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Rob 930 Rob 930 is offline
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Join Date: Dec 2002
Location: Seattle
Posts: 661
In the let's-totally-beat-this-to-death department:


Quote:
Originally Posted by TimT View Post
+1

There was a little misinformation regarding springs posted above

Analogy to a spring... say you have a lever 4 feet long you are using to lift something....

If you cut that lever in half... so you now have a 2 foot lever... and it will take more force to lift that something..

A coil spring is just a big wound up lever.. cut a coil spring in half you have a shorter stiffer spring.... cut a coil of that same spring you still have a shorter stiffer spring of an unknown rate

Keep your .8 bar spring... and get a boost controller..
Tim, we agree with the basic message: that a shorter spring is a stiffer spring, but I bristled a bit at the suggestion that I had given misinformation. So, to nitpick a bit here, the helical spring is analogous to a torsion bar, not a lever. That can be seen from the formula for spring stiffness which is:

k = d^4 * G / (8 * D^3 * N)

Where
d = wire diameter
G = Torsional modulus of rigidity
D = mean diameter of coil
N = Number of active coils

It can be seen from the above equation that the spring stiffness is a function of G, which is a property of torsion. But more interesting (and relevant) is that this formula shows that for the same construction, a shorter spring will be stiffer. If the number of active coils (N) decreases, the spring stiffness k will increase. (Note that a shorter spring has a lower N).

But I was more concerned about the statement by 911nut, which had elements of truth but I think could be misleading:
"Hooke's law states that a spring applies a force that is linearly proportional to the distance of it's equilibrium length. In other words, take two different lengths of springs made from the same steel and wire diameter. Compress the springs to an equal distance the longer spring applies a greater force than a shorter one.
That's why longer and longer springs with the same wire diameter are used in the stock wastegate to raise the pressure at which the wastegate opens. The longer spring is crammed into the chamber above the diaphragm. Since the longer spring has been displaced a greater amount than a shorter one, the force applied by the spring to the diaphragm is greater. "
Hooke's law is:

F = k * x

Where
F = spring force (lb)
k = spring stiffness (lb/in)
x = spring deflection (in)

From Hooke's law, it is clear that the greater the deflection of a spring, the greater the force exerted by it. But if you compare the force exerted by two springs of different lengths, it's not clear which will exert more force. If the springs are of the same construction, then the longer one will have a lower stiffness, and if they're deflected the same amount, the shorter one will exert a greater force. If they're deflected different amounts, then Hooke's law will determine which exerts a greater force.

For the wastegate case, it is possible that the longer spring will exert more force than the shorter spring of the same construction, depending on how much more it must be deflected to fit into that space. But if you take that spring and shorten it, then it will be stiffer than it was originally, and Hooke's law must be used to determine the net effect.

Bottom line: shortening a spring increases its stiffness. That may be okay, as long as you understand that and account for it. But I wouldn't be doing that with a wastegate spring. And it sounds like you've already solved that issue.

I find this to be an interesting (and non-intuitive) artifact of physics. Geek, anyone?

Rob
Old 02-21-2009, 04:19 PM
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