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	<title>The other blog of Miguel Pereira &#187; rx8</title>
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		<title>Follow Up: Rotary Engines Illustrated!</title>
		<link>http://www.myrotarycar.com/blog/index.php/2009/02/06/follow-up-rotary-engines-illustrated/</link>
		<comments>http://www.myrotarycar.com/blog/index.php/2009/02/06/follow-up-rotary-engines-illustrated/#comments</comments>
		<pubDate>Fri, 06 Feb 2009 16:06:49 +0000</pubDate>
		<dc:creator>Miguel Pereira</dc:creator>
				<category><![CDATA[Cars]]></category>
		<category><![CDATA[Technology]]></category>
		<category><![CDATA[mazda]]></category>
		<category><![CDATA[rotary engine]]></category>
		<category><![CDATA[rx8]]></category>

		<guid isPermaLink="false">http://www.myrotarycar.com/blog/?p=235</guid>
		<description><![CDATA[<p style="text-align: justify;">I thought it would be prudent to throw in some material that can help human beings like me visualize the construction and cycles of a rotary engine. All I had to do was a quick search through YouTube.</p>
<p style="text-align: justify;">The rotary engine anatomy
</p>
<p>Here is a Car &#38; Driver TV segment talking about a Mazda [...]]]></description>
			<content:encoded><![CDATA[<p style="text-align: justify;">I thought it would be prudent to throw in some material that can help human beings like me visualize the construction and cycles of a rotary engine. All I had to do was a quick search through YouTube.</p>
<p style="text-align: justify;"><strong>The rotary engine anatomy</strong><br />
<object width="640" height="505" data="http://www.youtube.com/v/TxqCtXVJoEY&amp;hl=en&amp;fs=1" type="application/x-shockwave-flash"><param name="allowFullScreen" value="true" /><param name="allowscriptaccess" value="always" /><param name="src" value="http://www.youtube.com/v/TxqCtXVJoEY&amp;hl=en&amp;fs=1" /><param name="allowfullscreen" value="true" /></object></p>
<p><strong>Here is a Car &amp; Driver TV segment talking about a Mazda RX-8 Renesis&#8217; engine (13B)</strong><br />
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		<title>RX-8: Stock Air/Fuel Ratio</title>
		<link>http://www.myrotarycar.com/blog/index.php/2009/02/05/rx-8-stock-airfuel-ratio/</link>
		<comments>http://www.myrotarycar.com/blog/index.php/2009/02/05/rx-8-stock-airfuel-ratio/#comments</comments>
		<pubDate>Thu, 05 Feb 2009 19:39:50 +0000</pubDate>
		<dc:creator>Miguel Pereira</dc:creator>
				<category><![CDATA[Cars]]></category>
		<category><![CDATA[Technology]]></category>
		<category><![CDATA[mazda]]></category>
		<category><![CDATA[rotary engine]]></category>
		<category><![CDATA[rx8]]></category>
		<category><![CDATA[wankel]]></category>

		<guid isPermaLink="false">http://www.myrotarycar.com/blog/?p=222</guid>
		<description><![CDATA[<p style="TEXT-ALIGN: justify">Here is another oldie from the my old collection of Rotary Engine &#38; RX8 articles. Original publish date 1/24/2005</p>
<p style="TEXT-ALIGN: justify">A brief introduction to Air/Fuel ratios</p>

<p style="TEXT-ALIGN: justify">Definition: the mass of air supplied to the engine divided by the mass of fuel supplied in the same period of time. The Stoichiometric, or chemically correct, [...]]]></description>
			<content:encoded><![CDATA[<p style="TEXT-ALIGN: justify">Here is another oldie from the my old collection of Rotary Engine &amp; RX8 articles. Original publish date 1/24/2005</p>
<p style="TEXT-ALIGN: justify"><strong>A brief introduction to Air/Fuel ratios</strong></p>
<blockquote>
<p style="TEXT-ALIGN: justify">Definition: the mass of air supplied to the engine divided by the mass of fuel supplied in the same period of time. The Stoichiometric, or chemically correct, air-fuel ratio (A/F ratio) is the exact ratio necessary to burn all the carbon and hydrogen in the fuel to carbon dioxide and water with no oxygen remaining. The fuel-air ratio is the reciprocal of the air-fuel ratio.</p>
</blockquote>
<p style="TEXT-ALIGN: justify">Stable combustion conditions require the right amounts of fuel and oxygen. The combustion products are heat energy, carbon dioxide, water vapor, nitrogen, and other gases (excluding oxygen). In theory there is a specific amount of oxygen needed to completely burn a given amount of fuel. In practice, burning conditions are never ideal.</p>
<p style="TEXT-ALIGN: justify"><span id="more-222"></span></p>
<p style="TEXT-ALIGN: justify">When air and gasoline are mixed together and ignited, the chemical reaction requires a certain amount of air to completely burn all of the fuel. The exact amount is 14.7 lbs of air for every pound of fuel. This is called the &#8220;Stoichiometric&#8221; Air/Fuel ratio. It&#8217;s also referred to the Greek letter &#8220;lambda.&#8221;</p>
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<p style="TEXT-ALIGN: justify">When lambda equals one, you have a 14.7:1 Stoichiometric Air/Fuel ratio and ideal combustion. When the Air/Fuel ratio is greater than 14.7:1, lambda also will be greater than one and the engine will have a lean mixture.</p>
<p style="TEXT-ALIGN: justify">Lean mixtures improve fuel economy but also cause a sharp rise in oxides of nitrogen (NOX). If the mixture goes too lean, it may not ignite at all causing &#8220;lean misfire&#8221; and a huge increase in unburned hydrocarbon (HC) emissions. This can cause rough idle, hard starting and stalling, and may even damage the catalytic converter. Lean mixtures also increase the risk of spark knock (detonation) when the engine is under load.</p>
<p style="TEXT-ALIGN: justify">When the Air/Fuel ratio is less than 14.7:1, lambda also is less than one and the engine has a rich fuel mixture. A rich fuel mixture is necessary when a cold engine is first started, and additional fuel is needed when the engine is under load. But rich mixtures cause a sharp increase in carbon monoxide (CO) emissions. When the relative proportions of air and fuel are &#8220;just right,&#8221; the mixture burns clearly and produces the fewest emissions. The trick is balancing the mixture as driving conditions, temperatures and loads are constantly changing.</p>
<p style="TEXT-ALIGN: justify"><strong>Enough theory, let&#8217;s go to the beef&#8230;</strong><br />
Several owners -me included- who have been toying around with CANScan tools, and doing real time data logging of the operational parameters of their RX-8s during normal / aggressive driving conditions have concluded that the stock Air/Fuel maps provided by Mazda tend to provide a richer than desired overall mix.</p>
