Scooterotica

eden wrote: Carlos aka non believer

all im gona say here is...........


P2+R2-2 RP cos(q) = L2 (Law of Cosines)

So P = R cos(q) + (L2 - R2 sin (q)2) 1/2



simples............ maybe not

Eden, WTF ,its Sunday just got in after a session on the beer, the clocks have gone forward , me heads battered, and now im reading this LOL, is nt it great that we can take the p*SS and learn at the same time,mmmmm time for another beer methinks.... .....

yes, but can you explanation it in english.

Ok just did a few calcs .....so 110 rod 60mm stroke. big end 90 deg to bore .
if i rotate the crank 20 deg towards tdc the piston travels 19.54 mm
if i rotate the crank 20 deg towards bdc the piston travels 16.86 mm
its the devils work i say

lol:)

don't need to understand why,we just need to know its just the way it is.

chuck the timing disk and just use the calculators, its far easier and more accurate.

x is centa line; R half stroke 30mm; L rod 110mm; wen crank spins past 90 degrees the angle tween L and R is obtuse; cus top end of L moves on a fixed centa line X; wen the bottom end of L is connected to R; crank rotation creates different r to l angles; 'see the picture in y minds eye'

eden wrote: Carlos aka non believer

all im gona say here is...........


P2+R2-2 RP cos(q) = L2 (Law of Cosines)

So P = R cos(q) + (L2 - R2 sin (q)2) 1/2



simples............ maybe not

That's the one...

Draw the cosine graph, degrees 0 to 360 on the X axis, 0 to max stroke on the Y axis. See how many degrees = Max stroke/2.
If TDC is at 0 degrees.
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