Swaybar rate?
#5
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most papers ive read assume bars are just mild steel. there is a formula:
500,000 D^4
K (lbs/in) = -------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)
B
-----_________________
A| /----------------------------\ C
| /------------------------------\
A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)
then use:
WHEEL RATE = K * Mr ^ 2
but there are many factors you have to take into account like the mounting point flex etc. but i do think you could at least get a ball park figure to help you combine a bar/springs to get pretty close to what was desired.
realistically there is a whole bunch of other stuff that needs to be figured too like roll center and spring frequency...its a giant can of worms.
500,000 D^4
K (lbs/in) = -------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)
B
-----_________________
A| /----------------------------\ C
| /------------------------------\
A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)
then use:
WHEEL RATE = K * Mr ^ 2
but there are many factors you have to take into account like the mounting point flex etc. but i do think you could at least get a ball park figure to help you combine a bar/springs to get pretty close to what was desired.
realistically there is a whole bunch of other stuff that needs to be figured too like roll center and spring frequency...its a giant can of worms.
Last edited by killerpenguin21; Jan 20, 2013 at 07:42 PM.
#6
most papers ive read assume bars are just mild steel. there is a formula:
500,000 D^4
K (lbs/in) = -------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)
B
-----_________________
A| /----------------------------\ C
| /------------------------------\
A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)
then use:
WHEEL RATE = K * Mr ^ 2
but there are many factors you have to take into account like the mounting point flex etc. but i do think you could at least get a ball park figure to help you combine a bar/springs to get pretty close to what was desired.
realistically there is a whole bunch of other stuff that needs to be figured too like roll center and spring frequency...its a giant can of worms.
500,000 D^4
K (lbs/in) = -------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)
B
-----_________________
A| /----------------------------\ C
| /------------------------------\
A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)
then use:
WHEEL RATE = K * Mr ^ 2
but there are many factors you have to take into account like the mounting point flex etc. but i do think you could at least get a ball park figure to help you combine a bar/springs to get pretty close to what was desired.
realistically there is a whole bunch of other stuff that needs to be figured too like roll center and spring frequency...its a giant can of worms.
Last edited by flyingscot; Jan 21, 2013 at 05:26 AM.
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#9
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most papers ive read assume bars are just mild steel. there is a formula:
500,000 D^4
K (lbs/in) = -------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)
B
-----_________________
A| /----------------------------\ C
| /------------------------------\
A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)
then use:
WHEEL RATE = K * Mr ^ 2
but there are many factors you have to take into account like the mounting point flex etc. but i do think you could at least get a ball park figure to help you combine a bar/springs to get pretty close to what was desired.
realistically there is a whole bunch of other stuff that needs to be figured too like roll center and spring frequency...its a giant can of worms.
500,000 D^4
K (lbs/in) = -------------------------------------
(0.4244 x A^2 x B) + (0.2264 x C^3)
B
-----_________________
A| /----------------------------\ C
| /------------------------------\
A - Length of end perpendicular to B (torque arm - inches)
B - Length of center section (inches)
C - Length of end (inches)
D - Diameter bar (inches)
then use:
WHEEL RATE = K * Mr ^ 2
but there are many factors you have to take into account like the mounting point flex etc. but i do think you could at least get a ball park figure to help you combine a bar/springs to get pretty close to what was desired.
realistically there is a whole bunch of other stuff that needs to be figured too like roll center and spring frequency...its a giant can of worms.
Last edited by BEKevo; Jan 21, 2013 at 10:11 AM. Reason: spelling
#10
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Besides the 4 basic measures of the swaybar, itself (i.e., the A, B, C, & D from the Fred Puhn formula), you need the motion ratio of the bar as it attached to the car. Then you'll be able to calculate pounds/degree.
Out of curiosity, why do you want pounds for 5* of roll? The more typical ways to describe the effects of bars is to either add their effect to that of the main springs to compute the total wheel rate in single-wheel bump and/or to compute the amount of body roll for a particular lateral g. The latter, of course, requires even more missing information, such as cg height, track, etc, but is one of the ways that people talk to each other, as in, for example: 3.5* of roll at one g.
