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Old Sep 22, 2006 | 12:12 PM
  #16  
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Originally Posted by Noob4life
oh, flag on the play...
teeny tiny nitpick here, but I think helper springs are typically coils with negligible rates, enough to keep the main coil seated but not enough to contribute to the overall spring rate, cut flat so that when it completely compresses its basically a flat disk.

I think we're talking about tender spring shere. I'm also looking forward to more opinions on the effectiveness of progressive rate springs and some solid evidence of their effect on "predictability" in handling.
Agreed, helper or tenders are not there to absorb bumps they (as even Mark agreed) are there to keep the spring seated during droop conditions. They are not "progressive systems"

Last edited by chronohunter; Sep 22, 2006 at 12:27 PM.
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Old Sep 22, 2006 | 12:35 PM
  #17  
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Originally Posted by DMS_Mark
Why do you keep inferring that if it is not your setup or idea, it is no good?
Because I have the backing of the entire professional road racing community with my set-ups and ideas (because I am a product of that environment).

You and your ideas/set-ups/commentary on the other hand seem to operate and/or originate on an alternate Universe (I'm not referring to Canada )
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Old Sep 22, 2006 | 03:00 PM
  #18  
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Originally Posted by chronohunter
Because I have the backing of the entire professional road racing community with my set-ups and ideas (because I am a product of that environment).

You and your ideas/set-ups/commentary on the other hand seem to operate and/or originate on an alternate Universe (I'm not referring to Canada )
That is pretty ignorant of you, don't you think?
You know nothing about me, my experiences, or our company apparently.

Your right chrono hunter, I woke up one morning a few months ago and said to myself "I am a suspension guru and everyone should bow before me" after pondering that statement, making my morning coffee, I then proceeded to add "I think that everything I say will be the equivalent of blowing smoke out of my butt, hence nothing what I say will ever have any means, I just enjoy stirring the pot"

I am really looking forward to meeting you one day, you really are something to accuse someone of knowing nothing. It also goes to show you have not been keeping up to speed on the entire racing community you so speak of, seeing as at autosport 2006 this past spring for example, over 95% of all open wheel, GT, prototype, rally and touring cars (basically any form of motorsport) where running some sort of linear / tender (tender is not a helper spring for the record) spring combination. This is at the highest forms of motorsport in every discipline.

The advantage of running a progressive spring is you get rid of that "step" that was mentioned before. You also have to say "are you running a shock that has 7" of stroke, but only using 1", or are you running a shock that is using the most of it's stroke whether it is 14" or 1"

If anyone has any real questions, feel free to e-mail me, as I am tired of listening to Chronohunter pretending to be god as opposed to trying to look at all sides of a different approach.

Cheers everyone,
mark

www.dmsnorthamerica.com .... yes it is a real company with real people, and real experiences, no matter what universe you are from
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Old Sep 22, 2006 | 06:20 PM
  #19  
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Originally Posted by chronohunter
Agreed, helper or tenders are not there to absorb bumps they (as even Mark agreed) are there to keep the spring seated during droop conditions. They are not "progressive systems"
Are we sure about this? I mean, let's say your main spring is 10kg/mm and your helper is like 3.5kg/mm. The combined rate of the two placed end to end would be:

= (10x3.5)/(10+3.5) = 2.59kg/mm <- so the little helper has a big effect on overall stiffness

So, before the helper is solid, this is the rate your springs would be effectively.

I am assuming by 'droop' you mean when the dampers extend fully (or nearly). If the helpers are only useful here, that means they must be preloaded so that they are solid when the car is at normal height so the car sees 10kg/mm right off the bat.

[In fact, the more compliant the helpers the more it reduces the overall rate.]

Anyhow, if you had dampers with say 5" total stroke, would it not be sensible to install springs with 5" + a little bit compression range? If so, the springs would never be un-seated (?) and you won't need anything additional. [again, i might be wrong about what I think you mean by 'droop'.]
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Old Sep 22, 2006 | 07:00 PM
  #20  
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Hang on a second....

M,kay just been thinking about this in the shower. It's actually on topic for a change!

