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Wales
by Joske Vermeule, September 1
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Forums39
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Posts838,211
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Jun 14th, 2026
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Joined: Jul 2007
Posts: 28,914 Likes: 354
Salty Sea Dog Member of the Inner Circle
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Salty Sea Dog Member of the Inner Circle
Joined: Jul 2007
Posts: 28,914 Likes: 354 |
Graham (G4FUJ)
Sold L44FOR 4/4 Giallo Fly
'11 MINI Countryman Cooper D All4
'90 LR 90 SW
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Joined: Dec 2011
Posts: 381
Learner Plates Off!
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Learner Plates Off!
Joined: Dec 2011
Posts: 381 |
Surely if you cut the lower spring if it's not free you might find the static height higher?
Nick
You are right! But by cutting the rebound spring its stiffness increases. In my case I would have to shorten my 90mm rebound spring to 45mm multiplying its stiffness by 2, but compression stroke will be longer.
Regards
2005 PLUS 4
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Joined: Sep 2014
Posts: 2,272 Likes: 7
Talk Morgan Expert
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Talk Morgan Expert
Joined: Sep 2014
Posts: 2,272 Likes: 7 |
I was talking to a Morgan Racer the other day. "Bill, I do not expect Morgan handling to conform to Me. I make every effort to get the best performance as it is, not as I wish it was. I have raced against too many other Marques to not know that Morgan handling is very good and I am competitive against them." Bill That kind of makes the point. On a racetrack there is generally a smooth surface and the most important factor is handling. Ride quality is least important and road holding is much less important than on "normal" roads. These mods to the rebound spring (be they the SSL kit, or a more DIY approach) gives you much better ride quality, so a long trip is less tiring on the body, better road holding because the wheels spend more time in contact with the tarmac rather than in the air, and the handling is about the same. For me it is a no-brainer, but I fully accept others have a different opinion.
Andy G
1999 +8 , Indigo Blue.
Ex-John McKecknie/Mike Duncan 1955 +4 racer.
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Joined: Aug 2017
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Talk Morgan Regular
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OP
Talk Morgan Regular
Joined: Aug 2017
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Seems like it is time to get the hack saw. But how can shortening the spring make it stiffer? If the spring rate is say 275 lb/inch and is linear, would not this be the same with a spring of two yards as opposed to two inches? And I would believe it takes more - not less - total force to totally compress two yards, as you have more inches to compress. Or am I still being blonde?
Robbie the Norseman
2004 V6 Roadster
Sherwood green
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Joined: Apr 2008
Posts: 12,268 Likes: 324
Scruffy Oik Member of the Inner Circle
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Scruffy Oik Member of the Inner Circle
Joined: Apr 2008
Posts: 12,268 Likes: 324 |
Seems like it is time to get the hack saw. But how can shortening the spring make it stiffer? If the spring rate is say 275 lb/inch and is linear, would not this be the same with a spring of two yards as opposed to two inches? And I would believe it takes more - not less - total force to totally compress two yards, as you have more inches to compress. Or am I still being blonde? The stiffness of a coil spring is, among other things like coil diameter, a function of the number of coils. Imagine a straight length of spring steel with one end clamped in a bench vice. Push on the free end with a set force and it will deflect. Now shorten the length by half and apply the same force. Will it deflect by the same amount? Take the original length of spring steel and wind it into a coil around a central former. Count the number of coils. Now take the half length and wind that around the same former so the coils are the same diameter. Count the number of coils. You have now proved that for the same material, fewer coils = shorter length = higher force needed to deflect by a set amount.
Tim H.
1986 4/4 VVTi Sport, 2002 LR Defender, 2022 Mini Cooper SE
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Joined: Apr 2017
Posts: 1,799 Likes: 4
Talk Morgan Enthusiast
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Talk Morgan Enthusiast
Joined: Apr 2017
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Robbie
A lot of people struggle with that calculation but as Tim said when you reduce the number of coils you reduce the overal length of the spring wire and therefore the deflection for a given load reduces. I.e the rate goes up.
If you look at the formula for calculating coil spring rate it includes the number of coils and it is an inverse ratio.
So fewer coils equals higher rate.
Bob
2009 Black Roadster
1999 4/4 2 litre Zetec
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Joined: Jul 2011
Posts: 1,246 Likes: 30
Has a lot to Say!
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Has a lot to Say!
Joined: Jul 2011
Posts: 1,246 Likes: 30 |
If you have a linear coil spring of say 140lbs. It takes 140lbs to deflect it 1 inch. Question. How many lbs to deflect it 2 inches ?
I thought i’d Got my head around Peters Ballards explanatory write up on GoMoG. Now thanks to this thread I am not so sure.
4/4 Ivory 4.1:1 axle, Mercedes A200 AMG
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Joined: Dec 2011
Posts: 1,943
Talk Morgan Enthusiast
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Talk Morgan Enthusiast
Joined: Dec 2011
Posts: 1,943 |
Depending on your car, the rebound spring may be free or not at static state. On my PLUS 4 2005 the rebound spring was not free.
In this case the stiffness of the suspension was very hard on small bump and after becomes soft on big bump, the opposite of a conventional suspension.
When the rebound spring is not free at static state, the gobal stiffness of the suspension is the sum of the both spring stiffness and determines the static height of the car. If you cut the rebound spring and get it free, the stiffness will be lower because only the main spring acts and you may get a lower static height.
I have installed the SSL front kit, in this case you can preload the main spring like in the conventional cars and get a free rebound spring. The suspension on small bumps is soft and increasing stiffness on bigger bump. You get quite a conventional suspension.
During the assembly with the SSL kit the global preload of the springs is lower, preload being adjusted and increased after assembly. Assembly is easier, main spring being free at the beginning. Surely if you cut the lower spring if it's not free you might find the static height higher? Nick Not much. Doubt You could measure it.
Button
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Joined: Aug 2017
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Talk Morgan Regular
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OP
Talk Morgan Regular
Joined: Aug 2017
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Think I understand now. If you compress a linear spring you have to add the same amount of force for each inch of compression. So in IcePack's case, 140 lbs for one inch, and another 140 for the next and so on. So that is the property of that given spring. But if you shorten it but otherwise leave it unchanged, you will get another spring, with different inherent properties. Still linear, but at another rate. Right?
Robbie the Norseman
2004 V6 Roadster
Sherwood green
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Joined: Apr 2008
Posts: 12,268 Likes: 324
Scruffy Oik Member of the Inner Circle
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Scruffy Oik Member of the Inner Circle
Joined: Apr 2008
Posts: 12,268 Likes: 324 |
If you have a linear coil spring of say 140lbs. It takes 140lbs to deflect it 1 inch. Question. How many lbs to deflect it 2 inches ?
I thought i’d Got my head around Peters Ballards explanatory write up on GoMoG. Now thanks to this thread I am not so sure. Part of the confusion comes from people (not just you) being a bit sloppy in their use of terminology. A "140 lb" spring is actually a "140lb/inch" spring. For every inch of deflection requires a force of 140lb. So 140 lb will compress it 1 inch, 280lb will compress it 2 inches, and so on.
Tim H.
1986 4/4 VVTi Sport, 2002 LR Defender, 2022 Mini Cooper SE
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