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Quote:
Originally Posted by gsmith660 View Post
but I am still going to get the tial.
Good move.

Then for less than 30 bucks you can get a second spring calibrated from Tial.

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Glenn
87' 930TT
Old 02-21-2009, 04:48 AM
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Quote:
if you can point me to a site that will refute that I am willing to look.
quick search regarding coiled springs

http://www.eatonsprings.com/atqCuttingCoilSpringsCalculations.htm

and further, a java app where you input data from a spring, and find out k or spring rate.. we dont know the properties of a waste gate spring though...
note that if you remove coil from the spring system, k increases..

http://www.engineersedge.com/calculators/comp_spring_k_pop.htm

and

http://www.engineersedge.com/spring_comp_calc_k.htm

When you cut a coil spring down, you increase the spring rate.

I was at the shop earlier, and found a .7 bar spring for a Tial.
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Last edited by TimT; 02-21-2009 at 08:58 AM..
Old 02-21-2009, 08:55 AM
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Tial has spring options for the 38 and 40mm gates all the way down to .25 bar


.25bar is the small diameter yellow spring
.4 bar is small red
.5 bar is small green
.6 is small blue

.7 to 1 bar use larger diameter springs of same color
1.1+ bar use dual spring groups to achieve desired boost pressure w/o a boost controller

Hope that helps.
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Old 02-21-2009, 09:02 AM
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Yes talked to Ben earlier and I have a 38mm tial with adapters to make it bolt in to my stock euro exhaust with a .4 bar spring on the way and I am going to set up a manual boost controller in the engine compartment to fine tune the boost later we must remember this is a training exercise. I have spent the morning in the shop modifing my engine tin for the turbo and still give a heat shield for the distributor. I also worked out my oil delivery and return to the motor so I think I have all the bases covered now, I am sure I will have more questions later but for now thanks to all and I hope I don't have to post pics of a blown engine. I am using a EFI that will and has worked on turbo apps. a MSD programmable ignition (so I am sure I will have question of the timing curve to use) and a wide band O2 setup to adjust AFR's. What do you guys think do I have all the bases covered and remember this is all to learn before I build a 3.2 for this purpose. And I will have heat for the first time since I put the car together, all this and I'm going to Disneyworld.
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Old 02-21-2009, 10:26 AM
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Sounds good to me. don't forget a support brace for the turbocharger. It's alot of weight to be hanging off a hot set of headers, let alone twelve tiny header flange studs.

What EFI system are you using? Some sort of EIC/AIC(auxillary injector controller, like SDS or 034) or a full blown EMS(Tec2/3, MS2, etc)

Just curious
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Old 02-21-2009, 02:19 PM
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Using Megasquirt (Bitzracing setup) and I have both the turbo and muffler bracket to order
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" Porsche there is no substitute" I always liked that saying. Air cooled is the only way to go!
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76 Blazer also restored by me
Old 02-21-2009, 02:21 PM
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Quote:
Originally Posted by gsmith660 View Post
Using Megasquirt (Bitzracing setup) and I have both the turbo and muffler bracket to order
Good deal, you'll have fun with that EMS. It's very easy to work with and upgradable down the road to fit whatever needs you may have on the 3.2L
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Old 02-21-2009, 02:30 PM
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Thanks the whole thing has been a dream since I built the car in 82 and I finally get to realize a dream I cant wait to fire it up the first time.
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" Porsche there is no substitute" I always liked that saying. Air cooled is the only way to go!
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76 Blazer also restored by me
Old 02-21-2009, 03:04 PM
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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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Quote:
Originally Posted by gsmith660 View Post
I get the impression that I have worn out my welcome here your responses are getting terse I am only trying to learn and I thought that this was the right place to ask questions.
Please do not feel that way. You came to the right place and got the answer...physics is physics.

Cutting spring will make it harder. Harder spring will increase your boost. It will also make spring work over narrower region before it bottoms or decouples. So answer is: you need equally long spring wound with thinner material in order to lower your base boost.

EBC only allows you to raise boost over "base boost". It cannot lower it. Thus, make sure base boost is on safe level and work from there.
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Last edited by beepbeep; 02-21-2009 at 04:27 PM..
Old 02-21-2009, 04:24 PM
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Quote:
but I bristled a bit at the suggestion that I had given misinformation
I never quoted or said which information was wrong... but there is incorrect information given in this thread...

Quote:
if you compare the force exerted by two springs
why introduce more variables? this is a simple system.. a force acting on a spring. i.e. the waste gate diaphragm acting on a spring.. If you take the spring, and make it shorter... the spring rate goes up! nothing suspicious or tricky... its what happens... so now you have your waste gate diaphragm acting on a spring with a higher rate... your low boost threshold is higher..

Quote:
The longer the wire is the lower the spring rate. As the wire get shorter, such as when cutting the coil, the spring rate increases.
Quoted from Eaton... spring mfg..

