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03102014, 01:20 PM


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Location: Ventura County CA
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BTW it would be nice to know what your unsprung weights are (axles, wheels, tires) so the above can be refined some. You will get slightly higher ride frequencies in practice since the sprung weight is less than the total weight.
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03112014, 05:51 PM


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Join Date: Feb 2014
Location: Taylor Bay Washington
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I think some clarification is needed.
Picture #1 is the rear of a car I built about 12 years ago.
Picture #2 is the front of the same car.
Picture # 3 is the one I am working on now.
Picture #4 is a car I built 67 years ago.
The Bandit
I ordered a "hanging scale" today 550# ,should be here by Fri.
So for a "flatride" could I just remove the short spring from the front since the front is a little stiff.
The car is just 98% cruising and maybe a "track day" if it passes tech.
The Goodguys AutoCross looks like a lot of fun.
Fred

03112014, 11:13 PM


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I honestly don't know what removing the short leaf will do to your spring rate. The beauty of this is it's easy to change and you can pull a leaf from just the upper and play around. Shocks and how they are mounted will play a big role in this too. Suspension is a system comprised of springs, shocks, swaybars, tires, chassis, bushings, etc so all the math and physics above wont mean squat (antisquat?) in the real world if the whole system doesn't work together. A bit of cut & try and iteration fortunately isn't too hard and it sounds like you've already built a few of these so you certainly have a better idea of what does and doesn't work than I do. I really should have qualified everything I posted above as classroom learnings; I have very little experience applying this in the real world.
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03112014, 11:32 PM


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Quote:
Originally Posted by lakesmod
I found this on Pirate4X4.com
I'm not questioning the knowledge here, just wondering what the tech behind it is
the moment of a simple span beam (your leafspring, L inches long) with a load in the middle (your axle, P) is PL/4
the moment for a cantilever beam with the load at the end is PL
so if you cut the spring in half, use it as a 1/4 eliptic, you should be doubling the spring rate.
I could have missed something, that's a gross simpification, but I think it'll get you in the ballpark.

The deflection of a simply supported beam with a point load is x=PL^3/48EI, therefore the stiffness is P/x=48EI/L^3
The deflection of a cantilevered beam with a point load is x=PL^3/3EI, therefore the stiffness is P/x=3EI/L^3. However the dimension L for the cantilevered beam in question is half the dimension of the simply supported beam (since we cut our leaf in half), so if we want "L" to be the same, the stiffness then becomes P/x=3EI/(L/2)^3. That my friend is P/x=24EI/L^3 which is exactly half the stiffness of the simply supported beam above.
Therefore I conclude with math and science that one half of the leaf spring mounted as you've shown, will produce a stiffness that is half of the original full length spring.
Another way to think about this is through superposition. If you were to cut a normal leaf spring in half, but mount both halves exactly as they originally were mounted, you would get the same stiffness you originally had. Take one half away and you get half the stiffness since each half of the spring contributed equally.
(okay to be technically accurate, the leaf spring is not a simple beam and can not be described by the formula above because it is arching and has a variable cross section, but superposition still holds!)
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Last edited by TheBandit; 03122014 at 02:06 AM.

03122014, 12:22 AM


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Join Date: Dec 2007
Location: Maple Ridge, B.C. Canada
Posts: 3,152


Quote:
Originally Posted by TheBandit
I honestly don't know what removing the short leaf will do to your spring rate. The beauty of this is it's easy to change and you can pull a leaf from just the upper and play around. Shocks and how they are mounted will play a big role in this too. Suspension is a system comprised of springs, shocks, swaybars, tires, chassis, bushings, etc so all the math and physics above wont mean squat (antisquat?) in the real world if the whole system doesn't work together. A bit of cut & try and iteration fortunately isn't too hard and it sounds like you've already built a few of these so you certainly have a better idea of what does and doesn't work than I do. I really should have qualified everything I posted above as classroom learnings; I have very little experience applying this in the real world.

I gotta say... I luv this comment..

03122014, 01:00 AM


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Join Date: Jan 2006
Location: Ventura County CA
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Quote:
Originally Posted by FrogDog
I gotta say... I luv this comment..

The part where I admit I don't know what the hell I'm talking about?
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Last edited by TheBandit; 03122014 at 01:35 AM.

03122014, 01:44 PM


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Quote:
Originally Posted by TheBandit
The part where I admit I don't know what the hell I'm talking about?

