GT3 Gurus - quick Q
11-10-2010, 01:05 AM
#31
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FWIW, I just measured the free length of the oem GT3 springs:
front: 9-3/16" (233 mm)
rear: 9-5/8" (244 mm)
Not sure how thick the spring spacer is for helper/main springs or how relevant this info may be to you. Figured I'd measure them before they go into long term storage.
Cheers,
front: 9-3/16" (233 mm)
rear: 9-5/8" (244 mm)
Not sure how thick the spring spacer is for helper/main springs or how relevant this info may be to you. Figured I'd measure them before they go into long term storage.
Cheers,
11-10-2010, 01:36 AM
#32
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Vic I translated from my language so I don't know if you have that difference in the term there for sure but to simplify it two types of "secondary" springs exist the ones that are designed to completely compress at static load, and usually used to maintain the main spring in its place at full travel and the ones that are part of a dual rate spring system or prgressive, these are still open at normal static load and close at a predetermined load after maybe depending on the N/mm already cause the main springs to compress too.
It's easy to see what you have, look under your car when on the ground resting if the secondary spring is fully compressed or not.
Also I have to say that all this is an over simplification for the shake of discussion here, one might argue that even a very soft helper spring plays a role on how long and at what force an unloaded wheel will touch the ground when a car is turning etc etc.
It's easy to see what you have, look under your car when on the ground resting if the secondary spring is fully compressed or not.
Also I have to say that all this is an over simplification for the shake of discussion here, one might argue that even a very soft helper spring plays a role on how long and at what force an unloaded wheel will touch the ground when a car is turning etc etc.
11-10-2010, 02:07 AM
#34
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^ if fully compressed at rest, then it is very very soft rate, thus a HELPER spring, used to keep your main string tight b/n perches.
a tender is like progressive spring when used with main spring. tender have higher rate than helper
a tender is like progressive spring when used with main spring. tender have higher rate than helper
11-10-2010, 02:30 AM
#35
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When you combine two linear rate springs like we do with our cars, it works this way: the car will use the combined (effective) rate until the soft spring is compressed and then will use the higher spring rate.
Using the example above of 143lb/in helpers with 627lb/in mains, you need to first calculate the effective rate of these two combined springs. That's done with this formula:
effective rate = (main spring x tender spring) / (main spring + tender spring)
So we have:
effective rate = (627 x 143) / (627 + 143)
effective rate = 89661 / 770
effective rate = 116lb/in
So the spring stack of 143/627 springs will act like a 116lb/in spring until the helper spring is fully compressed and then the main spring will take over at a rate of 627lb/in.
Consider the weight on each wheel and you'll see that under most cases the helper springs will be fully compressed at rest and when you are driving over bumps where the spring stack can get a bit unloaded, you will feel the low rate of 116lb/in in this example.
Cheers,
Using the example above of 143lb/in helpers with 627lb/in mains, you need to first calculate the effective rate of these two combined springs. That's done with this formula:
effective rate = (main spring x tender spring) / (main spring + tender spring)
So we have:
effective rate = (627 x 143) / (627 + 143)
effective rate = 89661 / 770
effective rate = 116lb/in
So the spring stack of 143/627 springs will act like a 116lb/in spring until the helper spring is fully compressed and then the main spring will take over at a rate of 627lb/in.
Consider the weight on each wheel and you'll see that under most cases the helper springs will be fully compressed at rest and when you are driving over bumps where the spring stack can get a bit unloaded, you will feel the low rate of 116lb/in in this example.
Cheers,
11-10-2010, 06:40 AM
#36
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You may also use this math for calculating two springs in series : http://scienceworld.wolfram.com/phys...l?affilliate=1
And talking about it we have to introduce the wheel rate and motion ratio into the discussion:
FYI For GT3 is around 1 for the front and about 0.75 for the rear end.
Spring and Wheel Rates
The relationship is not always as you'd expect...
by Julian Edgar
People modifying their car suspension often talk about spring rates. They might say: “I’ve got 600 pound springs in the front and 400 pound springs in the back.”
At this point someone else might well say: “Hell, that’s hard. I run only 300 pound springs back and front and my car’s heavier than yours.”
Trouble is, comparing spring rates between different cars is useless. That’s because it’s the wheel rates that matter, not the spring rates. Confused? Let’s take a look at the relationship between spring and wheel rates.
