Technical Really hard pressure plate. Need help!

The only issue is when the clutch pedal is too close to the floor

He can lengthen the lever for travel , then use a smaller bore M/C
But a larger bore can need less travel so the lever can be shortened

Floor mounted pedals on Formula cars are only about 8" long from the pivot.[due to space] yet we can get them to work with a 4:1 ratio [the M/C is about 1-1/2" off the floor with vertical bolt pattern]
here is a Lotus FF


here is a Lola [the driver's feet go under the R & P which is where the infamous "Lola Limp" comes from]


At the end of it all, it is just a ratio between feet movement/ force vs pressure plate movement and resistence.
There are many ways to alter the ratios M/C to Slave bores pedal leverage, clutch fork leverage etc.

 

@Kerrynzl not arguing with you (ha!) but the spec on the Wilwood slave requires a 3/4" master, as does the Ram htob I have (well 3/4" + actually), so going smaller on the master to build pressure isn't going to fly imo. Those skinny, fit, muscular (ha!) racers will have no problems with a heavy clutch for racing, but not the regular guy driving on the street, queuing at lights, manouvering, parking etc. Wilwoods on the floor setup is 6:1 but the pendulum pedal is 7:1, just saying that as a reasonableness test.

Chris

 

You don't go smaller on the M/C to build up pressure.
You go smaller on the M/C because you have lengthened the lever [which reduces the pedal ratio]

To get the 1.12" of piston travel [which you mentioned in post #60] would require 6.72" pedal travel with a 6:1 pedal ratio [so the pedal needs to be 6.72" off the floor]
A 3/4" [0.75"] M/C moves 0.495 in3 of fluid at full 1.12" travel
50 lbs pressure at a pedal = 300 lbs pressure at the pushrod ÷ 0.441 in2 [piston area] = 680 psi line pressure

Now if the pedal ratio was reduced to 4:1 [due to a longer arm] to get the same pressure at the line would require a smaller M/C
50 lbs pressure at a pedal now = 200 lbs pressure at the pushrod , so to get 680psi the piston area now = 0.294 in2
A 0.294 in2 bore M/C = 0.611" bore [slightly under 5/8" or 0.625"]
Now here is the irony..... We have managed to balance the line pressure to 680psi in both scenarios but what about the volume?
For a 0.611" bore to move the same volume as the 0.75" bore M/C [at 1.12" travel] the smaller bore M/C would need to move 1.68" of travel [0.56" further]
For this M/C to have 1.68" of travel multiplied by the 4:1 pedal ratio = 6.72" movement at the pedal [which is exactly the same as the 6:1 ratio longer pedal]
This is why most aftermarket M/C's have longer travel [Willwood etc]

The critical measurement is how far the pedal can move before it contacts the floor.

To round off the above calculations to the nearest commercially available bore M/C , going from a 3/4" bore 6:1 ratio to a 5/8" bore M/C would require a 4.17:1 ratio.
The 4.17:1 ratio would be easy to get correct with the lever length

 

That upper bell crank is playing a big part in the resistance. The ratio built into it is really working against you since it is reversing the mechanical advantage you would have had if it was reversed or at a 1:1 ratio. Ideally the bolt holes would be the same distance away from the pivot point so that nothing was gained or lost it would just be redirecting the direction of force (1:1).

 

That is how I see it also.
Firstly I prefer simplifying things vs overcomplication , But with the correct measurements the whole situation could be remedied with one part [the bellcrank]

If you look at the photo below the ratio of the bellcrank is the Yellow vs Blue lines But the directional force changes the ratio to the Yellow vs Red lines [and it gets worse as the bellcrank arcs inward further]

This below is how McLaren engineered "rising rate suspension" on F1 cars in the 1970's [they had the spring perpendicular to the A-Arm at maximum compression which is the reverse scenario to below]



This is how I would modify the existing situation to save it [but @farmer013 will need to calculate motion ratios to get the correct pedal ratio to the M/C size]



The M/C pushrod needs to be lengthened so the bellcrank goes through being perpendicular throughout it's range of movement [so if there is 20° of bellcrank movement, the bellcrank needs to be offset 10° before perpendicular and it finishes 10° past perpendicular ]
The same applies to the upper lever part of the bellcrank [eg 10° before and 10° past]
But the upper lever part needs to be perpendicular to the "pull rod" halfway through it's range. Because of this the bellcrank doesn't necessarily need to be exactly 90°

If the upper lever on the bellcrank interferes with the mounting bolt in the photo simply notch the lever and use it as a stopper.
Geometry is an imaginary straight line between 2 pivot points.

It will require some thought and calculations, but it can be a way to save what has already been fabricated by simply replacing one component .

 

I still wonder how much friction is in the bellcrank...is there a bushing at each end of it?