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Length without load, Length with specified load, type of action (compression/traction), OD and composition (steel, stainless, bronse...). You can specify Preload, if any and shut length if it matters.

The parameters available are Length unloaded. Rate of Compression in pounds per inch. Length Under Load. OD, ID and Wire Diameter.

I guess it’s the Rate of Compression parameter I don’t understand. Do I specify the length unloaded and loaded and figure out the Rate of Compression from there. If I did that how do I know the spring won’t compress so the coils touch each other and ‘bottom out’ with the specified amount of compression?

"Length unloaded. Rate of Compression in pounds per inch. Length Under Load", there's one redundant parameter here; At least they are interactive, as long as the elasticity limits are not exceeded.The parameters available are Length unloaded. Rate of Compression in pounds per inch. Length Under Load. OD, ID and Wire Diameter.

Length under load= Length unloaded-(load.Rate) (for a compression spring - for a traction spring use sign plus instead of minus).

If "shut length" is not spec'd, use number of turns x wire diameterhow do I know the spring won’t compress so the coils touch each other and ‘bottom out’ with the specified amount of compression?

sounds like math (algebra).Does anyone here know how to spec a spring? Say I want a spring that has 2” static compression under a 1 lb load. I also want the spring to be 3” long when under load?

looking at mcmaster carr about the best specified vendor for onsey-twosey.

I didn't see anything close to your spec.

JR

Hookes law.

For TMI, beware that Hooke's law is significantly inaccurate for helicoidal springs when the pitch changes under strain.Hookes law.

The force is the elastic torque divided by the sinus of the angle between a turn and the axis.

As the turns get closer, the sinus of the angle increases, so the compression rate also decreases.

From a basic lab exercise i had in school last year

A spring was elongated (expanded) by different weigths. I havnt recorded how long the unloaded spring was but i think it was around 5-10cm long..

The angle of the graph tells us what the stiffness is for every wight/force applied to the spring.

A spring was elongated (expanded) by different weigths. I havnt recorded how long the unloaded spring was but i think it was around 5-10cm long..

The angle of the graph tells us what the stiffness is for every wight/force applied to the spring.

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What is TMI btw?

"Too Much Information", internet forums live in the world of TMI because everybody posts everything they know about a given subject, not just the answer to the OP's (original poster) question.What is TMI btw?

JR

Aahhhaaa...

If your goal is sound isolation don't use a reverb spring...

JR

JR

They don’t have anything called a reverb spring at McMaster so I should be okayIf your goal is sound isolation don't use a reverb spring...

JR

I’m planning on using 1” PVC sheet for the plinth. Seems appropriate. I’m not sure why manufacturers use resonant material like wood or high transmissibility material like stone. PVC is dead. If you tap it you get nothing. If you put a speaker on it with mid to high frequencies you still get nothing. If its good enough to make records out it should work as a plinth.

I like plastic. I have a bunch of Kartel furniture. High end Italian plastic furniture with classic designs back to the 1960’s. It’s mostly sold as commercial furniture but i love it. People don’t generally like to pay a lot for plastic though. 1” PVC sheet isn’t cheap but probably looks it.

well... a mass + spring system is also resonant.

Right. Hopefully the resonant frequency will be less than 1Hz. That’s the point.well... a mass + spring system is also resonant.

F = k x delta l

Where F = the static force applied to a spring by a wieght in newtons

k = stiffness konstant of the spring in Newtons per meter

Delta l = the leghth or elongation, of which the spring has been alterd by the applied force, in meters.

The angle velocity gives

2 x pi x f = sq root(k / m)

Where f = the resonant freq of the system.

m = the mass of the system in kg.

This is equivalent to LCR circuits

Armed with this information I have roughly a 0% chance of getting it right. As best I can tell from manufacturer tear sheets a static compression of 2" has a resonant frequency around 1Hz. I haven't found any spring isolators with more than 2" of static compression. Using more than one equally specified spring doesn't seem to change the resonant frequency. Using 10 springs with 2" static compression has the same cutoff frequency as using 100 springs with 2" static compression. I'm flying blind like I do with Ohm's Law. I can rarely apply that correctly.

F = k x delta l

Where F = the static force applied to a spring by a wieght in newtons

k = stiffness konstant of the spring in Newtons per meter

Delta l = the leghth or elongation, of which the spring has been alterd by the applied force, in meters.

The angle velocity gives

2 x pi x f = sq root(k / m)

Where f = the resonant freq of the system.

m = the mass of the system in kg.

This is equivalent to LCR circuits

JR

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