<p style="TEXT-ALIGN: justify">Please take a look at the chart below. The RED horizontal line represents what -at least- I would consider the ideal Air/Fuel ratio for the given Engine RPM range and acceleration conditions. Notice that this sample was taken from 5,000rpms through 8,300rpms -for reference, peak power output on the Renesis engine is advertised @ 8,500rpms but, we&#8217;ll talk about that later on. It is also worth noting that this sample was recorded while the engine management was operating under an open loop state -I will touch on that as well on a later article.</p>
<p style="TEXT-ALIGN: justify">Now lets take a look at the PURPLE trace, which represents the actual AF measured by the RX-8&#8242;s own Wide Band O2 sensor (WBO2). It is stuck @ 11.xx AF, which is an extremely rich mix. Nearly as rich as required by some turbocharged / supercharged engines running stupid amounts of boost. The YELLOW trace represents the Stoichiometric AF ratio -which is observed, mostly, under closed loop operation. As explained before, the Stoichiometric ratio indicates the necessary mix of Air &amp; Fuel to eliminate the emission of combustion byproducts into the atmosphere. However, under high load conditions, it is desirable to maintain a semi-rich mix to prevent detonation and knock. Rotary engines are particularly sensitive to such phenomenon.</p>
<div id="attachment_323" class="wp-caption aligncenter" style="width: 843px"><a href="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/2005-11-22_231958_2004_mazda_rx8_air_fuel_011705.gif"><img class="size-full wp-image-323" title="Open loop A/F curve on a stock '04 Mazda RX-8." src="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/2005-11-22_231958_2004_mazda_rx8_air_fuel_011705.gif" alt="Open loop A/F curve on a stock '04 Mazda RX-8." width="833" height="569" /></a><p class="wp-caption-text">Open loop A/F curve on a stock &#39;04 Mazda RX-8.</p></div>
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<p style="TEXT-ALIGN: justify"><strong>Open Loop / Closed Loop briefly&#8230;</strong><br />
To briefly clarify two concepts, the definition of OPEN &amp; CLOSED loop, here is what they entail to the best of my knowledge:</p>
<p style="TEXT-ALIGN: justify"><strong>CLOSED LOOP</strong>: observed under light loads, partial throttle &amp; cruising conditions. The RX-8&#8242;s ECU is constantly monitoring the A/F state (Rich or Lean) reported by the Narrow Band Oxygen sensor located on the exhaust and adjusting fuel injector cycle (fuel delivery) to maintain a ratio close to Stoichiometric (14.7 A/F). These adjustments are simply leaning or richening the mix by a fixed amount, until the ECU detects an opposite condition, thus it &#8220;steps&#8221; back on its last adjustment. Generally, the A/F ratio is maintained @ 14.7 to maximize catalytic converter efficiency.</p>
<p style="TEXT-ALIGN: justify"><strong>OPEN LOOP</strong>: observed under wide open throttle (WOT) or high loads. The ECU will ignore the Oxygen Sensor and will receive information from Mass Air Flow (MAF), RPM, vehicle speed, load, and potentially other sources. It then determines the appropriate amount of fuel based on a lookup table stored in memory. Under this condition, the ECU no longer attempts to keep a Stoichiometric A/F ratio, but rather a richer mix. Such mix is intended to protect the engine from knock / pinging.</p>
<p style="TEXT-ALIGN: justify">Lets take a look at the closed loop operation of the Mazda RX-8. Notice that during the interval of steady acceleration -light load- the A/F ratio remains pretty close to the Stoichiometric 14.7; however, upon close examination, you will notice that the ratio fluctuates above/below &#8220;stoich.&#8221; That corresponds to the ECU reading values from the narrow band oxygen sensor, and applying a correction value (fuel trim) to the injector duty cycle. Yes, it will rarely ever stay put on 14.7. Why? Because the driving conditions rarely remains constant, and in order to maintain a neatly &#8220;stoich&#8221; ratio, the ECU has to constantly adjust. This adjustments, could be explained by the following kindergarten logic:</p>
<p style="TEXT-ALIGN: justify">Is there too much fuel on the mix? Then decrease fuel supply by X.<br />
Is there too little fuel on the mix? Then increase fuel supply by X.<br />
or&#8230;</p>
<p style="TEXT-ALIGN: justify">If the mix is too rich, decrease fuel injector duty cycle by X.<br />
If the mix is too lean, increase fuel injector duty cycle by X.</p>
<p style="TEXT-ALIGN: justify">As I mentioned earlier, the ECU is constantly asking and reacting to those 2 parameters. The value represented by X corresponds to the Fuel Trim stored in memory. Fuel Trim can be Short Term or Long Term (more on that some other day!)</p>
<div id="attachment_324" class="wp-caption aligncenter" style="width: 921px"><a href="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/2005-11-22_232004_2004_mazda_rx8_air_fuel_011705a.gif"><img class="size-full wp-image-324" title="Closed loop A/F curve on a stock '04 Mazda RX-8." src="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/2005-11-22_232004_2004_mazda_rx8_air_fuel_011705a.gif" alt="Closed loop A/F curve on a stock '04 Mazda RX-8." width="911" height="623" /></a><p class="wp-caption-text">Closed loop A/F curve on a stock &#39;04 Mazda RX-8.</p></div>
<p style="TEXT-ALIGN: justify"><strong>A word of caution&#8230;<br />
</strong>There is one last thing I&#8217;d like to make clear before wrapping up this article.</p>
<p style="TEXT-ALIGN: justify">We all know engines run @ high temperatures. The leaner the A/F ratio, the less fuel the injectors will supply, the higher the temperature of the combustion will be. On the other hand, the more fuel in the mix the cooler the combustion will be (too rich will foul plugs, etc)</p>
<p style="TEXT-ALIGN: justify">Running under lean conditions all the time will raise the combustion chamber temperature. The higher the temprature, the higher the likelyhood of detonation. What is detonation? Let&#8217;s just keep it at this: in a piston engine, detonation generaly means that the mix is combusted before the piston has reached maximum compression. Or even worse, even before the intake cycle is complete, having the intake valve(s) still open. This allows the flame to travel outside of the cylinder&#8230;..and&#8230; you can make out the rest.</p>
