Return-Path: Received: from [24.25.9.100] (HELO ms-smtp-01-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 723425 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 08:32:42 -0500 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-185-127.carolina.rr.com [24.74.185.127]) by ms-smtp-01-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id j1ADVrbo017253 for ; Thu, 10 Feb 2005 08:31:54 -0500 (EST) Message-ID: <001601c50f74$e45277c0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece Date: Thu, 10 Feb 2005 08:31:56 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0013_01C50F4A.FB375070" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0013_01C50F4A.FB375070 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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.=20 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. =20 Ed=20 ----- Original Message -----=20 From: Tom=20 To: Rotary motors in aircraft=20 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,=20 >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? TIA Tom Ed Anderson wrote: Tom, I have no experience with the Ellison, but the answer may be not in = the total air consumed, but in how it is sucked in. Producing the same = HP requires essentially the same air/fuel regardless - its how it gets = there that may make a difference regarding the Ellison. The aircraft engine gulps in air in large chunks. The four large = cylinders running at say 2800 rpm and only two cylinder "suck" each = revolution. So there air flow characteristic is different than a = rotary. With the rotary you have six faces (piston analogs) of less = displacement rotating (the rotors not the eccentric shaft) at approx = 2000 rpm (for 6000 rpm eccentric shaft). The rotary sips smaller chunks = of air. The total amount of air would have to be the same for both engines = (same HP), however, "Average" covers a multitude of difference in the = actual air flow pattern. I see 2 large masses of air in the intake for = the aircraft engine each revolution. The rotary would have 4 smaller = airmass packages ( Yes, the rotary has six faces but only four have come = around in a 720 deg revolution) in the intake. So the interval between = the center of mass for each package is roughly 1/2 that of the Lycoming. = =20 For a specific example let see what numbers may tell us. Lets take a Lycoming of 360 CID turning at 2800 rpm and a rotary of = 80 CID with the rotors turning at 2100 rpm (6300 E shaft ). This will = have both engines sucking (assuming 100 % Ve for both) approx 291.67 = Cubic Feet/Minute. And assuming the same BSFC they would be producing = the same HP. =20 But, lets see where there are differences. 1st a 360 CID Lycoming at 2800 rpm has a period of revolution of = 2800/60 =3D 46.6666 Revs/Sec or a rotation period of 1/rev-sec =3D 1/ = 46.666 =3D .021428 seconds or 21.428 milliseconds. During that time its = sucking intake air for 2 cylinders in 360 deg of rotation. The rotary = however, has its rotors spinning at 2100 rpm (to draw the same amount of = air) which gives it a rotation period of 2100/60 =3D 33.3333 Revs/Sec or = a period of 1/35 =3D 0.02857 seconds or 28.57 ms. The rotary is also = drawing in two chambers of air in 360 deg of rotation.=20 Here the rotary e shaft is spinning at 6300 rpm to give the rotor a = rotation rate of 2100 rpm 6300/3 =3D 2100. Eshaft rpm =20 Displacement rpm =20 CFM =20 360 2800 291.67 =20 =20 =20 =20 =20 80 6300 291.67=20 =20 =20 =20 =20 =20 6300 40 2100(rotor) 291.67 =20 =20 So right there we have a difference of approx 25% difference in the = rotation time of the pumps pulling in the same average amount of air. = The rotary takes approx 25% more time than the Lycoming to complete a = revolution.. A 360 CID Lycoming (forgetting compression ratios for this = discussion) has 360/4 =3D 90 cid displacement per cylinder or 180 CID = for on rev. 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. 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. The air velocity of the area in the intake for the rotary would = appear to have to be much higher than the Lycoming. If my assumptions = and calculations are correct that would imply (at least to me) that to = minimize air flow restriction a larger opening would be required on the = rotary compared to the same HP Lycoming. Its not that one is taken in = more air its that the rotary has less time and smaller displacement pump = so must take in the air at a higher velocity. The fact that the rotary has no valves to block the flow of air = may be one reason that it can over come what would appear to be less = favorable parameters for sucking air. An additional factor that may play = a role is the fact that air mass pulsation in the rotary intake is less = than the Lycoming. This would mean less starting and stopping of air = movement, so the velocity would seem to remain steadier and on an = average higher than for the air pulses for the Lycoming which if you = factor the start/slowing/start of air flow may lower its overall = velocity compared to the rotary. In summary, while the total air intake in equal for engines = producing equal HP. It is likely that the air flow to the rotary may be = considerably higher in order to ingest the same amount of air over the = same time. This may be why there is a perception that the Ellison = model that may work well for a Lycoming may not work as well for a = rotary. =20 Well, anyhow, that's my best shot - if its incorrect perhaps = somebody can take it from here, but I think the answer lies in the = different pumping configuration of the two engines. Best Regards Ed =20 I'm under the impression I have an answer. Isn't there a law of motor performance that says that two motors = putting out the same horsepower are consuming the same amount of = air&fuel, assuming efficiency differences were not significant? So if you had a 13b and a O-360 putting out the same horsepower = for a single given 1 revolution of the propeller, they should be = consuming the same amount of air and fuel during that 1 propeller = revolution. (I THINK chosing 1 propeller rpm is a correct standard) Bill pointed out that the 13b operates at a higher rpm, and we = know that there's more combustion charges consumed by the 13b to make = that 1 prop rpm. =20 The difference, the missing piece, each 13b combustion charge = consumes a SMALLER amount of fuel/air than the piston powerplants less = frequent combustion charge. ??? So the 13b burns a smaller amount = more frequently. ??? If this is all true, then the Ellison isn't on the trash heap yet. = Tom WRJJRS@aol.com wrote: Group, I want to remind everyone about how much a priority the large = volume inlets are to us. I believe Ed Anderson was mentioning in one of = his posts how difficult it can be to get a MAP signal in the airbox of = one of our PP engines. This is a perfect indication of why the smaller = throttle bodies used on some of the slow turning engines will kill our = HP.=20 -------------------------------------------------------------------------= - Do you Yahoo!? Yahoo! Search presents - Jib Jab's 'Second Term' -------------------------------------------------------------------------= ----- Do you Yahoo!? Yahoo! Search presents - Jib Jab's 'Second Term' ------=_NextPart_000_0013_01C50F4A.FB375070 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Good question, Tom.
 
