Spring Rates and Spring Length Correlation?
12-02-2004, 05:56 PM
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Spring Rates and Spring Length Correlation?
I'm just pondering at how greater spring rates are achieved. If I understand correctly, spring rates are not a function of length? Meaning if you have a 10" long spring and a 20" long spring, they could share the same spring rates? So, really the spring rates are achieved by the way the coils are bent? So, does spring length have any correlation with spring rates?
12-02-2004, 05:59 PM
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Re: Spring Rates and Spring Length Correlation? (Understeer)
No, it doesn't have a significant effect on spring rate, unless the spring is too short for the travel that suspension needs and it binds up, then you get a nice big jump to infinity once the spring is blocked and bound up completely.
A 6" 600 lb/in spring has the same spring rate as an 8" spring with the same wire diameter and number of coils per inch or ramp angle.
A 6" 600 lb/in spring has the same spring rate as an 8" spring with the same wire diameter and number of coils per inch or ramp angle.
12-02-2004, 06:06 PM
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Length does affect it... If you took a 20" long spring and cut it in half, each half would be stiffer than the original. The wire is still the same strength, but there are less coils for the weight to compress. When you put weight on a spring, it will compress each coil a certain amount... take away half of the coils and the spring will only compress half as much overall. The coils don't add strength... they add the capacity to absorb energy, which makes it softer.
That's my understanding of it at least...
That's my understanding of it at least...
12-02-2004, 07:41 PM
#4
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Re: Spring Rates and Spring Length Correlation? (TeamSlowdotOrg)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by TeamSlowdotOrg »</TD></TR><TR><TD CLASS="quote">No, it doesn't have a significant effect on spring rate, unless the spring is too short for the travel that suspension needs and it binds up, then you get a nice big jump to infinity once the spring is blocked and bound up completely.
A 6" 600 lb/in spring has the same spring rate as an 8" spring with the same wire diameter and number of coils per inch or ramp angle. </TD></TR></TABLE>
To illustrate why this is wrong:
Take 2 ideal 500 lb/in springs, put them on top of each other. Put a 500 lb weight on the top.
(500 lbs)
/////////// <--- spring
/////////// <--- spring
-----------
The bottom spring has 500 lbs pushing down on it (plus 500 lbs pushing up from the ground) and compresses 1 inch.
The top one has 500 lbs of weight on top and 500 lbs pushing back from the bottom spring. another 1 inch compression.
The 500 lb weight has 500 lbs of gravity pulling down, and 500 lbs pushing back from the top spring.
Hooray, everyone's net force is zero, and the "spring system" has compressed 2 inches. This system has a rate of 250 lb/in.
So if you double the spring length, without changing the coils, the spring rate is cut in half.
A 6" 600 lb/in spring has the same spring rate as an 8" spring with the same wire diameter and number of coils per inch or ramp angle. </TD></TR></TABLE>
To illustrate why this is wrong:
Take 2 ideal 500 lb/in springs, put them on top of each other. Put a 500 lb weight on the top.
(500 lbs)
/////////// <--- spring
/////////// <--- spring
-----------
The bottom spring has 500 lbs pushing down on it (plus 500 lbs pushing up from the ground) and compresses 1 inch.
The top one has 500 lbs of weight on top and 500 lbs pushing back from the bottom spring. another 1 inch compression.
The 500 lb weight has 500 lbs of gravity pulling down, and 500 lbs pushing back from the top spring.
Hooray, everyone's net force is zero, and the "spring system" has compressed 2 inches. This system has a rate of 250 lb/in.
So if you double the spring length, without changing the coils, the spring rate is cut in half.
12-02-2004, 08:16 PM
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Re: Spring Rates and Spring Length Correlation? (ccfries)
Aye, so what am I thinking of with ramp rate?
Oh, snap, I totally said coils per inch would be the same for the different length and same rate, woops. Yeah, didn't mean to say that at all.
Oh, snap, I totally said coils per inch would be the same for the different length and same rate, woops. Yeah, didn't mean to say that at all.
12-02-2004, 11:19 PM
#6
Re: Spring Rates and Spring Length Correlation? (TeamSlowdotOrg)
I'ts much easier to think of a spring as a straight bar. This puts binding/progressive rates out of the picture. If you have two bars of equal diameter but different lengths, the longer one will be easier to bend, period. A coil spring is the same as a straight bar just in a different shape.