<p style="TEXT-ALIGN: justify">Guess what? A way of preventing detonation is by using fuels that have higher flame thresholds. Thus being less prone to being ignited without direct exposure to spark. Now, ever wondered why 87, 89, 91, 93 &amp; 100 octane fuels exist? You got it! The higher the octane rating, the more difficult to ignite the fuel is. Sounds counterintuitive? You&#8217;ll get used to it.</p>
<p style="TEXT-ALIGN: justify">Alright, by now you are wondering where am I going with this? For now, just a friendly piece of advise: DO NOT ASSUME THAT BY LEANING YOUR A/F RATIOS WITH ONE OF THOSE NICE ELECTRONIC TOYS (eManage, SAFC &amp; SAFC-II, etc&#8230;) YOU WILL BE GAINING POWER WITHOUT ANY TRADEOFFS. It is very important that you have the necessary tools and mechanisms to detect knock, detonation or pinging on your engine before it is too late. Those three conditions are caused mainly by excessive combustion temperatures from a lean mix condition.</p>
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		<title>Rotary Engines: Volumetric Efficiency</title>
		<link>http://www.myrotarycar.com/blog/index.php/2009/02/05/rotary-engines-volumetric-efficiency/</link>
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		<pubDate>Thu, 05 Feb 2009 19:26:53 +0000</pubDate>
		<dc:creator>Miguel Pereira</dc:creator>
				<category><![CDATA[Cars]]></category>
		<category><![CDATA[Technology]]></category>
		<category><![CDATA[mazda]]></category>
		<category><![CDATA[rotary engine]]></category>
		<category><![CDATA[rx8]]></category>
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		<description><![CDATA[<p class="wp-caption-text">Renesis Rotary Engine</p>
<p>Yet another one of my old Rotary Engine technical articles (originally published on 2/2/2005) I managed to recover from my old site. Enjoy it!</p>
<p> </p>
<p style="text-align: justify;">Characteristics of a rotary engine compared to a 4 stroke piston engine:</p>
<p style="text-align: justify;">A rotary engine&#8217;s rotor completes one stroke for every 270º of crank rotation:</p>


Intake 270º of [...]]]></description>
			<content:encoded><![CDATA[<div id="attachment_204" class="wp-caption alignright" style="width: 160px"><img class="size-thumbnail wp-image-204 " title="Mazda RX-8" src="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/rx8-tx_vred_2-150x150.jpg" alt="Renesis Rotary Engine" width="150" height="150" /><p class="wp-caption-text">Renesis Rotary Engine</p></div>
<p>Yet another one of my old Rotary Engine technical articles (originally published on 2/2/2005) I managed to recover from my old site. Enjoy it!</p>
<p> </p>
<p style="text-align: justify;"><strong>Characteristics of a rotary engine compared to a 4 stroke piston engine:</strong></p>
<p style="text-align: justify;">A rotary engine&#8217;s rotor completes one stroke for every 270º of crank rotation:</p>
<ul>
<li>
<div style="text-align: justify;">Intake 270º of crank rotation.</div>
</li>
<li>
<div style="text-align: justify;">Compression 540º of crank rotation.</div>
</li>
<li>
<div style="text-align: justify;">Combustion 810º of crank rotation.</div>
</li>
<li>
<div style="text-align: justify;">Exhaust 1080º of crank rotation.</div>
</li>
</ul>
<p style="text-align: justify;">In other words, it takes a rotary engine 1080º of crankshaft rotation to complete an intake, compression, combustion &amp; exhaust cycle. Or 3 crankshaft rotations per cycle.</p>
<p style="text-align: justify;">A piston completes one stroke every 180º of crank rotation:</p>
<ul>
<li>
<div style="text-align: justify;">Intake 180º of crank rotation.</div>
</li>
<li>
<div style="text-align: justify;">Compression 360º of crank rotation.</div>
</li>
<li>
<div style="text-align: justify;">Combustion 540º of crank rotation.</div>
</li>
<li>
<div style="text-align: justify;">Exhaust 720º of crank rotation.</div>
</li>
</ul>
<p style="text-align: justify;">A piston engine requires 720º of crankshaft rotation to complete a cycle. In other words, 2 complete revolutions of the crankshaft.</p>
<p style="text-align: justify;">A rotor rotates @ 1/3 of the rate of the crankshaft. In other words, for every 1 revolution of a rotor, the crankshaft has done 3 revolutions. For example when the tachometer on a vehicle indicates 9000rpms, one rotor is turning at 3000rpms.</p>
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<p style="text-align: justify;"><span id="more-216"></span></p>
<p style="text-align: justify;">On two rotor engines, the front &amp; rear rotors are offset 180º from each other. A rotation of 360º of the crankshaft will bring the 2 rotors through the combustion stroke. Since each combustion chamber is -in the case of a 13B- 654cc, every 360º of crankshaft rotation will displace a total of 1308cc.</p>
<p style="text-align: justify;">In order to interpolate the cycles and volume displaced by a rotary engine vs. a 4 piston engine, we can use the following logic:</p>
<ul>
<li>
<div style="text-align: justify;">A piston engine takes 720º of crank rotation to complete a cycle.</div>
</li>
<li>
<div style="text-align: justify;">In a rotary engine, 720º will produce 4 combustion strokes:</div>
</li>
<li>
<div style="text-align: justify;">360º of crank rotation =&gt; 2 combustion strokes.</div>
</li>
<li>
<div style="text-align: justify;">720º = 360º x 2</div>
</li>
<li>
<div style="text-align: justify;">720º of crank rotation =&gt; 4 combustion strokes.</div>
</li>
<li>
<div style="text-align: justify;">4 strokes = 654cc x 4</div>
</li>
<li>
<div style="text-align: justify;">4 strokes = 2616cc</div>
</li>
</ul>
<p style="text-align: justify;">For simplicity reasons, we can stipulate that a 1.3L, two rotor rotary engine is similar to a 2.6L 4 piston, 4 stroke piston engine. It may not be academically correct, but it is a relatively simple way of visualizing how the rest of this write up and the formulas that are normally applied to piston engines can be applied to a rotary engine.</p>
<p style="text-align: justify;">Furthermore, applying the same calculations utilized to determine Volumetric Efficiency (VE) on a piston engine, but for a rotary engine will yield optimistic results. If we were to consider the rotary engine, a 4 stroke engine with a displacement of 1.3L, the results would render a VE of over 100% in more than one instance, which is very unrealistic.</p>
