Going lower

[noE30]

The drop looks good.

[1990_325i]

BUMP okay, would these be a good buy? http://cgi.ebay.com/ebaymotors/Parts...=p4506.c0.m245 I am new to e30s btw and if i get that kit on ebay do I need anything else to it? How much does the auto shop charges for springs installation or its easy to install them my self?

[JK_E30]

BUMP okay, would these be a good buy? http://cgi.ebay.com/ebaymotors/Parts...=p4506.c0.m245 I am new to e30s btw and if i get that kit on ebay do I need anything else to it? How much does the auto shop charges for springs installation or its easy to install them my self?

go to the suspension section a check out some set ups..

learn which sets ups and which mfg are legit and have good products.. and yes you can do suspension yourself.. just need tools and use a write up.

depending on how low you go you might need new shocks as well.

[E30_pilot]

IE's are nice. I have the same springs but with koni's

BUMP okay, would these be a good buy? http://cgi.ebay.com/ebaymotors/Parts...=p4506.c0.m245 I am new to e30s btw and if i get that kit on ebay do I need anything else to it? How much does the auto shop charges for springs installation or its easy to install them my self?

link no work. but if it is like the one you posted earlier,Cosmo "coilovers", they are trash.

get real suspension like this OP did.
And yes you can install the suspension yourself.
Use the Search feature, its wonderful.

[kid325e]

If you have to ask if you can do it yourself you're prolly better off having a friend or someone that knows how to do it for sure do it to your car. Suspension is not to be fucked with.

Anyway, the drop did really good things to your car. I really like it, those rims are dope btw, my buddy has a set of them on his IX.

[JK_E30]

Im looking into getting new tires since these are a bit much... what do you guys think would look better... 225/45/r16 are the same overall size as stock. Or should I go with a little smaller and get 225/40/r16?


What do you guys think would look best?

[der affe]

205/50-16 or 205/55-16 still ride nice and cheap to buy!

http://www.discounttiredirect.com/direct/searchTiresBySize.do?cs=205&rd=16&ar=55

they are PLUS 1 sizing

[alpinaclone]

I've never really heard it to be dangerous... just its still gonna be a softer ride with more body roll and if you're lower to the ground that could cause trouble...

other than that I haven't heard anything else.

Cutting a spring will raise the spring rate. I can't remember the formula for it right now, but it does.

[JK_E30]

Cutting a spring will raise the spring rate. I can't remember the formula for it right now, but it does.

Oh really?

huh I guess you learn something new everyday.

[z31maniac]

225/45/16 is the perfect looking tire for an E30 I think.

Or a 215/40/17

Car looks good!

[kronus]

Oh really?

huh I guess you learn something new everyday.

Yes, it does. The formula for the force a spring exerts is F = k*dx, where F is in newtons, k is the spring constant (usually in N/m), and dx is the change in the spring length.

For a clear example, consider a compression spring with a k of 10 N/m. Applying a force of 1 N to it, then, decreases its length by 10 cm. Consider one half of the spring, though. With the same 1N force applied, it has only been compressed by 5 cm. (both halves contract by 5cm) Therefore, k for half of the spring is double of what it is for the entire spring.

From this, it follows that if we reduce the length of a spring to x% of its original length, it will become 100*(100/x) % stiffer. For example, reducing the spring to 80% of its length makes it 100/80 = 10/8 = 5/4 = 125% as stiff as it was. (increases the spring ratio by 25%)

Sorry, I've been doing math proofs for the last 6 hours and couldn't resist... I'm in the proof zone.

[lostininglewood]

drop looks sweet!! nice setup!

[rodrigo]

:loco:

hmmm. . .

nevermind. no use

[JK_E30]

Yes, it does. The formula for the force a spring exerts is F = k*dx, where F is in newtons, k is the spring constant (usually in N/m), and dx is the change in the spring length.

For a clear example, consider a compression spring with a k of 10 N/m. Applying a force of 1 N to it, then, decreases its length by 10 cm. Consider one half of the spring, though. With the same 1N force applied, it has only been compressed by 5 cm. (both halves contract by 5cm) Therefore, k for half of the spring is double of what it is for the entire spring.

From this, it follows that if we reduce the length of a spring to x% of its original length, it will become 100*(100/x) % stiffer. For example, reducing the spring to 80% of its length makes it 100/80 = 10/8 = 5/4 = 125% as stiff as it was. (increases the spring ratio by 25%)

Sorry, I've been doing math proofs for the last 6 hours and couldn't resist... I'm in the proof zone.

lol well thanks for clearing this up..

I always hated proofs... you just boggled my mind.

[alpinaclone]

Yes, it does. The formula for the force a spring exerts is F = k*dx, where F is in newtons, k is the spring constant (usually in N/m), and dx is the change in the spring length.

For a clear example, consider a compression spring with a k of 10 N/m. Applying a force of 1 N to it, then, decreases its length by 10 cm. Consider one half of the spring, though. With the same 1N force applied, it has only been compressed by 5 cm. (both halves contract by 5cm) Therefore, k for half of the spring is double of what it is for the entire spring.

From this, it follows that if we reduce the length of a spring to x% of its original length, it will become 100*(100/x) % stiffer. For example, reducing the spring to 80% of its length makes it 100/80 = 10/8 = 5/4 = 125% as stiff as it was. (increases the spring ratio by 25%)

Sorry, I've been doing math proofs for the last 6 hours and couldn't resist... I'm in the proof zone.

Thanks for typing all of that up. I could not be assed to go dig out my notebook or go looking through the bookshelves. :)