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Rotary engine = 2-cycle engine?

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Old 08-09-02 | 11:47 AM
  #26  
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Originally posted by AlmightyRX3
not to sound stupid, but they really arent any type of "Stroke" considering a 4 stroke and 2 stroke engine have the term STROKE derived fromt eh stroke of a crankshaft moving the piston up and down. Rotaries dont have a stroke .
qualitativey, it does have 4 cycles or 'strokes of the oar'.

quantitatively, it does not have a linear stroke of a piston engine. it does have an effective stroke, based on the displaced volume rating of .65L pre face for 13B. the effective stroke would be the .65L/A, where A is area per face. But as is the case with the piston engine, the eccentric shaft offset would be used in any bmep calculation, and not the effective stroke.
Old 08-10-02 | 12:24 AM
  #27  
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Originally posted by KevinK2
5 events? why not include intake and exh open and close events for a total of 9?
I don't know. Maybe you should talk to SAE about that?
http://www.sae.org/servlets/index

Originally posted by KevinK2
main problem is too much emphasis on words like cycle and stroke, which are non specific terms. ex, you can have 'power' and 'return' strokes when rowing a boat, no pistons needed. One can buy 2-cycle oil for a 2-stroke heat engine. Just words.
This is true. Lots of words, but no defined standards. I'm rather surprised that some of you are not familiar with the "event" term, though.

Originally posted by KevinK2
Hopefully u know Brayton refers to 'turbine cycles' that continuously produce power with no crankshaft needed. and the sterling engines use external heat and no combustion products.
LOL, yes, I have 8 years of experience with open-cycle Brayton engines.

Apparently you are not aware of the Brayton/Otto comparison to the Wankel. Not that I agree with the reasoning, but it is based on the mechanical aspects of each engine type. The Otto uses a shaft with a full crank for linear motion of the pistons for compression, a Brayton cycle gas turbine uses a straight shaft for rotational compression, while a Wankel uses a combination of the two for compression. This is also where the Wankel "stroke" theory comes from, as the eccentric shaft may be said to have a partial "stroke" (ie offset from centerline), while a piston engine crankshaft has a full stroke (or half stroke, depending on how you look at it), and a gas turbine shaft has no offset from centerline which would be measured as 0 stroke. As you wrote, it all comes down to one's definition of "stroke", and where it is measured.

Originally posted by KevinK2
The 'otto cycle' can refer to the P-V behavior of any heat engine with the same 4 thermodynamic steps. The wankel engine has an otto cycle P-V curve, the 4 stroke diesel does not.
Hey, what's this P-V diagram? Hmm, it looks familiar somehow That T-s diagram looks funky, though. Why, it's a Stirling cycle, showing why cycles are not classified solely on their P-V diagram.


Hey, this Stirling cycle engine somehow looks familiar, too, but I can't quite place the resemblance.
Old 08-10-02 | 05:45 AM
  #28  
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Originally posted by KevinK2


qualitativey, it does have 4 cycles or 'strokes of the oar'.

quantitatively, it does not have a linear stroke of a piston engine. it does have an effective stroke, based on the displaced volume rating of .65L pre face for 13B. the effective stroke would be the .65L/A, where A is area per face. But as is the case with the piston engine, the eccentric shaft offset would be used in any bmep calculation, and not the effective stroke.
SHEESH OK Mr. Wizard. is the atomic weight of Cobalt .58???

lol i was just beign a smartass, to get people to laugh
Old 08-10-02 | 05:15 PM
  #29  
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aviator ....

sae .... I can't even find an sae spec for displacement. Best soft spec I could find was 'swept volume per engine cycle' which puts 13b rotary at 3.9L.

Like u, I see no link in mechanical aspects of turbines vs wankels. turbines are continuous processes with fans, wankels and boingers are batch with cranks.

Regarding PV's, as your example shows, the complete diagram includes event specifications, which are idealized as adiabatic for the ideal otto cycle.

The eccentric does create torque based on offset and the tangential vector load at the rotor bearing. This is just like piston engines, and unrelated to turbines.

Regarding sterling wankle, I suspect it's a lot easier to animate than build, esp with .6" crank offset like an fd. Any real working versions?
Old 08-10-02 | 11:19 PM
  #30  
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Originally posted by KevinK2
The eccentric does create torque based on offset and the tangential vector load at the rotor bearing. This is just like piston engines, and unrelated to turbines.
Yes, the eccentric does create torque, but so does the rotation of the rotor. Turbine engines also produce torque despite a centerline shaft with 0 stroke.