.. Naw... Not that part .. It was the kerfuffle part
r'z .......
Last edited by FrogDog; 03132014 at 02:52 AM.
Reason: Cause I Can

03122014, 01:59 PM


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Join Date: May 2006
Location: Denver, NC
Posts: 1,724


I've been avoiding this thread because I have an aversion to leaf springs. They're usually the simplest, cheapest way to go, but tough to analyze and model. We've done a couple of simulation projects at work on vehicles with leaf springs and it's always a tough deal to get right.
In Milliken's "Race Car Vehicle Dynamics" there is a section about leaf springs. They site the Bosch Automotive Handbook for their calculations, but the gist of it is that for the same geometry, a semielliptic and quarter elliptic spring have the same rate. The reason is because you're dividing the load between the two ends, so the stress level in the semielliptic spring is half of a quarterelliptic. This means that if you cut a semielliptic spring in half, the rate will be the same, but the allowable maximum load before you overstress the spring is cut in half. (On edit: This isn't right...got confused because the authors of the book redefined spring rate from one configuration to the other...see post below)
For a monoleaf spring, they use the equation c = (E * b * t^3) / (4 * l^3), where c is the spring rate, E is Young's modulus, b is the width of the leaf, t is the thickness, and l is the length from the axle centerline to the spring eye(s). It assumes the axle is centered between the eyes for the semielliptic.
Just for grins, a 24" long leaf, 2" wide by 0.25" thick mounted as a quarter elliptic would have a rate of:
c = (30,000,000 * 2 * 0.25^3) / (4 * 24^3) = 17 lb/in
If there are multiple leaves, the equation changes to:
c = ((2 + (n'/n)) * E * n * b * t^3) / (6 * l^3)
Where n is the number of leaves, and n' is the number of leaves at the spring eye(s).
So, say 5 leaves, 2" wide x 0.25" thick, 24" long, with two leaves going the full length of the spring would result in the rate being:
c = ((2 + (2/5)) * 30,000,000 * 5 * 2 * 0.25^3) / (6 * 24^3) = 135.6 lb/in
Which does not account for friction between the leaves. The configuration of the shackles also has an effect on the rate and how it changes vs. deflection.
So that's the theory. I haven't actually taken a leaf pack and compared the measured rate to the calculated values for it's dimensions, but these numbers seem pretty reasonable.
Last edited by Graham08; 03122014 at 03:37 PM.

03122014, 02:07 PM


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Quote:
Originally Posted by Graham08
IFor a monoleaf spring, they use the equation c = (E * b * t^3) / (4 * l^3), where c is the spring rate, E is Young's modulus, b is the width of the leaf, t is the thickness, and l is the length from the axle centerline to the spring eye(s). It assumes the axle is centered between the eyes for the semielliptic.

That agrees with my equation; it just includes "I" for a rectangular cross section.
Quote:
Originally Posted by Graham08
In Milliken's "Race Car Vehicle Dynamics" there is a section about leaf springs. They site the Bosch Automotive Handbook for their calculations, but the gist of it is that for the same geometry, a semielliptic and quarter elliptic spring have the same rate. The reason is because you're dividing the load between the two ends,

Well shit that just blows up my assertion, but for the moment I disagree because the P when calculating deflection for the semielliptical is defined at the center (not P/2 at each end) and the P for the quarterelliptical is defined at the end; both of those positions agree with where the vehicle mass/weight is reacted and therefore I believe my proof above still holds water.
I need to wrap my head around it because I'm sure Miliken knows what he's talking about.
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Last edited by TheBandit; 03122014 at 02:13 PM.

03122014, 03:15 PM


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Join Date: May 2006
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Quote:
Originally Posted by TheBandit
I need to wrap my head around it because I'm sure Miliken knows what he's talking about.

Here's the diagram. Now as I'm rereading it, I'm questioning my original post.
The equate the spring rate "c" for the semielliptic as being 2F/s (force over displacement). The quarter elliptic is just F/s...meaning cutting a spring pack in half also cuts the rate in half because you're only deflecting one quarter elliptic spring instead of two...
This is probably my least favorite chapter in the book. The other spring equations are just as murky. As someone trying to tune a racecar, I want to know what the stiffness of the spring or torsion bar is. Stress is important if I'm designing a spring, or double checking someone else's design, but not a primary factor in making the car turn.

