Sway Bar Motion Ratios
08-04-2010, 06:53 PM
#2
My front view approximation is:
F: 0.51 mmARB/mmWheelTravel
with a ~10.25" lever arm yields 2.86 degARB/degRoll
R: 0.45 mmARB/mmWheelTravel
with a ~8.25" lever arm yields 3.26 degARB/degRoll
My estimates.
Then I used theoretical bar in bending and torsion in series to estimate bar stiffness (Nm/degARB). I have not incorporated mount stiffness, though.
F: 0.51 mmARB/mmWheelTravel
with a ~10.25" lever arm yields 2.86 degARB/degRoll
R: 0.45 mmARB/mmWheelTravel
with a ~8.25" lever arm yields 3.26 degARB/degRoll
My estimates.
Then I used theoretical bar in bending and torsion in series to estimate bar stiffness (Nm/degARB). I have not incorporated mount stiffness, though.
08-05-2010, 02:21 AM
#3
Very close to my estimates. 
F: 0.53 mm/mm
36.5" straight length
10.5" total lever arm (9.5" perpendicular)
R: 0.42 mm/mm
38" straight length
8.5" total lever arm (8.25" perpendicular)
Do you like Puhn or Milliken's stiffness formula?
F: 0.53 mm/mm
36.5" straight length
10.5" total lever arm (9.5" perpendicular)
R: 0.42 mm/mm
38" straight length
8.5" total lever arm (8.25" perpendicular)
Do you like Puhn or Milliken's stiffness formula?
08-05-2010, 02:19 PM
#5
Milliken's formula is theoretically correct, but only for the case of perpendicular lever arms. Puhn's formula is an attempt to account for non-perpendicular lever arms, but it's an approximation, not a true generalization. By which I mean, when you apply Puhn's formula to a bar with perpendicular arms, it does NOT reduce to Milliken's.
The basic idea:
If the arms are perpendicular, then the usual formula given by Milliken et al. applies, which is just the rate of a cylindrical torsion spring:
S = (pi*d^4*G)/(32*L*R^2)
where G is constant (shear modulus of steel, ~11,000ksi).
Now for a bar with angled arms, Puhn's formula works out to be:
SP = (pi*d^4*G)/(30*L*R^2 + 16*C^3)
So obviously if you let R=C (same as assuming that the arms in Puhn's diagram are perpendicular), it does NOT reduce to the correct expression for a torsion spring.
That said, Milliken's formula is clearly not quite right for a bar with angled ends, and presumably Puhn had some real-world data that suggested this equation was pretty good. So if R/C is very close to 1 (like on our rear bar), I use Milliken, but if there's a substantial difference (like on the front bar) I use Puhn.
The basic idea:
Code:
L
_________________
R| / \ C
| / \
L = Length of center section (inches)
R = Perpendicular length of arm (torque arm - inches)
C = Total length of arm (inches)
d = Bar diameter (inches)
S = (pi*d^4*G)/(32*L*R^2)
where G is constant (shear modulus of steel, ~11,000ksi).
Now for a bar with angled arms, Puhn's formula works out to be:
SP = (pi*d^4*G)/(30*L*R^2 + 16*C^3)
So obviously if you let R=C (same as assuming that the arms in Puhn's diagram are perpendicular), it does NOT reduce to the correct expression for a torsion spring.
That said, Milliken's formula is clearly not quite right for a bar with angled ends, and presumably Puhn had some real-world data that suggested this equation was pretty good. So if R/C is very close to 1 (like on our rear bar), I use Milliken, but if there's a substantial difference (like on the front bar) I use Puhn.
08-06-2010, 06:47 AM
#6
Thanks for the information. I was oblivious to the possibility that one formula might be more accurate than another under certain conditions.
I've used this formula in the past. Both with the online calculator or the math itself.
How much angle is required to switch between the two without running the math on both? I don't have the Gendron bar in my hands yet, but visually it is angled. How much is yet to be determined.
