Mugen MTX-3
06-06-2003, 04:50 PM
#4876
Originally posted by eddiethefish
Hello TSR6,
There is no altered on the spring in any ways. I would just like to know the differences among the two spring brand, see attached pictures--left is FiveStars, the other is Mugen. They are the same in length/height, have the same thickness. The only difference is the number of coils. Please advise.
Regards,
eddie
Hello TSR6,
There is no altered on the spring in any ways. I would just like to know the differences among the two spring brand, see attached pictures--left is FiveStars, the other is Mugen. They are the same in length/height, have the same thickness. The only difference is the number of coils. Please advise.
Regards,
eddie
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
06-06-2003, 04:55 PM
#4877
Originally posted by TSR6
It's the same thing.
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
It's the same thing.
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
Thank you. Here's the picture of my twins!
06-06-2003, 05:02 PM
#4878
Tech Apprentice
Originally posted by TSR6
It's the same thing.
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
It's the same thing.
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
This can also be used as a tuning aid. You can use stock springs and cut 1/2 to full coils off of them to alter the stiffness of them.
06-06-2003, 05:04 PM
#4879
Tech Apprentice
Sorry, you guys snuck in there before I hit submit.
06-06-2003, 05:19 PM
#4880
Originally posted by PMT
Eddie, TSR6 is right on this one. Basicaly if a spring is the same length/height as the stock one but less coils it means that the wire that it is made out of is shorter. So even though the diameters of the wires are the same it is going to be stiffer because it is shorter. Example, take a pen, hold it at both ends and you can probably bend it easily. Hold it in the middle with your fingers about and inch apart and it is going to be hard to bend. Same principale applies to the wire that the springs are made out of, longer/more coils=softer (easier to bend), shorter/less coild=stiffer (harder to bend).
This can also be used as a tuning aid. You can use stock springs and cut 1/2 to full coils off of them to alter the stiffness of them.
Eddie, TSR6 is right on this one. Basicaly if a spring is the same length/height as the stock one but less coils it means that the wire that it is made out of is shorter. So even though the diameters of the wires are the same it is going to be stiffer because it is shorter. Example, take a pen, hold it at both ends and you can probably bend it easily. Hold it in the middle with your fingers about and inch apart and it is going to be hard to bend. Same principale applies to the wire that the springs are made out of, longer/more coils=softer (easier to bend), shorter/less coild=stiffer (harder to bend).
This can also be used as a tuning aid. You can use stock springs and cut 1/2 to full coils off of them to alter the stiffness of them.
I'll have to bring my softer one, [COLOR=red]RED, just in case I need to soften the rear.
06-06-2003, 09:08 PM
#4881
Originally posted by TSR6
It's the same thing.
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
It's the same thing.
With a two springs of the same dia. wire, the spring with the less coils will be stiffer.
You could work it out - a bit fiddly though. The thickness of the paint makes it practically impossible to get an accurate reading of the wire diameter.
Measure the diameter of the wire, the number of free coils, the outside diameter of the coils and apply this:
p = (G * F * d^4 ) / ( 8 * N * ( D - d )^3 )
p = rate in lbf/in
G = torsional modulus of elasticity = 11.5 * 10 ^ 6 for most spring steels
d = wire diamter (inches)
N = number of working coils
D = outside diameter (inches)
06-06-2003, 09:55 PM
#4882
Originally posted by Taylor-Racing
Yep.
You could work it out - a bit fiddly though. The thickness of the paint makes it practically impossible to get an accurate reading of the wire diameter.
Measure the diameter of the wire, the number of free coils, the outside diameter of the coils and apply this:
p = (G * F * d^4 ) / ( 8 * N * ( D - d )^3 )
p = rate in lbf/in
G = torsional modulus of elasticity = 11.5 * 10 ^ 6 for most spring steels
d = wire diamter (inches)
N = number of working coils
D = outside diameter (inches)
Yep.
You could work it out - a bit fiddly though. The thickness of the paint makes it practically impossible to get an accurate reading of the wire diameter.
Measure the diameter of the wire, the number of free coils, the outside diameter of the coils and apply this:
p = (G * F * d^4 ) / ( 8 * N * ( D - d )^3 )
p = rate in lbf/in
G = torsional modulus of elasticity = 11.5 * 10 ^ 6 for most spring steels
d = wire diamter (inches)
N = number of working coils
D = outside diameter (inches)
06-06-2003, 09:56 PM
#4883
eddiethfish
can i order this part by phone (kawahara sway bar, five star alum shock stopper) cause i dont have that paypal
thank you
nelly
can i order this part by phone (kawahara sway bar, five star alum shock stopper) cause i dont have that paypal
thank you
nelly
06-06-2003, 10:08 PM
#4884
in the formula mentioned, what is meant by the wire diameter and outside diameter? If I understand it correctly, you want the actual wire diameter and the overall diameter (including the paint)? What does the overall diameter have to do with it since the only difference between it and the wire diameter is the thickness of teh paint and last time I checked, paint isn't significant enough to change the spring rate.
06-06-2003, 10:23 PM
#4885
Spring Rate = (Gd^4) / (8ND^3)
as said before, G is 11.25x10^6 for most steel springs
Wire Diameter, or D is measured with a caliper, and should be the same for the whole spring ( unless it is a progressive spring, but you cant use this equation on a progressive spring anyways... so if it is progressive...stop here. )
N - Active coils.. only the ones that are doing the compressing. You don't want to count the "flat" coil that sits on the sping cup.
