Today's Message Index:
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1. 06:05 AM - Rotary engine displacement (Gary Casey)
2. 08:05 AM - Re: Rotary engine displacement (Russell Duffy)
3. 08:50 AM - Re: Rotary engine displacement (Tedd McHenry)
4. 09:30 AM - Re: Rotary engine displacement (Tedd McHenry)
5. 10:37 AM - Re: Rotary engine displacement (Tracy Crook)
6. 05:03 PM - Re: Rotary engine displacement (Royce Wise)
Message 1
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| Subject: | Rotary engine displacement |
--> Engines-List message posted by: "Gary Casey" <glcasey@adelphia.net>
<<Can you point to a source for these numbers? It's my understanding that
each chamber of a 13B displaces 654 cc's, which gives 1308 cc's for a single
rotation of the e-shaft. I've believed that for about 20 years, so I've got
a couple decades of unlearning to do if it's wrong :-)
Cheers,
Rusty
Turbo 13B RV-3 >>
I'm afraid I can't provide an official reference and I vaguely remember 654
being the number told by some people, although the "rated" displacement is
1,308 and I do know that is the displacement of 4 of the 6 chambers. I
think the 654 number came from the early days when some were saying the
engine is really a 2-stroke (The ported intake and exhaust ports do remind
one of a piston 2-stroke. One power stroke per rev, ported breathing
events, ergo 654 cc two-stroke.). It sounded more impressive to get all
that power from only 654 cc's. It doesn't sound impressive if you count all
the chambers and say it is almost 2 liters of displacement. Do you count
only one chamber in the rotor or all three? I say you have to count all
three as they are truly 3 independent chambers. There is no rationale in my
mind for counting only 2 of the 3 chambers swept by each rotor. It was just
a method created by the early race organizers to put them in a competitive
category. Call them 654 and they'll win every race. 2-liter and they will
lose. 1308 - just right. So 1308 it was. Just my opinion, observing the
events.
Gary Casey
Message 2
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| Subject: | Rotary engine displacement |
--> Engines-List message posted by: "Russell Duffy" <13brv3@bellsouth.net>
although the "rated" displacement is 1,308 and I do know that is the
displacement of 4 of the 6 chambers.
----------------
This is our basic disagreement. I'm quite certain that 1308 is the
displacement for one revolution of the e-shaft (2 of 6 chambers). Each
chamber is 654 cc, which is the number that's listed in the Mazda workshop
manual. Just as a sanity check, find a good picture of a rotor in a
housing, and do a rough calculation of the intake and compression areas. It
will be very clear that 654 is the correct number.
I do agree that there has been lots of confusion over the displacement.
Most of this is because folks are trying to make the rotary fit the 4-cycle
conventions.
Cheers,
Rusty
Message 3
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| Subject: | Re: Rotary engine displacement |
--> Engines-List message posted by: Tedd McHenry <tedd@vansairforce.org>
On Wed, 8 Oct 2003, Gary Casey wrote:
--snip--
> There is no rationale in my mind for counting only 2 of the 3 chambers swept
> by each rotor. It was just a method created by the early race organizers to
> put them in a competitive category. Call them 654 and they'll win every
> race. 2-liter and they will lose. 1308 - just right. So 1308 it was. Just
> my opinion, observing the events.
Gary:
You make it sound arbitrary, but it's not. The displacement of an internal
combustion engine is, by definition, the swept volume of the combustion
chambers for each revolution of the output shaft. This makes engineering
sense, since it is the number of combustion cycles per revolution of the output
shaft (times the power of each cycle) that determines the output power of the
engine.
Since the output shaft of a rotary engine turns at 1/3 of the rotor speed, it
sweeps only one of the three chambers of each rotor for each revolution of the
output shaft. Hence, the displacement of a two-rotor engine is 2/3 of the
displacement of all the chambers.
Tedd McHenry
Surrey, BC
Message 4
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| Subject: | Re: Rotary engine displacement |
--> Engines-List message posted by: Tedd McHenry <tedd@vansairforce.org>
On Wed, 8 Oct 2003, Tedd McHenry wrote:
> Since the output shaft of a rotary engine turns at 1/3 of the rotor speed,
Sorry, I said that backwards. The rotor turns at 1/3 the speed of the output
shaft.
