spring rates

[GuzziMoto]

I didn't think such a simple question would cause such a drama I only asked because I'm going to try a new spring on the sachs shock as I was browsing this site http://teknikracing.rtrk.com.au/?scid=28689&kw=4315973. But it appears I'll have to measure my spring myself to get an accurate answer or strip it and take it to the experts( the real ones not the virtual ones)

Measuring the stock shock is the best way to get an accurate spring rate from it. You can have pro's do it with real measuring equipment or you can do it yourself. All you need to do is measure the force it takes to compress the spring a measured distance. The standard for rear springs (in the US) is how many lbs to compress the spring 1", but that is a high number and going for 1/2" or even 1/4" would likely do just fine although it would double or quadruple your error factor.

[raz]

Sorry if I'm being ultra daft now, but won't the force multiply with the leverage ratio too? So 200 lbs weight on the seat would be 400 (or 100 ) lbs on the shock spring? Or am I confusing force with weight where I'm not allowed to?

[GuzziMoto]

You guys confuse me. If we're talking travel (and assuming 2:1), the shock will compress 2 cm from a 1 cm wheel movement. And 20 mm more preload gives 10 mm higher rear. Right?

 

No the shock travel to wheel travel is not a 2:1 ratio but a 1:2 ratio.

Any change of preload or length at the shock will result in a change at the wheel of twice the change at the shock, ie, if the shock is lengthened 10mm or 10mm preload is added to the shock then the rear wheel will move down (actually the rest of the bike will move up) 20mm.

 

To compress a 400# spring 1 inch, we apply 400 lbs at the spring. That is the same as 200 lbs at the wheel (or rather the seat). But as the swing arm only move 0.5 inch when we do this, the "effective spring rate" is still 400#. Is that right?

 

Yes, a 400lb spring takes 400 lbs to compress it 1". But if you apply the force at the rear wheel 200lbs would only compress the spring 1/2", but as the rear wheel moves twice as far as the shock it would appear that the rate of the rear spring is only half what it really is. That is why a change in rate at the spring will only have half the effect at the wheel as it did at the shock.

That is part of the reason rear springs are sold with much higher rates then fork springs and the increments of change between the different rates is much larger. It takes a larger rate change at the rear to accomplish a given change in how the suspension works then it does at the front, where springs are typically sold in steps of 12.7 lbs/inch. Of course since you have two springs holding up the front of the bike a 12.7lb change equates to 25.4lbs at the wheel (not taking the rake of the forks into it).

 

Edit: Whoops, sorry. I had a decimal one place off. Fork springs are actually sold in steps of 2.799lbs/inch.

 

I am not sure what you mean by "effective spring rate" but if you mean the rate that it will feel like then no, a 400lb spring with a 1:2 leverage ratio between the shock and the wheel will feel like a 200lb spring.

[GuzziMoto]

This is interesting and maybe I'm over thinking this.

I understand the ratio and if you lengthen the shock by 1cm you would realize 2cm at the wheel but wouldn't adding 1cm preload only change the wheel by 1cm assuming it is carrying the same load and working against a 2:1 ratio?

No, it is a 1:2 shock to wheel ratio, not a 1:2 ratio. It works like a lever. A given force will move the rear suspension twice as far as the shock moves but it is only doing half the work. In this case the work is compressing the rear shock. And a given change in shock length ( that is all preload is really going, is setting the length of the shock under the weight of the bike)will have double the change at the rear wheel as it moves 2" for every 1" of movement at the shock.

[raz]

To compress a 400# spring 1 inch, we apply 400 lbs at the spring. That is the same as 200 lbs at the wheel (or rather the seat)

Correct^.

Yes, a 400lb spring takes 400 lbs to compress it 1". But if you apply the force at the rear wheel 200lbs would only compress the spring 1/2", but as the rear wheel moves twice as far as the shock it would appear that the rate of the rear spring is only half what it really is. That is why a change in rate at the spring will only have half the effect at the wheel as it did at the shock.

OK, here's what I'm trying to say. Surely I'm wrong but please tell me what is wrong with the following:

 

1. To compress a 400# spring 1 inch, we apply 400 lbs at the spring. With a 2:1 ratio we can instead apply 200 lbs at the seat. So when applying 200 lbs at the seat, the spring compresses 1 inch.

