Lightened Flywheel
#1
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Lightened Flywheel
It's time for a clutch (62k miles) and I remember a post about a lightened flywheel. Are there any downsides. Which manufacturers? Are there any other mods that should be done now. Any specific clutch or clutch/flywheel combos recommended. It's a 97' C4S. Thanks for your input.
#3
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in '97 stalling is not a problem, only in '95 (OBD I) cars usually have it.
Other "problems" (I don't see them as problems) are higher noise from tranny/clutch. IMO just better that way but like I said, some people don't like it.
Definitely worthy upgrade, car revs more freely and this is the way it should've always been. Dual mass flywheel erally shouldn't be there in a first place.
Definitely get the factory parts.
Engine does not to be dropped so there's not much other things you should do that benefit from having tranny out (check all the seals and driveshaft & components).
I'd recommend RS tranny mount but some people don't like it.
You can search and find out more details about this, lot of discussions in the past.
Other "problems" (I don't see them as problems) are higher noise from tranny/clutch. IMO just better that way but like I said, some people don't like it.
Definitely worthy upgrade, car revs more freely and this is the way it should've always been. Dual mass flywheel erally shouldn't be there in a first place.
Definitely get the factory parts.
Engine does not to be dropped so there's not much other things you should do that benefit from having tranny out (check all the seals and driveshaft & components).
I'd recommend RS tranny mount but some people don't like it.
You can search and find out more details about this, lot of discussions in the past.
#6
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Originally Posted by Flying Finn
Engine does not to be dropped so there's not much other things you should do that benefit from having tranny out (check all the seals and driveshaft & components).
Engine doesn't need to be dropped?
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#9
King of Cool
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Originally Posted by mborkow
i thought that with the C4(S) you had to drop the engine...learn something new here every day.
#10
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Dont forget to get the RMS replaced - if for no other reason than it makes 996 owners feel better
LWF - lightning reactions from your car - slight increase in noise - well worth the smile!!!
LWF - lightning reactions from your car - slight increase in noise - well worth the smile!!!
#12
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I don't this that is can possibly accelerate faster, from the conservation of energy standpoint.
Yes, I have a LFW in my race car and will be putting one in my 993. But not for faster acceleration.
Yes, I have a LFW in my race car and will be putting one in my 993. But not for faster acceleration.
#13
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See my sig for perceived HP gain from LWF. It does work, it does make a bit more acceptable noise. Find someone who has one.
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I have a 95 C4 w/ LWF and get the occasional stall. One of those who have it and experience the stall from time to time. The noise is faint and sounds like a bunch a marbles in can rolling around. Not loud, annoying, or unbearable-I dont think I even hear it any more because its just a part of my car. People riding the car would never even notice it.
#15
Difference in HP calc
A lightened flywheel (and/or cover plate) can "free up" a significant amount of HP in first gear acceleration.
This is how it goes (check my math).
E=1/2 * I * w^2 where
E=energy in joules
I=moment of inertia of of rotating mass (i.e. flywheel) in kg*m^2
w=the angular velocity of the rotating mass in radians/second (2*Pi radians in one revolution)
^2 is squaring the value
Modeling the flywheel as a cylindrical disk of constant density is a good approximation, and the moment of inertia of a cylindrical disk rotating about the cylindrical axis is:
I=1/2 * m * r^2 where
m=mass of disk in kg
r=radius of disk in meters
Now the principle of a lightened flywheel is to reduce the rotating mass, and thereby, reduce the amount of power required to accelerate the rotating mass.
So power is given by P=E/t where
P=power in hp (746 joules/second=1hp)
E=energy in joules
t=time to accelerate rotating mass from velocity time(1) to velocity at time(2)
After substituting for I in the first equation and reducing the result of calculating the difference in power required to accelerate a cylindrical disk of m1 and m2 is Pdiff:
Pdiff=(m1-m2)/2*(r^2/2*(w2^2-w1^2)) where
m1=the mass of the more massive flywheel in kg
m2=the mass of the lightened flywheel in kg
r=radius of the flywheel in meters
w2=the angular velocity of the higher rotational speed (i.e. 6000 rpm)
w1=the angular velocity of the lower rotational speed (i.e. 4000 rpm)
Scenario
m1=26lbs*.454kg.lb=11.8kg
m2=13lbs*.454kg.lb=15.9kg
r=260mm/2=.13 meters
w2=6000 rpm
w1=4000 rpm
t=.5 seconds (i.e. time to get from 4000 rpm to 6000rpm in first gear level ground)
Pdiff=14.66hp
Now anyone who can feel the difference in engine response when the AC compressor is engaged should be able to notice the difference in top end power from a lightened flywheel.
I did guess at these numbers, but they seem reasonable to me. Anyone have any actual numbers?
One note for the careful reader(s) is that if the time to accelerate the original flywheel is .5 seconds, the lightened flywheel will accelerate in less time, but I used the same time for both. This is an approximation, but this error gives a lower than actual hp difference, so the actual difference is more than the value calculated.
This is how it goes (check my math).
E=1/2 * I * w^2 where
E=energy in joules
I=moment of inertia of of rotating mass (i.e. flywheel) in kg*m^2
w=the angular velocity of the rotating mass in radians/second (2*Pi radians in one revolution)
^2 is squaring the value
Modeling the flywheel as a cylindrical disk of constant density is a good approximation, and the moment of inertia of a cylindrical disk rotating about the cylindrical axis is:
I=1/2 * m * r^2 where
m=mass of disk in kg
r=radius of disk in meters
Now the principle of a lightened flywheel is to reduce the rotating mass, and thereby, reduce the amount of power required to accelerate the rotating mass.
So power is given by P=E/t where
P=power in hp (746 joules/second=1hp)
E=energy in joules
t=time to accelerate rotating mass from velocity time(1) to velocity at time(2)
After substituting for I in the first equation and reducing the result of calculating the difference in power required to accelerate a cylindrical disk of m1 and m2 is Pdiff:
Pdiff=(m1-m2)/2*(r^2/2*(w2^2-w1^2)) where
m1=the mass of the more massive flywheel in kg
m2=the mass of the lightened flywheel in kg
r=radius of the flywheel in meters
w2=the angular velocity of the higher rotational speed (i.e. 6000 rpm)
w1=the angular velocity of the lower rotational speed (i.e. 4000 rpm)
Scenario
m1=26lbs*.454kg.lb=11.8kg
m2=13lbs*.454kg.lb=15.9kg
r=260mm/2=.13 meters
w2=6000 rpm
w1=4000 rpm
t=.5 seconds (i.e. time to get from 4000 rpm to 6000rpm in first gear level ground)
Pdiff=14.66hp
Now anyone who can feel the difference in engine response when the AC compressor is engaged should be able to notice the difference in top end power from a lightened flywheel.
I did guess at these numbers, but they seem reasonable to me. Anyone have any actual numbers?
One note for the careful reader(s) is that if the time to accelerate the original flywheel is .5 seconds, the lightened flywheel will accelerate in less time, but I used the same time for both. This is an approximation, but this error gives a lower than actual hp difference, so the actual difference is more than the value calculated.