Return-Path: Received: from fed1rmmtao05.cox.net ([68.230.241.34] verified) by logan.com (CommuniGate Pro SMTP 4.3c1) with ESMTP id 725005 for flyrotary@lancaironline.net; Fri, 11 Feb 2005 10:59:04 -0500 Received-SPF: none receiver=logan.com; client-ip=68.230.241.34; envelope-from=ALVentures@cox.net Received: from BigAl ([68.7.14.39]) by fed1rmmtao05.cox.net (InterMail vM.6.01.04.00 201-2131-117-20041022) with ESMTP id <20050211155816.LNAJ11322.fed1rmmtao05.cox.net@BigAl> for ; Fri, 11 Feb 2005 10:58:16 -0500 From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Last Word on Displacement was Re: [FlyRotary] Solved!!!! => Rotary AirFlow Equation? Help? Date: Fri, 11 Feb 2005 07:58:18 -0800 Message-ID: <001801c51052$813abc40$6400a8c0@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0019_01C5100F.73177C40" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.6626 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 Importance: Normal In-Reply-To: This is a multi-part message in MIME format. ------=_NextPart_000_0019_01C5100F.73177C40 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: quoted-printable Thank you; Leon. That's how I had always looked at it; but with all the math spinning = about, I thought maybe I had overlooked something. =20 Al =20 Subject: [FlyRotary] Last Word on Displacement was Re: [FlyRotary] Solved!!!! =3D> Rotary AirFlow Equation? Help? =20 Hi Guys, =20 I have that feeling of "de ja vue" I've been here before. Here's how = to understand it all without a million pages of marthematics. =20 =20 The face of a rotary piston presents itself to the inlet port each revolution. (forget about the fact that the rotor has 3 faces - it only confuses the issue!!). Each piston of a 4 stroke piston engine go = through an "Otto Cycle" every TWO revolutions. =20 =20 With a rotary, you get ONE suck, per ROTOR, per REV. With a 4 = stroke, you get ONE suck per PISTON, per (get this) TWO revs!!! =20 The same goes for Power Pulses. A rotary gets ONE power pulse per REV, = per ROTOR. With a 4 stroke piston engine, you get ONE power pulse, per PISTON, per TWO revs. =20 1. The "SWEPT VOLUME" of a 6.5B/13B/20B rotary is 654 cc per rotor. 2. For EACH revolution of the E-Shaft, EACH rotor sucks 654 cc. 3. So, for a 6.5B, it sucks 654 cc, for a 13B, it sucks 1308 cc = PER REV, and for a 20B it sucks 1962 cc. =20 I'll leave you to do the conversion into Cu.In.=20 =20 Now, a 4 stroke engine completes it's Otto Cycle in TWO revolutions. = So to equate a rotary to a 4 stroke "Otto Cycle" piston engine, we must = compare how much a rotary sucks in TWO revolutions. =20 According to my limited mathematical ability, (I could never "run the numbers" with the alacrity of the blokes on ACRE! (}:>)), =20 =20 A 6.5B is equivalent to a 1308 cc 4 stroke piston (Otto Cycle) engine. = (two times 654 cc per rev) A 13B is equivalent to a 2616 cc 4 stroke piston (Otto Cycle) engine. = (two times 1308 cc per rev) A 20B is equivalent to a 3924 cc 4 stroke piston (Otto Cycle) engine. = (two times 1962 cc per rev) =20 It is THAT simple. Compare Granny Smith Apples with Granny Smith Apples (not mangoes or bananas!). =20 Now there are a couple of caveats. =20 1. The rotary, when ported, has the ability to easily run 125% to = 130% volumetric efficiency. So it has a slightly higher ability to draw air = than a piston engine. Runners, throttle bodies, and injectors must be = sized accordingly. =20 2. This being so, it will put out slightly (up to 20%) more power = than the equivalent swept volume of a 4 cycle piston engine of the same SWEPT capacity. =20 So there you have it. It's THAT simple! =20 Enjoy! =20 Leon ----- Original Message -----=20 From: Ed Anderson=20 To: Rotary motors in aircraft=20 Sent: Friday, February 11, 2005 8:18 AM Subject: [FlyRotary] Solved!!!! =3D> Rotary AirFlow Equation? Help? =20 > Ed, I think you've blinded yourself with science 8*) >=20 > Each face of the rotor will cycle 40cid of air on each revolution. = You=20 > can ignore the eshaft. Every time the rotor turns around, each face=20 > will have processed 40cid. If a rotor is turning at 2000 RPM, then = three=20 > faces will each cycle 2000*40cid of air. You bring in referencing off = > the eshaft, and all of a sudden only 4 faces are processing air. I=20 > think that is the mistake. >=20 > You reference 720 degrees of eshaft rotation. But all that means is=20 > that the rotors haven't completed their cycle. You're just ignoring=20 > every third rotor face. >=20 > =20 Ernest, I think you are making my very point.