Estimate springs for use on coil-overs

Mrblaine, Thank you for taking the time to post this info; I'll be using it to sort out the springs on my LJ . So far this info is the most straight forward and succinct I have come across, and I appreciate seeing/learning how this is figured out. The tips and insight you provide, like accounting for the small amount of lift from the reservoir, are invaluable.

Thanks again for this and all the information you share with this forum.

Someone smarter than me needs to work out the math for that one. I think I understand it but I'm not 100 % sure that I do.
I know how to calculate pressure and in the case of the reservoir it would be the charge pressure x the surface area of the piston.
If we take the theoretical of 200 psi charge pressure x the piston area of a 2.5" reservoir which is roughly 5 square inches, the math says that is 1000 lbs of force. That would extend the shock fully and not let it move. My theory is the force is acting against the diameter of the fluid equal to the cross section of the hose. In that case, the area of that is .19 square inches or 39 lbs additional lift which is about what we see.
 
Someone smarter than me needs to work out the math for that one. I think I understand it but I'm not 100 % sure that I do.
I know how to calculate pressure and in the case of the reservoir it would be the charge pressure x the surface area of the piston.
If we take the theoretical of 200 psi charge pressure x the piston area of a 2.5" reservoir which is roughly 5 square inches, the math says that is 1000 lbs of force. That would extend the shock fully and not let it move. My theory is the force is acting against the diameter of the fluid equal to the cross section of the hose. In that case, the area of that is .19 square inches or 39 lbs additional lift which is about what we see.


For a shock the "acting piston" is the shaft. Reservoir/hose size doesn't matter. Shock piston has holes in it so it doesn't matter. Hose diameter may be similar to shaft diameter so that's why it happens to be close.
 
@mrblaine this is very clear. The formula for combined spring rate is just like capacitors connected in series, and if you think about it, that's exactly how the two CO springs connected in series in the system function in terms of how they support the weight individually if the rates are different. The differential equations governing the systems have the same form, and the math works out exactly the same. There is a broader methodology that is used for both understanding and problem solving using mechanical / electrical analogues.

https://en.wikipedia.org/wiki/Impedance_analogy

The methodology you described is exactly what I would use if I were designing a circuit. I entirely understand the tuner wanting top and bottom spring rates not to deviate from each other too much. However the preload requirements are still somewhat a mystery to me. I guess this more a question for your tuner, but if you know .. I would like to understand. Why the 1" vs 2" difference between front and rear? To account for the weight differences? What happens when you use say 0.75" and 1.5", or the other side 1.5" and 3"? It's seems that this will have an effect on the eventual shock tune, correct?

:LOL: Leave it to an electrical engineer to compare simple springs to circuit design like that makes it easier :ROFLMAO:
 
For a shock the "acting piston" is the shaft. Reservoir/hose size doesn't matter. Shock piston has holes in it so it doesn't matter. Hose diameter may be similar to shaft diameter so that's why it happens to be close.

That puts it at .6 square inches x 200 psi or 120 lbs of extra lift x 2 for a pair which is 240 lbs. So, that seems too high for what we see. Generally the difference between a charged and not charged reservoir is about 1/4 -1/2" of lift.

I haven't had enough coffee yet this morning to try and figure out how much additional spring rate it would take to lift 1650 lbs 1/2".
 
For a shock the "acting piston" is the shaft. Reservoir/hose size doesn't matter. Shock piston has holes in it so it doesn't matter. Hose diameter may be similar to shaft diameter so that's why it happens to be close.

Sadly, I know that since that is how the piston in a proportioning valve works. There is a spring on one side of it but it is the diameter of the piston exiting the block to atmosphere that determines how much it moves under pressure.
 
That puts it at .6 square inches x 200 psi or 120 lbs of extra lift x 2 for a pair which is 240 lbs. So, that seems too high for what we see. Generally the difference between a charged and not charged reservoir is about 1/4 -1/2" of lift.

I haven't had enough coffee yet this morning to try and figure out how much additional spring rate it would take to lift 1650 lbs 1/2".


Hmmmm.... that's a puzzle. I want to say friction but rocking the car should take care of that.

I'd consider the the shock pressure force more of a static number than a spring rate. Technically it'd be more like a very low spring rate with a lot of preload. Depends on the reservoir gas volume vs the shaft volume.