Originally posted by KevinK2
Regarding sterling wankle, I suspect it's a lot easier to animate than build, esp with .6" crank offset like an fd. Any real working versions?
Hehehe, as far as I know it's just someone's dream, but it looks much more feasible than an RX-7 air pump "supercharger". Here is the link where I found the picture:
http://www-ifkm.mach.uni-karlsruhe.d.../stirling.html
Old 08-11-02 | 07:37 PM
  #31  
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"Yes, the eccentric does create torque, but so does the rotation of the rotor...."

Don't think rotor rotation produces any significant torque, except mabe resistive torque due to friction.

Each rotor face projected area is centered on the rotor spin bearing, so no pressure induced tendency to spin rotor, assuming power cycle pressure does not vary across the rotor face. The fixed gear mesh is for rotor positioning, not torque generation, afaik.
Old 08-13-02 | 12:51 PM
  #32  
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Originally posted by KevinK2
"Yes, the eccentric does create torque, but so does the rotation of the rotor...."

Don't think rotor rotation produces any significant torque, except mabe resistive torque due to friction.

Each rotor face projected area is centered on the rotor spin bearing, so no pressure induced tendency to spin rotor, assuming power cycle pressure does not vary across the rotor face. The fixed gear mesh is for rotor positioning, not torque generation, afaik.
Anything that produces rotation will also produce torque. Here is a web page ref the torque of a gas turbine engine which has no crankshaft yet still produces torque.
http://www.turbinecar.com/turbook/pg02b.htm
Old 08-14-02 | 01:34 AM
  #33  
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Originally posted by Evil Aviator

Anything that produces rotation will also produce torque. Here is a web page ref the torque of a gas turbine engine which has no crankshaft yet still produces torque.
http://www.turbinecar.com/turbook/pg02b.htm
a rotating turbo hot wheel or a gas turbine, both produce shaft torque. This is irrelavant to the issue of whether the rotation of the wankel rotor, free of pressure induced torque, also create shaft torque, as does the eccentric as I noted. Need explanation on this, not gas turbines
Old 08-15-02 | 12:04 AM
  #34  
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I don't quite understand your point. Maybe it's because I'm just a dumb pilot an not an engineer or mechanic.

Originally posted by KevinK2
[BEach rotor face projected area is centered on the rotor spin bearing, so no pressure induced tendency to spin rotor, assuming power cycle pressure does not vary across the rotor face.[/B]
OK, I'm going to take a guess here based on this comment. Take a look at this animation and see where the leading plug fires. See the moment arm? There's your rotor torque.
http://www.monito.com/wankel/j-wankel.html

The eccentric shaft is for compression, not for torque production, although it does increase the moment arm of the rotor which adds to the torque. As far as I can tell, you could run a Wankel engine with a straight shaft and round rotor housing, but it would have no way to make compression. Hydrogen peroxide maybe?
Old 08-15-02 | 12:21 AM
  #35  
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dont compare rotary with piston engines its like comparing apples and kiwis
Old 08-15-02 | 12:25 AM
  #36  
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one more thing, this thing is getting too technical for me
Old 08-15-02 | 02:48 PM
  #37  
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Originally posted by Evil Aviator
I don't quite understand your point. Maybe it's because I'm just a dumb pilot an not an engineer or mechanic.


OK, I'm going to take a guess here based on this comment. Take a look at this animation and see where the leading plug fires. See the moment arm? There's your rotor torque.
http://www.monito.com/wankel/j-wankel.html

The eccentric shaft is for compression, not for torque production, although it does increase the moment arm of the rotor which adds to the torque. As far as I can tell, you could run a Wankel engine with a straight shaft and round rotor housing, but it would have no way to make compression. Hydrogen peroxide maybe?
the flame front propegation is almost instant, as compared with the main power cycle, so any related unbalanced load will generate trivial torque.

check this link:

http://web.ukonline.co.uk/Members/jr.marsh/wankel2.html

this shows quite well how the eccentric shaft is fundamental and critical in torque used in comprssion and generated in power cycle. remember the eccentric offset centers in the rotor, and the shaft spin axis centers on the fixed gear shown. Ignore their general cycle descriptions, and focus on the positions of the chambers shown in orange (fig 1 and 2) and red for fig 3 and 4:

fig1 oran, beginning of compr stroke (top chamber)

fig2 oran, mid compression ... note max torque position of eccentric, relative to projected area of active rotor face.

fig3 red, TDC-ignition no torque created, since no eccentric offset, just like piston eng.

fig4 red, mid power cycle ... eccentric positioned for best torque output (like piston engine).

no offset, no torque for mazda's.

hope this clears it up for u, if not we can agree to disagree.

I can't fly a kite, but am a practicing mechanical engineering consultant, board certified, and have lectured at the local university. very little significance except in discussions like this.
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