 

2. One inch on the spring is two on the wheel. So when applying 200 lbs at the seat, the wheel (or seat ) moves 2 inches.

 

3. This means the "effective rate" (is there a term for this?) is just 200/2 = it feels like a 100# spring.

 

Unless both GM and Ratch are wrong, there must be a flaw in my reasoning. I've tried googling but it only adds to the mess. I found a post claiming "the spring rate at the axle is the actual spring-rate divided by the SQUARE of the leverage ratio." but that doesn't fit any better. It does sound better in my ears though

 

Edit: um, wait. I read it as SQUARE ROOT but if they mean just square, it actually supports my reasoning. I am so confused.

[GuzziMoto]

Correct^.

 

OK, here's what I'm trying to say. Surely I'm wrong but please tell me what is wrong with the following:

 

1. To compress a 400# spring 1 inch, we apply 400 lbs at the spring. With a 2:1 ratio we can instead apply 200 lbs at the seat. So when applying 200 lbs at the seat, the spring compresses 1 inch.

 

No, the 200lbs would move the rear wheel 1" but the shock would only compress 1/2" with a 1:2 ratio.

 

2. One inch on the spring is two on the wheel. So when applying 200 lbs at the seat, the wheel (or seat ) moves 2 inches.

 

No, with a 400lb spring 200lb would move the seat 1" but the shock is only moving 1/2", exactly how far it would compress with 200lb applied to it.

 

3. This means the "effective rate" (is there a term for this?) is just 200/2 = it feels like a 100# spring.

 

If by "effective rate" you mean how it actually feels then with a 400lb spring (which compresses 1" when 400lbs is applied to it) then since because the rear suspension would only need 200lbs of weight to compress it 1" (although the shock would only compress 1/2" under this 200lb load) the suspension would feel like it had a 200lb spring even thogh the actual spring rate is 400lb.

Unless both GM and Ratch are wrong, there must be a flaw in my reasoning. I've tried googling but it only adds to the mess. I found a post claiming "the spring rate at the axle is the actual spring-rate divided by the SQUARE of the leverage ratio." but that doesn't fit any better. It does sound better in my ears though

 

Edit: um, wait. I read it as SQUARE ROOT but if they mean just square, it actually supports my reasoning. I am so confused.

 

Let me try some more.

First, it is a 1:2 ratio of shock travel to wheel travel. That is different then a 2:1 ratio. In fact it is the opposite of a 2:1 ratio. Think of a 1:2 ratio as 1" of shock travel equals 2" of wheel travel. So if 400lbs of force (weight) will compress the spring 1" then the wheel will move 2". So it takes 400lbs to compress the shock 1" but only 200lbs will move the rear wheel 1" (and 400lb would move the rear 2") So a 400lb rear spring with a 1:2 ratio would feel like a 200lb spring since 200lb of weight would move the rear suspension 1". If you had a straight 1:1 ratio then with a 400lb spring it would take 400lb of weight to move the rear wheel 1" and 200lb would only move the wheel 1/2". That would feel like you had a 400lb spring at the back. But with a 1:2 leverage ratio that same 400lb spring would only feel like a 200lb spring since 200lb would move the rear wheel that same 1" that the 400lb spring with a straight 1:1 ratio would require 400lb of weight to move.

Hope this helps.

[Guest ratchethack]

Hatchet Wacker, you seem to mostly understand the ratio of shock to wheel movement, do you still think that a rate change at the shock will have a larger effect at the wheel due to the ratio?

And do you still think a 2% increase in spring rate would deliver a 31% increase in ride height due to the 1:2 ratio?

You rode the short bus to school, didn’t you, Quazimodo? Uh-huh. . .

 

Not that there’s anything wrong with that. . .

 

But you mustn’t put expectations on yourself that you‘d ever be capable of functioning on the same level with the “long bus” kids when you became an adult. They should’ve taught you that in “special” school – if not then, certainly either in “remedial” or “reform” school. If not then, well, your "Ju-vee" classes should've covered it sufficiently. If not there, I reckon you should have picked it up by osmosis after a few years of "education" from the yardbirds at The Big House. . . after that, your parole officer should've made absolute certain that you understood the concept in full. . . But I digress. . .