That interpretation did occur to = me.  I think=20 the answer depends on your assumptions, IF using commonly accepted = formulas for=20 calculating air flow vs rpm and displacement (and considering both are = positive=20 displacement pumps) - then the 360 CID lycoming turning 2800 rpm and=20 the rotors in the rotary turning 2100 rpm (6300 rpm E shaft) ingest = the=20 same total quantity of air in one minute - approx 291 CFM.  In = comparing=20 the two engines, its accepted that you compare them over the standard = 720deg 4=20 stroke cycle - that means that 4 of the rotary faces have gone through = their=20 cycle in the same 720 deg of rotation.
 
But, assuming the formulas are correct, = then they=20 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=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 = Lycoming=20 and the time it takes to complete that rotation is slower AND you still = ingest=20 the same amount of  total Air then the only way I can see that = happening is=20 the velocity of the air in the rotary's intake has to be=20 considerably higher than in the Lycoming. 
 
 The only other alternative answer = I see if=20 that the commonly accepted formula for comparing the rotary to the = reciprocating=20 4 stroke is incorrect (I got beat about the = head mercilessly by a=20 number of respected rotary experts  challenging that formula , = so I=20 wont' go there again (at least not now {:>)).
 
 Air Flow =3D Total Displacement * = RPM/(2 -=20 accounting for only every other cylinder sucking on each rev * 1728 = (conversion=20 of cubic inches to cubic feet)  =3D TD*RPM/(2*1728)
 
For the 360 CID Lycoming at 2800 rpm, = Air Flow =3D=20 360*2800/(2*1728) =3D 291.66 CFM
 
Using the commonly accepted notion that = a rotary is=20 equivalent to a 160 CID 4 stroke reciprocating engine because of the 4 = faces of=20 40 CID that complete there cycle in 720 deg.
 