This also helps show you how a progressive rate spring works. You can always tell them from regular ones because they have some coils closer together than others. When the coils compress, the closer winds bind up befor the loose ones effectively making a shorter bar. This causes the spring to get much stiffer as the spring compresses.
To get a spring with a different length but the same rate, they change the wire diameter. The wires angle doesn't change anything, only the length and diameter. A sway bar is the same as your coils. It is just a spring that connects one side of your suspension w/ the other. This is why sway bars are listed as springs in the honda parts catalogs. This also shows why a 22mm sway that mounts to your lower rear strut bolts (suspension techniques) is not as stiff as a 22mm swaybar that mounts to the integra LCA swaybar bolt holes (comptech).
This also helps show you how a progressive rate spring works. You can always tell them from regular ones because they have some coils closer together than others. When the coils compress, the closer winds bind up befor the loose ones effectively making a shorter bar. This causes the spring to get much stiffer as the spring compresses.
To get a spring with a different length but the same rate, they change the wire diameter. The wires angle doesn't change anything, only the length and diameter. A sway bar is the same as your coils. It is just a spring that connects one side of your suspension w/ the other. This is why sway bars are listed as springs in the honda parts catalogs. This also shows why a 22mm sway that mounts to your lower rear strut bolts (suspension techniques) is not as stiff as a 22mm swaybar that mounts to the integra LCA swaybar bolt holes (comptech).
12-03-2004, 05:59 AM
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Re: Spring Rates and Spring Length Correlation? (ccfries)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ccfries »</TD></TR><TR><TD CLASS="quote">
To illustrate why this is wrong:
Take 2 ideal 500 lb/in springs, put them on top of each other. Put a 500 lb weight on the top.
(500 lbs)
/////////// <--- spring
/////////// <--- spring
-----------
The bottom spring has 500 lbs pushing down on it (plus 500 lbs pushing up from the ground) and compresses 1 inch.
The top one has 500 lbs of weight on top and 500 lbs pushing back from the bottom spring. another 1 inch compression.
The 500 lb weight has 500 lbs of gravity pulling down, and 500 lbs pushing back from the top spring.
Hooray, everyone's net force is zero, and the "spring system" has compressed 2 inches. This system has a rate of 250 lb/in.
So if you double the spring length, without changing the coils, the spring rate is cut in half.
</TD></TR></TABLE>
sorry your wrong
for 500 lbs each spring would move 1/2 inch , totaling 1" at 500 lbs
spring rate is a math equation ,, wire diameter ,, thickness of wire ,, centre to centre of wire to determine the coil diameter
,, this X the modulus of elasticity of the wire material will give the spring rate
do a search on google to "determine spring rate"
ian
To illustrate why this is wrong:
Take 2 ideal 500 lb/in springs, put them on top of each other. Put a 500 lb weight on the top.
(500 lbs)
/////////// <--- spring
/////////// <--- spring
-----------
The bottom spring has 500 lbs pushing down on it (plus 500 lbs pushing up from the ground) and compresses 1 inch.
The top one has 500 lbs of weight on top and 500 lbs pushing back from the bottom spring. another 1 inch compression.
The 500 lb weight has 500 lbs of gravity pulling down, and 500 lbs pushing back from the top spring.
Hooray, everyone's net force is zero, and the "spring system" has compressed 2 inches. This system has a rate of 250 lb/in.
So if you double the spring length, without changing the coils, the spring rate is cut in half.
</TD></TR></TABLE>
sorry your wrong
for 500 lbs each spring would move 1/2 inch , totaling 1" at 500 lbs
spring rate is a math equation ,, wire diameter ,, thickness of wire ,, centre to centre of wire to determine the coil diameter
,, this X the modulus of elasticity of the wire material will give the spring rate
do a search on google to "determine spring rate"
ian
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12-03-2004, 06:29 AM
#8
Honda-Tech Member
If the rate is 500 lb/in, why would the springs only compress 1/2"? Wouldn't that imply that it's only supporting 250 lbs? This would be true if the springs were in a parallel configuration (side by side), but aren't we talking about springs in series (end to end)? I've always thought that springs in series add like resistors in parallel (the formula is the same). Is this not right???