<p style="text-align: justify;">Did I make any sense? Mmm maybe not but, try to follow the next steps as I try to make sense out of what I have gathered so far.</p>
<p style="text-align: justify;"> </p>
<p style="text-align: justify;">According to <a href="http://web.archive.org/web/20050327123026/http://www.auto-ware.com/combust_bytes/eng_sci.htm" target="_blank">Auto-Ware</a>, volumetric efficiency is:</p>
<blockquote style="text-align: justify;"><p>&#8230;.used to describe the amount of fuel/air in the cylinder in relation to regular atmospheric air. If the cylinder is filled with fuel/air at atmospheric pressure, then the engine is said to have 100% volumetric efficiency. On the other hand, super chargers and turbo chargers increase the pressure entering the cylinder, giving the engine a volumetric efficiency greater than 100%. However, if the cylinder is pulling in a vacuum, then the engine has less than 100% volumetric efficiency. Normally aspirated engines typically run anywhere between 80% and 100% VE. So now, when you read that a certain manifold and cam combination tested out to have a 95% VE, you will know that the higher the number, the more power the engine can produce.</p></blockquote>
<p style="text-align: justify;"><strong>A little experiment&#8230;..</strong><br />
Well, today I finally gave up and decided to do a little experiment that I found while searching for an effective method to calculate Volumetric Efficiency on a vehicle without having to yank the engine from the car. I came across the following experiment: <a href="http://web.archive.org/web/20050327123026/http://www.installuniversity.com/install_university/installu_articles/volumetric_efficiency/ve_computation_9.012000.htm" target="_blank">Calculating your cars volumetric efficiency</a></p>
<p style="text-align: justify;">I will assume that you are either too lazy or tired to follow the link, so I will explain a bit on what does the experiment entail.</p>
<p style="text-align: justify;">The experience requires the following: (1)vehicle with an engine; (1)OBD-II Scan Tool (e.g. <a href="http://web.archive.org/web/20050327123026/http://www.ghg.net/dharrison" target="_blank">CANScan Harrison R&amp;D</a>); (1)stock air intake with a Mass Air Flow sensor (MAF) in stock trim -according to the author, a slight variance from factory stock, such as removing the screen or repositioning the sensor will give the experiment little value- (1)stretch of private, safe, deserted road.</p>
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<p style="text-align: justify;">Once you&#8217;ve acquired all those items the procedure is rather painless. Attach the scan tool to the vehicle and make sure it can report the following things: Engine RPM, Intake Air Temperature &amp; Air Flow. Using the deserted stretch of private road, run the vehicle from a low engine rpm (2500rpm) @ WOT to redline (or as far as you want your sample to go..) while recording Intake Air Temperature (IAT), Engine Speed (RPM) &amp; Intake Mass Air Flow (IMAF).</p>
<p style="text-align: justify;">Once you have your data logged, re-read the experiment from the link provided and start crunching numbers! Its principle seems straight forward: based on the calculated theoretical volumetric air flow for your engine (a Renesis in this case), and the data you logged, you can approximate the actual VE of your particular engine. I will provide the formulas I used at the end of this article. For now, let&#8217;s take a look at the chart below.</p>
<div id="attachment_326" class="wp-caption aligncenter" style="width: 847px"><a href="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/2005-11-23_220506_13b_msp_volumetric_efficiency_020205_a.gif"><img class="size-full wp-image-326" title="2004 Mazda RX-8 Actual Volumetric Efficiency" src="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/2005-11-23_220506_13b_msp_volumetric_efficiency_020205_a.gif" alt="2004 Mazda RX-8 Actual Volumetric Efficiency" width="837" height="569" /></a><p class="wp-caption-text">2004 Mazda RX-8 Actual Volumetric Efficiency</p></div>
<p style="text-align: justify;"><strong>Experiment results plotted.</strong><br />
Theoretical Volumetric Air Flow was calculated assuming that a 13B MSP rotary engine has similar displacement in 720º of crankshaft rotation as a 2.6 Liter, 4 stroke piston engine. Notice how VE climbs as Engine Speed increases, until it reaches 5500rpms. This is were the engine is rated to produce peak torque, therefore its safe to assume that VE will peak at or nearby 5500rpms. Furthermore, you can safely assume that Volumetric Efficiency plotted against engine speed will mimic the shape and characteristics of the torque curve produced by the engine.</p>
<p style="text-align: justify;">Note that the plotted VE is somewhat linear: starts @ 80% and climbs its way to a tad over 100%. If this experiment&#8217;s results could be validated and the parameters I used were accurate, it would signify that the Renesis engine -on my car at least- is in fact very efficient for a normally aspirated internal combustion powerplant -VE definition above.</p>
<p style="text-align: justify;"><strong>Calculating Volumetric Efficiency (VE) for the Renesis (13B MSP) rotary engine:</strong><br />
We will use the following values obtained during our data log:</p>
<p style="text-align: justify;">Data:</p>
<p style="text-align: justify;">Intake Air Temperature (IAT) = 82ºF<br />
Engine Speed (RPM) = 8561rpm<br />
AirFlow (MAF) = 27.3lb/minute</p>
<p style="text-align: justify;">&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;</p>
<p style="text-align: justify;"><strong>THEORETICAL AIRFLOW CALCULATION:</strong></p>
<p style="text-align: justify;">Formula:</p>
<p style="text-align: justify;">[(ED) x (rpm) x (VE)] / [(ES) x (C)] = TAF</p>
<p style="text-align: justify;">Variables:</p>
<p style="text-align: justify;">ED = Engine Displacement [in³]<br />
rpm = Engine Speed [RPMs]<br />
VE = Volumetric Efficiency [%]<br />
ES = Engine Stroke Coefficient [#]<br />
C = Conversion coefficient from in³ to ft³<br />
TAF = Theoretical Air Flow [ft³]</p>
<p style="text-align: justify;">Solving:</p>
<p style="text-align: justify;">[(159.64in³) x (8561rpm) x (1)] / [(2) x (1728 in³/ft³)] = TAF<br />
TAF = 395.42ft³</p>
<p style="text-align: justify;">Values:</p>
<p style="text-align: justify;">ED = 2.6 Liters (1308cc x 2) &gt;&gt; 159.64in³<br />
rpm = I chose 8561rpm arbitrarily.<br />
