I'm looking for some help on calculating the spring rates on a dual spring setup. I tried searching here and on the web for a calculator of some sort, but all I can find are calculators which give me the initial rate...

I'm looking to find the point which the tender spring is at coil-bind and the main spring rate takes over. I'm been playing with some calculations in excel, but I have a feeling I'm trying to recreate the wheel, someone has to have already done this.




Modified by Mugenlude at 2:26 PM 11/15/2007

 

Tender springs, or zero-rate springs, have such low rates that I imagine they'd coil bind with hardly any weight on them. They shouldn't be affecting your spring rate.

 

if you know the spring rate of both springs, its easy to calculate the combined spring setup. you inverse the sum of the inverse of both spring rates.

but then you need to calculate the compressed distance for each spring based on their respective spring rate.

then you set the limit of the shorter, most likely weaker, spring compressed length when it bottoms out. at that point, then its a single spring rate of the likely greater spring rate.

knowing all this itd be faster to write a formula in excel than waiting for someone else to spoon feed you it.

granted, thats theoretical because at the extreme ends of barely compressed and almost fully compressed, the spring rate isnt really constant anymore. but thats how you would calculate it mathematically.

 

I wasn't looking to have it 'spoon feed' to me, I'm just looking to make sure I'm calculating it correctly. You act like it's readily available someplace and it's as easy as looking it up...

I have some numbers which I formulated in excel already, which at this point I believe are right, but I have nothing to compare them to...

Maybe someone could take a look at them and let me know if they look correct, thanks.

http://www.teamundercoatracing...c.xls

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ThoseDarnKids &raquo;</TD></TR><TR><TD CLASS="quote">Tender springs, or zero-rate springs, have such low rates that I imagine they'd coil bind with hardly any weight on them. They shouldn't be affecting your spring rate.</TD></TR></TABLE>

zero-rate springs and helper springs are this way, tender springs are designed to provide a lower rates until they react coil bind, at which point the main spring takes over at it's designed spring rate.


Modified by Mugenlude at 12:04 PM 11/19/2007

 

I doubt you are going to find any solid information that you can translate to your set-up because it really changes a lot.

The rate is progressive and increases through the travel until the tender spring is dead, and then it picks up the rate of the main spring.

I used a spring rate testing machine, and figure out my set-up.

All in all i ended up taking them out and just running a main spring. I didn't like the progressiveness that was there. And my setup had about .5 inchs of compression untill i was on the main spring.

I looked at your spread sheet, but don't know if it is correct. Ill see if i can find the graph i made of my dual rate spring set-up.

Also if you want some tender springs i have a few sets that im not using any more.

 

Did you use your dual spring setup for street or track, at this point I'm looking for street.

I might be interested in your tender springs depending on the rates and length, let me know what you have.

I'm looking to get a little bit of travel out of the tender spring to eliminate harshness over bumps on the street (under normal driving conditions).

I'm keeping the weight of the rear of the car in mind to stop all of the compression from being taken up from vehicle weight alone. My Prelude is around 660# LR and 600# RR.

 

thanks for showing your spreadsheet. it helps a lot in seeing why you cant solve the math problem.

i have no clue where you got that formula for "lbs until coil bind" for the tender spring (C15). it should just be like you did for the main spring. = (Rt * Ltcb) = 350 lbs.

so basically before 350 lbs, the combined spring rate should be 138.5 lbs/inch. then right after 350 lbs, or 2.52 inches of shock travel (combined compressed lengths, =350/138.5), the tender spring will be fully compressed. and the spring rate is now just the main spring rate, with 2.84 inches of travel left in the main spring (2.52 -1.75 = compressed length already taken up by the main spring = Lx, then Lmcb-Lx = 2.84 inches ).

but like slammed said, the real results are not going to be straight forward like that. springs arent perfect, so there will be progressiveness and blending of the numbers, and perhaps total inaccuracy.

 

Not sure of my thinking in C15 at this point, I remember having a reason... obviously a wrong one.

Thanks for looking it over for me... I understand that it isn't going to be perfect, but I want to get in the ballpark of not having the tender springs at bind with vehicle weight only.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mugenlude &raquo;</TD></TR><TR><TD CLASS="quote">I understand that it isn't going to be perfect, but I want to get in the ballpark of not having the tender springs at bind with vehicle weight only.</TD></TR></TABLE>OK, now with this in mind, lets use 600# as the vehicle weight on this corner...