#11
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ahhh.. I remember this formula... You can figure out the swaybar rate in ftlbs, while not being mounted. But this information would be fruitless, as killerP pointed out. There are so many variables that affect the "swaybars" rate (ie mounting points, spring frequencies etc.. )
Again, assuming infinitely stiff bushings and no play in the end-links, once you have the effective K for the bar from Puhn's formula, you multiply by the square of the motion ratio and add this to the wheel-rate K from the main springs.
#13
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You're welcome. Swaybar math is both simple and complicated. The simple way to look at swaybars is that they are twice as effective against body roll as they are against single-wheel bump. So, if you want to stop the car from rolling so much but want to keep the suspension somewhat compliant, you use a lot of bar relative to the amount of spring. The complicated way to look at swaybars is to actually calculate the final wheel-rate for single-wheel bump when both the bars and main springs are included, as well as calculate the amount of roll for a given lateral g (or the amount of weight transfer required to roll the car one degree, if you prefer).
The reason why I have been so interested in the complicated approach is two-fold. First, I've been trying to figure out if it would be possible to create what might be called a multi-purpose duffer-mobile: rallycross with the swaybars detached and autocross with the swaybars connected. I'm worried that this will end up such a compromise that the car will stink at both, but I really want to try it for Ss & Gs.
Second, I cannot for the life of me figure out why no formula for desired initial damping ever includes the effect of the swaybars on single-wheel bump. The reason for wanting to know this relates to the above duffer-mobile. If I get shocks that are correct for, say, 4K springs on a rough surface, how badly will said shocks be when large swaybars are added for tarmac?
The reason why I have been so interested in the complicated approach is two-fold. First, I've been trying to figure out if it would be possible to create what might be called a multi-purpose duffer-mobile: rallycross with the swaybars detached and autocross with the swaybars connected. I'm worried that this will end up such a compromise that the car will stink at both, but I really want to try it for Ss & Gs.
Second, I cannot for the life of me figure out why no formula for desired initial damping ever includes the effect of the swaybars on single-wheel bump. The reason for wanting to know this relates to the above duffer-mobile. If I get shocks that are correct for, say, 4K springs on a rough surface, how badly will said shocks be when large swaybars are added for tarmac?
#14
EvoM Guru
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Material is irrelevant because most steels are all just about the same modulus of elasticity, though if you want numbers look up spring steel. It most likely what youll find the bars made from.
Absolute numbers are nearly useless (to 99.99%) of people. You are better off looking at change in stiffness. Roughly speaking, you can get it by:
Diameter change: 1 - Dold^4/Dnew^4. (Larger Diameter is stiffer, results in positive number)
Leg length change (If my math is correct): 1 - Anew^2/Aold^2. (Longer leg, softer bar) (This also assumes B is more than several times the length of A & C, and the results will be off a couple percent of the overall change depending on several bar specific factors).
Absolute numbers are nearly useless (to 99.99%) of people. You are better off looking at change in stiffness. Roughly speaking, you can get it by:
Diameter change: 1 - Dold^4/Dnew^4. (Larger Diameter is stiffer, results in positive number)
Leg length change (If my math is correct): 1 - Anew^2/Aold^2. (Longer leg, softer bar) (This also assumes B is more than several times the length of A & C, and the results will be off a couple percent of the overall change depending on several bar specific factors).
#15
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I totally agree that you can use the 500,000 constant for pretty much any material. I also agree that diameter (D in Puhn's formula) is raised to the fourth power. And I also agree that Puhn's formula is for standard-type swaybars made from one piece of bent bar with a body length (B to Puhn) that is very high relative to the other lengths. However...
Lever-arm length (A to Puhn) is the one that is squared. Leg length (C to Puhn) is actually cubed.
edit: I just noticed that you wrote "Anew" and "Aold," so we are not disagreeing about what power to use for A. It's merely a labeling disagreement. The correct name for A is "lever-arm length" or "effective lever-arm length"; the term "leg length" is used for C.
Lever-arm length (A to Puhn) is the one that is squared. Leg length (C to Puhn) is actually cubed.
edit: I just noticed that you wrote "Anew" and "Aold," so we are not disagreeing about what power to use for A. It's merely a labeling disagreement. The correct name for A is "lever-arm length" or "effective lever-arm length"; the term "leg length" is used for C.
Last edited by Iowa999; Jan 21, 2013 at 11:21 AM.