For the 160 CID Rotary at 6000 rpm, Air = Flow =3D 160=20 * 6300/(2*1728) =3D 291.66 CFM
 
So if both ingest the 291 CFM and the = rotary has=20 less total displacement (over 720 deg) then disregarding any of my math = on=20 rotation period differences you still have to account for why the rotary = can=20 ingest the same amount of air with less displacement.  (Now I must = admit I=20 have my suspicions about the commonly accepted (racing approved) formula = for the=20 rotary.  However, if my suspicions about the rotary formula = are=20 correct, it would make the rotary even more efficient at ingesting air - = so I=20 won't go there {:>)).
 
If my logic and calculations are = correct then this=20 implies 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 = rotary may=20 indeed be better in the previous message.
 
Now, its possible that the stories = about the=20 Ellison not working well on the rotary is just that - a story OR there = could be=20 a plausible physical reason as I have poorly attempted to present.  =
 
 
Ed
 
 
----- Original Message -----
From:=20 Tom
Sent: Wednesday, February 09, = 2005 11:54=20 PM
Subject: [FlyRotary] Re: Same = HP =3D Same=20 Air Mass <> same air Velocity [FlyRotary] Re: Ellison, the = missing=20 piece

Ed,
 
>The rotary has 40 CID displacement per face and 2 facesx = 2=20 rotors =3D  4*40 or 160 CID for one rev.  So the rotary = has 22%=20 less displacement per revolution and the longer rotation=20 period.<
 
and
 
>So if the rotary has less displacement of the sucking = component and=20 must take 25% longer for each revolution.  Therefore the only way = it can=20 obtain an equal amount of air is for the intake air to have a higher=20 velocity than the Lycoming does.<
 
Isn't 'displacement' equal to the amount of air needing to be=20 ingested?   So 22% less displacement equates to 22% less=20 air and the rotarys longer rotation period gives it more time=20 for air to push in?    And then the intake = air=20 velocity should be lower?
 
TIA
Tom

Ed Anderson <eanderson@carolina.rr.com>=20 wrote:
Tom,
 
I have no experience with the = Ellison, but the=20 answer may be  not in the total air consumed, but in how it is = sucked=20 in.  Producing the same HP requires essentially the same = air/fuel=20 regardless - its how it gets there that may make a difference = regarding the=20 Ellison.
 
 
  The aircraft engine gulps in = air in=20  large chunks.  The four large cylinders running at say = 2800 rpm=20 and only two cylinder "suck" each revolution.  So there air = flow=20 characteristic is different than a rotary.  With the rotary you = have=20 six faces (piston analogs) of less displacement rotating (the rotors = not the=20 eccentric shaft) at approx 2000 rpm (for 6000 rpm eccentric = shaft). =20 The rotary sips smaller chunks of air.
 
The total amount of air would have = to be the=20 same for both engines (same HP), however, "Average" covers a = multitude of=20 difference in the actual air flow pattern.  I see 2 large = masses of air=20 in the intake for the aircraft engine each revolution.  The = rotary=20 would have 4 smaller airmass packages ( Yes, the rotary has six = faces but=20 only four have come around in a 720 deg revolution) in the intake. = So the=20 interval between the center of mass for each package is roughly 1/2 = that of=20 the Lycoming. 
 
For a specific example let see what = numbers may=20 tell us.
 
Lets take a Lycoming of 360 CID = turning at 2800=20 rpm and a rotary of 80 CID with the rotors turning at 2100 rpm (6300 = E shaft=20 ).  This will have both engines sucking (assuming 100 % Ve for = both)=20 approx 291.67 Cubic Feet/Minute.  = And assuming=20 the same BSFC they would be producing the same HP. =20
 
But, lets see where there are=20 differences.
 
1st a 360 CID Lycoming at =  2800 rpm=20 has a period of revolution of 2800/60 =3D 46.6666 Revs/Sec or a = rotation=20 period of 1/rev-sec =3D 1/ 46.666 =3D .021428 seconds or 21.428=20 milliseconds.  During that time its  sucking = intake air=20  for 2 cylinders in 360 deg of rotation.  The rotary = however,=20 has its rotors spinning at 2100 rpm (to draw the same amount of air) = which=20 gives it a rotation period of 2100/60 =3D 33.3333 Revs/Sec or a = period of 1/35=20 =3D 0.02857 seconds or 28.57 ms. The rotary is also drawing = in two=20 chambers of air in 360 deg of rotation. 
 