 

The point is that somehow, you've apparently "dropped thru the cracks of the system" -- and that's truly a sad thing. But then, as you insist on demonstrating repeatedly here lately, your comprehension level has likely always been well below nitwit level from the get-go, so however unfortunately -- there’s the rub. And waddayagonna do?

 

Have you noticed yet that of all the points that you’ve attempted to call me out on in this thread, I never actually made any of them? Have you noticed that I pointed this out to you very specifically, point by point, once already? Surely you remember. I challenged you to find any evidence of each, quoting you word for word on each false claim, and you drew a complete, 100% blank on providing any evidence wotsoever on each and every one. But, No. . . no, by your posts since then, I reckon you never as much as remembered, if not simply failed to register any of this a-tall. . .

 

Now you’ve posted 2 new questions above, and you've addressed them to me -- sloppy, ignorant, and insincere as they both are. Your first question makes no sense in normal adult level parlance. It’s the equivalent of asking, ‘How long is a string?’ It’s the kind of question that might be expected of a 7 year-old. It is, of course, not a legitimate question, and, of course, it’s unanswerable.

 

Your second question makes marginally more sense, and might have been asked by a 12 year-old. However, it illustrates most spectacularly (again) your chronic, atrocious lack of reading comprehension, which I pointed out previously, not all that long ago in this thread. It’s also a dead give-away to both your appallingly shallow knowledge of the topic at hand, and your staggering level of confusion over many of the basic suspension concepts we’ve been talking about here.

 

But I’ll respond to it on the off-chance you might learn something important to you that's far, far more imortant than moto suspensions.

 

Maybe if I keep this *R - E - A - L - L - Y* *S - I - M - P - L - E*, you might be able to pick up something you can benefit from this time around.

 

I’ll put this in the same question format as last time, so you won’t be further confused:

 

Ready? Here it is:

 

Where, exactly, (without taking my words out of context) did I ever post anything (in this thread or any other, ever) that even hints at anything even remotely close to what you claimed I posted here:

. . .do you still think a 2% increase in spring rate would deliver a 31% increase in ride height due to the 1:2 ratio. . .

Please do take your time. No rush wotsoever. Calm your mind. Breathe deeply. . . . .

 

Search. Read. Concentrate. Search again. Read again. Think again.

 

Think hard. Think harder. Take a break to rest your mind.

 

Think some more. . . And then by all means, do get back to me.

[Guest ratchethack]

SPECIAL NOTE:

 

I really, truly, and seriously don't WANT to b'lieve there's anyone here (besides Quazimodo) who suffers from confusion over this one. Now Quazi clearly has a severe, chronic reading comprehension problem, compounded by some kind of cognitive development challenges on the basics here, he's apparently now in the manic phase of a bipolar cycle, and he can't (and/or won't) ever likely recover from any of that. So the evidence so far in this thread is that he's incapable of grasping this.

 

Now by all means, sincere Q's of all kinds most welcome here as always from anyone and everyone. At least some of us are always open to learning. But the rank kiddie stuff being broadcasted here lately by the sole. . . um, problematic and most insincere , and quite ignorant poster here lately has just gotta stop.

 

Now just in case anyone else hasn't got this one down,

 

On all V11's:

 

Shock travel to wheel travel is a 1:2 ratio.

 

Wheel travel to shock travel is a 2:1 ratio.

 

Anyone still confused about this^, you might as well ask Quazimodo. He'll straighten you right out.

[helicopterjim R.I.P.]

I am surprised no one has mentioned lambda factor with regards to spring rate ....!!

[raz]

Let me try some more.

First, it is a 1:2 ratio of shock travel to wheel travel. That is different then a 2:1 ratio. In fact it is the opposite of a 2:1 ratio. Think of a 1:2 ratio as 1" of shock travel equals 2" of wheel travel. So if 400lbs of force (weight) will compress the spring 1" then the wheel will move 2". So it takes 400lbs to compress the shock 1" but only 200lbs will move the rear wheel 1" (and 400lb would move the rear 2") So a 400lb rear spring with a 1:2 ratio would feel like a 200lb spring since 200lb of weight would move the rear suspension 1". If you had a straight 1:1 ratio then with a 400lb spring it would take 400lb of weight to move the rear wheel 1" and 200lb would only move the wheel 1/2". That would feel like you had a 400lb spring at the back. But with a 1:2 leverage ratio that same 400lb spring would only feel like a 200lb spring since 200lb would move the rear wheel that same 1" that the 400lb spring with a straight 1:1 ratio would require 400lb of weight to move.