Spring Rates
A car spring is specified according to how much it deflects for a given mass placed on top of it. For example, a 400 pounds/inch spring compresses by 1 inch when 400 pounds is placed on top. (Or a 20 kg/mm spring compresses by 1mm when 20kg is placed on top.)
Most springs are linear devices, so if you double this mass, the spring compresses by twice as much.
Wheel Rates
Wheel rates are expressed in the same units as spring rates. For example, say the car weighs 3000 pounds and this mass is divided equally between all four wheels, giving a static load of 750 pounds per wheel. When you lower the car onto its suspension, you find the wheel rises by 3 inches as it supports the weight. 750/3 gives a wheel rate of 250 pounds/inch.
(However, see the ‘Preload?’ box below.)
Motion Ratios
So what’s the relationship between spring and wheel rates? That depends entirely on the leverage ratio built into the suspension. Most suspension systems do not have a 1:1 relationship between wheel movement and spring movement. In semi-trailing arms, wishbones, multi-links and many others, the spring:wheel movement ratio is not 1:1. (In strut systems it is close to 1:1.) In most cases, the wheel moves further than the spring.
The relationship between the wheel movement and the spring compression is called the motion ratio. It can be measured in two ways. The first way is to measure the suspension arm, working out its leverage ratio by looking at where the pivot point is relative to the wheel and the spring. However, it’s easy to make measuring errors in locating the centre of pivot points and complex suspension designs get very hard to analyse like this.
A better way is to carefully measure the wheel travel and the spring travel, preferably over small increments. (Remove the spring and/or the damper to do this and use a jack to move the wheel through its travel.) If the wheel moves twice as far as the spring, the motion ratio is 2:1. Note that motion ratios are seldom neat numerical figures like this – it might be 2.2:1 or 1.87:1. The motion ratio might even change through the range of wheel travel and an average may need to be taken.
So how do you use this motion ratio to get the relationship in pounds/inch between the spring and the wheel?
Spring:Wheel Rate Relationship
Here’s the key point: the relationship between the wheel and spring rates is worked out by squaring the motion ratio.
So if the motion ratio is 2:1, the relationship between the spring rate (eg in pounds/inch) and the wheel rate (also in pounds/inch) is 4:1 (2 x 2 = 4). If the motion ratio is 1.87:1, the spring:wheel rate relationship is 3.5:1 (1.87 x 1.87 = 3.4969).
Taking as an example the motion ratio of 1.87:1 (giving a 3.5:1 spring:wheel rate relationship), a wheel rate of 250 pounds/inch will require a spring rate of 875 pounds/inch. But if the motion ratio is 2:1 (and so the spring:wheel rate relationship is 4:1), the required spring will be 1000 pounds/inch!
You can see that small changes in motion ratio make for large changes in the spring:wheel rate relationship. That’s why it’s pointless comparing the spring rates used across different cars – unless of course those different cars have exactly the same motion ratios in their suspensions.
Modifying Springs
To a great extent you don’t need to worry about motion ratios if you have a car that already has springs. Irrespective of the motion ratio, upgrading the rate of a spring by (say) 30 per cent will increase the wheel rate by 30 per cent also. (However, a 1 inch shorter spring will have a much greater affect on ride height if the suspension uses a high leverage ratio – again it’s best to use a percentage change.) But even a cursory understanding of the affect of motion ratios will give you two advantages:
*
- You won’t try to compare spring rates from different suspension designs
*
- You’ll know why some springs are so much stronger than other springs, even though the car weights and handling outcomes are similar
If you’re designing a suspension from scratch, keep in mind that while a high motion ratio has some advantages, it will require a very strong spring that will be heavy and, for a given diameter, have less available travel before coil-binding occurs (because of its use of thicker wire).
Preload?
Note that how much the car settles on its suspension will depend not only on wheel rate but also on the spring pre-load, if any. In some suspension systems (eg struts), the springs are compressed and then held captive in that compressed state. Pre-load holds the spring in place when the strut reaches full extension (which occurs when the car is jacked up or the wheels leave the road) but it also means that the car does not settle to the extent that you might first expect when the weight is on the springs.
