Top 3-5 cost efficient ways to decrease body roll? - Page 6 - PeachParts Mercedes-Benz Forum

turbobenz · #76 01-14-2011, 04:00 AM

Quote:

Originally Posted by MBeige

Even the AMG tuned 500CE had some body roll

thats my dream car btw. I actually have a red coupe parts car sitting. Just needs a lot of stuff. The interior is molding bad (steering wheel is growing) and needs lots of parts. Make it just like that one but with a 606 tuned to about 275 hp I think would be perfect. Or a black SEC

English Bulldog · #77 01-14-2011, 06:18 AM

Quote:

Originally Posted by MBeige

Even the AMG tuned 500CE had some body roll

Not much body roll though. How they do it? special sway bars???

this car is lowered about 2 1/2 inches or even 2.75 from w123 stock.

Euro lights

wheels 235 16s ?????

wheels painted two colors. if it wasnt amg, some would say two tone wheels are stupid.

English Bulldog · #78 01-14-2011, 06:32 AM

Quote:

Originally Posted by EDBSO

I plan on redoing the sway bar bushings and end links and then
My E300 is veI plan on simply doubling up the bars for a simple cost effective solution. As weather permits I will do the work and report back. I am realistic and expect less body roll.

What are the steps to do this.

What is a budget to do this.

winmutt · #79 01-14-2011, 10:09 AM

Quote:

Originally Posted by leathermang

Doing the front sway bar is a big job because lots of other things have to be taken out and put back in to replace it..
but where do you think something could be attached TO which would accomplish what the antisway bar does .... which basically uses the OTHER WHEEL position to modify with spring action the following of the body towards centrifugal force.

IIRC all that is needed is to remove the battery tray, separate the mbc from firewall and jack the rear of the trans up....

BTW I recently replaced my front spring pads some months after the springs. They really do smooth out the ride a surprisingly amount.

Some roll is good in these cars. If you look at the way the rear end geometry is setup. Toe adjusts with camber, its certainly a precursor to the multilink rear.

mhin1956 · #80 01-14-2011, 11:43 AM

SORRY, can't resist...

best way to reduce roll is to stay away from DONUTS.

MNBenz · #81 01-14-2011, 12:11 PM

My vote is for bilstein hd's, I experienced much less body roll on the freeway after putting them in.

t walgamuth · #82 01-14-2011, 12:36 PM

Quote:

Originally Posted by leathermang

The longer one will give more...
but you have not made it stiffer...
you have simply doubled the wire length and it is giving under the same load per increment of length.
Cutting the length will make it bottom out faster... but does not make it stiffer.

I don't understand your first three lines and I disagree with your third. Tonight I will try to decifer what Roy posted.

#83 01-14-2011, 12:57 PM

Let me try:

Think of each coil of the spring as a twisted torsion bar. Assuming you are within the design operating range of the spring, the force exerted by that coil is proportional to the displacement of that coil (Hooke's law). So if my spring is 10 inches long and has 10 coils, I can think of it like 10 small springs stacked up.

Lets say I depress the 10-inch spring by 1 inch and feel 100 pounds of force, that means each of the coils was displaced by 1/10-inch and exerted 10 ponds of force. If I depress the spring 2 inches, I will feel 200 pounds, etc.; until I exceed the operating range of the spring.

Now, if we remove one coil we have a 9-inch, 9 coil spring. If I depress this spring by 1 inch, each of the coils will be displaced by 1/9-inch and the resulting force will be 11.1 pounds. However, I only have 9 coils contributing to the total force so, 11.1 x 9 = 100 pounds.

The result is that the overall spring coefficient is the same for the shorter spring, but it has less usable travel. In both cases a total displacement of 1 inch results in a force of 100 pounds. Shortening the spring does not make it stiffer, just shorter.

moon161 · #84 01-14-2011, 02:54 PM

Wrong.

Shear does not accumulate like displacement through the length of the spring. If it did, there would be no reaction at one of the supports- either the spring perch or the LCA.