Hope this helps.

Thanks. So in order to put 400 lbs on the spring, I just put the same 400 lbs on the seat, regardless of ratio. I thought that would be multiplied too. How come I can lift things heavier than myself using a leverage (and longer travel) then? Not the same thing?

[GuzziMoto]

You rode the short bus to school, didn’t you, Quazimodo? Uh-huh. . .

 

Blah, blah, blah....

Have you noticed yet that of all the points that you’ve attempted to call me out on in this thread, I never actually made any of them? Have you noticed that I pointed this out to you very specifically, point by point, once already? Surely you remember. I challenged you to find any evidence of each, quoting you word for word on each false claim, and you drew a complete, 100% blank on providing any evidence wotsoever on each and every one. But, No. . . no, by your posts since then, I reckon you never as much as remembered, if not simply failed to register any of this a-tall. . .

 

Now you’ve posted 2 new questions above, and you've addressed them to me -- sloppy, ignorant, and insincere as they both are. <_ your first question makes no sense in normal adult level parlance. it the equivalent of asking long is a string kind that might be expected year-old. course not legitimate and unanswerable.>

 

Your second question makes marginally more sense, and might have been asked by a 12 year-old. However, it illustrates most spectacularly (again) your chronic, atrocious lack of reading comprehension, which I pointed out previously, not all that long ago in this thread. It’s also a dead give-away to both your appallingly shallow knowledge of the topic at hand, and your staggering level of confusion over many of the basic suspension concepts we’ve been talking about here.

 

But I’ll respond to it on the off-chance you might learn something important to you that's far, far more imortant than moto suspensions.

 

Maybe if I keep this *R - E - A - L - L - Y* *S - I - M - P - L - E*, you might be able to pick up something you can benefit from this time around.

 

I’ll put this in the same question format as last time, so you won’t be further confused:

 

Ready? Here it is:

 

Where, exactly, (without taking my words out of context) did I ever post anything (in this thread or any other, ever) that even hints at anything even remotely close to what you claimed I posted here:

 

Please do take your time. No rush wotsoever. Calm your mind. Breathe deeply. . . . .

 

Search. Read. Concentrate. Search again. Read again. Think again.

 

Think hard. Think harder. Take a break to rest your mind.

 

Think some more. . . And then by all means, do get back to me.

Sorry, I was asking you a serious question but I realized afterwards that it was a waste of time and tried to delete it before it was too late. Obviously I did not succeed. Sorry.

 

But the short answer to your question(?) is back in post #28 where you claimed to have gotten a 31% reduction in your "sag delta", "The sag delta dropped by a great whallopping 31%" while saying that you only increased your spring rate 11lbs (about 2% of the original spring rate, 532lb).

You followed that with with a bunch of fancy words that had little to do with the topic (mostly designed to insult the people who did not agree with you) and then a "Helpful reminder" which presumably was intended to explain the large change in your "sag delta" from a small change in spring rate. It went as follows... (from post #28)

"HELPFUL REMINDER: When changing rate on the shock spring, there's a ~1:2 multiplier at work WRT sag delta change vs. spring rate change, due to the 1:2 swingarm leverage (shock travel to wheel spindle travel)."

This statement, "there's a ~1:2 multiplier at work WRT sag delta change vs. spring rate change, due to the 1:2 swingarm leverage"is what I was asking about. You pointed out that your"sag delta" was 100% affected by/representative of spring rate. It is actually affected by other things as well but spring rate is the biggest influence on the difference between your free sag and your race sag (you call that number your "sag delta").

I am sorry I brought it up. I apologize to the rest of the board.

[GuzziMoto]

Thanks. So in order to put 400 lbs on the spring, I just put the same 400 lbs on the seat, regardless of ratio. I thought that would be multiplied too. How come I can lift things heavier than myself using a leverage (and longer travel) then? Not the same thing?

