Dear Paul,

With all due respect, the shaft of the Wankel needs to turn three (3)

!!!!!

times for one chamber to complete the four stroke cycle.!!

Again who cares how long it takes one chamber to complete the four
stroke cycle? The ENGINE completes the four stroke cycle in one
revolution of the output shaft. As I said before all three chambers
cooperate to achieve that end.

Since there are three chambers, one ignition occurs at every single

shaft

revolution. Three revolutions divided by three chambers = 1 revolution
per chamber.

Nobody is arguing that. Vance is saying the engine operates like a
three cylinder four cycle piston engine and nothing could be further
from the truth. Obviously when you focus on the three chambers of
the Wankel you can confuse yourself and be misled as Vance is confused
and misled.

Also,
The RX-7 compares to a 240 in 3 four cycle RPE, where it does not matter
how many cylinder are used with regard to total displacement and

throughput.

However, the shaft has to turn 1.5 times faster.

Using equal shaft speed, which is an incorrect comparison, we can only

talk

about "similarities" between the Wankel and the RPE. With equal shaft

speed,

the equivalent displacement is 160 in3. If the RPE should be a six

cylinder

or a four cylinder is another debatable issue.

I was working at NSU in Neckarsulm, in the 60's during the Wankel
development years.

Rolf P.

I am surprised. For somebody that worked at NSU you too appear to be

misled

and confused as well. Forget the RPM. RPM is only another way of keeping

time.

The number of explosions of a given size per second, per minute, per day
per year is all that counts. There is absolutely no doubt
the one rotor Wankel takes in one chambers worth of air and fuel and fires
once per output shaft revolution no matter how long that takes.

EXACTLY THE SAME AS A TWO CYLINDER FOUR STROKE PISTON ENGINE OF TWICE THE

DISPLACEMENT!!

EXACTLY THE SAME AS A ONE CYLINDER TWO STROKE PISTON ENGINE OF EXACTLY THE

SAME DISPLACEMENT!!

I AM GETTING TIRED OF WASTING MY TIME ON THIS SUBJECT.
END OF THREAD!!!!!!!!!!!

Paul Lamar

OK! 0ne last simple rule for the displacement of any positive displacement
IC engine
It will work for any engine - 4 stroke, two stroke, cam, vane - any thing
you can think of
"The displacement of any positive displacement IC engine is the total swept
volume equivalent
of all the power strokes ("explosions") that take place in the period
required for any single
chamber (cykinder etc) to complete the thermodynamic cycle"

WRONG. To quote Ansdale page 57 paragraph four "Evidently this comparison underlines the 
careful analytical approach to RC engines as mere counting of chambers
will certainly fail to yield the correct answer.."

This renders the remaining part of your message invalid.

What volume of air (or displacement) goes in and is processed in a given time
period is the ONLY criteria that works.

 END OF THREAD!!!!!!!!!!!        PLEASE!!!!!!!!!!!

Paul Lamar

You can try that on any engine you can think of, and please tell me if you
think it does not
work and why. For reference the "period" for a classic two stroke is one
revoloution
of the crank shaft, for a four stroke it is two revs of the crank -
regardless of the number
or arrangement of chambers. For a Wankel it is three turns of the eccentric,
or one
turn of the rotor. The firing chamber does NOT complete a thermodynamic
cycle
in one turn of the output  - starting with ignition it only completes it's
expansion
stroke and only half of the subsequent exhaust stroke, and it takes two full
turns
 of the eccentric before that chamber is ready to fire again.. I will leave
it to
you to compute the displace,ent of a 13B with this rule - Vance

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