For example, a spring might have a rate of 100 pounds/inch and a free length of 15 inches. When it is mounted on the strut it is compressed to (say) 11.6 inches, and so the pre-load is 3.4 inches. To compress the spring by each inch takes 100 pounds, and so the compressive force acting on the spring when it is on the strut must be 340 pounds (3.4 inches x 100 pounds per inch rate). If there are 665 pounds acting through that corner of the car when the car is stationary, the spring will settle only 3.25 inches, because the first 340 pounds of the car’s weight is needed just to overcome the pre-load. If there was no pre-load, the car would settle by 6.65 inches (665 pounds divided by the 100 pounds/inch spring rate).
And talking about it we have to introduce the wheel rate and motion ratio into the discussion:
FYI For GT3 is around 1 for the front and about 0.75 for the rear end.
Spring and Wheel Rates
The relationship is not always as you'd expect...
by Julian Edgar
People modifying their car suspension often talk about spring rates. They might say: “I’ve got 600 pound springs in the front and 400 pound springs in the back.”
At this point someone else might well say: “Hell, that’s hard. I run only 300 pound springs back and front and my car’s heavier than yours.”
Trouble is, comparing spring rates between different cars is useless. That’s because it’s the wheel rates that matter, not the spring rates. Confused? Let’s take a look at the relationship between spring and wheel rates.
Spring Rates
A car spring is specified according to how much it deflects for a given mass placed on top of it. For example, a 400 pounds/inch spring compresses by 1 inch when 400 pounds is placed on top. (Or a 20 kg/mm spring compresses by 1mm when 20kg is placed on top.)
Most springs are linear devices, so if you double this mass, the spring compresses by twice as much.
Wheel Rates
Wheel rates are expressed in the same units as spring rates. For example, say the car weighs 3000 pounds and this mass is divided equally between all four wheels, giving a static load of 750 pounds per wheel. When you lower the car onto its suspension, you find the wheel rises by 3 inches as it supports the weight. 750/3 gives a wheel rate of 250 pounds/inch.
(However, see the ‘Preload?’ box below.)
Motion Ratios
So what’s the relationship between spring and wheel rates? That depends entirely on the leverage ratio built into the suspension. Most suspension systems do not have a 1:1 relationship between wheel movement and spring movement. In semi-trailing arms, wishbones, multi-links and many others, the spring:wheel movement ratio is not 1:1. (In strut systems it is close to 1:1.) In most cases, the wheel moves further than the spring.
The relationship between the wheel movement and the spring compression is called the motion ratio. It can be measured in two ways. The first way is to measure the suspension arm, working out its leverage ratio by looking at where the pivot point is relative to the wheel and the spring. However, it’s easy to make measuring errors in locating the centre of pivot points and complex suspension designs get very hard to analyse like this.
A better way is to carefully measure the wheel travel and the spring travel, preferably over small increments. (Remove the spring and/or the damper to do this and use a jack to move the wheel through its travel.) If the wheel moves twice as far as the spring, the motion ratio is 2:1. Note that motion ratios are seldom neat numerical figures like this – it might be 2.2:1 or 1.87:1. The motion ratio might even change through the range of wheel travel and an average may need to be taken.
So how do you use this motion ratio to get the relationship in pounds/inch between the spring and the wheel?
Spring:Wheel Rate Relationship
Here’s the key point: the relationship between the wheel and spring rates is worked out by squaring the motion ratio.
So if the motion ratio is 2:1, the relationship between the spring rate (eg in pounds/inch) and the wheel rate (also in pounds/inch) is 4:1 (2 x 2 = 4). If the motion ratio is 1.87:1, the spring:wheel rate relationship is 3.5:1 (1.87 x 1.87 = 3.4969).
Taking as an example the motion ratio of 1.87:1 (giving a 3.5:1 spring:wheel rate relationship), a wheel rate of 250 pounds/inch will require a spring rate of 875 pounds/inch. But if the motion ratio is 2:1 (and so the spring:wheel rate relationship is 4:1), the required spring will be 1000 pounds/inch!
You can see that small changes in motion ratio make for large changes in the spring:wheel rate relationship. That’s why it’s pointless comparing the spring rates used across different cars – unless of course those different cars have exactly the same motion ratios in their suspensions.
