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 723549 for flyrotary@lancaironline.net; Thu, 10 Feb 2005 10:07:59 -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 j1AF7Cbo029775 for ; Thu, 10 Feb 2005 10:07:12 -0500 (EST) Message-ID: <000d01c50f82$34d29bf0$2402a8c0@edward2> From: "Ed Anderson" 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 10:07:15 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000A_01C50F58.4BCCF870" 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_000A_01C50F58.4BCCF870 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Be interested in seeing what you come up with, Tom. As best I recall I think the 13B is 1308 cc displacement and assume the = 360 lycoming really means 360 CID - but don't know for certain. Ed ----- Original Message -----=20 From: Tom=20 To: Rotary motors in aircraft=20 Sent: Thursday, February 10, 2005 9:12 AM Subject: [FlyRotary] Re: : Same HP =3D Same Air Mass <> same air = Velocity II [FlyRotary] Re: Ellison, the missing piece Thanks Ed,=20 I think I can see where some trouble is and I'll try to work it out = after doing taxi service and getting coffee. Bottom line, for the = smaller displacement rotary to match the airflow of the lycoming, it = should be as simple as 'burning smaller charges at a higher rate'. =20 Are the CID numbers accurate for the rotary and lycoming? Tom=20 Ed Anderson wrote: 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.=20 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' -------------------------------------------------------------------------= ----- Do you Yahoo!? Yahoo! Search presents - Jib Jab's 'Second Term' ------=_NextPart_000_000A_01C50F58.4BCCF870 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Be interested in seeing what you come = up with,=20 Tom.
 
As best I recall I think the 13B = is 1308 cc=20 displacement and assume the 360 lycoming really means 360 CID - but = don't know=20 for certain.
 
Ed
----- Original Message -----
From:=20 Tom
Sent: Thursday, February 10, = 2005 9:12=20 AM
Subject: [FlyRotary] Re: : Same = HP =3D Same=20 Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the = missing=20 piece

Thanks Ed,
 
I think I can see where some trouble is and I'll try to work it = out after=20 doing taxi service and getting coffee.    Bottom line, = for the=20 smaller displacement rotary to match the airflow of the lycoming, = it=20 should be as simple as 'burning smaller charges at a higher=20 rate'.   
 
Are the CID numbers accurate for the rotary and lycoming?
 
Tom 
 
 


Ed Anderson = <eanderson@carolina.rr.com>=20 wrote:
Good question, Tom.
 
That interpretation did occur to = me.  I=20 think the answer depends on your assumptions, IF using commonly = accepted=20 formulas for calculating air flow vs rpm and displacement (and = considering=20 both are positive displacement pumps) - then the 360 CID lycoming = turning=20 2800 rpm and the rotors in the rotary turning 2100 rpm (6300 = rpm E=20 shaft) ingest the same total quantity of air in one minute - approx = 291=20 CFM.  In comparing the two engines, its accepted that you = compare them=20 over the standard 720deg 4 stroke cycle - that means that 4 of the = rotary=20 faces have gone through their cycle in the same 720 deg of=20 rotation.
 
But, assuming the formulas are = correct, then=20 they both end up with the same amount of air in the engine to create = the=20 same HP.  I think my math is correct on the smaller/unit = displacement=20 and longer period of rotation for the rotary for the same intake of=20 air.  However, in both cases the air flow is pulsating and = pulsating=20 differently.  So if the total displacement for the rotary over = that 720=20 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=20 then the only way I can see that happening is the velocity of the = air in the=20 rotary's intake has to be considerably higher than in the=20 Lycoming. 
 
 The only other alternative = answer I see=20 if that the commonly accepted formula for comparing the rotary to = the=20 reciprocating 4 stroke is incorrect (I got beat about the=20 head mercilessly by a number of respected rotary=20 experts  challenging that formula , so I wont' go there = again (at=20 least not now {:>)).
 
 Air Flow =3D Total = Displacement * RPM/(2 -=20 accounting for only every other cylinder sucking on each rev * 1728=20 (conversion of cubic inches to cubic feet)  =3D=20 TD*RPM/(2*1728)
 
For the 360 CID Lycoming at 2800 = rpm, Air Flow=20 =3D 360*2800/(2*1728) =3D 291.66 CFM
 
Using the commonly accepted notion = that a=20 rotary is equivalent to a 160 CID 4 stroke reciprocating engine = because of=20 the 4 faces of 40 CID that complete there cycle in 720 = deg.
 