If you look at Eibach's website, they show some simple spring rate calculations for their dual spring (tender/main spring combo) setup. One is the initial spring rate, which is the effective spring rate of the two springs stacked together, before the tender spring is at full coil bind.
http://eibach.com/cgi-bin/html...23731
The equation is spring rate = (main spring rate * tender spring rate) / (main spring rate + tender spring rate)
In the above example, if you have two 500 lb/in springs stacked (say one is the main, and the other the tender, although it doesn't matter), the effective rate is 500 * 500 / (500 + 500) = 2500 / 1000 = 250 lb/in
I also found:
http://www.alloyspringproducts.com/ET_info.asp
If you look at Eibach's website, they show some simple spring rate calculations for their dual spring (tender/main spring combo) setup. One is the initial spring rate, which is the effective spring rate of the two springs stacked together, before the tender spring is at full coil bind.
http://eibach.com/cgi-bin/html...23731
The equation is spring rate = (main spring rate * tender spring rate) / (main spring rate + tender spring rate)
In the above example, if you have two 500 lb/in springs stacked (say one is the main, and the other the tender, although it doesn't matter), the effective rate is 500 * 500 / (500 + 500) = 2500 / 1000 = 250 lb/in
I also found:
http://www.alloyspringproducts.com/ET_info.asp
12-03-2004, 06:45 AM
#9
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Re: Spring Rates and Spring Length Correlation? (icarp)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by icarp »</TD></TR><TR><TD CLASS="quote">
sorry your wrong
for 500 lbs each spring would move 1/2 inch , totaling 1" at 500 lbs
spring rate is a math equation ,, wire diameter ,, thickness of wire ,, centre to centre of wire to determine the coil diameter
,, this X the modulus of elasticity of the wire material will give the spring rate
do a search on google to "determine spring rate"
ian </TD></TR></TABLE>
Yeah man, each spring has 500 lbs compressing it. Each one compresses one inch.
-Chris
sorry your wrong
for 500 lbs each spring would move 1/2 inch , totaling 1" at 500 lbs
spring rate is a math equation ,, wire diameter ,, thickness of wire ,, centre to centre of wire to determine the coil diameter
,, this X the modulus of elasticity of the wire material will give the spring rate
do a search on google to "determine spring rate"
ian </TD></TR></TABLE>
Yeah man, each spring has 500 lbs compressing it. Each one compresses one inch.
-Chris
12-03-2004, 07:15 AM
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Re: (117)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 117 »</TD></TR><TR><TD CLASS="quote"> I've always thought that springs in series add like resistors in parallel (the formula is the same). Is this not right???
</TD></TR></TABLE>
yes, it's right. You are correct, and I am afraid that Ian is wrong this time. Springs behave "opposite" of resistors. They add in parallel and divide in series.
</TD></TR></TABLE>
yes, it's right. You are correct, and I am afraid that Ian is wrong this time. Springs behave "opposite" of resistors. They add in parallel and divide in series.
12-03-2004, 08:46 AM
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Re: Spring Rates and Spring Length Correlation? (vtecvoodoo)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by vtecvoodoo »</TD></TR><TR><TD CLASS="quote">Yay we`re talking springs!
</TD></TR></TABLE>
It's amazing how much of a "black art" it is still. I worked for a spring manufacturer a few years back and you'd be amazed at how many engineers have no clue how springs work We had the neat SMI program that I could plug in a multitude of variables and find out what the spring would do.
There's a lot of miss-information in this post. In the original example the 20" spring would have to have thicker wire than the 10" spring to maintain the same rate. That just a general statement tho, other factors come into play.
Rate is how much weight it takes to compress (for compression springs) a linear spring 1". So a 200# spring rate with 400# on top of it will be compressed 2" (roughtly). So a 10" 200# spring will act the same as a 20" 200# spring up until they start to bind.
</TD></TR></TABLE>It's amazing how much of a "black art" it is still. I worked for a spring manufacturer a few years back and you'd be amazed at how many engineers have no clue how springs work
We had the neat SMI program that I could plug in a multitude of variables and find out what the spring would do. There's a lot of miss-information in this post. In the original example the 20" spring would have to have thicker wire than the 10" spring to maintain the same rate. That just a general statement tho, other factors come into play.