Modifying Springs
To a great extent you don’t need to worry about motion ratios if you have a car that already has springs. Irrespective of the motion ratio, upgrading the rate of a spring by (say) 30 per cent will increase the wheel rate by 30 per cent also. (However, a 1 inch shorter spring will have a much greater affect on ride height if the suspension uses a high leverage ratio – again it’s best to use a percentage change.) But even a cursory understanding of the affect of motion ratios will give you two advantages:
*
- You won’t try to compare spring rates from different suspension designs
*
- You’ll know why some springs are so much stronger than other springs, even though the car weights and handling outcomes are similar
If you’re designing a suspension from scratch, keep in mind that while a high motion ratio has some advantages, it will require a very strong spring that will be heavy and, for a given diameter, have less available travel before coil-binding occurs (because of its use of thicker wire).
Preload?
Note that how much the car settles on its suspension will depend not only on wheel rate but also on the spring pre-load, if any. In some suspension systems (eg struts), the springs are compressed and then held captive in that compressed state. Pre-load holds the spring in place when the strut reaches full extension (which occurs when the car is jacked up or the wheels leave the road) but it also means that the car does not settle to the extent that you might first expect when the weight is on the springs.
For example, a spring might have a rate of 100 pounds/inch and a free length of 15 inches. When it is mounted on the strut it is compressed to (say) 11.6 inches, and so the pre-load is 3.4 inches. To compress the spring by each inch takes 100 pounds, and so the compressive force acting on the spring when it is on the strut must be 340 pounds (3.4 inches x 100 pounds per inch rate). If there are 665 pounds acting through that corner of the car when the car is stationary, the spring will settle only 3.25 inches, because the first 340 pounds of the car’s weight is needed just to overcome the pre-load. If there was no pre-load, the car would settle by 6.65 inches (665 pounds divided by the 100 pounds/inch spring rate).
11-10-2010, 11:14 AM
#37
Three Wheelin'
Finally, a discussion of wheel rates and motion ratios for the GT3. I have been looking for this information for quite some time.
Thanks John! Isn't that the motion ratio you posted? (f=1/r=.75)
In another week we will be onto suspension frequency....
Thanks John! Isn't that the motion ratio you posted? (f=1/r=.75)
In another week we will be onto suspension frequency....
11-10-2010, 12:25 PM
#38
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You are welcome.
Yes it is approximately, and how nice it is to start a thread with "Guys I have a Quick suspension question" and to end up talking about suspension frequency.
Proceed with caution : )
11-10-2010, 01:10 PM
#39
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11-10-2010, 01:30 PM
#40
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An excellent book for petrolheads that like suspension beyond therapy is "Race Car Vehicle Dynamics" by William F. Milliken & Douglas L. Milliken.
It's expensive but worth every penny.
It's expensive but worth every penny.
11-10-2010, 03:21 PM
#41
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Seriously, all the extra info is great, but I'm still struggling with what springs I should opt for.
-td
11-10-2010, 04:47 PM
#42
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Unless you're doing this all yourself, your shop should take care of it for you.
...otherwise....
Like I said before, oem free lengths are as follows:
front: 9-3/16" (233 mm)
rear: 9-5/8" (244 mm)
Look at the way your car is right now and see where the lower spring perch is located on the threaded shock body. If you are near the top or bottom, then err on the side of caution to make sure your new spring stack will give you enough leeway to adjust the ride height.
Front: I'd try 160mm mains with 60mm tenders and assume the spacer will take up a few mm.
Rear: If you have the space, I'd also go with 160mm mains and 60mm tenders here.
...if I guessed wrong then it would be obvious after installation and I'd order new main springs.
Consolation: 160mm lengths are fine for the new Moton line so *when* you upgrade to Motons you'll be set. 
In fact, wouldn't Motons be much easier about now? No guesswork because many folks have done this.
Cheers,
...otherwise....
Like I said before, oem free lengths are as follows:
front: 9-3/16" (233 mm)
rear: 9-5/8" (244 mm)
Look at the way your car is right now and see where the lower spring perch is located on the threaded shock body. If you are near the top or bottom, then err on the side of caution to make sure your new spring stack will give you enough leeway to adjust the ride height.
Front: I'd try 160mm mains with 60mm tenders and assume the spacer will take up a few mm.
Rear: If you have the space, I'd also go with 160mm mains and 60mm tenders here.
...if I guessed wrong then it would be obvious after installation and I'd order new main springs.
Consolation: 160mm lengths are fine for the new Moton line so *when* you upgrade to Motons you'll be set. In fact, wouldn't Motons be much easier about now? No guesswork because many folks have done this.
Cheers,