VE = Since this corresponds to Theoretical VE, we assume VE = 100% (1)<br />
ES = Since we simplify a 13B engine to a 4 stroke piston engine -thus 2.6L- we use a coefficient of 2.<br />
C = 1728in³/ft³</p>
<p style="text-align: justify;">&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;</p>
<p style="text-align: justify;"><strong>AIR DENSITY &amp; TEMPERATURE CALCULATION:</strong></p>
<p style="text-align: justify;">Formula:</p>
<p style="text-align: justify;">[(t1) / (t2)] = [(d2) / (d1)]</p>
<p style="text-align: justify;">Variables:</p>
<p style="text-align: justify;">t1 = Temperature of air for a known density [ºR]<br />
t2 = Temperature of the intake air measured by the IAT sensor [ºR]<br />
d1 = Density of air for a known temperature [lb/ft³]<br />
d2 = Density of the intake air [lb/ft³]</p>
<p style="text-align: justify;">Solving for [d2]:</p>
<p style="text-align: justify;">[(t1) / (t2)] x (d1) = (d2)<br />
[(491.67ºR) / (541.67ºR)] x (0.0808lb/ft³) = d2<br />
d2 = 0.073341lb/ft³</p>
<p style="text-align: justify;">Values:</p>
<p style="text-align: justify;">t1 = 32ºF &gt;&gt; 491.67ºR<br />
t2 = 82ºF &gt;&gt; 541.67ºR<br />
d1 = 0.0808lb/ft³</p>
<p style="text-align: justify;">&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;</p>
<p style="text-align: justify;"><strong>VOLUMETRIC FLOW RATE CALCULATION:</strong></p>
<p style="text-align: justify;">Formula:</p>
<p style="text-align: justify;">[(MF) / (d2)] = AVF</p>
<p style="text-align: justify;">Variables:</p>
<p style="text-align: justify;">MF = Mass Flow rate taken from CANScan [lb/minute]/LI&gt;<br />
d2 = Density of the intake air [lb/ft³]<br />
AVF = Actual Volumetric Flowrate [ft³/minute][</p>
<p style="text-align: justify;">Solving:</p>
<p style="text-align: justify;">[(27.3lb/minute) / (0.073341lb/ft³)] = AVF<br />
AVF = 372.233ft³/minute</p>
<p style="text-align: justify;">Values:</p>
<p style="text-align: justify;">MF = 27.3lb/minute<br />
d2 = 0.073341lb/ft³</p>
<p style="text-align: justify;">&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;</p>
<p style="text-align: justify;"><strong>VOLUMETRIC EFFICIENCY CALCULATION:</strong></p>
<p style="text-align: justify;">Formula:</p>
<p style="text-align: justify;">[(AVF) / (TAF)] = VE</p>
<p style="text-align: justify;">Variables:</p>
<p style="text-align: justify;">AVF = Actual Volumetric Flow rate [ft³/minute]<br />
TAF = Theoretical Air Flow rate [ft³/minute]<br />
VE = Volumetric Efficiency [%]</p>
<p style="text-align: justify;">Solving:</p>
<p style="text-align: justify;">[(372.233ft³/minute) / (395.42ft³/minute)] = AVF<br />
AVF = 0.94 &gt;&gt; 94%</p>
<p style="text-align: justify;">Values:</p>
<p style="text-align: justify;">AVF = 372.233ft³/minute<br />
TAF = 395.42ft³/minute</p>
<p style="text-align: justify;">Is this remotely close to accurate? I truly don&#8217;t know! I simply took the time to do a little research through different channels &amp; gathered information. If you have any comments, or would like to make any suggestions or correct aything, feel free to make your voice heard!</p>
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		<title>Rotary Engines: Are they really just 1.3L?</title>
		<link>http://www.myrotarycar.com/blog/index.php/2009/02/05/rotary-engines-are-they-really-just-13l/</link>
		<comments>http://www.myrotarycar.com/blog/index.php/2009/02/05/rotary-engines-are-they-really-just-13l/#comments</comments>
		<pubDate>Thu, 05 Feb 2009 19:17:46 +0000</pubDate>
		<dc:creator>Miguel Pereira</dc:creator>
				<category><![CDATA[Cars]]></category>
		<category><![CDATA[Technology]]></category>
		<category><![CDATA[mazda]]></category>
		<category><![CDATA[rotary engine]]></category>
		<category><![CDATA[rx8]]></category>
		<category><![CDATA[wankel]]></category>

		<guid isPermaLink="false">http://www.myrotarycar.com/blog/?p=214</guid>
		<description><![CDATA[<p class="wp-caption-text">Renesis Rotary Engine</p>
<p>Here is an article I reproduced a long time ago from one of the most ungodly knowledgeable person I&#8217;ve ever heard regarding rotary engines.</p>
<p>Rotary engine displacement explained by Fred &#8220;rotarygod&#8221; Swain</p>
<p style="text-align: justify;">
The rotary engine is a 6 stroke internal combustion engine. I know, people will probably start screaming at me for this [...]]]></description>
			<content:encoded><![CDATA[<div id="attachment_204" class="wp-caption alignleft" style="width: 160px"><img class="size-thumbnail wp-image-204" title="Mazda RX-8" src="http://www.myrotarycar.com/blog/wp-content/uploads/2009/02/rx8-tx_vred_2-150x150.jpg" alt="Renesis Rotary Engine" width="150" height="150" /><p class="wp-caption-text">Renesis Rotary Engine</p></div>
<p>Here is an article I reproduced a long time ago from one of the most ungodly knowledgeable person I&#8217;ve ever heard regarding rotary engines.</p>
<p><strong>Rotary engine displacement explained by <span style="text-decoration: underline;">Fred &#8220;rotarygod&#8221; Swain</span></strong></p>
<p style="text-align: justify;"><strong><br />
</strong>The rotary engine is a 6 stroke internal combustion engine. I know, people will probably start screaming at me for this so let’s get into a little explanation as to why and how typical mathematical formulas for piston engines don&#8217;t work.</p>
<p style="text-align: justify;">First of all, lets get the terms &#8220;stroke&#8221; and &#8220;cycle&#8221; defined (Some of you get your heads out of the gutter!) since everyone commonly gets these terms interchanged. They are not the same thing. Every internal combustion engine whether it is a 2 stroke, 4 stroke, diesel, gasoline, propane injected, etc. is a 4 cycle engine. Why? All of these engines take in air (intake), compress the air (compression), ignite the air whether by spark plug or glow plug (ignition), and expel it out the tailpipe (exhaust). There you go 4 cycles. Simple isn&#8217;t it. The term &#8220;stroke&#8221; in this context refers to how many times the crankshaft or eccentric shaft makes a piston go up or down to complete the cycle.</p>
<p style="text-align: justify;">The connecting rods and pistons are just an extension of the offset lobes of the crankshaft. This is also true in regards to a rotor and eccentric shaft. When the lobe rotates upward, the piston goes up. When the lobe rotates down, the piston goes down. Every time it moves one way is considered a stroke. In a 2 stroke engine, all 4 phases or cycles of the combustion process are completed in only 2 strokes of the piston, 1 up and 1 down. This is only 1 complete revolution of the crankshaft. In a 4 stroke engine, it takes 4 strokes of the piston, up, down, up, down to go through the complete combustion process. This is 2 complete revolutions of the crankshaft. It&#8217;s all a very simple mathematical relationship.</p>
