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I did it like this: 23lbs/inch=.9055+lb/mm= .41kg/mm



As long as you know the number of mm in an inch..and the number of kg in a pound..and the piº radian of the hypotenuse, you're all set.

And YOU supplied the figures!!

..mostly...

Cheers!

there is 25.4mm in an inch... and there is 2.20 kg's in a pound. So if my next set of springs have 28 active coils then what is the rate?? ANd what about 27 active coils??

Figure it like you did last time.

OK..truth of the matter...I didn't check your figures..just converted your answer.

I'm assuming you did it right, but obviously arriving at a correct answer doesn't inidicate any such thing!

BTW..having read the link that I started this thread with, you know that with a given spring removing a coil will increase its rate...and versy vice-uh.

Note: Increasing the rate does NOT increase it's load-carrying cability.

BTW...double check your figures in your post above.

ya thats basically what i dont know how to do.... i dont know how to convert my answer. I know you did it like this "I did it like this: 23lbs/inch=.9055+lb/mm= .41kg/mm "


but can you explain it so i can do it? thanks

For starters..did you get the 'check' part? There are 2.2lb in a kg, NOT 2.2kg in a pound.

Again: 23lbs/inch=.9055+lb/mm= .41kg/mm

OK. You're starting with a spring that 'weighs' 23lb when you compress it one inch.

How many pounds is that per mm? Divide 23 by 25.4. You get .9055+lb. for each mm.

How many (how much of one) kg is .9055lb? Set that question up as a ratio. I don't have a math font on this machine, but it 'looks' like this:

2.2lb is to 1kg as .9055lb is to Xkg or...2.2 over 1 : .9055 over Xkg

Cross multiply (as you do with ratios when you know three out of the four numbers) and you get 2.2X=.9055 where 'X' is the number of kg in .9055lb.

Solve the equation for X: .9055 divided by 2.2 = .41(and change).

Sort'a like this:

WHICH...the clever amongst us will notice is a 220x220pixel graphic which is to say, I don't wanna have ANYthing to do with it agin!

Cheers!

Heck - I'am still trying to find out where the constant 11,500,000 came from?

IdahoCharley wrote:Heck - I'am still trying to find out where the constant 11,500,000 came from?

Is there a smilie after that?

It's 11.5 x Wire Diameter to the fourth power (D4).

Exponents aren't associative, natch. 11.5 raised to the 4th x WD isn't the same as 11.5 x WD to the 4th.

...and it's not '10 to the 4th' ..the BASE is not 10. 2 to the 5th is 2x2x2x2x2.... not 200,000



Actually, he didn't write it out...it could be (11.5 x WD) raised to the 4th, but you would have to intentionally misunderstand what he wrote to get that.

Doesn't mean HE knows what HE wrote...........

Actually, he didn't write it out...it could be (11.5 x WD) raised to the 4th, but you would have to intentionally misunderstand what he wrote to get that.

Doesn't mean HE knows what HE wrote........... [/quote]

i knew what i wrote . !!!!!


Acutally i kinda didnt. A little bit thought. But thanks for the other info, i just bought a caliper so now i can figure it out more exact.

well i used my digital caliper checked and triple checked every measurement and found the rate for each spring.

they came out to be

1. .419kg/mm
2. .416kg/mm
3. .380kg/mm

Im kinda bummed cause i was suppose to have a .36kg/mm set of springs. they turned out to be .419kg/mm. Well i guess i should try the .38kg/mm springs and see how i like them. Thanks for your help.

alright maybe someone can help me with this question. Which spring should be the lighest. The one with the most active coils or the one with the least active coils?? thanks

More synapses filled.

I'll bite!

Everything else being the same (material, diameters, mfg. methods and such), the more active coils you have, the lower the rate.

1. That's in the article I posted at the top.

2. Spring rate is increased by removing coils..which is ALSO in the article I posted at the top.

Exaggerate the situation to make the point. Take a spring 4" long and compare it to a spring 4' long.

For a given input (pressure), each coil on the longer spring will move less!

Right?

ya your right, i got the forks on today, they look really sweet. Thanks for your help.

In case you thought I was kidding...

From: here!

from where I said wrote:As the number of active coils increases, the spring rate decreases.

It's a pretty good write-up. Answers your questions from earlier, too.

Google is your friend!


BTW...I've whined more than once (read: stated most evocatively clear!) that springs are not linear..even linear springs! I've been called bad names and have suffered insult most large upon my person for my ludicrous idea(s).

Correctly stated, rate creep is not the same as a progressive type spring. My point has always been the linearity part of it...whether you call it rate creep or any other name you choose.

no i didnt think you were kidding, you really seem to know what you are talking about. Thanks for helping me.

AS far as the bike goes, it doesnt feel much differnt than my kdx forks, but the kx forks have brand new seals and bushings which need to break in, and i am sure i am way off with the adjusters. Whats a good starting point for the adjusters.

Also i have only been riding on my flat yard since i am just breaking in the bike. Come sunday i will be ( if no rain) ripping around on the trails.

Hopfully in agust enter some harescrambles.

On flat ground you won't be able to tell much of a difference. After your out on the trail a bit and get warmed up aim your bike straight for a section of trail that you previously had issues with. If you can't tell the difference between the stockers and the kx forks then something isn't right.

FWIW...posting links to what someone else said has nothing to do with knowing anything about it, and researching an answer doesn't require knowledge of the subject, either.

But, you're welcome anyway.

Yeah...flat ground doesn't tell you diddle.

What about Sunday?