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#1
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coilspring rates vs. torsionbar size
Tim;
In our discussion on torsionbar rates some time ago http://www.germanlook.com/Forums/sho...5&pagenumber=2 I noticed that your maths overdo mine greatly. Now I found this article of Porches torsion bar rates converted to inch/lbs http://tech.rennlist.com/911/pdf/torsionbars.PDF Maybe you can get something out of it with the information you have from the Baja Bug-book. If you get interested on it let me know of the results. Justin |
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#2
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Hi Justin,
I put the numbers from the Baja Bug book into the Rennlist formula. d = bar diameter = 22mm = 0.866in (beetle stock size) L= torsion bar length = 21.75in (effective length = 20.75in) A = arm length = 16.5in (torsion bar centre to wheel centre) The result is 116lb/in According to the formula from the Baja Bug book, the force to change 1 degree is 35.6lb. To move 1 inch of the suspension, the torsion bar will be preloaded approximately 3.4 degree. Therefore lb/in is 35.6lb/degree x 3.4degree/in = 121lb/in Looks like the two formulas are consistent . Hope that this is not too complicated to read. Tim |
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#3
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So the stock torsion bar rate is around 116-121 lbs/inch?
Thats not very much I think. But I guess its impossible to compare as such linear coil spring to torsion bar strength. I wanted to know this because I read that if the spring rates are equal front and back, then you might end up driving a rocking chair. Thanks again Tim, I owe you another one. Justin |
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#4
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The example which I was using is the torsion bar length of pre-'69 swing axle. The bar length of Supers is 26-9/16in and the spring rate is around 94 lb/in by using the Rennlist formula.
Tim |
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#5
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What is your opinion Tim about rating torsion bars; If we decide that the stock rate on super is that 94 lbs/inch, then is it linear or rather progressive?
You might understand my interest on this if you think of the rockingchair effect I was warned of.. Justin |
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#6
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I think a torsion bar spring rate is linear because a linear coil spring is basically a long torison bar wound up into a uniform spiral. In both cases, the load is absorbed by twisting the metal. Correct me if I am wrong.
Tim |
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