Let's say you have really stiff springs. Say 12kg/mm at the front (670.56lbs/in). Assuming your damper has 8" of stroke (200mm, I don't know actual evo figure) and you want to have +/-4" of travel for some reason. If you want the spring to just touch the seats when the dampers are fully extended, then the static load (when you're parked) needs to be 4 x 670.56 x [suspension linkage rate] = 2,682.24 x [suspension linkage rate] lbs.

Because of the strut arrangement in the evo, I'd say the suspension rate is almost 1 or thereabouts. Now it's clear we can't get that sort of weight on one wheel cos the evo doesn't weigh that much. However, it would still be a good idea to have the springs seated at all times. This, I'd suppose, is why the rates are progressive. So that the static weight of the car gives enough preload to allow for sufficient outward travel of the wheel. Obviously it is possible to add pre-load the spring by just winding up the spring seat, but I guess too much of that compromises the availably compression travel of the spring.

Going back slightly, can anyone confirm that the helpers are fully compressed under static load?

Again, I'm no expert....
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Old Sep 22, 2006 | 07:51 PM
  #21  
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Originally Posted by x838nwy
Are we sure about this? I mean, let's say your main spring is 10kg/mm and your helper is like 3.5kg/mm. The combined rate of the two placed end to end would be:

= (10x3.5)/(10+3.5) = 2.59kg/mm <- so the little helper has a big effect on overall stiffness

So, before the helper is solid, this is the rate your springs would be effectively.

I am assuming by 'droop' you mean when the dampers extend fully (or nearly). If the helpers are only useful here, that means they must be preloaded so that they are solid when the car is at normal height so the car sees 10kg/mm right off the bat.

[In fact, the more compliant the helpers the more it reduces the overall rate.]

Anyhow, if you had dampers with say 5" total stroke, would it not be sensible to install springs with 5" + a little bit compression range? If so, the springs would never be un-seated (?) and you won't need anything additional. [again, i might be wrong about what I think you mean by 'droop'.]
Well the point is a helper spring has a rate so low that it can be effectively left out of the overall spring rate equation at that corner. They might be on the order of 5lbs or so, just enough to keep the spring fully seated even when the suspension is at full droop.

I know you posed a hypothetical situation, but consider that a 5" spring is too short, and would require too thick a coil diameter/wire thickness in order to maintain a useable spring rate.

Indeed longer springs are used when longer stroke is called for, so in a sense I think manufacturers DO try and do what you are saying, to an extent, using free length to keep the spring seated. But I think other constraints in the design of the coilovers (notably height adjustment range, suspectibility of the spring to bowing, actual shock travel, available space) limit spring sizes that can be used. And so helper springs become necessary.

By definition, a helper spring will be fully compressed under static load, and only come into effect under droop.
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Old Sep 22, 2006 | 09:10 PM
  #22  
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Originally Posted by DMS_Mark
That is pretty ignorant of you, don't you think?
You know nothing about me, my experiences, or our company apparently.

Your right chrono hunter, I woke up one morning a few months ago and said to myself "I am a suspension guru and everyone should bow before me" after pondering that statement, making my morning coffee, I then proceeded to add "I think that everything I say will be the equivalent of blowing smoke out of my butt, hence nothing what I say will ever have any means, I just enjoy stirring the pot"

I am really looking forward to meeting you one day, you really are something to accuse someone of knowing nothing. It also goes to show you have not been keeping up to speed on the entire racing community you so speak of, seeing as at autosport 2006 this past spring for example, over 95% of all open wheel, GT, prototype, rally and touring cars (basically any form of motorsport) where running some sort of linear / tender (tender is not a helper spring for the record) spring combination. This is at the highest forms of motorsport in every discipline.

The advantage of running a progressive spring is you get rid of that "step" that was mentioned before. You also have to say "are you running a shock that has 7" of stroke, but only using 1", or are you running a shock that is using the most of it's stroke whether it is 14" or 1"

If anyone has any real questions, feel free to e-mail me, as I am tired of listening to Chronohunter pretending to be god as opposed to trying to look at all sides of a different approach.