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<p style="text-align: justify;"><span id="more-214"></span></p>
<p style="text-align: justify;">Now lets go look at the workings of a rotary engine. If we look at a rotary engine eccentric shaft and compare it to a piston engine crankshaft, we see essentially the same piece. Both have lobes and because of this both engines will have a stroke length, even the rotating rotary. It doesn&#8217;t matter if it is a piston going back and forth or a rotor going round and round. The crankshaft motion remains the same. On a rotary engine, the rotors are spinning at exactly 1/3 the speed of the eccentric shaft. From the time that the air entering one chamber goes through the combustion phases to the time it leaves the engine from the same chamber (rotor face), the eccentric shaft has gone around 3 complete times unlike a 4 strokes 2 times or a 2 strokes 1 time. If we do the math we see that the lobes of the eccentric shaft must have gone up and down 6 times (up, down, up, down, up, down). Since it does this process the exact same way every time for every rotor face, it is a 6 stroke engine. That’s right the rotary engine is a 6 stroke! Do not confuse these strokes with the 4 internal cycles that every engine has!</p>
<ol>
<li>Let&#8217;s sum this up in a simple chart to visually explain how this works:</li>
<li>2 stroke engine (up, down) &#8211; 1 complete crankshaft revolution.</li>
<li>4 stroke engine (up, down, up, down) &#8211; 2 complete crankshaft revolutions.</li>
<li>6 stroke (rotary) engine (up, down, up, down, up, down) &#8211; 3 complete crankshaft (eccentric shaft) revolutions.</li>
</ol>
<p style="text-align: justify;">See a pattern? All of these engines though are still 4 cycle engines! They are different stroke engines though so the amount of work they do per time is very different. A 2 stroke engine does twice the work per amount of time that a 4 stroke does. Don&#8217;t believe me? Go race 2-80cc motorcycles, 1-2 stroke and 1-4 stroke and see who wins! This must mean that the rotary engine does the least amount of work per time than both other engine types. Yes it does. But, unlike a piston engine, it uses 3 sides of it&#8217;s piston (rotor) at a time. In reality it makes no difference if we have 1 rotor with 3 usable faces or 6 rotors with 1 usable face each as in a piston engine.</p>
<p style="text-align: justify;">Here&#8217;s a little info on how to properly figure out displacement on a rotary engine. Everyone argues that it is really a 1.3 liter while others argue that it is really a 2.6 liter engine. They are both wrong! If we look at how a piston engines volume is calculated we arrive at a displacement based on total swept volume of every piston added together. It is not based on rpm. On a rotary, displacement is figured using one rotor face in one complete revolution then multiplied by 2. This only leaves the total for 2 combustion chambers though and the rotary has 6! Since the volume of a 13b rotary is rated at 1.3 liters (only 2 combustion chambers) it really adds up to 3.9 liters!!! I can hear it now, &#8220;&#8230;but we only have 2 rotors!&#8221; So what! Like I said it makes no difference if there are 2 rotors with 6 faces or 6 rotors with one face each. the total is always 6 and the base numbers are only based on 2 chambers. The rotary merely does 3 times the work in a package 1/3 the size. It&#8217;s just a 3.9 liter engine crammed into a 1.3 liter body. Just so none of you start a fight over this, I will explain this later so don&#8217;t chastise me yet!!!</p>
<p style="text-align: justify;">In case anyone is curious I did some math to determine what the 13B rotary would be sized at if it were a piston engine. The results are pretty neat. First of all the rotary would be a 3.9 liter, 6 cylinder engine. It would be a 6 stroke. Each cylinder would be 6.54&#8243; across (damn big piston!) but the stroke length would only be 1.18&#8243; in length peak to peak. Not much there. Interesting isn&#8217;t it. Now just imagine a way to make all this work with only 2 intake runners!</p>
<p style="text-align: justify;">In all fairness to the terms I have used, the word &#8220;stroke&#8221; can be interchanged with the word &#8220;cycle&#8221; since both technically have the same definition. The terms &#8220;periods&#8221;, &#8220;quarters&#8221;, or &#8220;phases&#8221; can also be used correctly. I merely wrote it the way I did to get a certain mental picture going.</p>
<p style="text-align: justify;">I have already dealt with why the rotary engine is really a 6 stroke engine and why displacement is really 3.9 liters and not 1.3 liters. Now I need to explain why the rotary engine doesn&#8217;t have the torque or horsepower of a good 3.9 liter engine or why it doesn&#8217;t get the gas mileage of a 1.3 liter engine. The world has always wondered so here&#8217;s why.</p>
<p style="text-align: justify;">Remember that I stated that the true displacement of the rotary engine, if figured out according to the way piston engine volumes are calculated, is according to the total number of rotor faces and not the number of rotors, nor does it have anything to do with rpm. This added up to 3.9 liters for a 2 rotor 13B engine and not the published spec of 1.3 liters. They just crammed all 3.9 liters into a 1.3 liter body. If the engine is really a 3.9 liter engine then why doesn&#8217;t it have the low end torque of a 3.9 liter engine? This has a very simple answer. Lack of leverage. OK, what the hell does that mean?</p>
<p style="text-align: justify;">First of all we must figure out what a lever is. It is a device that multiplies mechanical advantage over an object to do the same amount of work with a smaller amout of effort. Another way to look at it is to do a greater amount of work with the same amount of effort. It&#8217;s the same thing. Let&#8217;s look at leverage differences as an example in a piston engine.</p>