Rate is how much weight it takes to compress (for compression springs) a linear spring 1". So a 200# spring rate with 400# on top of it will be compressed 2" (roughtly). So a 10" 200# spring will act the same as a 20" 200# spring up until they start to bind.
12-03-2004, 08:49 AM
#13
Senior Member
Re: Spring Rates and Spring Length Correlation? (vtecvoodoo)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by vtecvoodoo »</TD></TR><TR><TD CLASS="quote">Yay we`re talking springs!
</TD></TR></TABLE>
Thanks for the insight
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nonsense »</TD></TR><TR><TD CLASS="quote">Rate is how much weight it takes to compress (for compression springs) a linear spring 1". So a 200# spring rate with 400# on top of it will be compressed 2" (roughtly). So a 10" 200# spring will act the same as a 20" 200# spring up until they start to bind. </TD></TR></TABLE>
Yes! But a 20" 200 lb/in spring is not the same as two 200 lb/in springs end-to-end, it's two 400 lb/in springs end to end. I think that was the confusion.
Not that any of us would ever cut a spring in half, or put two springs end-to-end.
-Chris
</TD></TR></TABLE>Thanks for the insight
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nonsense »</TD></TR><TR><TD CLASS="quote">Rate is how much weight it takes to compress (for compression springs) a linear spring 1". So a 200# spring rate with 400# on top of it will be compressed 2" (roughtly). So a 10" 200# spring will act the same as a 20" 200# spring up until they start to bind. </TD></TR></TABLE>
Yes! But a 20" 200 lb/in spring is not the same as two 200 lb/in springs end-to-end, it's two 400 lb/in springs end to end. I think that was the confusion.
Not that any of us would ever cut a spring in half, or put two springs end-to-end.
-Chris
12-03-2004, 10:28 AM
#14
Re: Spring Rates and Spring Length Correlation? (ccfries)
I have read some write up in a book that told of a suspension setup, using multiple springs, that would alow the suspension to soften up when an extreeme load was applied (similar to digressive shock valveing). The idea was to be able to hit curbs, if necisarry, to maintain a line. The writer was saying it would be really difficult to figure out how to do it w/o the springs softening under heavy cornering loads. I remember him saying something about a combonation of a regular coil along with a pre-compressed coil (not sure how that works).
Does anybody know of this?
Does anybody know of this?
12-03-2004, 11:09 AM
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Re: Spring Rates and Spring Length Correlation? (94eg!)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 94eg! »</TD></TR><TR><TD CLASS="quote">I have read some write up in a book that told of a suspension setup, using multiple springs, that would alow the suspension to soften up when an extreeme load was applied (similar to digressive shock valveing). The idea was to be able to hit curbs, if necisarry, to maintain a line. The writer was saying it would be really difficult to figure out how to do it w/o the springs softening under heavy cornering loads. I remember him saying something about a combonation of a regular coil along with a pre-compressed coil (not sure how that works).
Does anybody know of this?</TD></TR></TABLE>
Yeah, that'd work, but sounds hacky! Take a short 500 lb/in spring with 3" of travel before it bottoms out, and a tall 250 lb/in spring with 9" of travel, and pre-load it 6 inches (1500 lbs).
For the first 3 inches of travel, the 500 in-lb spring takes 1500 lbs to compress. The 250 lb/in spring doesn't move.
For the next 3 inches of travel, it only takes 750 lbs to finish compressing the 250 lb/in spring, so the effective spring rate dropped from 500 to 250.
You'd have to make sure the 1st heavy spring (500 in-lb) has enough travel to never bottom out while cornering.
I've never heard of an automotive application for this. Sounds like fun!
Does anybody know of this?</TD></TR></TABLE>
Yeah, that'd work, but sounds hacky! Take a short 500 lb/in spring with 3" of travel before it bottoms out, and a tall 250 lb/in spring with 9" of travel, and pre-load it 6 inches (1500 lbs).
For the first 3 inches of travel, the 500 in-lb spring takes 1500 lbs to compress. The 250 lb/in spring doesn't move.
For the next 3 inches of travel, it only takes 750 lbs to finish compressing the 250 lb/in spring, so the effective spring rate dropped from 500 to 250.
