---------------------------------------------------------- Engines-List Digest Archive --- Total Messages Posted Wed 10/08/03: 6 ---------------------------------------------------------- Today's Message Index: ---------------------- 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 _____________________________________ Time: 06:05:05 AM PST US From: "Gary Casey" Subject: Engines-List: Rotary engine displacement --> Engines-List message posted by: "Gary Casey" <> 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 _____________________________________ Time: 08:05:24 AM PST US From: "Russell Duffy" <13brv3@bellsouth.net> Subject: RE: Engines-List: 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 _____________________________________ Time: 08:50:10 AM PST US From: Tedd McHenry Subject: Re: Engines-List: Rotary engine displacement --> Engines-List message posted by: Tedd McHenry 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 _____________________________________ Time: 09:30:10 AM PST US From: Tedd McHenry Subject: Re: Engines-List: Rotary engine displacement --> Engines-List message posted by: Tedd McHenry 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 _____________________________________ Time: 10:37:57 AM PST US From: "Tracy Crook" Subject: Re: Engines-List: Rotary engine displacement --> Engines-List message posted by: "Tracy Crook" > > < 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 _____________________________________ Time: 05:03:54 PM PST US From: Royce Wise Subject: Re: Engines-List: Rotary engine displacement --> Engines-List message posted by: Royce Wise Tedd McHenry wrote: >--> Engines-List message posted by: Tedd McHenry > >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 > > > >