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1.3L calculation

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Old Nov 12, 2006 | 04:39 PM
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1.3L calculation

just wondering how our rotary engines' displacement were calucated. i recieved this comment.

Mazda is full of ****. The rotary is NOT a 2-stroke engine. When they set out the parameters of designating the displacement of a rotary, they only calculated it for one crank revolution, the same way a 2-stroke displacement is calculated. Since a rotary is a 4-stroke by all means of the definition, displacement should be calculated for two complete revolutions of the crank. And since the rotors only spin 1/3 the speed of the eccentric shaft due to the gear drive, it really isn't a 1.3L. The only way it could be a 1.3L is if it was a 2-stroke engine. They just decided to calculate it this way because it fires every revolution like a 2-stroke. But it still has 4 complete cycles which means Mazda engineers were either stupid when they did this, of just didn't give a damn at getting the facts right

please tell me.
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Old Nov 12, 2006 | 05:04 PM
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Arrow

https://www.rx7club.com/rotary-car-performance-77/rotary-capacity-yet-again-269803/
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Old Nov 12, 2006 | 10:17 PM
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It's too bad the link provided in that thread lacks some very fundamental facts and makes assumptions based off of others. That's what happens when someone with zero rotary experience tries to write a paper on how it works. It's correct in many aspects but he didn't get it completely correct. As the old saying goes, "close but no cigar". He ignores some obvious important facts and brushes them off as "coincendences". After a while I just gave up in that thread and am done with it.
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Old Nov 13, 2006 | 02:40 AM
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ha-ha- FUNNY

Originally Posted by pinkrx7
just wondering how our rotary engines' displacement were calucated. i recieved this comment.

Mazda is full of ****. The rotary is NOT a 2-stroke engine. When they set out the parameters of designating the displacement of a rotary, they only calculated it for one crank revolution, the same way a 2-stroke displacement is calculated. Since a rotary is a 4-stroke by all means of the definition, displacement should be calculated for two complete revolutions of the crank. And since the rotors only spin 1/3 the speed of the eccentric shaft due to the gear drive, it really isn't a 1.3L. The only way it could be a 1.3L is if it was a 2-stroke engine. They just decided to calculate it this way because it fires every revolution like a 2-stroke. But it still has 4 complete cycles which means Mazda engineers were either stupid when they did this, of just didn't give a damn at getting the facts right

please tell me.
Hi.

I couldn't help noticing this text : they only calculated it for one crank revolution, the same way a 2-stroke displacement is calculated.

Hello ! Aren't all engines calculated for one rev, regarding 2 or 4 stroke?

What a NUT.

He-He
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Old Nov 13, 2006 | 12:35 PM
  #5  
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Actually no. Displacement is calculated based on total swept volume of all cylinders in a piston engine. They didn't use the same standard in a rotary. When we talk about piston engine size we may say we have a 2 Liter 2 stroke or a 2 Liter 4 stroke engine. That doesn't tell us how much air they flow over time. It's just swept volume of all cylinders. A rotary technically should be called a 3.9 L 6 stroke but let's not get into that here. Crank rotation is important to know when you are trying to calculate airflow through an engine as you need a frame of reference over time. That's what rpm's are, revolutions per minute. That is the issue that the poster in the other thread is disagreeing with and he's flat wrong.
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