adnoh wrote:
Your welcome, here is some read and pic
Fl 350 spring change:
Here is some information on changing the spring in the rear stock shock. Based from what Terry H posted the stock spring has a free length of 11-5/8 inch. He has used a new 10” spring with a solid spacer of 1-¾”. This brings the new install to 11-3/4 inch total free length. The stock spring according to Terry H has a 210 lbs/in rate. He also has provided measurements for an increase in pre load poundage in fraction of inch.
In order to make the math easy we will change the fraction to a decimal number which easier to work with. To do this is pretty straight forward. You divide the first number by the second:
3/ (divided by) 4 = .75. Now we can take the rate per inch of 210 X .75 (of an inch) = 157.50 lbs, to get poundage. The following information is based from this math. Now everyone can do the math easy.
Please keep in mind the numbers provided by Terry H to get the correct number for the installed spring. With the set up at 11-3/4” and the installed length of 11-1/4” there is a pre-load of ½”or .50 on the shock. Take whatever spring rate times .50 (185 X .50 = 92.5) and you get the poundage of static preload on the shock before installed. The weight/force applied to this shock now must overcome this poundage before the shock/rate becomes usable. At a 1 to 1 rate you have to take the rate plus static pre-load to start with. At zero preload the shock will move 1” for every 185 lbs applied+/- 1% (guaranteed rate by MFG). With pre-load of 92.5 you now have to have 185+ 92.5 = 277.5 lbs to move it the first inch. This not a big deal as long as you understand the impact.
An example of this is the balance or weight applied to the shock to get sag or ride height. As mentioned before this shock is not at a 1 to 1 so there is a lever ratio involved. This takes us to a wheel rate versus spring rate which we can deal with later. Let’s look at the book rated balance of 396 lbs. if we divide that by half 396/2=198 lbs. If we use a 1 to 1 (for now) we find that with a static preload of 92.5- 198=105.5. So how far will the shaft move at that weight or sag (ride height), we can find this pretty easy. We take 105.5/185=.57 of an inch. If the shaft is six inches long (Book says 5.9 round up to 6) the sag in a percentage would be 6 -.57=5.43/ (divided by) 6=.905, 1-.905=.095 or 9.5% not even 10%.
Most say, a good sag or ride height number is 30% so we need at least 20% more which is controlled by the balance and spring rate. In this example a red flag flashes don’t do it. The 10” spring on a 6” shaft the spring most likely will go solid or bind. I know there is a lever ratio and the shaft is not 6” so this is why the information is important. Back to the example! We need too determine the amount of shaft and spring rate including the static pre load to get 30% sag. 30% of 6” shaft would be 6 x .30 =1.8”Remember the 105.5lbs rear balance for that shock, 105.5 / 1.8= 58.61 rate. Long way from 185, now what, first let’s look at the rest of the shaft or 6-1.8=4.2 remaining shaft. If at the 185 pound rate this would provide a bottom out poundage of 185 x 4.2=777 lbs or 1554 for the rear if they bottom out tighter not likely, this is why we look at each wheel. If you’re following correctly you should be asking how we got to 30% if we used the 185 with the spacer!
The correct number using the above 185 with spacer and a static pre load of 92.5 would provide us with a left over shaft of 6-.57=5.43” x 185=1004.55 bottom out poundage ( BOP). This is where we can use the old spring and do the math to get an idea of what to do. Once we get that we can match up a dual rate spring set up to get both 30% and correct bottom out poundage with more control.
The old spring has a 210 rate with static preload of 78.75. Using the math above time for some fun, I will leave it up to the board to get the right answer for BOP using old spring and the above example. Then we will look a dual rate set up to do the job.
We can discuss the three options for a dual rate set up for dirt, sand and woods. Each has a different set up for god tire traction.
FYI: The Terry H spring has a Travel height of 5.64” and a block height (coil bind) of 4.36” as a safe rule use no more 70% for the spring height. See the red flag ½ static plus 6” shaft is 6.5”. The travel for that spring is 5.64. ooops. This means it will go solid at a shaft length of 5.14”. Travel Data from Eibach(see pic). Remember we know it’s not a 1 to 1 however the shaft travel better not be longer than 5.14. Oh ya 70% of the 10” spring travel height of 5.64 x .70=3.948-.5 static pre-load= shaft travel of 3.448”
As always someone check the math
Andoh
adnoh I don't want to sound like an Ahole but now you have to do this all again

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Honda lied when they said the oddy has 5.9" of rear travel. This wouldn't be the first time a sales team fudged the numbers a bit would it ??
The actual rear wheel travel as shown above is actually 5.6" and that's WITHOUT a snubber. The sales team may have "over looked" that part as well.
Still the above info is good stuff --- I like it.
Looking foreword to the spread sheet you post with the adjusted numbers and the motion ratio.
Edit: Can you also do a sheet with the corrected motion ratio and a 15% sag ??
The reason I ask is because a 30% sag number takes away to much of the travel for an oddy.
The suspension is suspect at best and now we rob 30%
Now what do we do about the snubber ?? I actually cut some of my snubbers to get some extra travel.
Thx in advance for any info.
Edit again: Hey Terry if you want me to move this crap just PM me and I will contact hoser. I just want this stuff in one place. Still looking foreword to your final spring rate and ride report -- CO.