For the 160 CID Rotary at 6000 rpm, = Air Flow =3D=20 160 * 6300/(2*1728) =3D 291.66 CFM
 
So if both ingest the 291 CFM and = the rotary=20 has less total displacement (over 720 deg) then disregarding any of = my math=20 on rotation period differences you still have to account for why the = rotary=20 can ingest the same amount of air with less displacement.  (Now = I must=20 admit I have my suspicions about the commonly accepted (racing = approved)=20 formula for the rotary.  However, if my suspicions about the=20 rotary formula are correct, it would make the rotary even more=20 efficient at ingesting air - so I won't go there = {:>)).
 
If my logic and calculations are = correct then=20 this implies the Ve of the rotary is considerably better than the = Lycoming=20 and is 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 possible that the stories = about the=20 Ellison not working well on the rotary is just that - a story OR = there could=20 be a plausible physical reason as I have poorly attempted to = present. =20
 
 
Ed
 
 
----- Original Message -----
From:=20 Tom=20
To: Rotary motors in = aircraft=20
Sent: Wednesday, February = 09, 2005=20 11:54 PM
Subject: [FlyRotary] Re: = Same HP =3D=20 Same Air Mass <> same air Velocity [FlyRotary] Re: Ellison, = the=20 missing 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=20 22% less displacement per revolution and the longer rotation=20 period.<
 
and
 
>So if the rotary has less displacement of the sucking = component=20 and must take 25% longer for each revolution.  Therefore the = only way=20 it can obtain an equal amount of air is for the intake air to have = a=20 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=20 velocity should be lower?
 
TIA
Tom

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

Eshaft rpm   =

   Displacement

     = rpm   =20

CFM

360

2800

291.67

80

6300

291.67

6300

40

 2100(rotor)

291.67

 
 
 So right there we have a = difference=20 of approx 25% difference in the rotation time of the pumps = pulling in=20 the same average amount of=20 air. The rotary takes  approx = 25% more=20 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=20 per cylinder or 180 CID for on rev.  The rotary has 40 CID=20 displacement per face and 2 facesx 2 rotors =3D  4*40=20 or 160 CID for one rev.  So the rotary has 22% =  less=20 displacement per revolution and the  longer rotation=20 period.
 
So if the rotary has less = displacement of=20 the sucking component and must take 25% longer for each=20 revolution.  Therefore the only way it can obtain an equal = amount=20 of air is for the intake air to have a higher velocity than = the=20 Lycoming does.
 
The air velocity of the area in = the intake=20 for the rotary would appear to have to be much higher than the=20 Lycoming.  If my assumptions and calculations are correct = that=20 would imply (at least to me) that to minimize air flow = restriction a=20 larger opening would be required on the rotary compared to = the same=20 HP Lycoming.  Its not that one is taken in more air = its that=20 the rotary has less time and smaller displacement pump so must = take in=20 the air at a higher velocity.
 
  The fact that the rotary = has no=20 valves to block the flow of air may be one reason that it can = over come=20 what would appear to be  less favorable parameters for = sucking air.=20 An additional factor that may play a role is the fact that air = mass=20 pulsation in the rotary intake  is less than the = Lycoming. =20 This would mean less starting and stopping of air movement, so = the=20 velocity would seem to remain steadier and on an average higher = than for=20 the air pulses for the Lycoming which if you factor the=20 start/slowing/start of air flow may lower its overall velocity = compared=20 to the rotary.
 
In summary, while the total air = intake in=20 equal for engines producing equal HP. It is likely that the air = flow to=20 the rotary may be considerably higher in order to ingest = the same=20 amount of air over the same time.  This may be why =  there=20 is a perception that the Ellison model  that may work well = for a=20 Lycoming 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=20 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=20 significant?
 
So if you had a 13b and a O-360 putting out the same = horsepower=20 for a single given 1 revolution of the propeller, they should = be=20 consuming the same amount of air and fuel during that 1 = propeller=20 revolution. (I THINK chosing 1 propeller rpm is a correct=20 standard)
 
Bill pointed out that the 13b operates at a higher rpm, = and we=20 know that there's more combustion charges consumed by the 13b = to make=20 that 1 prop rpm. 
 
The difference, the missing piece, each 13b combustion = charge=20 consumes a SMALLER amount of fuel/air than the piston = powerplants less=20 frequent combustion charge.   ???   So the = 13b=20 burns a smaller amount more frequently.   ???
 
If this is all true, then the Ellison isn't on the trash = heap=20 yet.
 
Tom

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

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

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

Do you Yahoo!?
Yahoo! Search presents - Jib=20 Jab's 'Second Term' ------=_NextPart_000_000A_01C50F58.4BCCF870--