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Anybody have an 64-66 factory swaybar?
https://slantsix.org/forum/viewtopic.php?t=17235		 Page 2 of 3
Author:  emsvitil [ Wed Apr 26, 2006 1:44 pm ]
Post subject: 
Quote:
does anyone know an equation that will determine swaybar rates? -james

I have something somewhere....... There's the geometry part (length of center section,arm length, diameter) and then there's the spring part.

Would also be applicable for the torsion bars as I've only seen torsion bar diameter discussed and not their rates.
Author:  Dart270 [ Wed Apr 26, 2006 5:05 pm ]
Post subject: 
There must be a web calculator for this. I could whip up a spread sheet to calculate this easily enough (few hours). Anyone else?

James, ask me about it next time we meet...

Lou
Author:  NewLancerMan [ Wed Apr 26, 2006 5:38 pm ]
Post subject: 
Quote:
I have a factory swaybar from a 1966 Barracuda Formula S if you are still looking for one. I'll have a 66 k member and suspension if you need/want it for your 62, its from a v8 car and has 10" drums. I would also be interested in a lighter swaybar.
Argh! I completely forgot you had parted that 66 out! DUH! You're here in my backyard and I'm thinking of having people send them all over the country. Email me directly cause I don't think I have your number anymore.

how you feeling? Getting better?
Author:  emsvitil [ Wed Apr 26, 2006 6:12 pm ]
Post subject:  Mathematically Springy/swayey rateys.......
Coil Springs:

K = ((W^4)*G)/(8*N*(D^3))

K= spring rate lb/in
W= diameter of spring wire
G = 12,000,000 for steel springs
N = number of active coils
D = diameter of coil (center to center)


Torsion Bars:

K= 1,178,000 * (D^4)/(L*(A^2))

K= spring rate lb/in
D = diameter of bar
L = length of bar
A = length of lever arm (assumed to not bend)


Sway Bars (solid):

K = (500,000 * (D^4))/((.4244*(A^2)*B) + (.2264 * (C^3)))

K= spring rate lb/in
D = diameter of bar
B = length of bar (center section i.e major bend to major bend)
A = Perpendicular length of lever arm (major bend to end)
C = Actual length of lever arm (major bend to end, unless bend is exactly 90 degrees, longer than A)

Formula considers twist of center section and bending of end sections.


Tubular (hollow bars) will have a similiar formula, just don't know what it is.
Author:  Dart270 [ Wed Apr 26, 2006 7:19 pm ]
Post subject: 
I think the "actual length of lever arm" should be defined as C.

You can do this for the tubular arms/bars by "the superposition principle."

For example,

K= 1,178,000 * (D^4-E^4)/(L*(A^2)) for torsion bars.

where D is the outer diameter and E is the inner diameter. I don't have time at the moment to work this out for sway bars, but it will be similar.

Lou
Author:  emsvitil [ Wed Apr 26, 2006 7:51 pm ]
Post subject: 
Quote:
I think the "actual length of lever arm" should be defined as C.

Whoops.......

fixed it...
Author:  james longhurst [ Wed Apr 26, 2006 9:00 pm ]
Post subject: 
Quote:
Quote:
I think the "actual length of lever arm" should be defined as C.

Whoops.......

fixed it...
i just read both of those posts twice and thought "wha?..." :D

-james
Author:  emsvitil [ Wed Apr 26, 2006 9:23 pm ]
Post subject: 
Quote:
You can do this for the tubular arms/bars by "the superposition principle."

For example,

K= 1,178,000 * (D^4-E^4)/(L*(A^2)) for torsion bars.

where D is the outer diameter and E is the inner diameter. I don't have time at the moment to work this out for sway bars, but it will be similar.

Lou

Thinking about this a little, and I'm not sure it's just the cross-sectional area of metal that's resisting the twisting force.......

It's a combination of area and how far the mass of metal is from the centerline.

i.e. 2 tubes that have the same cross sectional area, the larger diameter tube resists the twisting motion more even though it's thinner walled
Author:  emsvitil [ Wed Apr 26, 2006 9:23 pm ]
Post subject: 
Quote:
Quote:
Quote:
I think the "actual length of lever arm" should be defined as C.

Whoops.......

fixed it...
i just read both of those posts twice and thought "wha?..." :D

-james

I had my definition of terms wrong...
Author:  Dart270 [ Thu Apr 27, 2006 3:10 am ]
Post subject: 
Ed,

You're right, it's not just the cross-sectional area, since it goes as the 4th power of bar radius/diam, so metal farther out contributes DRAMATICALLY more than towards the axis of the bar. That's why a tubular bar with a thin wall doesn't need to be much bigger diam than a solid bar for the same rate, and why you save mass with a tubular bar.

The mass of the torsion bar material is a tiny effect.

Lou
Author:  Al T [ Thu Apr 27, 2006 7:42 pm ]
Post subject:  4th power 1/2 life decay rates emperical coefficient degree
Damn scientific types!! Can the rest of us join in? :wink: :wink:

Actually its great you guys bringing theory to the hobby.

BTW Lou, where you moving too?
Author:  Hyper'72Valiant [ Thu Apr 27, 2006 10:28 pm ]
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I was planning on getting a sway bar for my '72 Valiant later on, but, if these get made soon, I will have to buy one. :D
I never knew that the ones that are being sold by addco and other places had ground clearance problems. Is that because of where they are mounted?
Cory
Author:  sandy in BC [ Fri Apr 28, 2006 8:13 am ]
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The ground clearance thing is silly. You would pound your front valance flat before you would bottom the bar mounts.
If you saw some of the places my Valaint has been ......and never has the mount touched down. My valance however....
Author:  slantvaliant [ Fri Apr 28, 2006 8:13 pm ]
Post subject: 
I'm not worried about bottoming out, per se, but I am concerned about damage from some obstacle sticking up, waiting to go between the front wheels and grab a low-hanging, non-shielded, important something. I drive in some rough places, and have to deal with some pretty primitive parking situations as well!
Author:  sandy in BC [ Fri Apr 28, 2006 8:35 pm ]
Post subject: 
doh!
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