You'd have to make sure the 1st heavy spring (500 in-lb) has enough travel to never bottom out while cornering.
I've never heard of an automotive application for this. Sounds like fun!
12-03-2004, 01:09 PM
#16
Junior Member
Re: Spring Rates and Spring Length Correlation? (ccfries)
isn't the dual spring setup performing to the same effect as a progressive spring? If I am not mistaken, a progress would have a spring rate that increased as the spring becomes compressed, the coils which are farther apart would compress under a lighter load, and would not compress the tighter wound coils until a sufficient load has been met. Of course this would not give you two seperate linear rates, unless you had a really sharp change in the spring coils, btu this has the same effect right?
As for my question, is the unit pounds/inch, dependent on the length of spring?
Hypothetically, you have a 20 inch spring, it compresses 1 inch under 200 pounds (a rate of 200lb/in), then you cut it in half, would the resulting spring compress 0.5 inches under 200 pounds?(a new rate of 400lb/inch) I'm sure I am ignoring factors such as spring constants and such, but is that basically what happends?
As for my question, is the unit pounds/inch, dependent on the length of spring?
Hypothetically, you have a 20 inch spring, it compresses 1 inch under 200 pounds (a rate of 200lb/in), then you cut it in half, would the resulting spring compress 0.5 inches under 200 pounds?(a new rate of 400lb/inch) I'm sure I am ignoring factors such as spring constants and such, but is that basically what happends?
12-03-2004, 01:18 PM
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Re: Spring Rates and Spring Length Correlation? (Understeer)
All spring move when any magnitude of force is applied to them. The spring rate for a linear spring is simply the slope of the force/displacement curve. The "lb/in" designation is only for conveinient units. So you can't say a spring won't move when a force is applied to it.
Length of spring is involved in calculating the spring rate.
Length of spring is involved in calculating the spring rate.
12-03-2004, 01:39 PM
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Re: (SJR)
SJR
500lbs per inch means just that
if you have 500 lbs and you stack two 500lb per " springs the total compression
will = one inch
the 10" spring plus another 10"spring = a 20" spring its still a 500lb rate as long as the two springs are the same rate .
ian
500lbs per inch means just that
if you have 500 lbs and you stack two 500lb per " springs the total compression
will = one inch
the 10" spring plus another 10"spring = a 20" spring its still a 500lb rate as long as the two springs are the same rate .
ian
12-03-2004, 01:42 PM
#19
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Re: Spring Rates and Spring Length Correlation? (GSpeedR)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by donz »</TD></TR><TR><TD CLASS="quote">isn't the dual spring setup performing to the same effect as a progressive spring? </TD></TR></TABLE>
Its 100% backwards. Instead of having a soft spring take the small bumps, then bottom out and hit the stronger spring... we're keeping the car flat with a strong spring, but if we hit a curb we compress then next 3 inches easily.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR »</TD></TR><TR><TD CLASS="quote">All spring move when any magnitude of force is applied to them. The spring rate for a linear spring is simply the slope of the force/displacement curve. The "lb/in" designation is only for conveinient units. So you can't say a spring won't move when a force is applied to it.
Length of spring is involved in calculating the spring rate.</TD></TR></TABLE>
You replying to me or the guy above?
Anyhow-- the light (250 in-lb) would be preloaded by 1500 lbs, by a perch on both sides that held it compressed by 6 inches, but would allow it to compress more. So untill you applied more than 1500 lbs, you'd just be taking weight off of the collar holding the perch compressed, the spring wouldn't notice the difference until you exceeded 1500 lbs.
I suck at teh words, let me draw a pic.
0 load: no compression String has 1500 lbs of tension holding the 250 lb spring compressed.
500 lbs: 1 inch deflection.
1000 lbs: 2 inch deflection
1500 lbs: 3 inch deflection. String has 0 tension anymore, the 1500 lbs of compression force are from the 500 lb spring and the top of the car.
The 500 lb-inch spring has bottomed out, as well.
The next inch of travel will compress the 250 in-lb spring by, well, 250 lbs. The effective spring rate went down. By the time the whole system has bottomed out (6") the total weight is only 1500 + 750 = 2250.
Voila.
