Snips

VJ - I will repeat again! the out put shaft is not the definative time base
for a
Wankel - the reason that it takes 3 turns to complete a four stroke cycle is
because it is chasing the the working srufaces - the rotor which is turning
that
extra turn - the crank (and it is a crank) makes the classic 2 revs per four
stroke
RELATIVE to the rotor which is the pumping/expanding element for the
extraction
of the power. 

snips.

 - AGAIN the cycle is the time line
!-
not the rotation of some mechanical part of the machine!
Vance - if someone missed the web page address it is
www.geocities.com/vjaqua

HP is explosions of a given size per second. We all agree on that do we not?

We do agree that a one rotor fires once per output shaft rev do we not?

OK one explosion per rev of the one rotor Wankel output shaft.
at 6000 RPM or 100 revolutions per second or 100 explosions per second. 
One explosion in .01 second.

We do agree that the rotor is turning at 1/3rd output shaft speed do we not?

OK at 6000 RPM the rotor is turning at 2000 RPM or 33.3333 revs per second.
Three explosions in one rev of the rotor. 3 X 33.3333 is
100 explosions per second. One explosion in .01 second.

Pick any moving part of the rotary engine you wish as your definative 
time base. The time line is the time line.
 
Paul Lamar
 
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