It is a leverage ratio, so it works the same. The way levers work in this case is they move something with half the force that should be required but they only move it half as far. So that 200lb applied at the seat will compress a shock with a 400lb spring 1/2", even though the seat drops down 1".

[GuzziMoto]

Dave,

Blah, blah, blah.... Edited for content.

 

PRE-EMPTIVE CAVEAT/REMINDER: Many on this board still don't seem to grasp the fact that a laden sag measurement ALONE is 100% useless information when it comes to upgrading spring rates, as I've been whingeing on about like a broken record, lo these many years. Once again -- you ALSO need a correct UNLADEN SAG measurement to be of any use wotsoever when re-springing and/or setting up suspension.

 

I'm presently running 38 mm laden sag and 20 mm unladen sag, with a Wilbers 641 custom order shock and 95 N/mm (543 lbs/in) spring. (sag delta = 18 mm).

 

With the previous OE Sachs shock and OE 532 lbs/in (93 N/mm) spring, I was last running a 40 mm laden sag and a 14 mm unladen sag. (sag delta = 26 mm).

 

Q: BUT RATCHET, RATCHET! You only went up a measly 2 N/mm (11 lbs/in) on a shock re-spring rate?!?!?

 

A: Yes, Glasshoppel. . . Please pay attention now, and observe carefully how this allowed me to hit my sag delta target with the spring rate upgrade -- not quite, but very nearly *dead-nuts*:

 

The sag delta dropped by a great whallopping 31%.

 

Yeah, I know. Here’s^ where eyes start to glaze over, and all the insincere and less than adequately motivated (and less than adequately equipped) peel off. . . For all such as these, the above and all that follows below will be gibberish and nonsense. By all means -- rather than even attempt an understanding, far better to continue to let hearsay, any GROUPTHINK that tickles your fancy, the latest popular delusions, anybody who's delirious about wotever they did (or wotever somebody else said somebody else was delirious about wot they did), and old wives’ tales be your guide. The Hooter’s thread beckons seductively, and, ‘Bye Now!

 

For those still interested and capable, but possibly not accustomed to the term, ‘sag delta’ (laden sag minus unladen sag), it’s the ONLY single measurement number that directly relates spring rate to load. When tuning suspension, it’s a very handy number to use for comparison purposes when setting up and balancing the chassis, since it cuts all the usual unnecessarily confusing, irrelevant, and useless foolishness and folderol of preload considerations OUT of the picture.

 

Now if you absolutely can’t quite wrap your mind all the way around this simple concept^, this might be the best time to call in a suspension Pro, place all your faith and trust in whoever you can find that you imagine might fit the bill, and/or whoever has the flashiest looking Web site, pay the man as handsomely as he can convince you it’s worth for wotever he sells you, and vaya con Dios, mi compadre.

 

HELPFUL REMINDER: When changing rate on the shock spring, there's a ~1:2 multiplier at work WRT sag delta change vs. spring rate change, due to the 1:2 swingarm leverage (shock travel to wheel spindle travel). With fork springs, it's a direct 1:1 ratio (but also keep in mind there's 2 of 'em ).

 

HELPFUL REMINDER (Part II): Laden sag alone can be set anywhere you like at any time, REGARDLESS OF SPRING RATE. That’s right – you could go so far as to replace your existing shock spring with a solid pipe, and adjust preload to set your laden sag exactly where it was before, or wherever it strikes your fancy. But since the unladen sag would then be the same as laden sag (sag delta = ZERO), your ride experience after a few hours on back roads might well warrant the professional services of Docc, and/or a team of ortho surgeons. . .

 

Blah, blah, blah... Edited for space/time constraints

 

Hope this helps.

Here is post #28 that I reference so you do not have to go looking for it.

[Dan M]

No, it is a 1:2 shock to wheel ratio, not a 1:2 ratio.....

 

 

What???

 

As far as the rest of it. No shit, really?

[GuzziMoto]

What???

 

As far as the rest of it. No shit, really?

Sorry, I typo'd that. I meant to say it is a 1:2 ratio of shock travel to wheel travel, not a 2:1 ratio as you said. And yes, really...

And if it is that simple, why do you not understand it???

WTF???