-Chris
Its 100% backwards. Instead of having a soft spring take the small bumps, then bottom out and hit the stronger spring... we're keeping the car flat with a strong spring, but if we hit a curb we compress then next 3 inches easily.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR »</TD></TR><TR><TD CLASS="quote">All spring move when any magnitude of force is applied to them. The spring rate for a linear spring is simply the slope of the force/displacement curve. The "lb/in" designation is only for conveinient units. So you can't say a spring won't move when a force is applied to it.
Length of spring is involved in calculating the spring rate.</TD></TR></TABLE>
You replying to me or the guy above?
Anyhow-- the light (250 in-lb) would be preloaded by 1500 lbs, by a perch on both sides that held it compressed by 6 inches, but would allow it to compress more. So untill you applied more than 1500 lbs, you'd just be taking weight off of the collar holding the perch compressed, the spring wouldn't notice the difference until you exceeded 1500 lbs.
I suck at teh words, let me draw a pic.
0 load: no compression String has 1500 lbs of tension holding the 250 lb spring compressed.
500 lbs: 1 inch deflection.
1000 lbs: 2 inch deflection
1500 lbs: 3 inch deflection. String has 0 tension anymore, the 1500 lbs of compression force are from the 500 lb spring and the top of the car.
The 500 lb-inch spring has bottomed out, as well.
The next inch of travel will compress the 250 in-lb spring by, well, 250 lbs. The effective spring rate went down. By the time the whole system has bottomed out (6") the total weight is only 1500 + 750 = 2250.
Voila.
-Chris
12-03-2004, 01:43 PM
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Re: Spring Rates and Spring Length Correlation? (ccfries)
ccfries
it dose not matter if you pre load a 200lb spring to 2000lb the next inch will take 200 lbs to compress it
that is what spring rate means
ian
it dose not matter if you pre load a 200lb spring to 2000lb the next inch will take 200 lbs to compress it
that is what spring rate means
ian
12-03-2004, 01:43 PM
#21
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Re: (icarp)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by icarp »</TD></TR><TR><TD CLASS="quote">SJR
500lbs per inch means just that
if you have 500 lbs and you stack two 500lb per " springs the total compression
will = one inch
the 10" spring plus another 10"spring = a 20" spring its still a 500lb rate as long as the two springs are the same rate .
ian</TD></TR></TABLE>
no ian youre wrong.
springs added in series create a spring rate less than the average spring rates.
1/x1 + 1/x2 = 1/X
x1 = spring rate 1
x2 = spring rate 2
X = new spring rate.
look up the formulas yourself.
springs added in parallel simply ADD together.
x1 + x2 = X
its ok to be wrong, its confusing at first, but it makes sense once you do the free body diagram as described before.
500lbs per inch means just that
if you have 500 lbs and you stack two 500lb per " springs the total compression
will = one inch
the 10" spring plus another 10"spring = a 20" spring its still a 500lb rate as long as the two springs are the same rate .
ian</TD></TR></TABLE>
no ian youre wrong.
springs added in series create a spring rate less than the average spring rates.
1/x1 + 1/x2 = 1/X
x1 = spring rate 1
x2 = spring rate 2
X = new spring rate.
look up the formulas yourself.
springs added in parallel simply ADD together.
x1 + x2 = X
its ok to be wrong, its confusing at first, but it makes sense once you do the free body diagram as described before.
12-03-2004, 01:47 PM
#22
Honda-Tech Member
Re: Spring Rates and Spring Length Correlation? (icarp)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by icarp »</TD></TR><TR><TD CLASS="quote">ccfries
it dose not matter if you pre load a 200lb spring to 2000lb the next inch will take 200 lbs to compress it
that is what spring rate means
ian </TD></TR></TABLE>
that also is quite wrong in a certain case if you preloaded it with say something like a common spring compressor. if you held it in place, you would need to first overcome the tension the spring compressor is holding the spring, then it will move.
now if you just put 100 lbs on a spring and called that preloading it, then another 50 lbs would move it more. but thats not what ccfires is talking about.
btw, ive seen pictures of ppl doing this to tune their progressive suspension. but for like really big shocks ppl use for desert rallies on buggies or trucks that have huge suspension travel.
it dose not matter if you pre load a 200lb spring to 2000lb the next inch will take 200 lbs to compress it
that is what spring rate means
ian </TD></TR></TABLE>
that also is quite wrong in a certain case if you preloaded it with say something like a common spring compressor. if you held it in place, you would need to first overcome the tension the spring compressor is holding the spring, then it will move.
now if you just put 100 lbs on a spring and called that preloading it, then another 50 lbs would move it more. but thats not what ccfires is talking about.
btw, ive seen pictures of ppl doing this to tune their progressive suspension. but for like really big shocks ppl use for desert rallies on buggies or trucks that have huge suspension travel.
