X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from [24.25.9.100] (HELO ms-smtp-01-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 5.0.4) with ESMTP id 892364 for flyrotary@lancaironline.net; Thu, 22 Dec 2005 08:46:40 -0500 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.100; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-025-165.carolina.res.rr.com [24.74.25.165]) by ms-smtp-01-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id jBMDjVWe023565 for ; Thu, 22 Dec 2005 08:45:32 -0500 (EST) Message-ID: <000f01c606fd$fe20e8e0$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Props Date: Thu, 22 Dec 2005 08:45:38 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000C_01C606D4.14FE95A0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.2180 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2900.2180 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_000C_01C606D4.14FE95A0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable I really do not know, George. I have known folks with aircraft speeds = somewhat faster than the "screw" calculations would have predicted - = perhaps this "lift" factor played a role. Most figures I have seen puts = efficient of prop around 80-85% mostly around 80. Ed ----- Original Message -----=20 From: george lendich=20 To: Rotary motors in aircraft=20 Sent: Thursday, December 22, 2005 4:45 AM Subject: [FlyRotary] Re: Props Thanks Ed, I like to reduce things down to the simplest formula, provided one = doesn't lose track of what's relevant. So I guess I should look at this as 100 % theoretically efficient - = whereas we may be looking at only 80 % efficiency ( in real terms), Do = you agree with 80%? However your saying that the lift may give more speed ( MPH, in real = terms) - So are you suggesting that these two (variables) may balance = out? Or should we use a fudge factor, and that could be what? I'm not looking for a finite answer, just close enough to be able to = access the experts opinion, and we all know what an expert is - " A drip = under pressure". George (down under) Yes, George. If we assumed no slip in the prop then theoretically = the formula will give you the pitch of the prop you would need to screw = through the air and achieve your MPH at your RPM input. This is based = simply on the screw principal. I am certainly not a prop expert, but I = also believe that since the prop is a spinning wing (airfoil), that = there is a lift component in the direction of flight which may result = in more speed than the formula could account for. =20 Your second equation can be reduced further from = (Inches/Minute)/(RPM) =3D (Inches/Minute)/(Revs/Minute) =3D inches/Rev Ed A ----- Original Message -----=20 From: Bulent Aliev=20 To: Rotary motors in aircraft=20 Sent: Wednesday, December 21, 2005 7:08 PM Subject: [FlyRotary] Re: Props Don't ask me George. I'm not an engineer or good with mathematics. = Ed A will know better.=20 Buly On Dec 21, 2005, at 6:20 PM, george lendich wrote: Thanks Buly, So, Pitch" =3D MPH x12 x 5280 RPM x 60 I figure 12 (inches in a foot) and 5280 ( feet in a mile) and 60 = ( minute in an hour). So the top line is converted into inches per minute and the = bottom line RPM. So, Pitch " =3D inches per minute RPM Does that look right to you ? George ( down under) Thanks George. my mistake was that i did not convert it to MPH. I was also given the formula for cruise speed: Pitch" x RPM x 60 =3D MPH 12 x 5280 Buly On Dec 21, 2005, at 4:39 PM, george lendich wrote: Buly, Looking at the formula given by members on this list, 2 out = of 3 were spot on with Hoverhawk and the third was close. Examples: 1. Prop Dia x 3.1416x RPM/ 720 - Correct! 2. Prop RPM x Dia x .00436 - Correct ( 3.14/ 720 =3D.00436) 3. 3.14 x 64 {given prop dia} ( 5500/2.17 {given RPM})=3D = 508346/707 - slightly incorrect ( as compared to the other 3 examples, = which include Hoverhawk) i.e 3.14/ 707 =3D .004809 whereas all the others are 3.14/ = 720 =3D .00436. Now that I have that sorted - can someone tell me how to = calculate pitch, easily!! George ( down under) Thanks Perry. This makes me feel much better. Looks like = the formulas given to me were not good. Buly On Dec 19, 2005, at 2:00 PM, Perry Casson wrote: I think your math is off a bit - Here's a online = calculator http:// www.hoverhawk.com/propspd.html I think 6900 puts a 72" prop close to mach 0.9 but I've = never ran mine above 6500 Perry Casson -- Homepage: http://www.flyrotary.com/ Archive and UnSub: = http://mail.lancaironline.net/lists/flyrotary/ Buly atlasyts@bellsouth.net -- Homepage: http://www.flyrotary.com/ Archive and UnSub: = http://mail.lancaironline.net/lists/flyrotary/ -- Homepage: http://www.flyrotary.com/ Archive and UnSub: = http://mail.lancaironline.net/lists/flyrotary/ Bulent "Buly" Aliev = http://home.bellsouth.net/p/s/community.dll?ep=3D16&ext=3D1&groupid=3D164= 323&ck=3D Bulent "Buly" Aliev = http://home.bellsouth.net/p/s/community.dll?ep=3D16&ext=3D1&groupid=3D164= 323&ck=3D ------=_NextPart_000_000C_01C606D4.14FE95A0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
I really do not know, George.  I have known = folks=20 with aircraft speeds somewhat faster than the "screw" calculations would = have=20 predicted - perhaps this "lift" factor played a role.  Most figures = I have=20 seen puts efficient of prop around 80-85% mostly around 80.
 
