Cliff Help please
06-13-2011, 03:13 PM
#11
3rd Gear Member
Join Date: Mar 2009
Location: NOLA
Posts: 665
the t45 didn't come with a hydraulic actuator. the t3650 didn't come with one until 2005+. i'm just curious why you decided to have that added on.. unless its an 05+ t3650 in there??? IMO, its just something else that can fail and that needs fluid changed/added. i love having the feel of a fully mechanical clutch, gives it more of an "authentic" feeling, i guess
06-13-2011, 06:59 PM
#12
2nd Gear Member
Thread Starter
Join Date: Nov 2009
Location: NJ
Posts: 251
The Hyrdraulic setup was custom made with a tilton set up. I really dont know all the details but it did take off some pressure but is still not desirable. I only did it because i hated how hard the clutch was and like i said it did help a bit. I just feel there is something wrong in general. The car never consistantly got out of first without being jumpy and its annoying. I am not sure if they used a new drive shaft or not or if its required for that conversion. I would put the cable setup back if i knew i can get the feel better. I really would like to know why its never been smooth getting out of first. Could they of set something up wrong on the conversion
06-13-2011, 07:15 PM
#13
1st Gear Member
Join Date: May 2011
Location: NC
Posts: 133
06-13-2011, 08:09 PM
#14
TECH SAVANT
Join Date: Dec 2006
Location: Saint Augustine, FL
Posts: 10,938
Do you know which Tilton release bearing, and what master cylinder (particularly the bore diameter of the m/c) was used?
Most of the Tilton release bearings have a 1.221in² piston area, so used with a 5/8" bore (0.307in² area) m/c would give you a bit less than 4:1 mechanical advantage between the master cylinder and the release bearing.
One thing that must be understood is that for any given clutch assembly the work (force times distance) needed to release the clutch is a constant--no actuator; mechanical, hydraulic, or otherwise--can change that. They are just different ways of transmitting the force, none having any significant advantage over the other.
With the stock cable operated clutch the release lever provides about a 3.5:1 mechanical advantage and the pedal lever and quadrant another 3.5:1 or so, for a total advantage of 12.25:1.
I.e 20lbs of force applied to the pedal over a distance of 6.0" will create 240lbs of force over a distance of 0.5" at the TOB.
So in order to mimic the pedal pressure of the stock cable actuator, and assuming a 5/8" bore master cylinder, your pedal assembly would have to have a mechanical advantage of 3.1:1. If it is less than that more force over a lesser distance will be required at the pedal, more than that and less force (over a greater distance) would be required.
The other dimensions needed to calculate the pedal force would be the length of the pedal lever from the pivot point to the center of the foot pad, and the distance from the pivot to the point at which the master cylinder operating rod connects to the pedal lever.
As to the master cylinder bore; a smaller bore will require less force over a greater distance, and a larger bore greater force over a lesser distance.
Most of the Tilton release bearings have a 1.221in² piston area, so used with a 5/8" bore (0.307in² area) m/c would give you a bit less than 4:1 mechanical advantage between the master cylinder and the release bearing.
One thing that must be understood is that for any given clutch assembly the work (force times distance) needed to release the clutch is a constant--no actuator; mechanical, hydraulic, or otherwise--can change that. They are just different ways of transmitting the force, none having any significant advantage over the other.
With the stock cable operated clutch the release lever provides about a 3.5:1 mechanical advantage and the pedal lever and quadrant another 3.5:1 or so, for a total advantage of 12.25:1.
I.e 20lbs of force applied to the pedal over a distance of 6.0" will create 240lbs of force over a distance of 0.5" at the TOB.
So in order to mimic the pedal pressure of the stock cable actuator, and assuming a 5/8" bore master cylinder, your pedal assembly would have to have a mechanical advantage of 3.1:1. If it is less than that more force over a lesser distance will be required at the pedal, more than that and less force (over a greater distance) would be required.
The other dimensions needed to calculate the pedal force would be the length of the pedal lever from the pivot point to the center of the foot pad, and the distance from the pivot to the point at which the master cylinder operating rod connects to the pedal lever.
As to the master cylinder bore; a smaller bore will require less force over a greater distance, and a larger bore greater force over a lesser distance.
Thread
Thread Starter
Forum
Replies
Last Post