Viewing each coil as a free body is useful, and spring body connects the equal and opposite (in a static case) forces at the spring perch and LCA via shear.

Consider the spring again 10" 10 coils, 100lb/inch. For a 100lb load, each coil sees 100lb, and deflects .1". The spring constant of each coil is the force over the deflection, or 100/.1 = 1000lb/in.

Cut a coil off, 100lb load, deflection is 9x.1" or .9". To achieve 1" deflection, each coil deflects 1/9" = .111", the reaction is 1000*.111=111 lb. The new spring constant is 111 lb/in.

Another way to look at it is each coil being in series w/ the next.

http://en.wikipedia.org/wiki/Hooke%27s_law#Multiple_springs

Springs in series add like resistors in parallel:
If added in series, 1/k = Sum(1/k1 + 1/k2 ....)

10 coils 1/k = 10 * 1/1000 =.01
k = 1/.01 = 100

9 coils 1/k = 9 * 1/1000 = .009
k = 1/.009 = 111

t walgamuth · #85 01-14-2011, 03:42 PM

Quote:

Originally Posted by moon161

Wrong.

Shear does not accumulate like displacement through the length of the spring. If it did, there would be no reaction at one of the supports- either the spring perch or the LCA.

Viewing each coil as a free body is useful, and spring body connects the equal and opposite (in a static case) forces at the spring perch and LCA via shear.

Consider the spring again 10" 10 coils, 100lb/inch. For a 100lb load, each coil sees 100lb, and deflects .1". The spring constant of each coil is the force over the deflection, or 100/.1 = 1000lb/in.

Cut a coil off, 100lb load, deflection is 9x.1" or .9". To achieve 1" deflection, each coil deflects 1/9" = .111", the reaction is 1000*.111=111 lb. The new spring constant is 111 lb/in.

Another way to look at it is each coil being in series w/ the next.

http://en.wikipedia.org/wiki/Hooke%27s_law#Multiple_springs

Springs in series add like resistors in parallel:
If added in series, 1/k = Sum(1/k1 + 1/k2 ....)

10 coils 1/k = 10 * 1/1000 =.01
k = 1/.01 = 100

9 coils 1/k = 9 * 1/1000 = .009
k = 1/.009 = 111

So you are saying cutting a coil off increases rate by 10%, right?

leathermang · #86 01-14-2011, 03:42 PM

Moon161, Do you want to correct the sentence fragment which says 'each coil sees 100 lbs' ... you have the math in the next clause correct for your initial example ..but each coil only sees 10 lbs... and does deflect .1 inch
before we go on ?

leathermang · #87 01-14-2011, 03:47 PM

Tom, he is trying to say that ..but that is not correct.
Given the same load.. the 100 lbs... spread between either 10 coils or 9...
the deflection per coil is the same.. but with 9 as the load is increased it will encounter the limits of its ability to compress.. thus feeling stiffer... and since we are assuming the load of the car in question remains the same.... each coil still represents a deflection of one inch per 100 lb load .. so the short one is not stiffer.. only shorter and has less travel to offer.

( the load is increased in corners for instance ...and less roll will feel like stiffer.. but will just be less travel available.)

moon161 · #88 01-14-2011, 04:20 PM

Pretty sure my example is correct. Get comfortable with free body diagrams, & refer to a physics, machine design or solid mechanics book for further explanation.

leathermang · #89 01-14-2011, 04:27 PM

"For a 100lb load, each coil sees 100lb,"
Your choice... but for 10 coils and one inch deflection per 100 lbs your math of .1 inch deflection is correct.. but your statement that EACH coil sees 100 lbs is not.. your own math does not match your wording...

t walgamuth · #90 01-14-2011, 04:59 PM

Quote:

Originally Posted by leathermang

"For a 100lb load, each coil sees 100lb,"
Your choice... but for 10 coils and one inch deflection per 100 lbs your math of .1 inch deflection is correct.. but your statement that EACH coil sees 100 lbs is not.. your own math does not match your wording...

Coils don't have eyes, right?