=20 =20 1. Each rotor has 3*40 =3D 120 cid of volume displace per 360 deg = rotation and with two rotor that is 2*120 =3D 240 CID each revolution. so we have 2000*240 cid/1728 =3D 277.77 CFM exactly what quantity the = formula gives when treating the rotary as a 160 CID 4 stroke engine with e = shaft at 6000 rpm - but that is for a FULL 360 deg rotation of the rotor. See = part 2 next {:>). =20 =20 2. Example =20 160*6000/(2*1728) =3D 277.77 CFM which agrees with the above but if the = 160 CID figure is correct then that means that 160/4 =3D 4 , only 4 rotor = faces worth of displacement have been considered in the formula. Which if = that is correct then the rotor have only turned 240 degs and not 360 deg. But = if the rotor has only turned 240 deg (240*3 =3D 720Deg Checks!) then that = means that to make a valid comparison with 1 above, we need to reduce the = amount of rotation in statement 1 to 240 deg instead of 360 which means that we only have 277.77*4/6 =3D 185 CFM which leaves me back where I started. =20 There is no question that if the two rotors have completed their 360 deg rotation - then all six faces have completed the cycle. =20 That gives 40*6*2000/1728 =3D 277.77 CFM, so lets not debate that point = - we both agree! . =20 My problem has been the use of the 160 CID equivalent in the formula. I understand how it was arrived at ( I believe) its simply the amount of displacement occurring in the rotary at 720 deg of e shaft rotation = which is the standard of comparison with other engines. =20 At 720 deg of e shaft rotation then indeed 160 CID of displace has = occurred. If I calculate using the 240 deg of rotor rotation I again get 240/360 * = 6 =3D 4 rotor faces. But if that is indeed the actually amount of = displacement of the rotor then according to our calculations in 1 above we need to use = 240 deg not 360 and that gives 185 cfm not 277.77. =20 I mean I don't care if we use 240 CID (the total displacement across = 1080) or 160 CID (the total displacement across 720 deg - the standard for comparison) but there should be some logical consistency between the = two. =20 WAIT! Stop the presses I Finally Understand! The key is the 240 CID = total displacement for 360 rotor degrees or for 1080 deg of e shaft rotation. = =20 It was not a math problem or logic problem, my problem was a reference transformation problem! =20 =20 Using the rotor representation we have Air flow =3D 240 * 2000/1728 =3D 277.77 this is rotary based on 360 deg of rotation. Now = if we take that formula (which I think we agree on), I then want to convert = it into a formula that conforms to the 720 deg standard. So I can compare apples and apples. =20 Here's what happens.=20 =20 1. First the e shaft rpm is the e shaft rpm and is the standard = reference point for rpm of engines. So we can simply multiply our 2000 rpm rotor = rpm by 3 to reference it to the eshaft. So 3* 2000 =3D 6000. That "transformation" gives us the rpm figure in the "Standard" equation. =20 2. However, to reference the 240 CID displacement to the 720 deg = standard we must use the ratio of the transformation from 1080 to the 720 = reference system. Because the amount of displacement IS affected by the choice of reference. This gives 720/1080 * 240 =3D 160 CID referenced to the = standard. =20 So now the formula should migrate from rotor reference of 240*2000/1728 = =3D 277.77 to e shaft reference of 160*6000/1728 =3D 277.77 by reference = system transformation. To do that: =20 1. We multiply the rotor rpm by 3 - the rpm does not care whether its = 720 or 1080 referenced, one revolution/min is one revolution/min. 2. The 720 deg standard DOES affect the amount of displacement so = 720/1080 *240 =3D 160 CID. =20 Taking the rotor reference formula above we can directly transform it to = the e shaft reference standard. Doing so gives us =20 [720/1080*240]*[3*2000]/1728 =3D [0.666*240]*[6000]/1729, taking = the first factor 0.6666* 240 =3D 159.9999 =3D 160 and we have =20 Airflow =3D 160CID*6000/1728 =3D 277.77 So the rotor reference is = transformed to the e shaft reference of 720 deg. =20 Thanks all for helping me out of my problem area. I just knew there had = to be a connection someway. If this is not correct - please refrain from informing me {:>). =20 =20 Be heartened, Ernest, I have been faithfully using the 160 formula - = just wanted to understand what appeared to me to be a difference when looking = at it from two different perspectives. If the "reference" transformation = is applied to the rotor equation then it translates into the 720 reference state. =20 Best Regards and thanks for your input =20 Ed =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 ------=_NextPart_000_0019_01C5100F.73177C40 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Thank you; Leon.

