http://www.ebay.com/itm/CoilOverSprin ... rmvSB=true
Thinking about buying this spring and cutting it in half to use on the rear. If cutting the spring in half changes the spring rate, should I get the 450lbs/in? so it would be 900lbs/in if cut in half?
alternative rear springs
alternative rear springs
Last edited by toylet on 29 Mar 2016 20:28, edited 1 time in total.
Re: alternative rear springs
That's correct. Ride height might be an issue. That spring at 9" may need no shimming for height, but it depends on what you're looking at for ride height. I only have data for a cut D50 spring, which was much softer.
Because when you spend a silly amount of money on a silly, trivial thing that will help you not one jot, you are demonstrating that you have a soul and a heart and that you are the sort of person who has no time for Which? magazine. – Jeremy Clarkson
Re: alternative rear springs
ordered will report back
Last edited by toylet on 29 Mar 2016 20:26, edited 1 time in total.
Re: alternative rear springs
UPDATE
I bought this 450 lb/in spring from Ebay in color red to match the front Eibach springs. $99 plus shipping
http://www.ebay.com/itm/CoilOverSprin ... SwQItUIwLe
Cut the 18" spring in half than cut them to 8.5" each and installed on car. Test drove and all I have to say is WOW. The car is a pleasure to drive now. It sits 6" ground to pinch in front and 6.5" from ground to rear pinch. Nice ride height and very little squat on accel. The best mod for the money on this project.
I bought this 450 lb/in spring from Ebay in color red to match the front Eibach springs. $99 plus shipping
http://www.ebay.com/itm/CoilOverSprin ... SwQItUIwLe
Cut the 18" spring in half than cut them to 8.5" each and installed on car. Test drove and all I have to say is WOW. The car is a pleasure to drive now. It sits 6" ground to pinch in front and 6.5" from ground to rear pinch. Nice ride height and very little squat on accel. The best mod for the money on this project.
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 SteveEdmonton
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 Location: Edmonton, AB
Re: alternative rear springs
Glad this worked out well for you.
However, I've been thinking about the way you described your plan, ever since you posted it. It doesn't seem logical to me that cutting a spring in half will change its rate.
I'm starting with the premise that spring rates are expressed in terms of pounds of tension per inch of spring material. Therefore, whether the spring is 8.5 inches long or 18 inches long, that rate remain constant. The total amount of energy stored within the entirety of the spring will indeed be halved by cutting the spring in half, but total energy isn't what we're talking about. The rate per inch will remain exactly the same.
Am I wrong?
However, I've been thinking about the way you described your plan, ever since you posted it. It doesn't seem logical to me that cutting a spring in half will change its rate.
I'm starting with the premise that spring rates are expressed in terms of pounds of tension per inch of spring material. Therefore, whether the spring is 8.5 inches long or 18 inches long, that rate remain constant. The total amount of energy stored within the entirety of the spring will indeed be halved by cutting the spring in half, but total energy isn't what we're talking about. The rate per inch will remain exactly the same.
Am I wrong?
'71 4door
'74 MGBGT
'04 Miata
'74 MGBGT
'04 Miata
Re: alternative rear springs
Cutting them does change the spring rate  as far as everything I've read. Unfortunately I do not have enough knowledge in physics to be able to tell you why  I'm sure someone on here has the answer….
Last edited by James on 19 Apr 2016 06:49, edited 1 time in total.
Finished is better than perfect......
Re: alternative rear springs
Finished is better than perfect......
Re: alternative rear springs
I just went off what I read.. it was a experiment and it worked out good. Any stiffer and It would not be very streetable car.
 SteveEdmonton
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 Posts: 399
 Joined: 27 Aug 2010 13:20
 Location: Edmonton, AB
Re: alternative rear springs
Aha thanks, James, that link helps a great deal, especially the final explanation by "Manny." Until then I didn't see the flaw in my logic but now I surely do. Guess that's what happens when I think too much about stuff I know too little about...
Here's his explanation:
In problems like this, it's easier to work with a specific case as opposed to the general one. Let's assume that the original spring has 12 coils. Then each of the cut springs would have 3 coils.
Let's say that applying a 12 N force to the original spring would compress it 12 cm. Now let's apply some force, Y, to the spring that was cut (has 3 coils). This force compresses the cut spring by 12 cm as well.
Here, our intuition tells us that Y should be much greater than 12 N since compressing a small spring by 12 cm should require more force than compressing a large spring by 12 cm. So how does this work and what is this force, Y?
Essentially we are compressing each coil by 1 cm in the original spring but we are compressing each coil by 4 cm in the cut spring.
Force, however, is transmitted across the entire spring, something that we frequently forget (we also forget this a lot when looking at a rope under tension and that the tension is the same at each at some point on the rope). So the force on the spring is the same at every point on the spring.
This means that 12 N of force is being applied to each of the coils in the original spring to compress each coil by 1 cm. In other words, 12 N of force are required to compress 12 coils by 12 cm in total and also to compress 1 coil by 1 cm in total. Yes, that's right, applying 12 N will cause 12 coils to compress by 12 cm but applying 1 N to 1 coil will not compress the coil by 1 cm; 12 N will compress 1 coil by 1 cm (this does not seem intuitive so let me rephrase it this way: 12 N compresses the original spring by 12 cm and one coil of the spring by 1 cm).
Then 48 N of force are required to compress 12 coils by 48 cm in total and also to compress 1 coil by 4 cm in total. Using this logic, 48 N of force would be required to compress 3 coils by 12 cm in total. So Y=48 N. When we disregard direction, the equation F=kx becomes F=kx. So k=F/x. Plugging in the values from the original spring gives k=1 N/cm. Plugging in the values from the cut spring gives k=4 N/cm. So yes, your teacher was wrong because his logic would give 1/4 N/cm not 4 N/cm but your friends were wrong too because k is not constant regardless of the length of the spring. And it makes sense intuitively: it's harder to compress a short spring by some distance than it is to compress a longer spring by that same distance.
