Question about Lowering Springs
Thats not what I'm saying.
I'm saying that because the spring has less time and one less coil of resistance to it, the suspension becomes easier to bottom out, less restistance.
Your saying that by cutting the one spring the spring rate increases a proportional ammount by some magic, and the car only bottoms out because you hitting bumps harder. (Proportionally higher spring rate would mean the suspension would move the same ammount as before over same bump.)
I'm saying that because the spring has less time and one less coil of resistance to it, the suspension becomes easier to bottom out, less restistance.
Your saying that by cutting the one spring the spring rate increases a proportional ammount by some magic, and the car only bottoms out because you hitting bumps harder. (Proportionally higher spring rate would mean the suspension would move the same ammount as before over same bump.)
Last edited by DriftFB; Nov 6, 2006 at 01:08 AM.
Originally Posted by GavinJuice
it's magic 

open the calculator he posted there, and change the ammount of coils from 10 to 5, then hit the "modulus of rigidity" button and the number will be cut in half. Pure, unbelievable magic!
That means that with half the coils, it takes half the work to fully compress the spring.
Thank you for posting the link, I now have science on my side. Do any of you guys feel like rewriting science?
The modulus of rigidity is a property of the material used, and doesn't change when you cut the spring.
However, I generally agree with you that cutting the springs makes a car more prone to bottoming. I don't think arguments about spring rates alone hold the key to unravelling this mystery.
-Max
However, I generally agree with you that cutting the springs makes a car more prone to bottoming. I don't think arguments about spring rates alone hold the key to unravelling this mystery.
-Max
I made the same bonehead error this time that I made the last time the subject of spring cutting came up. It wasn't until I realised that a coil spring is simply a torsion bar that is wound. As the spring compresses, it twists through it's center line, the same way a torsion bar does. If you shorten a torsion bar, it takes more force to twist it, same goes for the spring, only this translates into compression, through the length of it.
according to the machinist handbook (25th edition, pg 299-301 for those who are wondering), the formulas do support that the deflection per coil does not change as the length changes for a said spring under a given load. therefore...as the spring is shortened, the springrate does increase.
im not about to write out the formulas and method on figuring this out though...
im not about to write out the formulas and method on figuring this out though...
Originally Posted by DriftFB
But you see the springs are doing the same amount of work, thats why cars with cut springs bottom out so easily. A spring, reguardless of its lb rating can only support so much weight before the sping is compressed flat and it has no where else to go. And you guys are saying that if you take out 1/2 spring it won't take 1/2 the weight to get the spring flat..
Very true REVHED. I had to trim my bump stops after installing the Eibachs. Remove the bump stops, one stands a chance of blowing out the shocks, not to mention the hard abrupt hit the body takes without the stops.
Just a wild guess, but wouldn't a car bottom out after cutting springs because the travel is so much shorter... and while the springs are now stiffer, they're not stiffer enough to make up for the lost range of travel?
Alright, so you guys get one result with the calculator and I get another, I checked and they're both right. But I still can't believe that this will happen with a spring.
So, the constant of delflection changes in your favor, but the modulus of rigidity changes in mine. Which one represents what we're talking about and why?
So, the constant of delflection changes in your favor, but the modulus of rigidity changes in mine. Which one represents what we're talking about and why?
Wow, Still trying to tell us were wrong.
Your calculation simply says that for a spring half the size to compress the same amount, it would have to be twice as easy to compress.
This doesnt prove your point, Just ours.
What you want to calculate is how much the spring will compress with half the coils, and as gavinjuice said much earlier it will take twice the force. This only works if the material stays the same, and in our case it does. This is where you lost track.
Look at K- which is the constant (which is pounds of load per inch of deflection.)
You were looking at a factor that wont change when you cut the spring in half. It wont magically become easier to compress.
Read over this thread again and maybe youll figure it out.
Your calculation simply says that for a spring half the size to compress the same amount, it would have to be twice as easy to compress.
This doesnt prove your point, Just ours.
What you want to calculate is how much the spring will compress with half the coils, and as gavinjuice said much earlier it will take twice the force. This only works if the material stays the same, and in our case it does. This is where you lost track.
Look at K- which is the constant (which is pounds of load per inch of deflection.)
You were looking at a factor that wont change when you cut the spring in half. It wont magically become easier to compress.
Read over this thread again and maybe youll figure it out.
Last edited by 1badFB; Nov 6, 2006 at 01:31 PM.
To get some practical use out of this discussion
-- my son and I are working on a '66 Mustang. He replaced the stamped A-arms with the welded steel tubing ones, and put in new springs. The front end is sitting a little high, so we have to cut the springs (in 1/4 coil increments) until we get it right. Will this make the front end stiffer or not?
Thanks - David
-- my son and I are working on a '66 Mustang. He replaced the stamped A-arms with the welded steel tubing ones, and put in new springs. The front end is sitting a little high, so we have to cut the springs (in 1/4 coil increments) until we get it right. Will this make the front end stiffer or not?Thanks - David
Originally Posted by DavidMyers53
To get some practical use out of this discussion
-- my son and I are working on a '66 Mustang. He replaced the stamped A-arms with the welded steel tubing ones, and put in new springs. The front end is sitting a little high, so we have to cut the springs (in 1/4 coil increments) until we get it right. Will this make the front end stiffer or not?
Thanks - David
-- my son and I are working on a '66 Mustang. He replaced the stamped A-arms with the welded steel tubing ones, and put in new springs. The front end is sitting a little high, so we have to cut the springs (in 1/4 coil increments) until we get it right. Will this make the front end stiffer or not?Thanks - David
Thats what this whole discussion is about.
Wow! I didn't think it was going to be a big subject. Well guys I was looking more toward handeling and performance. I notices a few sprints for a good price now my question is this. Should I upgrade the entire suppension system? Shocks, Sprints, etc..? If so then I will have to wait until I am more finacialy set to do a job like that.
Originally Posted by Secondmessiah
i got a five on my AP physics test and i can say that cutting a spring shorter will not appreciably increase the spring rate
The spring rate comparason is listed at the bottom of this page:
http://www.mazdatrix.com/h4579-85.htm
Originally Posted by Secondmessiah
i got a five on my AP physics test and i can say that cutting a spring shorter will not appreciably increase the spring rate
hahah.... guess that shows the true value of AP courses....
...good luck in university...
lmao
*For those who are still can't follow, k is the spring constant (ie. constant for THE SPRING) G is a MATERIAL PROPERTY. So, keeping G, d, and D constant, and halving the number of coils will double k, the SPRING RATE.... simple enough, right?
...don't worry about how it works... just know that it does...
*And for Secondmessiah, here's an example for physics class:
380lb/in spring with 2in D, 0.3002in d, 10 coils
now you cut off one spring (just one coil... thats it), and the spring rate goes to 423lb/in... ...I dunno what they call appreciable in your physics classes, but thats sure as hell appreciable in engineering....
Matt
Originally Posted by DJAngelicon
Wow! I didn't think it was going to be a big subject. Well guys I was looking more toward handeling and performance. I notices a few sprints for a good price now my question is this. Should I upgrade the entire suppension system? Shocks, Sprints, etc..? If so then I will have to wait until I am more finacialy set to do a job like that.