<p style="text-align: justify;">What happens to a piston engine when we make it a &#8220;stroker&#8221;? Ignoring a host of other variables, it gains torque. It also gains horsepower but they are both a fixed mathematical ratio between each other and you can&#8217;t increase or decrease one without the other. Why did it gain torque? Greater mechanical advantage or leverage over the crankshaft. The reason being is that on a “stroker” crankshaft as opposed to the stock crankshaft, the lobe centerline is farther out from the rotational centerline of the crankshaft. This increases the leverage that the piston has over the crankshaft. Don&#8217;t believe me? Try this. Get a short pole and hold it at the end straight out away from your body. Attach a 10 lb weight to it exactly 1 foot away from your hands. The weight is exerting exactly 10 ft. lbs. of torque on your hands. Now move that weight out away from you to 2 feet away from your hands. Now the same weight is exerting 20 ft. lbs. of torque on your hands. You have just in essence made a &#8220;stroker&#8221;. Now let&#8217;s get back to the engine.</p>
<p style="text-align: justify;">Now we know that the greater the stroke length, the greater the engine torque. As I stated, the rotary engine only has an effective stroke length of 1.18&#8243;. My weed eater has that! There is not very much mechanical advantage over the eccentric shaft. This still doesn&#8217;t explain everything though.</p>
<p style="text-align: justify;">Remember, I stated that if the 13B rotary were a piston engine it would have pistons 6.54&#8243; across. Now we just discovered another enemy of efficiency, flame front speed. When the spark plug ignites the mixture in the engine, it doesn&#8217;t just ignite everything all at once. The spark ignites at the plug and then has to travel outward away from the plug at a certain rate of speed. While this only takes milliseconds, this amount of time gets more critical the higher the rpm gets due to the shorter amount of available time. The result is that as rpm&#8217;s rise efficiency decreases. The larger the area of the piston, the farther the flame front has to travel and the greater the chance that all of the mixture does not get ignited when it should. Just can&#8217;t go far enough fast enough. Today’s rotaries have 2 sparkplugs per chamber to help combat this problem. Varying their ignition time in relation to each other even helps somewhat with power and emission. That&#8217;s right they don&#8217;t necessarily fire together even though they are in the same chamber. This can get complex so I will not deal with it at this time. Some race engines even have 3 plugs per chamber to improve efficiency and ignition wave front speed. On piston engines, Mercedes has capitalized on this and uses 2 plugs per cylinder in some of their higher end cars. Do they know something others don&#8217;t?</p>
<p style="text-align: justify;">There is also one more aspect that affects it. Remember that the rotary is a 6 stroke engine. A 2 stroke engine does twice the amount of work per amount of time that a 4 stroke engine does. A 4 stroke engine does 50% more work per amount of time that a 6 stroke does. The rotary engine does less work per eccentric shaft rotation than your typical 4 stroke counterpart. All of these characteristics combine to make an engine that has relatively little low end power and needs to be revved up to be truly powerful.</p>
<p style="text-align: justify;">I make it sound like we should have less torque than a 1.3 liter engine due to the above reasons. This isn&#8217;t true though. Remember that we still have a 3.9 liter engine even though it only uses 2 lobes on the eccentric shaft. We should not expect to develop the torque numbers of a 1.3 liter engine. It should settle in somewhere around 50% less than a 3.9 liter engine which would put it around equal to a 2.6 liter engine in power.</p>
<p style="text-align: justify;">These traits of the rotary engine are also why the engine gets worse gas mileage than your typical 1.3 liter engine. Hell it gets worse gas mileage than your typical 2.6 liter engine. Another aspect that affects this is port timing and duration. If we had a piston engine of 2.6 liters in size that had the same intake and exhaust timing as the rotary then it would get comparable gas mileage to the rotary. The 12A/13B rotary though have much more exhaust duration than intake duration due to the peripheral exhaust port location. This contributes to several factors which decrease efficiency. Exhaust gas dilution is one of them. For each stroke there is a small amount of overlap. The exhaust ports and intake ports are open to the same chamber at the same time for a short amount of time as measured in degrees of eccentric shaft rotation. The higher the rpm&#8217;s the less important this becomes since air velocity will generally keep the gasses where we want them to go. At lower rpm&#8217;s though, the intake and exhaust air velocity is not very high. This will cause some exhaust to go back through the combustion chamber again. When this happens volumetric efficiency decreases and there is less room for fresh air to fit inside the combustion space. Also this re-circulated exhaust gas is very hot. A hotter air molecule is larger than a cold one which means a fewer number of molecules can fit in the same area per amount of pressure exerted on them. Another aspect of the rotary&#8217;s peripheral exhaust port configuration that contributes to less low end power and greater fuel consumption is its incredibly long duration or time it is open for. Unfortunately when we make the port bigger we also change it&#8217;s timing. We don&#8217;t have the luxury of being able to mill out a head to accept a larger valve while still being able to use the same cam. The timing is really only optimized for high rpm use. We are leaving it open for too long which gets back to the whole overlap problem. Again, all of this is just a generalization and can be affected by how well the intake and exhaust flow and how well they can scavenge. The affects of scavenging, intake design, Helmholtz effect, and proper exhaust design are all out of the scope of this article. So just assume it is an even world.</p>
