Return-Path: Received: from [65.54.168.112] (HELO hotmail.com) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 724067 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 14:37:47 -0500 Received-SPF: pass receiver=logan.com; client-ip=65.54.168.112; envelope-from=lors01@msn.com Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Thu, 10 Feb 2005 11:36:00 -0800 Message-ID: Received: from 4.174.7.95 by BAY3-DAV8.phx.gbl with DAV; Thu, 10 Feb 2005 19:35:21 +0000 X-Originating-IP: [4.174.7.95] X-Originating-Email: [lors01@msn.com] X-Sender: lors01@msn.com From: "Tracy Crook" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece Date: Thu, 10 Feb 2005 14:35:17 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0151_01C50F7D.BD78B3E0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: MSN 9 X-MIMEOLE: Produced By MSN MimeOLE V9.10.0009.2900 Seal-Send-Time: Thu, 10 Feb 2005 14:35:17 -0500 X-OriginalArrivalTime: 10 Feb 2005 19:36:00.0906 (UTC) FILETIME=[C03C4EA0:01C50FA7] This is a multi-part message in MIME format. ------=_NextPart_000_0151_01C50F7D.BD78B3E0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Doesn't prove anything conclusive, but in the SUN 100 race my NA 2nd gen = 13B powered RV-4 did beat a bunch of 0 - 360 powered planes of the same = type (RV-4s & 6s). I think it is safe to say that a *properly tuned* = 13B of any generation will equal or exceed the power of an O - 360. = Where did your understanding come from? FWIW, It is the definition of *properly tuned* that we are really = discussing here. Tracy It's my understanding that NA non-renesis rotary installations produce = less power than 360s, Perry Mick might have a word on this.=20 Eric Ruttan = > wrote: 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 =3D velocity * carb area. If the velocities are higher, the rotary consumes more air, and = makes more power. Eric ----- Original Message -----=20 From: "Ed Anderson"=20 To: "Rotary motors in aircraft"=20 Sent: Thursday, February 10, 2005 8:31 AM Subject: [FlyRotary] : Same HP =3D 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 =3D Total Displacement * RPM/(2 - accounting for only every = other cylinder sucking on each rev * 1728 (conversion of cubic inches to = cubic feet) =3D TD*RPM/(2*1728) For the 360 CID Lycoming at 2800 rpm, Air Flow =3D 360*2800/(2*1728) = =3D 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 =3D 160 * 6300/(2*1728) = =3D 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 -----=20 From: Tom To: Rotary motors in aircraft Sent: Wednesday, February 09, 2005 11:54 PM Subject: [FlyRotary] Re: Same HP =3D 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 = =3D 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? >> Homepage: http://www.flyrotary.com/ >> Archive: http://lancaironline.net/lists/flyrotary/List.html -------------------------------------------------------------------------= ----- Do you Yahoo!? Yahoo! Search presents - Jib Jab's 'Second = Term' ------=_NextPart_000_0151_01C50F7D.BD78B3E0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Doesn't prove anything conclusive, but in the SUN 100 race my = NA 2nd=20 gen 13B powered RV-4 did beat a bunch of 0 - 360 powered = planes of the=20 same type (RV-4s & 6s).   I think it is safe to say that = a =20 *properly tuned*  13B of any generation will equal or exceed the = power of=20 an O - 360.  Where did your understanding come from?
 
FWIW, It is the definition of  *properly tuned* that we are = really=20 discussing here.
 
Tracy
It's my understanding that NA non-renesis rotary installations=20 produce less power than 360s, Perry Mick might have a word on = this.=20


Eric Ruttan <ericruttan@chartermi.net>= =20 wrote:
Warning=20 top poster, who cuts the post size down.

A hopothises for = your=20 examination.

A 360 Lyc does not produce the same power as a=20 rotary.

If true, then the Ellison card may not get enough=20 air.

If not true, then there is no real reason why the = Ellison cannot=20 feed a
rotary.

Ed, I understand your math, but even if the = local=20 inlet velocity is much
higher, we dont care. the velocities = adverage out=20 to the same, as the
volume of air =3D velocity * carb = area.

If the=20 velocities are higher, the rotary consumes more air, and makes=20 more
power.

Eric

----- Original Message ----- =
From: "Ed=20 Anderson"
To: "Rotary motors in = aircraft"=20
Sent: Thursday, February 10, 2005 = 8:31=20 AM
Subject: [FlyRotary] : Same HP =3D Same Air Mass <> same = air=20 Velocity II
[FlyRotary] Re: Ellison, the missing = piece


Good=20 question, Tom.