600 / 138.5 = 4.33" of travel to support the vehicle weight (VW), meaning my example setup would be at bind w/ VW.

I'd estimate that I would want to have at least an inch of travel to absorb bumps in the road, maybe more. So I would need a spring with about 5.5-6" of travel, which is going to account for the entire spring assembly (standard spring that would come with the Prelude GC kit would be an 8" spring)... so that obviously isn't going to work. I need to up the tender spring rate and/or make it longer.

Going to play around with some numbers and available rates/length from the Eibach site to see what I can come up with.

 

when i was looking at the eibach tenders i couldn't find one that would go dead in the rear of my car.

the wheel weight is not what you need to using. It will be different, the leverage system at work, just like the spring rate doesn't equal wheel rate. If you know the motion ratio then you can figure out how much weight is actually on the spring.

I made this mistake when i first started looking at doing it.

I can't remember my exact rates but i bet the rears would be great for the heavier car, and the fronts you could use a little higher then what i ran.

For whatever reason when you put the tender with a main spring you get a progressive rate until the tender is bound, but the binding can't be figured by the calculations because of the progressiveness of the set-up.

I don't know when i will see my car again but ill try and remember to look at the tenders.

This was all on my race car.

 

Using the tenders is tricky. Unfortunetely their is not any sort of calculation you can use to determine exactly what you need. Most of it is just trial and error.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mugenlude &raquo;</TD></TR><TR><TD CLASS="quote">OK, now with this in mind, lets use 600# as the vehicle weight on this corner...

600 / 138.5 = 4.33" of travel to support the vehicle weight (VW), meaning my example setup would be at bind w/ VW.</TD></TR></TABLE>

you definately are omitting wheel rate in this now.

speaking of recreating the wheel, why not just get a regular spring kit? there are lots of options, many do have rear progressive rates, for many different ride heights..

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">the wheel weight is not what you need to using. It will be different, the leverage system at work, just like the spring rate doesn't equal wheel rate. If you know the motion ratio then you can figure out how much weight is actually on the spring.</TD></TR></TABLE>

Ah yes, I forgot the motion ratio in my example above, rear motion ratio is ~0.75. That changes things, for the worse I believe.

<u>Wheel Rate</u>
( 0.75 ^ 2 ) * 138.5 lb/in = 77.9 lb/in

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Using the tenders is tricky. Unfortunetely their is not any sort of calculation you can use to determine exactly what you need. Most of it is just trial and error. </TD></TR></TABLE>

I understand that there is going to be some trial and error involved, I just want to "get it in the ballpark" to start...

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">speaking of recreating the wheel, why not just get a regular spring kit? there are lots of options, many do have rear progressive rates, for many different ride heights..</TD></TR></TABLE>

I don't see any height adjustable applications for the Prelude. Not to mention I already have the Koni shocks... if you know of some progressive springs that will work with a coil-sleeve sytem like the GC's I'd be happy to check them out.


Modified by Mugenlude at 8:25 AM 11/26/2007

 

I came across this in another thread. Perhaps it'll be helpful:

The equation is spring rate = (main spring rate * tender spring rate) / (main spring rate + tender spring rate)

https://honda-tech.com/zerothread/1073767

EDIT: you really should go through the second page of that thread if haven't done so already.


Modified by vinuneuro at 9:05 PM 12/17/2007

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by vinuneuro &raquo;</TD></TR><TR><TD CLASS="quote">I came across this in another thread. Perhaps it'll be helpful:

The equation is spring rate = (main spring rate * tender spring rate) / (main spring rate + tender spring rate)

https://honda-tech.com/zerothread/1073767

EDIT: you really should go through the second page of that thread if haven't done so already.


Modified by vinuneuro at 9:05 PM 12/17/2007</TD></TR></TABLE>

That isn't right, at least not for tender springs.

there is no ONE rate for them. The rate of the springs is progressive, untill the tender is "dead"

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by slammed_93_hatch &raquo;</TD></TR><TR><TD CLASS="quote">

That isn't right, at least not for tender springs.

there is no ONE rate for them. The rate of the springs is progressive, untill the tender is "dead"</TD></TR></TABLE>

theoretically, this is incorrect if the tender spring is made linear.

empirically its probably true.