Cheers everyone,
mark

www.dmsnorthamerica.com .... yes it is a real company with real people, and real experiences, no matter what universe you are from
I do not and have not accused you of knowing nothing. I constantly accuse you of spreading what I call propaganda on these boards...and I stick with it.

You (remember this?) told every one that the DMS 40's we perfect for beginners because the had no low speed damping (that was a gem), I called you out and said that's a nice story but the real reason they had no low speed damping was because they were actually rally dampers valved for off tarmac use. Please explain how beginners need excessive body roll

You claim there are several ways to do suspension on a car yet I constantly ask you to show me some professional teams driving our types of cars on tarmac that follow your philosophy (and you come up with tenders that are fully compressed at static ride height) and you call that proof?

Mark I really think you believe what you are saying and I think you have experience in certain areas of motorsport and you are trying to expand your company into this arena (which is fine and cool). It is not cool to try to fly against what is proven time and time again out on the race track every day and say it isn't so praying on people who don't have much knowledge in this area. That's why I really feel I have a responsibility to everyone here to weigh in whenever something unconventional is stated to be true (and I will continue to do so).

BTW I would love to meet you talk and go drive some cars I think it would be fun considering all of this
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Old Sep 22, 2006 | 09:16 PM
  #23  
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Originally Posted by Noob4life
Well the point is a helper spring has a rate so low that it can be effectively left out of the overall spring rate equation at that corner. They might be on the order of 5lbs or so, just enough to keep the spring fully seated even when the suspension is at full droop.

I know you posed a hypothetical situation, but consider that a 5" spring is too short, and would require too thick a coil diameter/wire thickness in order to maintain a useable spring rate.

Indeed longer springs are used when longer stroke is called for, so in a sense I think manufacturers DO try and do what you are saying, to an extent, using free length to keep the spring seated. But I think other constraints in the design of the coilovers (notably height adjustment range, suspectibility of the spring to bowing, actual shock travel, available space) limit spring sizes that can be used. And so helper springs become necessary.

By definition, a helper spring will be fully compressed under static load, and only come into effect under droop.
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Old Sep 23, 2006 | 12:17 AM
  #24  
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Originally Posted by Noob4life
Well the point is a helper spring has a rate so low that it can be effectively left out of the overall spring rate equation at that corner. They might be on the order of 5lbs or so, just enough to keep the spring fully seated even when the suspension is at full droop.
The point here is that the lower the rate of the helper, the more it effects the overall rate, not less. Therefore it becomes more important to the overall rate.

If you have two springs, one rate 'k' and another 'e x k' where e < 1, then the combined rate of them in series is:

k x [e/(1+e)]

So, if your main spring is 700lbs/in and your helper is 7lbs/in then the overall rate becomes 700 x (0.01/1.01) = 6.93lbs/in which is far away from the 700lbs of the main spring. So its effects cannot be ignored - that is if it hasn't fully compressed under the static load. You can see that the smaller 'e' becomes the more (e/(1+e)) approaches e and therefore the overall rate becomes increasingly close to the helper and the main spring has less and less effect.

I think what might happen is that if the helper is really a very low rate, then it is highly likely that it is compressed fully (and some more) under static load and so does not have an effect on the actual wheel rate other than at the very end of the dampers extension where the main spring runs out of travel.

If it is the case that the helper spring is always fully compressed at static load, there won't be that transiton I was talking about and it would in fact behave like a constant rate spring, perhaps 90% or more of the time and would not be a bad solution at all.
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Old Sep 23, 2006 | 05:48 AM
  #25  
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chrono still can't tell the difference between a tender and a helper spring ... words get mixed up between his brain and typing fingers.
-mark signed off
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Old Sep 23, 2006 | 06:43 AM
  #26  
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Originally Posted by x838nwy
The point here is that the lower the rate of the helper, the more it effects the overall rate, not less. Therefore it becomes more important to the overall rate.

