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hustler 01-27-2012 01:17 AM

1 Attachment(s)
Got the new shocks on the daily, its insanely smooth over bumps.
https://www.miataturbo.net/attachmen...ine=1327645030
550/350 springs.

shuiend 01-27-2012 09:08 AM


Originally Posted by hustler (Post 827011)
Got the new shocks on the daily, its insanely smooth over bumps.
https://www.miataturbo.net/attachmen...ine=1327645030
550/350 springs.

How do you complain about a van being to expensive and then go and drop 2k on suspension for your daily?

Braineack 01-27-2012 09:26 AM

can you only adjust the ride height in .5" steps?

hustler 01-27-2012 10:00 AM


Originally Posted by shuiend (Post 827088)
How do you complain about a van being to expensive and then go and drop 2k on suspension for your daily?

lol @ thinking I've paid full price for anything in my life. This was old-stock on the "get it out of here" price.


Originally Posted by Braineack (Post 827098)
can you only adjust the ride height in .5" steps?

You have probrem? It was the right price. The fronts have 1/2" grooves.

Braineack 01-27-2012 10:02 AM

curious cat is all.

hustler 01-27-2012 10:14 AM

I should probably note that the rear shocks appear to be too short and I can't get the car higher than 4" so I have to come up with something, the spring is seated on the upper and lower perches at full droop.

BTW, full droop wheel position in the fender well is "lower" than factory Miata ride height, lol.

Braineack 01-27-2012 10:17 AM

just needs longer springs.

hustler 01-27-2012 10:20 AM


Originally Posted by Braineack (Post 827119)
just needs longer springs.

How will a longer spring increase ride height if the spring is snugged-up on the perches at full droop?

y8s 01-27-2012 10:24 AM

because the springs will be snugged up but longer.

hustler 01-27-2012 10:25 AM


Originally Posted by y8s (Post 827124)
because the springs will be snugged up but longer.

Won't that increase spring rate? The ride is insane-o smooth, but the springs are as firm as I want for the street. These shocks are so smooth over bump/RR tracks/speedbumps that I giggle.

y8s 01-27-2012 10:26 AM


Originally Posted by hustler (Post 827125)
Won't that increase spring rate?

think about that question for five minutes.

y8s 01-27-2012 10:27 AM

and furthermore, damn you for making me reply to this thread. I've been doing my best to not post in it for no good reason.

mgeoffriau 01-27-2012 11:09 AM


Originally Posted by mgeoffriau (Post 826705)
So at work we have the "UPS Computer" which is really just an aging Dell desktop that UPS supplied when we set up our shipper account. [...] Today I decided to take on the project of preparing a few basic upgrades and then wiping the hard drive and reinstalling Windows XP.

Well, poop. I managed to get the RAM bumped up from 256 MB to 768 MB which helped quite a bit, but I couldn't find an AGP video card. I was sure I had one in one of my boxes of old computer stuff. Must have tossed that old Geforce 2 I had kicking around.

I did find my old Matrox Millenium II PCI video card, however. Who knows why I kept that and tossed the AGP card.

So, now the computer has...

Celeron 2.4 GHz
768 MB DDR SDRAM @ 333 MHz
Onboard Intel vidchip

...and it's not terrible with a clean install of XP.

Yay.

hustler 01-27-2012 11:48 AM


Originally Posted by y8s (Post 827127)
and furthermore, damn you for making me reply to this thread. I've been doing my best to not post in it for no good reason.

What does "linear" mean?

y8s 01-27-2012 12:42 PM


Originally Posted by hustler (Post 827167)
What does "linear" mean?

remember back in school when you learned the equation of a line?

y=m*x+b ?

it's a first order polynomial. x is raised to the 1 power. the graph of the equation is a straight line. linear.

That means that for whatever input you stick in for x, you'll get a value of y that is "m times" more than x.

or in the case of springs, for every inch you compress, you get "k times" (since the spring equation is F=k*x) that amount of force in pounds (assuming your rate "k" is in lbs/inch).

and for most springs with equal spacing between coils (which your ASTs appear to have) they are constant rate. or "linear springs". That means that no matter what length they are at, the spring rate never changes. so if they are unloaded and sitting on a desk, they are still the same rate as they are just before they are coil binding. at coil bind, the spring rate is effectively the spring rate of solid steel. or rounding up: infinite.


References:
http://en.wikipedia.org/wiki/Degree_of_a_polynomial

Joe Perez 01-27-2012 12:48 PM


Originally Posted by hustler (Post 827167)
What does "linear" mean?

In this context, linearity describes the condition where the response produced by a system is always proportional to the input into the system, regardless of the magnitude of the input.

One example would be that applying 500 lbs of force to a spring causes it to compress by 1", applying 1,000 lbs causes it to compress by 2", applying 1,500 lbs for force causes it to compress by 3", etc. For every additional 500 lbs you add, the spring will compress by another inch.

In other words, the spring's effective rate does not change simply because it already partially compressed. It responds the same way (by compressing 1" per 500 lbs) regardless of how much input (weight) has already been applied to it, in much the same way that your mom's response to "input" does not vary throughout the course of the evening regardless of how much "input" she has already had.


edit: y8s beat me to it. And I got a chuckle out of "rounding up to infinity."

Doppelgänger 01-27-2012 03:20 PM

1 Attachment(s)
https://www.miataturbo.net/attachmen...ine=1327695603

Braineack 01-27-2012 03:42 PM

i think that chart is incorrect.

y8s 01-27-2012 03:49 PM


Originally Posted by Braineack (Post 827282)
i think that chart is incorrect.

you are correct.
the chart is incorrect.

the ordinate label should be "Force".

Braineack 01-27-2012 04:02 PM

or if the "linear" was horizontal to the travel axis.


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