump and your suspension compresses to the bumpstops, the spring IS compressed more than it normally would be without a spacer (like I said ealier with the 2in spacer example). BUT, when driving normally around town, you most likely will not use this region of suspension travel and will not contribute to a stiffer ride.


No, actually, ideally a spring will not deform at all. When Frank and I were discussing stress relaxation, that was outside the scope of this preload article and doesn't have much to do with it.

Ok, I'm gonna do this again but more basic....

Did you ever take Algebra? Do you remember the equation of a line Y=MX+B? It just happens that the spring rate number is the "M" or slope in that equation. Spring force F=KX, K being "M", the spring rate constant. Now pay close attention to the words you are using especially the word, "rate". The spring force increases with a LINEAR RATE. This means two things. First, it is true that to hold that bow stretched at a farther distance requires more force. (The X value is larger) Second, and very importantly, if takes X amount of force to stretch it say 1", it still takes that SAME amountof force to stretch it another inch even when it is pulled taught.

Your example of the motorcycle shock may or may not apply. For example, when it is in the bike, is the shock still maxed out? This is very important if you are to try to draw a parallel. If it is maxed, is is similar to taking torsion bar and adjusting it until at rest, the suspension is fully extended, then adjusting more. You are simulating more weight by smashing the bar against the bumpstops. We typically don't do this to our trucks so it would not apply. However, if the motorcycle shock is not fully extended (it isn't, I have one too, if it were you eat crap on corners when the tire leaves the ground because it can't extend into pavement depressions) then the shock will behave similar to our trucks. You move the adjustment point so that if say you add a heavy rider, the bike returns to a proper ride height as opposed to sagging. The extra load is from the RIDER!!!!!!!!!!!!

Anyway, all of this theory was the result of real world testing and has been proven millions of times and is at work in your truck, bike, scale, bow.. whatever as we speak.

Your leaf spring analogy is really applicable. The arching of the spring in fact strain hardens the metal, among other things, increasing the spring rate. You could do that to any spring if you wanted.

Springs are springs because they return to their original shape. Bubble gum deforms but it ain't too springy is it? Solder resists deformation but it isn't too springy either. Harden steels can resist deformation like crazy but when pushed too hard break instead, not very springy. Spring steel resists deformation but, as long as you don't exceed the yield strength, returns to it's original shape thus making it a spring.

And lastly... this is to explain WHY the torsion bar suspension feels stiffer when the angle inceases (it's not because of the spring, it's the difference in torque)!!!!! Not to say that it doesn't.!!!

Frank.

Well, it eventually stops at time=infinity. <img src="/forums/images/graemlins/smile.gif" alt="" />

For our experiments we projected out for a couple years life and it doesn't stop, just slows way down. However, I'm not sure how that would transfer over to room temperature and much less stress applications. I would have to guess that for practical purposes the tbars do not stop relaxing, but I could be wrong.



We didn't do any tests like this, but I would guess that there would be no difference unless the tbar has strain hardened. But then again, maybe the tbars are strain hardening as they stress relax?

Just to elaborate on this and prevent confusion, what frank means when he says, "it still takes that SAME amountof force to stretch it another inch even when it is pulled taught" he means it will take 2X of force to pull it back to that 2nd inch.

In my explanation I was assuming the spring came like that from the factory, not deformed into the position. Just wanted to clarify that.

That would be true for a K value of 2. Let's say you have a K value of 1 for simplicity's sake. Let's say you stetch 1 inch you have 100lb of force. (or vise versa). You stretch 3 inches, 300lb and so on. Even if you are at 1000lb of force, you add 100lb and you get another inch of flex.

Frank.

Hehe, I think we're trying to say the same thing here. From your original statement it sounds like the force is going to be the same to pull the bow from 0-1" as it is to pull it from 0-2". For example it'll take 10lbs to pull the bow from 0-1" and it'll still take 10lbs to pull from 1-2". At least that's what it sounds like to me (but I know what you mean). I can also see how mine doesn't quite sound right either...sigh...it's that communication thing agian.