04-20-2004, 01:56 PM
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CC Member
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Join Date: Jan 2004
Location: Shawbury,
Posts: 325
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Not Ranked
Bruce, I'm a little confused. Maybe I got this wrong, but I think the formula for calculating leverage is:
F1 x d1=F2 x d2
The pic below shows my clutch pedal (I hope it does  ).
If I assume d1 is one foot long and F2 is 100 Pound, and I want to have only 10Pound pedal effort , then d2 is calculated like this:
d2= F1 x d1/F2, in this case d2=10lbs x 1ft/100lbs=0.1ft
So the closer the cable is attached to the pivot the easier it is to push...or am I completely dumb  ? As far as I understand the closer it's attached to the pivot the more pedal travel is necessary.
Please tell me if I'm wrong or kind of stupid-I can take it as long as I find a solution for my clutch problem  |
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