D = Mean coil Diameter ( Racing Evo, this is for you ) - The mean coil diameter is the Outside diameter, minus the wire diameter (D).
as said before, G is 11.25x10^6 for most steel springs
Wire Diameter, or D is measured with a caliper, and should be the same for the whole spring ( unless it is a progressive spring, but you cant use this equation on a progressive spring anyways... so if it is progressive...stop here. )
N - Active coils.. only the ones that are doing the compressing. You don't want to count the "flat" coil that sits on the sping cup.
D = Mean coil Diameter ( Racing Evo, this is for you ) - The mean coil diameter is the Outside diameter, minus the wire diameter (D).
06-06-2003, 11:38 PM
#4886
Re: Question re Spring
Originally posted by eddiethefish
Hello,
I'm now using the FiveStars BLUE spring. When compare to the Mugen's Blue, I've noticed that the FiveStars has less coils. They both have almost the same thichness at 1.74mm in diameter and of course the same in height.
What would the less coils spring perform when compare to the Mugen's one?
Any better or worse?
Thank you,
Hello,
I'm now using the FiveStars BLUE spring. When compare to the Mugen's Blue, I've noticed that the FiveStars has less coils. They both have almost the same thichness at 1.74mm in diameter and of course the same in height.
What would the less coils spring perform when compare to the Mugen's one?
Any better or worse?Thank you,
06-06-2003, 11:44 PM
#4887
Re: Re: Question re Spring
Originally posted by Data
just go by the spring rate of the springs you got (if it is specified). since you are not going to modify the spring, don't worry about # of coils or wire diameter. not enough information is not good, too much information is not going to help either.
just go by the spring rate of the springs you got (if it is specified). since you are not going to modify the spring, don't worry about # of coils or wire diameter. not enough information is not good, too much information is not going to help either.
I used the cut-spring example to show that less coils = harder spring. I really wasn't suggesting to cut the springs, although it is an option...
06-07-2003, 04:31 AM
#4888
Originally posted by Manticore
great information. can you illustrate this with an example?
great information. can you illustrate this with an example?
http://faq.f650.com/FAQs/SpringRateFAQ.html
06-07-2003, 04:50 AM
#4889
Originally posted by TSR6
Spring Rate = (Gd^4) / (8ND^3)
as said before, G is 11.25x10^6 for most steel springs
Wire Diameter, or D is measured with a caliper, and should be the same for the whole spring ( unless it is a progressive spring, but you cant use this equation on a progressive spring anyways... so if it is progressive...stop here. )
N - Active coils.. only the ones that are doing the compressing. You don't want to count the "flat" coil that sits on the sping cup.
D = Mean coil Diameter ( Racing Evo, this is for you ) - The mean coil diameter is the Outside diameter, minus the wire diameter (D).
Spring Rate = (Gd^4) / (8ND^3)
as said before, G is 11.25x10^6 for most steel springs
Wire Diameter, or D is measured with a caliper, and should be the same for the whole spring ( unless it is a progressive spring, but you cant use this equation on a progressive spring anyways... so if it is progressive...stop here. )
N - Active coils.. only the ones that are doing the compressing. You don't want to count the "flat" coil that sits on the sping cup.
D = Mean coil Diameter ( Racing Evo, this is for you ) - The mean coil diameter is the Outside diameter, minus the wire diameter (D).
oww no don't enter the realm of maths and physics please!
sometimes it's easier if you just say to yourself: "it is, becouse it is"
get a pen spring compress it. soft. now stretch it out. it has now become harder. same principle.
06-07-2003, 05:06 AM
#4890
Originally posted by Racing4Evo
in the formula mentioned, what is meant by the wire diameter and outside diameter? If I understand it correctly, you want the actual wire diameter and the overall diameter (including the paint)? What does the overall diameter have to do with it since the only difference between it and the wire diameter is the thickness of teh paint and last time I checked, paint isn't significant enough to change the spring rate.
in the formula mentioned, what is meant by the wire diameter and outside diameter? If I understand it correctly, you want the actual wire diameter and the overall diameter (including the paint)? What does the overall diameter have to do with it since the only difference between it and the wire diameter is the thickness of teh paint and last time I checked, paint isn't significant enough to change the spring rate.
The "outside diameter" is the overall diameter of the whole spring.
The reason the formula wants these two measurements is that it calculates the "mean" diameter of the spring - that is, it calculates the spring diameter as measured from the centre of the wire on one side to the centre of the wire on the other side.
. . . and the reason it does that is because springs work with torsional loads - that is, as the spring compresses, the wire is actually twisted about the certre point of the wire's cross-sectional diameter. This is where the modulus of elasticity comes into it. As you might imagine, aluminium springs, having a lower modulus of elasticity, would not be as strong for the same diameter of material.
No, paint is not a significant factor in the spring rate.
The issue here, is that the paint on some springs is quite thick, and measurement, therefore becomes problematic. When we are talking about wire diameters of about 1.7mm, I wouldn't be surprised to see paint thicknesses of .5mm if you include both sides of the wire - that's a significant error being introduced.
Hope this helps.