I may be tail-end-of-the-cow here, since I haven't read
all the responses to this, but ...
Re:
> From: "Ed Anderson" <eanderson@carolina.rr.com>
> First, we need to agree on total displacement of an engine. My understanding is that the total displacement of an engine is the displacement that ALL its power producing volume. Not just for 3/4 of its cylinders or chambers, not 2/3rds but ALL! Now, If I am incorrect about that definition then I guess I missed that class in auto school {:>). (See definitions below this message)
>
Actually, the only displacement that counts is that
which comes into play in a single rotation of the output
shaft. Any components not adding power during that time
are actually parasitic. That is why four-stroke engines
put out so little power for their stated displacement.
Thus:
> The rotary has six power producing faces each which "Scoops" up its "share" of the total volume displacement of the rotary. Now if you want to throw out two of the chambers and claim what remains is the total displacement then we should throw out two cylinders of a V8 when we consider its displacement {:>)
>
should read "throw out *half* its cylinders".
my $.02
Dale R.
Well, Tracy, I've been wrong before and this
would not likely be the last time {:>). Lets see, the two rotors have
equal volume (I think that's safe to say).
First, we need to agree on total displacement of an
engine. My understanding is that the total displacement of an engine is the
displacement that ALL its power producing volume. Not just for 3/4 of its
cylinders or chambers, not 2/3rds but ALL! Now, If I am incorrect about
that definition then I guess I missed that class in auto school {:>).
(See definitions below this message)
The rotary has six power producing faces each which
"Scoops" up its "share" of the total volume displacement of the rotary.
Now if you want to throw out two of the chambers and claim what remains is the
total displacement then we should throw out two cylinders of a V8 when we
consider its displacement {:>)
I think we can agree that the two rotors provide
1308 CC displacement in 720 degrees that means 4 rotor faces have scooped
out their share of the 1308 displacement (which is only 2/3 of the rotors 1080
deg cycle). But sticking with the 1308 cc for four chambers worth we have
each chamber = 1308/4 = 327 cc/rotor face/chamber. Ok so shoot me - I was
6 cc off.
But if we take the FULL engine cycle of the
rotary which is 1080 deg not the reciprocating 4 strokes
720 degrees, that means there are two more chambers worth
before we have exercised all rotor faces/chambers once for a complete engine
cycle. Then we add 2 more 327 cc chamber cycles to the 1308 we
already have and we have a grand total of approx 1952 cc or approx 120
CID
As I pointed out, if you truncate the rotor's 1080
degree cycle, then you indeed come up with the 1308 cc displacement. But
the definition of displacement of an engine generally means the total
displacement of all its power producing surfaces (like piston top,rotary faces,
etc), not just some of them.
I've listed several defintions for engine
displacement I found on the web below: I think you will find that the
definition of total engine displacement call for ALL the power
producing volumns be included - not just 2/3 of them. But, then
that is my interpretation of the definitons below.
So until future evidence to the contrary, I'll
stick by the 120 CID
See! I told you this would invoke discussion
{:>)
Engine displacement (or size)
of the engine is the number of cylinders times
the displacement of one of the
pistons.
In an engine, displacement is the total volume of air and fuel that an engine
is theoretically capable of drawing into all its cylinders
in one cycle. Since an engine is three-dimensional (volume),
displacement is measured in cubic inches for U.S. cars. Some cars may have
displacements measured in liters.
Displacement
The total volume of air displaced by all the
pistons within an engine block. In general, the greater an engine's
displacement, the more power it can produce.
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