That’s how I had always looked at it; but = with all the math spinning about, I thought maybe I had overlooked = something.

 

Al

 

Subject: [FlyRotary] Last Word on Displacement was Re: [FlyRotary] Solved!!!! =3D> Rotary = AirFlow Equation? Help?

 

Hi Guys,

 

I have that feeling of = "de ja vue"  I've been here before.  Here's how to understand it = all without a million pages of marthematics. 

 

The face of a rotary piston = presents itself to the inlet port each revolution.  (forget about the = fact that the rotor has 3 faces - it only confuses the issue!!).  Each = piston of a 4 stroke piston engine go through an "Otto Cycle" every = TWO revolutions. 

 

With a rotary,  you = get ONE suck,  per ROTOR,  per REV.  With a 4 stroke,  you = get ONE suck per PISTON,  per  (get this) TWO = revs!!!

 

The same goes for Power Pulses.  A rotary gets ONE power pulse per REV,  per = ROTOR.  With a 4 stroke piston engine,  you get ONE power pulse,  per PISTON,  per TWO revs.

 

1.   The = "SWEPT VOLUME" of a 6.5B/13B/20B rotary is 654 cc per = rotor.

2.   For EACH = revolution of the E-Shaft,  EACH rotor sucks 654 cc.

3.   So,  = for a 6.5B,  it sucks 654 cc,  for a 13B,  it sucks 1308 cc PER REV,  and for a 20B it sucks 1962 cc.

 

I'll leave you to do the = conversion into Cu.In.

 

Now,  a 4 stroke = engine completes it's Otto Cycle in TWO revolutions.  So to equate a = rotary to a 4 stroke "Otto Cycle" piston engine,  we must compare how much a rotary sucks in TWO revolutions.

 

According to my limited = mathematical ability, (I could never "run the numbers" with the = alacrity of the blokes on ACRE! = (}:>)), 

 

A 6.5B is equivalent to = a 1308 cc 4 stroke piston (Otto Cycle) engine.  (two times 654 cc per = rev)

A 13B is equivalent to a = 2616 cc 4 stroke piston (Otto Cycle) engine.   (two times 1308 = cc per rev)

A 20B is equivalent to a = 3924 cc 4 stroke piston (Otto Cycle) engine.   (two times 1962 cc per = rev)

 

It is THAT simple.  = Compare Granny Smith Apples with Granny Smith Apples (not mangoes or = bananas!).

 

Now there are a couple of = caveats.

 

1.   The = rotary,  when ported,  has the ability to easily run 125% to 130% volumetric efficiency.  So it has a slightly higher ability to draw air than a = piston engine.  Runners,  throttle bodies,  and injectors must = be sized accordingly.

 

2.   This being = so,  it will put out slightly (up to 20%) more power than the equivalent = swept volume of a 4 cycle piston engine of the same SWEPT = capacity.

 

So there you have = it. It's THAT simple!

 

Enjoy!

 

Leon

=

----- Original Message = -----

From: Ed Anderson

Sent: Friday, February 11, 2005 8:18 AM

Subject: [FlyRotary] Solved!!!! =3D> Rotary AirFlow Equation? Help?

 


> Ed, I think you've blinded yourself with science 8*)
>
> Each face of the rotor will cycle 40cid of air on each = revolution.  You
> can ignore the eshaft.  Every time the rotor turns around, = each face
> will have processed 40cid. If a rotor is turning at 2000 RPM, then = three
> faces will each cycle 2000*40cid of air.  You bring in = referencing off
> the eshaft, and all of a sudden only 4 faces are processing = air.  I
> think that is the mistake.
>
> You reference 720 degrees of eshaft rotation.  But all that = means is
> that the rotors haven't completed their cycle.  You're just = ignoring
> every third rotor face.
>
>

 

Ernest, I think you are = making my very point. 