<p style="text-align: justify;">Luckily there is a cure for this. It is called Renesis! It is the new 13B based rotary engine in the new Mazda RX-8. The exhaust ports are no longer in the periphery of the chamber but have rather been moved to the side housings. This allowed the designers to more appropriately optimize the port timing duration. The location also allows more port area leaving the engine. So now we have more area to flow air out of faster. This new location also completely got rid of the port overlap. There is actually 64 degrees of dwell. This amount of dwell was originally greater in the early test engine called the MSP-RE since it had the intake timing of the &#8217;84-&#8217;91 n/a RX-7&#8242;s 6 port engine. However dwell is only useful if you just have enough to get the job done but not so much that you are getting losses from it. Because of this Mazda engineers learned that they could open the intake earlier than previously and still maintain all of the other good aspects of the new exhaust layout.</p>
<ol>
<li>
<div style="text-align: justify;">A bigger intake port = more time for air to enter and a greater CFM rating through the port.</div>
</li>
<li>
<div style="text-align: justify;">Less turbulence through the port as well.</div>
</li>
<li>
<div style="text-align: justify;">Less overlap gives us less dilution of the intake air and a cooler intake charge.</div>
</li>
<li>
<div style="text-align: justify;">More available room for incoming air.</div>
</li>
<li>
<div style="text-align: justify;">Volumetric efficiency increases.</div>
</li>
<li>
<div style="text-align: justify;">Since efficiency goes up, our use of gas gets more efficient. In other words it takes less fuel to do the same amount of work.</div>
</li>
</ol>
<p style="text-align: justify;">What’s the result? Better gas mileage. With today’s gas prices this is a very welcome thing. The efficiency increase also means that emissions characteristics are also improved -another bonus with today’s laws concerning air quality.</p>
<p style="text-align: justify;">So after reading this you are probably wondering why in the world anyone would want to use one of these engines. First and most obvious is size. They crammed a 3.9 liter engine, or more appropriately a 2.6 usable liter engine into a 1.3 liter body. Second, it is just such a simple design. There are only 3 moving parts. Fewer moving parts have less frictional losses. Also fewer moving parts have less chance statistically of failure. The more it moves the more chances you have for failure. Third, nothing moves back and forth. So what? A piston stopping and changing direction exerts a lot of stress on everything from the crankshaft to the connecting rods, to the pistons, to the wristpins, etc. Let’s not also forget the stresses on the valves for being slammed open and shut as well as the temperature extremes they see during the combustion cycle. A body in motion tends to stay in motion. It is a very unnatural act to change direction suddenly or at all for that matter. A rotary just spins away in the same direction. Yes the lobes of the eccentric shaft do see stress but remember that we don&#8217;t have very much leverage over them. The rotors are also exerting some of their rotational stress on the stationary gears as well so some stress is never transmitted to the eccentric shaft from the rotors. The lack of stroke length and pure rotational motional do make it very naturally adapted to high rpm use. If we look at really high horsepower piston race engines, their stroke length has been shortened to reduce the stresses to all of the engine components at high rpms. The last and most important reason why the rotary engine is still a popular engine despite its shortcomings is because it is different. There is always something to be said for individuality and uniqueness. If you own a piston engine it doesn&#8217;t matter how big it is or if it is made by Chevrolet or Honda. It is still the same device.</p>
<p style="text-align: justify;">Just to shoot down right now any arguments on displacement think about this:</p>
<ol>
<li style="text-align: justify;">The 13B rotary engine is a 1.3 liter. Yes.</li>
<li style="text-align: justify;">The 13B rotary engine is a 2.6 liter. Yes.</li>
<li style="text-align: justify;">The 13B rotary engine is a 3.9 liter. Yes.</li>
</ol>
<p style="text-align: justify;">Notice that all of these statements are TRUE!!! That&#8217;s right there is a truth to all of those statements. Go read the whole thing again. To understand why this is so, lets define truth. Truth can be defined in a couple of ways: Anything that is not false (none of those statements is) or it can be defined as: One&#8217;s individual interpretation of presented facts. This herein is the source of our debate. We can&#8217;t change the facts no matter how hard we try. Arguing won&#8217;t do it. What is debatable however, is each individual&#8217;s interpretation of facts. If your interpretation doesn&#8217;t match someone else&#8217;s, you argue about it.</p>
<p style="text-align: justify;">Here are the facts: The rotary engine as rated by Mazda is 1.3 liters because each individual rotor, following one face of one rotor through the complete cycle, has a swept displacement of 654cc or .65 liters. Multiply this times 2 rotors to achieve 1.3. Since this only accounts for 2 of the total of 6 rotor faces, we multiply our answer by 3 to get an actual displacement of 3.9 liters. However since the rotary engine is a 6 stroke engine and not a 4 stroke engine since it takes 3 complete eccentric shaft revolutions to fire all faces instead of the typical engine&#8217;s 2, it only does 66% the work of a 4 stroke 3.9 liter engine. Calculating for this we divide 3.9 by 1.5 to get a total of 2.6 liters equivalent work to a 4 stroke piston engine. All of these, from a 1.3 liter in physical size package.</p>
<p style="text-align: justify;">No one can argue that this is not correct and any response saying otherwise will have been explained by what I just said. Any debate will only focus on one aspect and not the total facts.</p>
<p style="text-align: justify;">Just to put a cap on this whole thing: If at any time you try to calculate proper sizing for a turbo, intake manifold runners, intake plenum size, exhaust size, etc, and you try to use the 1.3 liter number in your equations, you will be way, way, way off!!!!!!!!! There are only 2 ways to flow more air: increase displacement or increase rpm. A 1.6 liter Honda engine doesn&#8217;t flow anywhere even remotely near what a 13B (1.3 liter) flows per the same rpm. Just some food for thought.</p>
<p style="text-align: justify;">Author: Fred &#8220;rotarygod&#8221; Swain</p>
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