Yes me again - Tyres :-)

I cant speak about any of those tyres as I have never used or know someone that has.
I can tell you anyway that I have a set of Malatesta Kaiman 30/9.5/15 and I love them . . .

Great grip all around specially in mud, but I have to say they are noise on the road.

They are only a few months old and so signs of wear yet...

Usually, if I want to change the springs on a vehicle - I'll just get some custom wound to spec. Generally they cost around £50 per spring (don't know if that's expensive?)

To work out what length & weight you need - you do this:
Weigh the vehicle. A stock Jimny is about 1200kg - so it may not be necessary unless you've added winches etc. Wheels, tyres & axles are about 200kg (or maybe a little less), so each wheel is supporting 250kg.

When you go over a bump, the effective weight of the vehicle is multiplied (because of it's inertia). With this multiplied weight, the axle should just touch the bump-stop. I'll say more about the multiple to use later.

You need to measure the separation of the axle & bump stop then add the amount of lift you want. For a 4.5" lift, that's going to be about 7" - Call this the 'Travel Length'

When the spring has compressed those 7", the force must equal the weight multiple. Take the weight multiple for that wheel (In Lbs) minus the un-multiplied weight and divide it by the travel. That gives you the spring rate.

When the vehicle is stationary, the springs are only supporting the un-multiplied weight. Using the rate you have calculated above, you can say that the compression of the spring will be the un-multiplied weight / rate. - call this the 'Stationary Length'

You need to measure the length of the compressed spring when the axle is sat on the bump stop. With the vehicle on level ground, measure the length of a spring and subtract the separation of the axle and bump stop. This is the 'unused length'

Add the 'unused length' + 'travel length' + 'Stationary Length' and that equals the uncompressed length of the spring.

For a soft ride - something like a Rolls Royce, use a weight multiple of 1.6. For an average road vehicle, 1.8 and for an off roader, 1.8 to 2.

So, a worked example - with a few guesses.
Each wheel has 250kg or 550Lbs sitting on it. (Un-Multiplied weight)
This is an off roader, so multiply by 2 = 1100Lbs - 550Lbs = 550Lbs (Multiplied weight)
For 7" travel, the rate needs to be 550/ 7 = 78 Lb/in

The stationary length is 550Lb / 78 Lb/in = 7"
The unused length is likely to be about 3"
So the uncompressed spring length wants to be 7 + 3 + 7 = 17"

The spec for the spring is therefore 78 Lb/in, 17" Free length with a maximum compressed length of 4" (rounded to round numbers).

For the above, I have assumed that each of the wheels has equal loading - which is not the case.
If you take your vehicle over a weigh bridge, it will tell you the axle weights. It's reasonable to assume that each wheel on one axle will have the same weight and can be the same.
If you want to be even more accurate, because vehicles are mostly driven with only the driver on board, they tend to lean slightly towards the driver. Vehicle manufacturers compensate by taking an average weight for a person - say 150Lbs, divide it equally between the front & rear and make the drivers side springs a little longer to compensate. 1" in this case.

It might be worth calculating the rate and free length of an original spring using the formula here (clicky)
This gives a poundage of about 100 (based on guesses) which means they have used a multiplier of 1.8. Also, using that formula, you can work out the spec of any after market spring to have your own made, tweaked to your spec.

Si

Thanks Si, for the technical explanation

Though I must admit. When it comes to that, buying suspension parts for my future Jimny, I think I'll just go for a complete kit. Perhaps from here on bigjimny.com or jimnybits.co.uk - I don't know yet

But maybe I'll just stick with a Silverstone MT117 Extreme or something similar