 

1.  Each rotor has = 3*40 =3D 120 cid of volume displace per 360 deg rotation and with two rotor that is = 2*120 =3D 240 CID each revolution.

 so we have 2000*240 = cid/1728 =3D 277.77 CFM exactly what quantity the formula gives  when treating = the rotary as a 160 CID 4 stroke engine with e shaft at 6000 rpm - but that = is for a FULL = 360 deg rotation of the rotor.  See part = 2 next {:>). 

 

2.  = Example

 

160*6000/(2*1728)  =3D = 277.77 CFM which agrees with the above but if the 160 CID =  figure is correct then that means that 160/4 =3D 4 , only 4 rotor faces = worth of displacement have been considered in the formula.  Which if that is correct then the rotor have only turned 240 degs  and not 360 = deg.  But if the rotor has only turned 240 deg (240*3 =3D 720Deg Checks!) then = that means that to make a valid comparison with 1 above, we need to reduce = the amount of rotation in statement 1 to 240 deg instead of 360 which means = that we only have 277.77*4/6 =3D 185 CFM which leaves me back where I = started.

 

There is no question that = if the two rotors have completed their 360 deg rotation - then = all six faces have = completed the cycle. 

That gives 40*6*2000/1728 = =3D 277.77 CFM, so lets not debate that point - we both agree! = .

 

My problem has been the use = of the 160 CID equivalent in the formula.  I understand how it was arrived = at ( I believe) its simply the amount of displacement occurring in the rotary = at 720 deg of e shaft rotation  which is the standard of comparison with = other engines.

 

At 720 deg of e shaft = rotation then indeed 160 CID of displace has occurred.  If I calculate using = the 240 deg of rotor rotation I again get 240/360 * 6 =3D 4 rotor = faces.  But if that is indeed the actually amount of displacement of the rotor then = according to our calculations in 1 above we need to use 240 deg not 360 and that = gives 185 cfm not 277.77.

 

I mean I don't care if we = use 240 CID (the total displacement  across 1080) or 160 CID (the total displacement across 720 deg - the standard for comparison) but there = should be some logical consistency between the two.

 

WAIT! Stop the presses I  Finally Understand!  The key is the 240 CID total = displacement for 360 rotor degrees or for 1080 deg of e shaft = rotation. 

 

It was not a math problem = or logic problem, my problem was a reference transformation = problem!

 

 

    &nbs= p;        Using the rotor representation we have Air flow =3D 240 * 2000/1728 = =3D 277.77 this is rotary based on 360 deg of rotation.  Now if we  take that formula (which I think we agree on), =  I then want to convert it = into a formula that conforms to the 720 deg = standard.  So I can compare apples and apples.

 

Here's what = happens. 

 

1.   First the e = shaft rpm is the e shaft rpm and is the standard reference point for rpm of engines.  So we can simply multiply our 2000 rpm rotor rpm by 3 = to reference it to the = eshaft. So 3* 2000 =3D 6000.  That "transformation" gives us the rpm = figure in the "Standard" equation.

 

2.  However, to  reference the 240 CID displacement to the 720 deg standard we must = use the ratio of the transformation from 1080 to the 720 reference = system.  Because the amount of displacement IS affected by = the choice of reference.   This gives 720/1080 * 240 =3D 160 CID = referenced to the standard.

 

So now the formula should = migrate from rotor reference of 240*2000/1728 =3D 277.77 to e shaft = reference of 160*6000/1728 =3D 277.77 by reference system transformation.  = To do that:

 

1.   We = multiply the rotor rpm by 3 - the rpm does not care whether its 720 or 1080 = referenced, one revolution/min is one revolution/min.

2.   The 720 deg = standard DOES affect the amount of displacement so 720/1080 *240 =3D 160 = CID.

 

Taking the rotor reference = formula above we can directly transform it to the e shaft reference = standard.  Doing so gives us

    =

     [720/1080*240]*[3*2000]/1728 =3D [0.666*240]*[6000]/1729, taking the = first factor 0.6666* 240 =3D 159.9999 =3D 160 and we have

 

  Airflow =3D = 160CID*6000/1728 =3D 277.77  So the rotor reference is transformed to the e shaft = reference of 720 deg.

 

Thanks all for helping me = out of my problem area.  I just knew there had to be a connection = someway.  If this is not correct - please refrain from informing me = {:>).

 

 

Be  heartened, Ernest, =  I have been faithfully using the 160 formula - just wanted to understand = what appeared to me to be a difference when looking at it from two different perspectives.  If the "reference" transformation is = applied to the rotor equation then it translates into the  720 reference = state.

 

 Best Regards and = thanks for your input

 

Ed

 

 

 

 

 

 

 

 

 

 

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