By replacing my stock front bar with a Gendron bar, my initial math says I need to run the Gendron on the softest setting and add a rear bar to preserve the balance.
I've used this formula in the past. Both with the online calculator or the math itself.
How much angle is required to switch between the two without running the math on both? I don't have the Gendron bar in my hands yet, but visually it is angled. How much is yet to be determined.
By replacing my stock front bar with a Gendron bar, my initial math says I need to run the Gendron on the softest setting and add a rear bar to preserve the balance.
08-06-2010, 09:45 AM
#7
Originally Posted by twohoos,Aug 5 2010, 02:19 PM
The basic idea:
Now for a bar with angled arms, Puhn's formula works out to be:
SP = (pi*d^4*G)/(30*L*R^2 + 16*C^3)
Code:
L
_________________
R| / \ C
| / \
L = Length of center section (inches)
R = Perpendicular length of arm (torque arm - inches)
C = Total length of arm (inches)
d = Bar diameter (inches)
SP = (pi*d^4*G)/(30*L*R^2 + 16*C^3)
L= 36.5"
R= 9.5"
C= 10.5"
d= 28.2mm/25.4mm= 1.11023622
The formula you listed is giving a result I'm not expecting:
(pi*1.11023622**4*11)/(30*36.5*9.5**2 + 16*10.5**3)
0.00044744
The formula I listed feels closer but still not quite there:
500000*1.11023622**4/((0.4244*9.5**2*36.5)+0.2264*10.5**3)
457.60834217 lb/in
I'm told it should be ~300lb/in for the front bar of 28.2mmx5mm.
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08-09-2010, 08:08 PM
#8
There's a few things going on here.
1) My bad, sorry, should have said G = 11,000 ksi = 1.1e7 psi. So multiply your first answer by a million. I've corrected my post above.
2) The link you posted just uses Puhn's "A/B/C/D" formula, which has G embedded in it. I converted that to the more conventional form by factoring out G. (Turns out Puhn actually used G=11,250 ksi as I recall.)
3) To get the rate of a hollow bar, you have to subtract the rate of the hollowed-out part. So the '00 front 28.2mmx5mm bar is computed as (28.2mm solid bar rate) - (18.2mm solid bar rate) = 447 - 77 = 370 lb/in.
Hope this helps.
1) My bad, sorry, should have said G = 11,000 ksi = 1.1e7 psi. So multiply your first answer by a million.
I've corrected my post above.2) The link you posted just uses Puhn's "A/B/C/D" formula, which has G embedded in it. I converted that to the more conventional form by factoring out G. (Turns out Puhn actually used G=11,250 ksi as I recall.)
3) To get the rate of a hollow bar, you have to subtract the rate of the hollowed-out part. So the '00 front 28.2mmx5mm bar is computed as (28.2mm solid bar rate) - (18.2mm solid bar rate) = 447 - 77 = 370 lb/in.
Hope this helps.
08-09-2010, 08:56 PM
#9
Just to finish the thought:
The formulas here are the "spring rate" S of a torsion spring (in lb/in). The actual roll stiffness U = S*(M*K)^2*[pi/(12*180)] (lb-ft/deg), where M = motion ratio and K=track (in.).
For the '00 front bar, using 0.53 motion ratio and 57.9in for front track, I get a roll rate of 506 lb-ft/deg.
The formulas here are the "spring rate" S of a torsion spring (in lb/in). The actual roll stiffness U = S*(M*K)^2*[pi/(12*180)] (lb-ft/deg), where M = motion ratio and K=track (in.).
For the '00 front bar, using 0.53 motion ratio and 57.9in for front track, I get a roll rate of 506 lb-ft/deg.
08-10-2010, 11:47 AM
#10
When comparing the handling bias of bars, is it better to compare a wheel rate or a roll rate? Since I haven't modeled the chassis, either number appears to be quite theoretical in its affect on the balance.