Return-Path: Received: from front4.chartermi.net ([24.213.60.106] verified) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 723459 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 09:11:51 -0500 Received-SPF: none receiver=logan.com; client-ip=24.213.60.106; envelope-from=ericruttan@chartermi.net X-Virus-Scanned: by cgpav Received: from [24.236.229.73] (HELO [192.168.2.2]) by front4.chartermi.net (CommuniGate Pro SMTP 4.0.6) with ESMTP id 57528728 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 09:10:56 -0500 Received: from 127.0.0.1 (AVG SMTP 7.0.300 [265.8.6]); Thu, 10 Feb 2005 09:18:33 -0500 Message-ID: <002501c50f7b$670cbfd0$0202a8c0@eric> From: "Eric Ruttan" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece Date: Thu, 10 Feb 2005 09:18:04 -0500 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1478 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1478 Warning top poster, who cuts the post size down. A hopothises for your examination. A 360 Lyc does not produce the same power as a rotary. If true, then the Ellison card may not get enough air. If not true, then there is no real reason why the Ellison cannot feed a rotary. Ed, I understand your math, but even if the local inlet velocity is much higher, we dont care. the velocities adverage out to the same, as the volume of air = velocity * carb area. If the velocities are higher, the rotary consumes more air, and makes more power. Eric ----- Original Message ----- From: "Ed Anderson" To: "Rotary motors in aircraft" Sent: Thursday, February 10, 2005 8:31 AM Subject: [FlyRotary] : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece Good question, Tom. That interpretation did occur to me. I think the answer depends on your assumptions, IF using commonly accepted formulas for calculating air flow vs rpm and displacement (and considering both are positive displacement pumps) - then the 360 CID lycoming turning 2800 rpm and the rotors in the rotary turning 2100 rpm (6300 rpm E shaft) ingest the same total quantity of air in one minute - approx 291 CFM. In comparing the two engines, its accepted that you compare them over the standard 720deg 4 stroke cycle - that means that 4 of the rotary faces have gone through their cycle in the same 720 deg of rotation. But, assuming the formulas are correct, then they both end up with the same amount of air in the engine to create the same HP. I think my math is correct on the smaller/unit displacement and longer period of rotation for the rotary for the same intake of air. However, in both cases the air flow is pulsating and pulsating differently. So if the total displacement for the rotary over that 720 deg is less than the Lycoming and the time it takes to complete that rotation is slower AND you still ingest the same amount of total Air then the only way I can see that happening is the velocity of the air in the rotary's intake has to be considerably higher than in the Lycoming. The only other alternative answer I see if that the commonly accepted formula for comparing the rotary to the reciprocating 4 stroke is incorrect (I got beat about the head mercilessly by a number of respected rotary experts challenging that formula , so I wont' go there again (at least not now {:>)). Air Flow = Total Displacement * RPM/(2 - accounting for only every other cylinder sucking on each rev * 1728 (conversion of cubic inches to cubic feet) = TD*RPM/(2*1728) For the 360 CID Lycoming at 2800 rpm, Air Flow = 360*2800/(2*1728) = 291.66 CFM Using the commonly accepted notion that a rotary is equivalent to a 160 CID 4 stroke reciprocating engine because of the 4 faces of 40 CID that complete there cycle in 720 deg. For the 160 CID Rotary at 6000 rpm, Air Flow = 160 * 6300/(2*1728) = 291.66 CFM So if both ingest the 291 CFM and the rotary has less total displacement (over 720 deg) then disregarding any of my math on rotation period differences you still have to account for why the rotary can ingest the same amount of air with less displacement. (Now I must admit I have my suspicions about the commonly accepted (racing approved) formula for the rotary. However, if my suspicions about the rotary formula are correct, it would make the rotary even more efficient at ingesting air - so I won't go there {:>)). If my logic and calculations are correct then this implies the Ve of the rotary is considerably better than the Lycoming and is great than 100%. I mentioned a few of the reasons why the Ve of the rotary may indeed be better in the previous message. Now, its possible that the stories about the Ellison not working well on the rotary is just that - a story OR there could be a plausible physical reason as I have poorly attempted to present. Ed ----- Original Message ----- From: Tom To: Rotary motors in aircraft Sent: Wednesday, February 09, 2005 11:54 PM Subject: [FlyRotary] Re: Same HP = Same Air Mass <> same air Velocity [FlyRotary] Re: Ellison, the missing piece Ed, >The rotary has 40 CID displacement per face and 2 facesx 2 rotors = 4*40 or 160 CID for one rev. So the rotary has 22% less displacement per revolution and the longer rotation period.< and >So if the rotary has less displacement of the sucking component and must take 25% longer for each revolution. Therefore the only way it can obtain an equal amount of air is for the intake air to have a higher velocity than the Lycoming does.< Isn't 'displacement' equal to the amount of air needing to be ingested? So 22% less displacement equates to 22% less air and the rotarys longer rotation period gives it more time for air to push in? And then the intake air velocity should be lower?