Nobsound springs - load range

The Rate I calculated for @hm9000 was right at 20lb/inch of travel for the spring measured.

Based on your specs, @ryder I calculate about 12.6lbs per one inch travel for the Rate of Compression.

As I understand it, so long as you are in the travel range (not fully compressed) then no matter how large the gap is, the force to move the object further remains the same or constant. I'd guess leave it in the middle but no one seems to have any idea why. Can the 'different audible results' be identified other than 'pre-knowing' how many springs are being used...  

I'd still love a thought on best Rate of spring to use. I ordered 100 springs .75" long with a 9.3lb/inch compression rate. These will allow a lot more travel of the component (or more soft 'floating'). 

Of course I do not know whether one wants one's equipment on a very stiff, barely springy bed, or on a highly springy and very flexible bed (assuming stability). This is the question I'd love someone to weigh in on with knowledge of resonances.

Is there any difference in spring tension between silver or blue end caps

@musicaddict I try to put in some physics about spring mass systems (have to dig a little since this is long ago since I had to deal with that in my engineering study).

A spring extends or contracts in a linear relation with the force you are applying, or the other way round, the force F the spring excerts is proportional to the distance x that you are forcing the spring from its neutral (unloaded) position:

F = kx , where k is the spring constant. Your assumption that the force stays the same is wrong, the force gets higher the more you compress the spring. That is the reason why it settles in at a compression that depends on the mass you put on top.

From the formula above I can calculate k for my springs: k = F/x. Putting 2.5kg and 7mm gives me:  k = 3500 kg/s^2 (sorry for international units, but as continental European I cannot get used to imperial ;-)

Now let's calculate the resonance frequency, it is given by formula

f = 1/2Pi  sqrt(k/m), where m is the mass you put on the spring. You can see that for the same spring (or amount of springs), the frequency gets lower with higher mass. At the same time if you put two springs in parallel, you double the effective spring constant, and, with the same mass, increase the resonance frequency.

Putting the values for my DAC and spring configuration in the equation (mass ~3kg per spring), I get a resonance frequency of a little more than 5Hz, which is about what I would estimate when exciting the DAC and watching it swing.

Which resonance frequency does one want to get? The springs should decouple/isolate the component from its base, so that no vibrations are transmitted either way. The spring mass is a low pass system, it decouples above the resonance frequency, but kind of transmits vibrations below. What you want to avoid is transmitting vibrations for audio frequencies, so the resonance frequency should be below the audio band.

@hm9001, I think we agree but that I did not state it clearly enough. If a single spring has a 4kg rate per 4cm, then from the resting point of a 1kg wt at 1cm, it would take another 1kg of weight to depress 1 more cm additional. It's what I have heard referred to as a 'constant rate' compression spring.

(Different is a 'progressive rate spring' where the rate changes, and increases as you depress. A common example was fork spring replacements in motorcycles for a nicer ride.)

Thanks very much for the above info on frequencies. It makes sense and I will work to further digest the resonant frequency issue. This sure gets me started. It will be interesting, after I do the math conversion to see where the 9.3 lb springs may get me or not.  Thanks again.

My point on 'constant force' should have been 'constant additional force' for movement of a constant size, additionally. 

With a constant rate spring the additional force (or weight) from (equidistant positions) position 3-4 should be the same as the additional required to move from 5-6, for example.

My wondering was that if the equipment (wt or force) sits in the operating range of linear motion of the linear spring will require a similar push, etc. to move it whether it sits up high or low (while remaining in range).

With a static application what changes (or not) from riding high in the range to riding low in the range. I guess those were the results from changing out numbers of springs. Interesting for sure. Thx.