The change in height describes the force change with spring rate. 1/2" lift with 86lb springs means 43lb from the charge. I don't know why that's so different from the calculated 120. Maybe someone else here can chew on it.
 
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Hmmmm.... that's a puzzle. I want to say friction but rocking the car should take care of that.

I'd consider the the shock pressure force more of a static number than a spring rate. Technically it'd be more like a very low spring rate with a lot of preload. Depends on the reservoir gas volume vs the shaft volume.

The change in height describes the force change with spring rate. 1/2" lift with 86lb springs means 43lb from the charge. I don't know why that's so different from the calculated 120. Maybe someone else here can chew on it.

Which is why I've mostly given up on calculating it or trying to and just know that the charge in the reservoir is going to add some small amount of lift so your target rate should be slightly lower to compensate and then use small bits of preload adjustment to get back to DRH.
 
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Hmmmm.... that's a puzzle. I want to say friction but rocking the car should take care of that.

I'd consider the the shock pressure force more of a static number than a spring rate. Technically it'd be more like a very low spring rate with a lot of preload. Depends on the reservoir gas volume vs the shaft volume.

The change in height describes the force change with spring rate. 1/2" lift with 86lb springs means 43lb from the charge. I don't know why that's so different from the calculated 120. Maybe someone else here can chew on it.

I think to start we'd want to take the springs out of the equation for simplification. We should be able to validate the pressurized nitrogen's contribution to preload. Take a pressurized shock with no spring, mark 1" on the shaft, set it on a bathroom scale and have someone read approximate weight it takes to start moving the shaft.

Compare that with Blaine's previous explanation. The key being make sure the chamber is pressurized at full extension since that's where we'll be pushing from. My shocks are 5/8" shaft diameter and at 200 psi should require 61lbs before the shaft begins to move. Judging from my experience on compressing a shock by hand to assemble that seems about the right ballpark. Sounds like Blaine is using 7/8 shaft, so it will be higher at 120lbs.
 
copied and pasted the first post for safekeeping. Thanks for the valuable resource.
 
I think to start we'd want to take the springs out of the equation for simplification. We should be able to validate the pressurized nitrogen's contribution to preload. Take a pressurized shock with no spring, mark 1" on the shaft, set it on a bathroom scale and have someone read approximate weight it takes to start moving the shaft.

Compare that with Blaine's previous explanation. The key being make sure the chamber is pressurized at full extension since that's where we'll be pushing from. My shocks are 5/8" shaft diameter and at 200 psi should require 61lbs before the shaft begins to move. Judging from my experience on compressing a shock by hand to assemble that seems about the right ballpark. Sounds like Blaine is using 7/8 shaft, so it will be higher at 120lbs.

I've not gone too deep into for the highest level of nuance. Were we to start down that path, we wouldn't use 2.5" but instead would use the exact piston diameter inside the 2.5" OD reservoir. I just want folks to know it exists so they aren't surprised when they miss the target rate slightly.
 
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I just put together a spreadsheet to calculate the gas force on the shock for a given compressed length using the Ideal Gas Law to do the calculation. See attached. Fill in the highlighted cells, and the bold cell at the bottom gives the calculated gas force on the shock rod. @srimes, do you mind checking my math? It's a busy day today, so I did it in a hurry.


Quick glance and it looks good to me. Nice job!

Still need to explain why observation doesn't seem to match. @B00mb00m's bathroom scale test would be a great start
 
I think that's where the inaccuracy of manufacturing a spring with a given spring rate mentioned in my earlier post manifests itself. If you have ±10% error band (that's on the order of what I observed) in each spring's rate (there are three of them - the 200 lb/in initial spring and then the 150 lb/in and 200 lb/in final configuration), the range of actual rates could account for why a calculated gas force is 120 lb and the displacement works out to 43 lb. Just thinking "out loud," if the actual combined spring rate is 10% low, that puts it at 77.1 lb/in, and with a compression at ride height of 6.75", the spring force error is 6.75 in x (85.7 lb/in - 77.1 lb/in), or 58 lb, which is close to the difference we're looking for (120 lb - 43 lb, or 77 lb). That's oversimplified, I know, but I think there's something to it.

Ya'll can quit anytime now. Things we don't know-
Piston area
Gas charge pressure
Sticktion

This is not a rocket, it doesn't need rocket science. For the last time, this is a method to slow down the number of springs you toss into the corner.