If you have two springs, one rate 'k' and another 'e x k' where e < 1, then the combined rate of them in series is:

k x [e/(1+e)]

So, if your main spring is 700lbs/in and your helper is 7lbs/in then the overall rate becomes 700 x (0.01/1.01) = 6.93lbs/in which is far away from the 700lbs of the main spring. So its effects cannot be ignored - that is if it hasn't fully compressed under the static load. You can see that the smaller 'e' becomes the more (e/(1+e)) approaches e and therefore the overall rate becomes increasingly close to the helper and the main spring has less and less effect.

I think what might happen is that if the helper is really a very low rate, then it is highly likely that it is compressed fully (and some more) under static load and so does not have an effect on the actual wheel rate other than at the very end of the dampers extension where the main spring runs out of travel.

If it is the case that the helper spring is always fully compressed at static load, there won't be that transiton I was talking about and it would in fact behave like a constant rate spring, perhaps 90% or more of the time and would not be a bad solution at all.
That more or less sums it up. I use a helper spring for the purpose of being able to run a high rate main spring, yet not limit myself to only an inch or so of droop travel before unloading the spring. The idea is to get a spring that is fully compressed at static height, yet gives you an inch and a half or two more travel in droop, all be it at a somewhat low rate on the droop side of travel. (not 6 or 7lb/in but more like 130-150lbs/in) On the loaded side of travel, you will always be fully in the high rate spring. (Side benefit in running over the curbs)
Just my 2c

Mark

Last edited by markdaddio; Sep 23, 2006 at 06:46 AM.
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Old Sep 23, 2006 | 08:08 AM
  #27  
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Originally Posted by x838nwy
The point here is that the lower the rate of the helper, the more it effects the overall rate, not less. Therefore it becomes more important to the overall rate.

If you have two springs, one rate 'k' and another 'e x k' where e < 1, then the combined rate of them in series is:

k x [e/(1+e)]

So, if your main spring is 700lbs/in and your helper is 7lbs/in then the overall rate becomes 700 x (0.01/1.01) = 6.93lbs/in which is far away from the 700lbs of the main spring. So its effects cannot be ignored - that is if it hasn't fully compressed under the static load. You can see that the smaller 'e' becomes the more (e/(1+e)) approaches e and therefore the overall rate becomes increasingly close to the helper and the main spring has less and less effect.

I think what might happen is that if the helper is really a very low rate, then it is highly likely that it is compressed fully (and some more) under static load and so does not have an effect on the actual wheel rate other than at the very end of the dampers extension where the main spring runs out of travel.

If it is the case that the helper spring is always fully compressed at static load, there won't be that transiton I was talking about and it would in fact behave like a constant rate spring, perhaps 90% or more of the time and would not be a bad solution at all.
I see what you are saying, and you're correct (although I still would like to make the distinction between tender and helper springs... later). Calculating the rate before either spring binds (or bottoms out) will yield the rate you showed above in the simple formula.

Which is why helper spring *will* be fully compressed under static load, and only come into effect under full extension, thereby not contributing anything to the spring rate at that corner under any useful operation or range of travel (given spring rate means nothing when the tire is not in contact with the ground). edit: I see you hinted at this in your other posts.

As soon as a minimal amount of load (10lbs, 12lbs, 20lbs) is applied on a helper spring, it will collapse into a flat coil as is intended by its very short deflection distance between coils, and low rate. The purpose then, is only to keep the main spring fully seated when the corner is at full droop, nothing else.

In the case of a tender spring, the same thing will happen, but at a point after static load is applied. The tender spring will still collapse fully onto itself, binding, as is the purpose of its design as well. But the difference here is it will occur at some point during the useful range in which the shock operates, affecting the spring rate of the series at that corner enough to produce a useable, progressive rate.

Of course then, tender springs will have a higher rate than helper springs, with coils spaced further apart and longer free length. The purpose is to create a progressive rate.

Last edited by Noob4life; Sep 23, 2006 at 08:21 AM.
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Old Sep 23, 2006 | 04:11 PM
  #28  
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ok my turn...