Here's his explanation:
In problems like this, it's easier to work with a specific case as opposed to the general one. Let's assume that the original spring has 12 coils. Then each of the cut springs would have 3 coils.
Let's say that applying a 12 N force to the original spring would compress it 12 cm. Now let's apply some force, Y, to the spring that was cut (has 3 coils). This force compresses the cut spring by 12 cm as well.
Here, our intuition tells us that Y should be much greater than 12 N since compressing a small spring by 12 cm should require more force than compressing a large spring by 12 cm. So how does this work and what is this force, Y?
Essentially we are compressing each coil by 1 cm in the original spring but we are compressing each coil by 4 cm in the cut spring.
Force, however, is transmitted across the entire spring, something that we frequently forget (we also forget this a lot when looking at a rope under tension and that the tension is the same at each at some point on the rope). So the force on the spring is the same at every point on the spring.
This means that 12 N of force is being applied to each of the coils in the original spring to compress each coil by 1 cm. In other words, 12 N of force are required to compress 12 coils by 12 cm in total and also to compress 1 coil by 1 cm in total. Yes, that's right, applying 12 N will cause 12 coils to compress by 12 cm but applying 1 N to 1 coil will not compress the coil by 1 cm; 12 N will compress 1 coil by 1 cm (this does not seem intuitive so let me rephrase it this way: 12 N compresses the original spring by 12 cm and one coil of the spring by 1 cm).
Then 48 N of force are required to compress 12 coils by 48 cm in total and also to compress 1 coil by 4 cm in total. Using this logic, 48 N of force would be required to compress 3 coils by 12 cm in total. So Y=48 N. When we disregard direction, the equation F=kx becomes F=kx. So k=F/x. Plugging in the values from the original spring gives k=1 N/cm. Plugging in the values from the cut spring gives k=4 N/cm. So yes, your teacher was wrong because his logic would give 1/4 N/cm not 4 N/cm but your friends were wrong too because k is not constant regardless of the length of the spring. And it makes sense intuitively: it's harder to compress a short spring by some distance than it is to compress a longer spring by that same distance.
'71 4door
'74 MGBGT
'04 Miata
'74 MGBGT
'04 Miata

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Re: alternative rear springs
The diameter and the number of coils and the thickness of the wire, all affect the spring rate. The formula is...
11,250,000 X wire thickness (in inches)4
__________________________________
8 X number of coils X coil diameter (center to center) 3
11,250,000 X wire thickness (in inches)4
__________________________________
8 X number of coils X coil diameter (center to center) 3
"Nissan 'shit the bed' when they made these, plain and simple." McShagger510 on flattop SUs
Re: alternative rear springs
Here is an even simpler way to think about it.
Whether a spring is wound in a coil or remains a strait piece of bar stock, it will react basically the same. So a spring with more coils in longer. A longer bar required less force to deflect it a given amount, which is the same as a lower rate.
Another way to think about it; two springs each with a free height of 9" wound from the exact same piece of heat treated bar stock. Once spring has 7 coils, the other has 5. The one with 5 coils will effectively have shorter piece of bar stock and will effectively be stiffer that the one with 7 coils.
Make sense?
So in this case, you 450 lb spring at 18" will become a 900 lb spring at 9". Trimmed a bit more to 8.5" and lets say it's a 925 lb spring.
that friends and neighbors is a good auto X spring rate for a driven, weekend warrior car that needs to traverse real roads in between.
Nice work Toy, the only thing that could have made this better would be an actual verification of the spring rate as installed.
Thanks for sharing this with us.
Byron
Whether a spring is wound in a coil or remains a strait piece of bar stock, it will react basically the same. So a spring with more coils in longer. A longer bar required less force to deflect it a given amount, which is the same as a lower rate.
Another way to think about it; two springs each with a free height of 9" wound from the exact same piece of heat treated bar stock. Once spring has 7 coils, the other has 5. The one with 5 coils will effectively have shorter piece of bar stock and will effectively be stiffer that the one with 7 coils.
Make sense?
So in this case, you 450 lb spring at 18" will become a 900 lb spring at 9". Trimmed a bit more to 8.5" and lets say it's a 925 lb spring.
that friends and neighbors is a good auto X spring rate for a driven, weekend warrior car that needs to traverse real roads in between.
Nice work Toy, the only thing that could have made this better would be an actual verification of the spring rate as installed.
Thanks for sharing this with us.
Byron
Love people and use things,
because the opposite never works.
because the opposite never works.
 SteveEdmonton
 Supporter
 Posts: 399
 Joined: 27 Aug 2010 13:20
 Location: Edmonton, AB
Re: alternative rear springs
They say you learn something every day. Not always true in my experience, because they also say it's hard for old dogs to learn new tricks. But I've certainly learned something here, and appreciate the patience of all of you teachers. Thank you.
'71 4door
'74 MGBGT
'04 Miata
'74 MGBGT
'04 Miata
Re: alternative rear springs
I actually did something similar for my rear springs a few years ago.
I purchased a single Hyperco Coil 3" ID x18" tall x 400lbs spring and cut it in half. They can still be found pretty cheap https://www.amazon.ca/Hyperco1818E0400 ... B006H1DYPO
Here's how it rests at 9" with no spacers.
I purchased a single Hyperco Coil 3" ID x18" tall x 400lbs spring and cut it in half. They can still be found pretty cheap https://www.amazon.ca/Hyperco1818E0400 ... B006H1DYPO
Here's how it rests at 9" with no spacers.
'72 Datsun 510  original Ottawa car
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