I don't need a pair of springs or even 4 of them to be rocket science race bike dead on accurate. I need a pair of spring or 4 made by a company that has the same amount of error across the line up. Given how often this method works to get us very close, their errors are consistent and we are plenty close.

The bottom line is the average recreational user will not hunt down and employ a spring rate tester, take it to the spring store and sit there and test springs to find those within some mythical margin of error. We are all going to do the same thing, look at the number on the box and use it.
 
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:LOL: Leave it to an electrical engineer to compare simple springs to circuit design like that makes it easier :ROFLMAO:

It helps me understand the concepts in what Blaine explained without having to make excel sheets :)
 
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The "average recreational user" is likely going to end up with a box of springs because they were lost before reading the first paragraph of your first post. The people still following the thread are not average recreational users. ;)

Engineers engineer. Sorry!

Guess what my inclination will be the next time I'm faced with the choice to try and make folks life easier by putting up a post like the first one or not because I consider it not worth the bullshit?
 
F.F.S. @Chris can we please just delete every post past the first one, lock the thread under "stickes" or "important shit to know"? This kind of information isn't readily available on the interwebs and some jackass always comes along and mucks it up.
 
I didn't realize my contributions were bullshit. I was honestly trying to contribute to a discussion.



If my posts and efforts to contribute are just jackassery, I'll stop. Again, I'm just a guy trying to contribute to a great forum. If y'all want me to just shut up - just ask me to. I am a big boy and can take constructive criticism.

You are more than welcome to contribute. Back it down and keep it in context. I can only say knock it off, we don't need rocket science levels of nuance here for this issue so many times before I start looking for the delete button. We are NOT taking a spring rate tester to the store, we are not going to buy 20 samples of the same rate and blueprint match them. We don't need to understand hysteresis, we don't need to take into account how the vagaries of spring wind up affect spring rate and whether or not we should equip our spring cups with thrust bearings.

We need something that takes the mystery out of figuring out spring rate at the recreational level that will produce something that works within fractions of what we are after. This method does that and there are enough qualifiers and disclaimers in enough posts that the average person trying to set up coil overs will be just fine and if not, a conversation with their tuner will get them on the right path.
 
F.F.S. @Chris can we please just delete every post past the first one, lock the thread under "stickes" or "important shit to know"? This kind of information isn't readily available on the interwebs and some jackass always comes along and mucks it up.

I could do that but I have to respect Blaine’s wishes. Often times he will ask me to take these threads down if he gets tired of the jackasses. I’ll have to run this by him. @mrblaine?
 
I've deleted all my posts from this thread. Sorry for the trouble. I'll create my own threads if I want to get into more esoteric discussions.
 
I've deleted all my posts from this thread. Sorry for the trouble. I'll create my own threads if I want to get into more esoteric discussions.

Nothing wrong with esoteric discussions .. just need to find the right place and audience. I'm an engineer too and I get wanting to get into details and the physics and the math. However .. what is more important is to understand the big picture. This is not at you but everyone in general.

There is a lot of myth and misconception about setting up coilovers. Heck, if you follow the main board discussions, there is much nonsense discussed about plain springs. It's easy to get overwhelmed with the choices, and one can as easily mess up a CO setup as a regular setup.

What Blaine is attempting here is to provide the right information in a simple, clear way to understand without needless jargon. Not only did his post explain how to derive the right combination of springs for a coilover setup for a desired ride height, but it also explains what is it that springs really do. Did everyone notice that never once the topic of ride quality enter the picture in the entire post? Because that's determined by the shock tune. If this post doesn't make that abundantly clear, I do not know what will.

It's never easy to present a complex topic in a way such that you read it and think "I could have figured that out". It takes years of real world experience and directed learning to tease that simplicity out in the open. We are lucky to have someone like Blaine, who has both the intuition and the experience, who is willing to come here participate, and then share such useful information in an easily digestible manner.

You'll often read Blaine saying he doesn't chase numbers. That doesn't mean he doesn't understand them, but rather the opposite. He understands what matters where and he won't chase numbers for it's own sake. Everything has to fit and work well as a whole. You don't need an excel sheet calculator with hundred formulas and linked cells for that.

You are welcome to post and talk about details as much as you like. There are folks in this forum that will engage and atleast try to see if what you are saying is useful. Just make sure to understand the thread context and then post your heart's content.

Edit - before you (Sab) gets too upset, a good bit of the "you" usage above is generic.
 
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