Helper or tender springs are always in coil bind under the car's weight, therefore they do nothing in terms of the overall spring rate. The reason why they're used is because of travel and ride height compromises. If you need a really stiff spring for grip, it won't compress much under static load. So, to get the car low enough, the spring has to be relatively short, often shorter than the shock travel, which leaves a gap between the spring and perch at full droop. Add the *helper* spring to take up the gap.

Now if there's a situation where the wheel droops so far that the main spring is not engaged, there is nothing pushing the wheel down other than it's own weight. Usually (but not always) this arises because of excessive body roll, meaning the main springs are too soft. Another situation is where the rear sway bar is so stiff that it lifts the inside rear wheel in tight turns, because the main spring is so short that it becomes unloaded. The sway bar is then stiff enough to support the weight of the inside wheel because it's not fighting the (unloaded) spring. A tender spring helps here because it will push the wheel back down to the ground so that you can apply power to accelerate out of the turn. If you have an open or Torsen rear diff and face this situation tender springs will help, otherwise you need a clutch pack style diff or soften the sway bar and get stiffer rear springs. Overly stiff rear sway bars are bad anyways because they make the suspension less "independent" and more like a solid rear axle which causes these problems.

The notion that tender springs help over small bumps is absolutely incorrect. This may be percieved if the tender spring rate is close to the main spring rate, meaning it is not in full coil bind, and then it is in series with the main spring, therefore lowering the overall spring rate, and making the car feel softer over bumps. Even at full droop, when you hit a bump, this is a high speed motion for the wheel. Shocks are velocity dependent meaning high piston velocities cause high damping force. So if you have a tender spring, a relatively soft spring rate, and combine this with a huge amount of damping, the bump hit is going to feel stupid harsh.

I have yet to see a progressive rate spring that doesn't have the softer coils at or near full coil bind. Other cars that have progressive or staged spring rates are aero cars or anything the jumps (rally, off-road). These are completely different situations. High aero formula cars have super soft tires (a spring as well) with big side walls. Tires are also progressive springs, and in F1, the tire is 60% of the suspension travel. Past that they have really stiff springs, often higher rates than the tires, and under aero conditions they ride on bump rubbers. In off-road, the main spring is the "tender" long travel spring and this is combined with a high rate short spring. This is to absorb landing after jumps, but the shocks also have high speed blow off so that the damping stays lows and lets the spring take the impulse. Not to take this off topic, but these cars put ours in perspective.

Hope this adds something to the topic.
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Old Sep 23, 2006 | 11:39 PM
  #29  
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Originally Posted by DMS_Mark
chrono still can't tell the difference between a tender and a helper spring ... words get mixed up between his brain and typing fingers.
-mark signed off
so that's all you came up with I guess that means you agree with everything else (what a relief...finally a breakthrough )

and yes I am a pitiful typist
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Old Sep 23, 2006 | 11:44 PM
  #30  
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Originally Posted by Noob4life
I see what you are saying, and you're correct (although I still would like to make the distinction between tender and helper springs... later). Calculating the rate before either spring binds (or bottoms out) will yield the rate you showed above in the simple formula.

Which is why helper spring *will* be fully compressed under static load, and only come into effect under full extension, thereby not contributing anything to the spring rate at that corner under any useful operation or range of travel (given spring rate means nothing when the tire is not in contact with the ground). edit: I see you hinted at this in your other posts.

As soon as a minimal amount of load (10lbs, 12lbs, 20lbs) is applied on a helper spring, it will collapse into a flat coil as is intended by its very short deflection distance between coils, and low rate. The purpose then, is only to keep the main spring fully seated when the corner is at full droop, nothing else.

In the case of a tender spring, the same thing will happen, but at a point after static load is applied. The tender spring will still collapse fully onto itself, binding, as is the purpose of its design as well. But the difference here is it will occur at some point during the useful range in which the shock operates, affecting the spring rate of the series at that corner enough to produce a useable, progressive rate.

Of course then, tender springs will have a higher rate than helper springs, with coils spaced further apart and longer free length. The purpose is to create a progressive rate.
Thanks for clearing that up, I promise not to interchange them ever again
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