12-03-2004, 01:51 PM
#23
Re: Spring Rates and Spring Length Correlation? (ccfries)
So how do I figure out my max cornering weight? This sounds like math might not work here since there are so many variable (tires, swaybars, chassis stiffness, shock damping, etc...). Would you have to use an on-car shock dyno or something? Or mabey you could use a measured G force from a perfomance accelerometer (G-Tech Pro) and calculate the corner weights under those loads, and tune accordingly?
Finally, how do you preload a spring like that and still fit it on your shocks?
Finally, how do you preload a spring like that and still fit it on your shocks?
12-03-2004, 01:52 PM
#24
Senior Member
Re: (icarp)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by icarp »</TD></TR><TR><TD CLASS="quote">SJR
500lbs per inch means just that
if you have 500 lbs and you stack two 500lb per " springs the total compression
will = one inch
the 10" spring plus another 10"spring = a 20" spring its still a 500lb rate as long as the two springs are the same rate .
ian</TD></TR></TABLE>
Bzzzzt! Wrong answer. If each spring is 500 lb/in, and each one has 500 lbs of force on it (they are stacked, not sharing the weight), then each one will compress 2 inches. The system is a 250 lb-in spring.
If you put the springs side-by-side, they will only compress 1/2 inch each, because they are sharing the load.
Series springs:
http://scienceworld.wolfram.co....html
The force is the same on each of the two springs
Spring constant = (1/k1 + 1/k2) ^-1, which is 1/(2/500), or 250 lb/in.
Parallel springs:
http://scienceworld.wolfram.co....html
Spring constant = k1 + k2 (500+500 = 1000 lb/in)
Just like resistor math. Springs in series == Resistors in parallel.
500lbs per inch means just that
if you have 500 lbs and you stack two 500lb per " springs the total compression
will = one inch
the 10" spring plus another 10"spring = a 20" spring its still a 500lb rate as long as the two springs are the same rate .
ian</TD></TR></TABLE>
Bzzzzt! Wrong answer. If each spring is 500 lb/in, and each one has 500 lbs of force on it (they are stacked, not sharing the weight), then each one will compress 2 inches. The system is a 250 lb-in spring.
If you put the springs side-by-side, they will only compress 1/2 inch each, because they are sharing the load.
Series springs:
http://scienceworld.wolfram.co....html
The force is the same on each of the two springs
Spring constant = (1/k1 + 1/k2) ^-1, which is 1/(2/500), or 250 lb/in.
Parallel springs:
http://scienceworld.wolfram.co....html
Spring constant = k1 + k2 (500+500 = 1000 lb/in)
Just like resistor math. Springs in series == Resistors in parallel.
12-03-2004, 01:52 PM
#25
Honda-Tech Member
Re: Spring Rates and Spring Length Correlation? (icarp)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by icarp »</TD></TR><TR><TD CLASS="quote">ccfries
it dose not matter if you pre load a 200lb spring to 2000lb the next inch will take 200 lbs to compress it
that is what spring rate means
ian </TD></TR></TABLE>
when you add two springs together, no matter what rates, youve essentially created a NEW spring, with a NEW spring rate.
two springs of equal spring rate in series does NOT equal a spring twice the length and the same spring rate. more like twice the length and half the spring rate of the original.
it dose not matter if you pre load a 200lb spring to 2000lb the next inch will take 200 lbs to compress it
that is what spring rate means
ian </TD></TR></TABLE>
when you add two springs together, no matter what rates, youve essentially created a NEW spring, with a NEW spring rate.
two springs of equal spring rate in series does NOT equal a spring twice the length and the same spring rate. more like twice the length and half the spring rate of the original.