That interpretation did occur to me. I think = the=20 answer depends on your
assumptions, IF using commonly accepted = formulas=20 for calculating air flow vs
rpm and displacement (and considering = both=20 are positive displacement
pumps) - then the 360 CID lycoming = turning 2800=20 rpm and the rotors in the
rotary turning 2100 rpm (6300 rpm E = shaft)=20 ingest the same total quantity of
air in one minute - approx 291 = CFM. In=20 comparing the two engines, its
accepted that you compare them = over the=20 standard 720deg 4 stroke cycle -
that means that 4 of the rotary = faces=20 have gone through their cycle in the
same 720 deg of=20 rotation.

But, assuming the formulas are correct, then they = both end=20 up with the same
amount of air in the engine to create the same = HP. I=20 think my math is
correct on the smaller/unit displacement and = longer=20 period of rotation for
the rotary for the same intake of air. = However, in=20 both cases the air flow
is pulsating and pulsating differently. = So if the=20 total displacement for
the rotary over that 720 deg is less than = the=20 Lycoming and the time it takes
to complete that rotation is = slower AND=20 you still ingest the same amount of
total Air then the only way I = can see=20 that happening is the velocity of the
air in the rotary's intake = has to=20 be considerably higher than in the
Lycoming.

The only = other=20 alternative answer I see if that the commonly accepted
formula = for=20 comparing the rotary to the reciprocating 4 stroke is = incorrect
(I got=20 beat about the head mercilessly by a number of respected = rotary
experts=20 challenging that formula , so I wont' go there again (at least = not
now=20 {:>)).

Air Flow =3D Total Displacement * RPM/(2 - = accounting for=20 only every other
cylinder sucking on each rev * 1728 (conversion = of cubic=20 inches to cubic
feet) =3D TD*RPM/(2*1728)

For the 360 CID = Lycoming=20 at 2800 rpm, Air Flow =3D 360*2800/(2*1728) =3D = 291.66
CFM

Using the=20 commonly accepted notion that a rotary is equivalent to a 160 = CID
4=20 stroke reciprocating engine because of the 4 faces of 40 CID that=20 complete
there cycle in 720 deg.

For the 160 CID Rotary at = 6000=20 rpm, Air Flow =3D 160 * 6300/(2*1728) =3D 291.66
CFM

So if = both ingest=20 the 291 CFM and the rotary has less total displacement
(over 720 = deg)=20 then disregarding any of my math on rotation period
differences = you still=20 have to account for why the rotary can ingest the same
amount of = air with=20 less displacement. (Now I must admit I have my
suspicions about = the=20 commonly accepted (racing approved) formula for the
rotary. = However, if=20 my suspicions about the rotary formula are correct, it
would make = the=20 rotary even more efficient at ingesting air - so I won't go
there = {:>)).

If my logic and calculations are correct then this = implies=20 the Ve of the
rotary is considerably better than the Lycoming and = is=20 great than 100%. I
mentioned a few of the reasons why the Ve of = the=20 rotary may indeed be better
in the previous message.

Now, = its=20 possible that the stories about the Ellison not working well on=20 the
rotary is just that - a story OR there could be a plausible = physical=20 reason
as I have poorly attempted to=20 present.


Ed


----- Original Message ----- =
From:=20 Tom
To: Rotary motors in aircraft
Sent: Wednesday, February = 09, 2005=20 11:54 PM
Subject: [FlyRotary] Re: Same HP =3D Same Air Mass = <> same=20 air Velocity
[FlyRotary] Re: Ellison, the missing=20 piece


Ed,

>The rotary has 40 CID displacement = per face=20 and 2 facesx 2 rotors =3D 4*40
or 160 CID for one rev. So the = rotary has=20 22% less displacement per
revolution and the longer rotation=20 period.<

and

>So if the rotary has less = displacement of=20 the sucking component and must
take 25% longer for each = revolution.=20 Therefore the only way it can obtain
an equal amount of air is = for the=20 intake air to have a higher velocity than
the Lycoming=20 does.<

Isn't 'displacement' equal to the amount of air = needing to=20 be ingested?
So 22% less displacement equates to 22% less air and = the=20 rotarys longer
rotation period gives it more time for air to push = in? And=20 then the
intake air velocity should be = lower?



>>=20 Homepage: http://www.flyrotary.com/
>> Archive:=20 http://lancaironline.net/lists/flyrotary/List.html

Do you Yahoo!?
Yahoo! Search presents - Jib=20 Jab's 'Second Term'
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