Suspension 101: Stiffer is Better? Text by TANABE U.S.A.

[85 Supra-DOHC]

85 supra suspension

does anyone know what suspension ould be good for drifting a 85 celica-supra??

[emission]

does anyone know what suspension would be good for drifting a 85 celica-supra??

coilovers would be the best thing for your ma61.
But since coilovers don't exist for this chassis.

Just replaceing the stock f-up bushings makes for a more controled ride.

replace the rear bushings, bilstein shocks, front and rear sway bars, and a set of stiff lowering springs and your done.



just replaceing worn out bushings makes alot of diffrence when Im out messing around in the rain, i notice the diffrences before and after replaceing them. no wheel hop and more power to the ground.

[jinxtacy]

Firstly, it's important to understand the differences between dampers (shocks) and springs and what they do. Yeah, it's obvious that a damper is a tube-looking thingy and a sping is that coil but, more importantly:
The spring is the suspension element that supports the weight of the car and determine how much load it will take to compress that corner a certian amount. Spring rates are usually shown as "k" which is a constant representing how much force is needed to compress the spring a given distance. Generally, k is in lb/in but can also be shown in kg/mm depending on who manufactured them. Duly, a spring rate of 150 lbs/in will mean that it will take 150 additional kg's of weight to compress the spring one inch.
The Damper is the suspension component that will prevent the wheel from moving up and down so as to remain in contact with the road surface. A wheel that can be moved up and down too easily will be pushed off of the road surface by bumps, while a wheel that does move up and down easily enough will not absorb those bumps effectively.

On the right track, but a few corrections.

K = kilograms (a unit)

Springs rates are generally shown as kg/mm or lbs/in (unless of course the spring is progressive in which it will have a variable rate).


Thanks to 'Vankuen' @ evolutionm (or you can just google) for this explanation of the conversion process explained.

1lb = 0.45359237kg
1in = 25.4mm

.45359237 / 25.4 = .017857967322834645669291338582677

So now we know that 1lb/in = .017857967322834645669291338582677.

So if we want a spring that's roughly 560lbs; we'd multiply both sides by 560, and come up with roughly a 10k spring rate. Alternatively if you already know the kg/mm rate and want to find out what the lbs/in is, just divide the spring rate by that reaaally long factoral above (the .0178...).

But just in case you all didn't want to go through all that, here's a chart:

kg/mm to lbs/in
-----------------------------
16 = 896
15 = 840
14 = 784
13 = 728
12 = 672
11 = 616
10 = 560
9.0 = 504
8.5 = 476
8.0 = 448
7.5 = 420
7.0 = 392
6.5 = 364
6.0 = 336
5.5 = 308
5.0 = 280
4.5 = 252
4.0 = 224
3.0 = 168
2.0 = 112

This is also available: http://www.hraefn.net/projects/spring_rates.php

So a spring rated at 150 lbs/in will take a 150 lbs to compress the spring one inch, 300 lbs to to compress the spring two inches, 450 lbs for 3 inches, and on and on.

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My apologize if that's what you meant, but I felt that the wording was off.

[white2.4]

this is more than 2 years old

[teambergenholtz]

i gotta bump this so I can read it later