03122014, 03:30 PM


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*EDIT* I will come back to this, but I still stand by my original proof. The book defines the loads differently (F in one case and 2F in another) which is likely why it looks like the stiffness is the same, but in reality the load should be F not 2F for the semielliptical because we are talking about the same wheel force in both cases.
I will bolster my original proof with evidence later :)
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Last edited by TheBandit; 03122014 at 03:53 PM.

03122014, 03:52 PM


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I'm looking at the Bosch book now. It's done the same way there. I don't understand why. I think they dropped a "2" from the denominator of the right hand side, though, because if you divide both sides by two to get "F/s" on the left side, the rate is now divided in half, which can't be right for two cantilever springs in parallel.

03122014, 03:56 PM


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Another possibility for the discrepancy is they say the semielliptical is "rolled out parabolically" and show t0 for the thickness at the center. That implies variable cross section, which isn't shown for the elliptical spring.
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03122014, 03:59 PM


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here is some experimental evidence that the quarterelliptical is definitely less stiff than the semielliptical. Same ruler constrained both ways with the same tape dispenser used as a load. The semiellpitcal is not touching the desk. The quarterelliptical is. If I had another ruler handy I would measure the deflection too haha. Well at least I have plenty of tape dispensers.
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Last edited by TheBandit; 03122014 at 04:02 PM.

03122014, 04:06 PM


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Quote:
Originally Posted by TheBandit
here is some experimental evidence that the quarterelliptical is definitely less stiff than the semielliptical. Same ruler constrained both ways with the same tape dispenser used as a load. The semiellpitcal is not touching the desk. The quarterelliptical is. If I had another ruler handy I would measure the deflection too haha. Well at least I have plenty of tape dispensers.

Nice!
I just glanced at the "other" Milliken book (Chassis Design, which is really a compilation of Maurice Olley's notes) and there's a whole chapter devoted to the subject of leaf springs. It's going to take a bit to digest, though, because it's not just simple equations.
On edit: Olley calculates the rate for a general semiellliptic spring from the rates of the two quarterelliptics that make it up and their respective lengths. If the semielliptic is symmetric, it's rate is exactly twice that of the two quarterelliptics that it is comprised of.
If you're interested, the equation is:
K = (K1 * K2 * (a + b) ^ 2) / (K1 * a^2 + K2 * b^2)
Where K1 and K2 are the two quarterelliptic springs, and a and b are their respective lengths.
The equation in my initial post for the rate of a quarterelliptic is a simplified version of what Olley was doing, but it's valid.
Last edited by Graham08; 03122014 at 04:23 PM.

03122014, 05:45 PM


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Quote:
Originally Posted by Graham08
If the semielliptic is symmetric, it's rate is exactly twice that of the two quarterelliptics that it is comprised of.

So it appears we have further confirmation that the quarterelliptic is half the rate of the semielliptic it was cut from. Agreed?
Now maybe lakeshoremod can get back to building cool shit while us engineers suffer from analysis paralysis.
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03122014, 06:02 PM


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Agreed.
I think the rate equation for the semielliptical spring in the Bosch and Milliken books is misstated at best.
Apologies to the OP for cluttering up his thread with this mental mastrubation session.
Last edited by Graham08; 03132014 at 07:45 AM.

03142014, 12:18 PM


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Join Date: Feb 2014
Location: Taylor Bay Washington
Posts: 33


Some numbers
Engine and 3speed manual trans..................590#
Front end ( I beam ) with disk brakes............140#
Front wheel and tire 175/65 R 15..45# X 2 =90#
12 bolt chevy rear end( no breaks )..............160#
1 rear wheel and tire 265/75R 15..60#..X 2 = 120#
Frame with floor.........................................190#
Body with 1" X 1" tubing...............................60#
Total .................................................. ......1350#
Unsprung weight.........................................510 #
Fred

03232014, 11:25 AM


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Join Date: Feb 2014
Location: Taylor Bay Washington
Posts: 33


Have the front 1/4 E's tacked.

04072014, 09:24 PM


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Join Date: Feb 2014
Location: Taylor Bay Washington
Posts: 33


Did some work on the 1/4 E's.
There is a video on my photo bucket with new pictures.
http://s210.photobucket.com/user/lak...?sort=3&page=1
Let me know if the pictures don't show.
Fred



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