What is a modular clutch and why do I have to buy a flywheel when I replace it?

Car: 2004 Dodge Neon SRT-4.


In regards to the SRT-4 what I'm reading is the modular clutch is an all-in-one setup. The pressure plate, clutch disc and flywheel are a pre-assembled piece that mounts onto a "flex-plate".

The flex-plate is mounted where on most cars the flywheel is (to the crank and teeth on the outer edge for the starter motor).

Apparently this is mainly just an ease of installation, the components are aligned during assembly by the manufacturer instead of during installation.

You are going to have to buy a flywheel if you are looking to replace the modular clutch with a normal aftermarket unit. You will remove the flex-plate, mount the flywheel in its place, then the disc and cover as normal.

The next time you need to replace the clutch, you shouldn't have to replace the flywheel unless it has been damaged.

Information from: http://www.modernperformance.com/blog/difference-between-modular-and-nonmodular-clutches-for-srt4

Response to "By not having a flex plate with an aftermarket clutch, does this give an advantage in reduced rotational inertia ?"

There could be a slight weight advantage as you are taking out the flex plate and whatever extra housing there is to make the modular package.

On top of that, when picking a flywheel, you could go with a lightened one.

On my vehicle, I went with a lightened flywheel. OEM was 20lbs and the lightened is 11lbs so nearly a 50% reduction in weight.

I don't remember the exact numbers, but the manufacturer was claiming something like 30-40 HP is "used" in the effort to spin the stock OEM flywheel on my car and so the weight reduction could "free up" 15-20 HP. How much of that is marketing fluff, I don't know.

Mine does seem to rev easier now and when rev matching for shifts it is a lot easier to put the revs right where I want them for a smooth transition.

  • Thanks for the explanation. By not having a flex plate with an aftermarket clutch, does this give an advantage in reduced rotational inertia ? – John Alexiou Jul 11 '11 at 19:54

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