Quote:
the helical spring is analogous to a torsion bar, not a lever
Yes kinda maybe yes..
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Rent a GT3RS from us!! Call or e-mail.
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Old 02-21-2009, 04:43 PM
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Quote:
Originally Posted by TimT View Post
I

why introduce more variables? this is a simple system.. a force acting on a spring. i.e. the waste gate diaphragm acting on a spring.. If you take the spring, and make it shorter... the spring rate goes up! nothing suspicious or tricky... its what happens... so now you have your waste gate diaphragm acting on a spring with a higher rate... your low boost threshold is higher..
Tim, the only reason I did that was because of 911nut's comment, which brought up a condition worth considering. Though it is simple and true that a shorter spring is stiffer (for the same wire characteristics), that oversimplifies what's going on in a wastegate. What also matters in the wastegate is the force on the diaphragm, not just the spring stiffness. The spring in the wastegate must be compressed to the same degree, regardless of what spring is installed. Therefore it isn't necessarily true that a shorter spring will provide more force, one has to consider both the stiffness and the deflection of whatever spring is being used. There's no question that the shorter one will be stiffer, but if the longer (and less stiff) spring is able to be compressed enough, it may exert more force. So this isn't as simple as it looks...

What I would take back from this is that if you cut down a spring, be careful. Both the spring rate and the force at a given deflection will be changed.

Rob
Old 02-21-2009, 05:15 PM
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I'm way not qualified for this conversation I just fix nuclear reactors for a living, thanks to everyone for the input you all convinced me to lay out the bucks for a wastegate with the proper spring and now I need to find someone to buy OEM wastegate that woks really good now and appears to open right at .8 bar or close to it.
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Old 02-21-2009, 05:36 PM
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Quote:
Originally Posted by Rob 930 View Post
There's no question that the shorter one will be stiffer, but if the longer (and less stiff) spring is able to be compressed enough, it may exert more force. So this isn't as simple as it looks...
Having installed a 1 bar spring in a stock wastegate, I know this to be true!
This is where Hooke's Law makes perfect sense. Two spring of different length that are compressed into the same space. When I say different length these different grades of springs varied by 1/4 inch in length per 1/10 bar rating. In this case I think we can safely discount the "shorter is stiffer" argument as being trivial.
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Old 02-21-2009, 07:05 PM
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Quote:
Originally Posted by fredmeister View Post
.........................................
You don;t want to cut the spring because like others said this increases spring rate..................................
+1

Cutting a coil of a spring increases spring rate pressure.

I don't know of any factory springs that come in that low a pressure rate. Aftermarket is the only option, I'm afraid.
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Old 02-22-2009, 05:18 AM
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Quote:
Originally Posted by WERK-I View Post
+1

Cutting a coil of a spring increases spring rate pressure.
All you experts are avoiding answering one question: If a shorter spring is a stiffer spring, why is the .8 bar factory spring longer than the .7 bar factory spring?

So, whoever thinks they can explain the in's and out's (and I don't mean providing links to other pages; do it in your own words) of spring mechanics to us all, have at it.
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Old 02-22-2009, 07:27 AM
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Answer to the first question;
The diameter of the coil wire is different.

The second is using the formula to determining spring rate pressure;

spring rate =

modulus of spring steel X wire diameter4
--------------------------------------------------------------------
8 X number of active coils X mean coil diameter3

Where all other variables are unchanged except for the "number of active coils" (number goes down), a reduction of the divisor, results in an increase in the result.
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Old 02-22-2009, 08:05 AM
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A simpler experiment would be taking the springs out of two identical retractable ball point pens. Cut one of the springs in half. Try compressing each of the springs between your thumb and index finger. You will notice an increase in the effort on the shorter spring to get the same amount of distance of compression compared to the longer spring.
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Old 02-22-2009, 08:15 AM
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Here is what I discovered today:

1) I can't measure the 0.7 bar (10.15psi) stock spring - it's on the car
Spring Length = inch
Outer diameter = inch (Measured across middle coils with calipers)
Number of effective coils = (Two less than total coils)
Wire diameter = 0. inch (Measured at top near spring cut)
Installed height in WG = 2.5 inch (Estimated)
Computed spring rate = lb/inch

2) Next the 0.8 (11.60psi) bar spring -
Spring Length = 6.687 inch
Outer diameter = 2.70 inch
Number of effective coils = 6.75
Wire diameter = 0.198 inch
Installed height in WG = 2.5 inch
Computed spring rate = 43.38 lb/inch

3) Now the 1.0 (14.50psi) bar spring -
Spring Length = 6.750 inch
Outer diameter = 2.670 inch
Number of effective coils = 6.75
Wire diameter = 0.210 inch
Installed height in WG = 2.5 inch
Computed spring rate = 56.76 lb/inch

4) And the 1.1 (15.95) bar spring (Andial, I think) –
Spring Length = 6.937 inch
Outer diameter = 2.80 inch
Number of effective coils = 5.0
Wire diameter = 0.210 inch
Installed height in WG = 2.5 inch
Computed spring rate = 66.44 lb/inch

5) Finally the 1.2 (17.40) bar spring (If I had one...) –
Spring Length = inch
Outer diameter = inch
Number of effective coils =
Wire diameter = 0. inch
Installed height in WG = 2.5 inch
Computed spring rate = lb/inch

Used:
http://www.engineersedge.com/calculators/comp_spring_k_pop.htm

Mark
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Old 02-22-2009, 09:06 AM
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Quote:
Originally Posted by WERK-I View Post
Answer to the first question;
The diameter of the coil wire is different.
But it's not different; think I mentioned that in an earlier post.

Mark, that's exactly what I was getting at. Thanks for the hard numbers.

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Old 02-22-2009, 09:23 AM
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