Here the rotary e shaft is spinning = at 6300 rpm=20 to give the rotor a rotation rate of 2100 rpm 6300/3 =3D = 2100.

Eshaft rpm   =

   Displacement

     rpm    =

CFM

360

2800

291.67

80

6300

291.67

6300

40

 2100(rotor)

291.67

 
 
 So right there we have a = difference of=20 approx 25% difference in the rotation time of the pumps pulling in = the same=20 average amount of = air. The rotary takes  approx=20 25% more time than the Lycoming to complete a = revolution..
 
 A 360 CID Lycoming = (forgetting=20 compression ratios for this discussion) has 360/4 =3D 90 cid = displacement per=20 cylinder or 180 CID for on rev.  The rotary has 40 CID = displacement per=20 face and 2 facesx 2 rotors =3D  4*40 or 160 CID for = one=20 rev.  So the rotary has 22%  less displacement per = revolution=20 and the  longer rotation period.
 
So if the rotary has less = displacement of the=20 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.
 
The air velocity of the area in the = intake for=20 the rotary would appear to have to be much higher than the = Lycoming. =20 If my assumptions and calculations are correct that would imply (at = least to=20 me) that to minimize air flow restriction a larger opening = would be=20 required on the rotary compared to the same HP Lycoming.  = Its not=20 that one is taken in more air its that the rotary has less time and = smaller=20 displacement pump so must take in the air at a higher = velocity.
 
  The fact that the rotary has = no valves=20 to block the flow of air may be one reason that it can over come = what would=20 appear to be  less favorable parameters for sucking air. An = additional=20 factor that may play a role is the fact that air mass pulsation in = the=20 rotary intake  is less than the Lycoming.  This would mean = less=20 starting and stopping of air movement, so the velocity would seem to = remain=20 steadier and on an average higher than for the air pulses for the = Lycoming=20 which if you factor the start/slowing/start of air flow may lower = its=20 overall velocity compared to the rotary.
 
In summary, while the total air = intake in equal=20 for engines producing equal HP. It is likely that the air flow to = the=20 rotary may be considerably higher in order to ingest the same = amount of=20 air over the same time.  This may be why  there is a=20 perception that the Ellison model  that may work well for a = Lycoming=20 may not work as well for a rotary. 
 
Well, anyhow, that's my best shot - = if its=20 incorrect perhaps somebody can take it from here, but I think the = answer=20 lies in the different pumping configuration of the two = engines.
 
Best Regards
 
Ed

 
I'm under the impression I have an answer.
 
Isn't there a law of motor performance that says that two = motors=20 putting out the same horsepower are consuming the same amount of=20 air&fuel, assuming efficiency differences were not = significant?
 
So if you had a 13b and a O-360 putting out the same = horsepower for a=20 single given 1 revolution of the propeller, they should be = consuming the=20 same amount of air and fuel during that 1 propeller revolution. (I = THINK=20 chosing 1 propeller rpm is a correct standard)
 
Bill pointed out that the 13b operates at a higher rpm, and = we know=20 that there's more combustion charges consumed by the 13b to make = that 1=20 prop rpm. 
 
The difference, the missing piece, each 13b combustion charge = consumes a SMALLER amount of fuel/air than the piston powerplants = less=20 frequent combustion charge.   ???   So the 13b = burns a=20 smaller amount more frequently.   ???
 
If this is all true, then the Ellison isn't on the trash heap = yet.=20
 
Tom

WRJJRS@aol.com wrote:
Group,
 I want to remind everyone about how much a priority = the large=20 volume inlets are to us. I believe Ed Anderson was mentioning in = one of=20 his posts how difficult it can be to get a MAP signal in the = airbox of=20 one of our PP engines. This is a perfect indication of why the = smaller=20 throttle bodies used on some of the slow turning engines will = kill our=20 HP.

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

Do you Yahoo!?
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