Tedd
Message 5
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| Subject: | Re: Rotary engine displacement |
--> Engines-List message posted by: "Tracy Crook" <lors01@msn.com>
>
> <<Can you point to a source for these numbers? It's my understanding that
> each chamber of a 13B displaces 654 cc's, which gives 1308 cc's for a
single
> rotation of the e-shaft. I've believed that for about 20 years, so I've
got
> a couple decades of unlearning to do if it's wrong :-)
>
> Cheers,
> Rusty
> Turbo 13B RV-3 >>
>
> I'm afraid I can't provide an official reference and I vaguely remember
654
> being the number told by some people, although the "rated" displacement is
> 1,308 and I do know that is the displacement of 4 of the 6 chambers. I
> think the 654 number came from the early days when some were saying the
> engine is really a 2-stroke (The ported intake and exhaust ports do remind
> one of a piston 2-stroke. One power stroke per rev, ported breathing
> events, ergo 654 cc two-stroke.). It sounded more impressive to get all
> that power from only 654 cc's. It doesn't sound impressive if you count
all
> the chambers and say it is almost 2 liters of displacement. Do you count
> only one chamber in the rotor or all three? I say you have to count all
> three as they are truly 3 independent chambers. There is no rationale in
my
> mind for counting only 2 of the 3 chambers swept by each rotor. It was
just
> a method created by the early race organizers to put them in a competitive
> category. Call them 654 and they'll win every race. 2-liter and they
will
> lose. 1308 - just right. So 1308 it was. Just my opinion, observing the
> events.
>
> Gary Casey
Back to my original "Sanity Check" formula. It takes about 1.59 CFM of
fuel/air mixture to make each and every HP coming out of the crank
regardless of what kind of gasoline engine we are talking about. If the
Mazda 13B Mazda pumps 2 chambers of mixture every revolution (I hope we can
at least agree on that) which would make 654 cc per rev according to your
figures. This equates to 40 cubic inches. Multiply this times 6000 and you
get 240000 cubic inches. Divide this by 1728 (CI per cubic foot) and we get
138.888 CFM. Divide this by 1.59 and we get 87.35 which is the amount of HP
the engine should make at 6000 rpm.
If the engine only made 87.35 HP, I never could have won the Sun 100 race
at Sun & Fun this year (Class 8, 160HP RV class). So, either my formula
is wrong or the displacement of the 13B is actually double what Gary says it
is. Using the latter assumption, I was making 174 HP, which would explain
why I was 10+ knots faster than the fastest 0 - 320 powered RV.
But really, all you have to do is take one look inside the 13B engine to see
that one chamber is way bigger than 327 cc (20 CI).
Tracy Crook
Message 6
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| Subject: | Re: Rotary engine displacement |
--> Engines-List message posted by: Royce Wise <royce@wayxcable.com>
Tedd McHenry wrote:
>--> Engines-List message posted by: Tedd McHenry <tedd@vansairforceorg>
>
>On Wed, 8 Oct 2003, Gary Casey wrote:
>
>--snip--
>
>
>>There is no rationale in my mind for counting only 2 of the 3 chambers swept
>>by each rotor It was just a method created by the early race organizers to
>>put them in a competitive category Call them 654 and they'll win every
>>race 2-liter and they will lose 1308 - just right So 1308 it was Just
>>my opinion, observing the events
>>
>>
>
>Gary:
>
>You make it sound arbitrary, but it's not The displacement of an internal
>combustion engine is, by definition, the swept volume of the combustion
>chambers for each revolution of the output shaft This makes engineering
>sense, since it is the number of combustion cycles per revolution of the output
>shaft times the power of each cycle that determines the output power of the
>engine
>
>Since the output shaft of a rotary engine turns at 1/3 of the rotor speed, it
>sweeps only one of the three chambers of each rotor for each revolution of the
>output shaft Hence, the displacement of a two-rotor engine is 2/3 of the
>displacement of all the chambers
>
>Tedd McHenry
>Surrey, BC
>
Ted is correct, it is the output shaft that you deal with
Take away one of the rotors and look at it that way
Royce
>
>
>
>
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