Ed
----- Original Message -----
From:=20 george=20 lendich
Sent: Thursday, December 22, = 2005 4:45=20 AM
Subject: [FlyRotary] Re: = Props

Thanks Ed,
I like to reduce things down to the = simplest=20 formula, provided one doesn't lose track of what's = relevant.
So I guess I should look at this as = 100 %=20 theoretically efficient - whereas we may be looking at only 80 % = efficiency (=20 in real terms), Do you agree with 80%?
However your saying that the = lift may give=20 more speed ( MPH, in real terms) - So are you suggesting that = these two=20 (variables) may balance out?
 
Or should we use a fudge factor, and = that could=20 be what?
 
I'm not looking for a finite = answer, just=20 close enough to be able to access the experts opinion, and we all know = what an=20 expert is - " A drip under pressure".
 
George (down under)
Yes, George.  If we assumed no slip in = the prop=20 then theoretically the formula will give you the pitch of the prop = you would=20 need to screw through the air and achieve your MPH at your RPM = input. =20 This is based simply on the screw principal.  I am certainly = not a prop=20 expert, but I also believe that since the prop is  a spinning = wing=20 (airfoil), that there is a lift component  in the direction of = flight=20 which may result in more speed than the formula could account=20 for.  
 
Your second equation can be reduced further = from=20   (Inches/Minute)/(RPM) =3D (Inches/Minute)/(Revs/Minute) =3D=20 inches/Rev
 
Ed A
 
 
----- Original Message ----- =
From:=20 Bulent Aliev
To: Rotary motors in = aircraft=20
Sent: Wednesday, December = 21, 2005=20 7:08 PM
Subject: [FlyRotary] Re: = Props

Don't ask me George. I'm not an engineer or good = with=20 mathematics. Ed A will know better.=20
Buly
On Dec 21, 2005, at 6:20 PM, george lendich wrote:
Thanks=20 Buly,
So, Pitch"  = =3D MPH=20 x12 x 5280
           =       =20  RPM x 60
I figure 12 = (inches in a=20 foot) and 5280 ( feet in a mile) and 60 ( minute in an=20 hour).
So the top line is = converted=20 into inches per minute and the bottom line = RPM.
 
So, Pitch " =3D = inches=20 per minute
           =         =20 RPM
 
Does that = look right to=20 you ?
George ( down=20 under)
 
 
Thanks George. my mistake was that i did not convert it to=20 MPH.
I was also given the formula for cruise speed:

Pitch" x = RPM x=20 60  =3D MPH
       12 x 5280

Buly
On Dec 21, 2005, at 4:39 PM, george lendich = wrote:
Buly,
Looking at the=20 formula given by members on this list, 2 out of 3 were=20 spot
on = with Hoverhawk=20 and the third was close.
Examples:
1. = Prop Dia x=20 3.1416x RPM/ 720 - Correct!
2. = Prop RPM x Dia=20 x .00436 - Correct ( 3.14/ 720 =3D.00436)
3. = 3.14 x 64=20 {given prop dia} ( 5500/2.17 {given RPM})=3D 508346/707 = -
slightly=20 incorrect ( as compared to the other 3 examples, which=20 include
Hoverhawk)
i.e = 3.14/ 707 =3D=20 .004809 whereas all the others are 3.14/ 720 =3D = .00436.
Now = that I have=20 that sorted - can someone tell me how to calculate=20 pitch,
easily!!
George ( down=20 under)
Thanks Perry.=20 This makes me feel much better. Looks like the=20 formulas
given to me=20 were not good.
Buly
On = Dec 19,=20 2005, at 2:00 PM, Perry Casson wrote:
I = think your=20 math is off a bit - Here's a online calculator=20 http://
www.hoverhawk.com/propspd.html
I = think 6900=20 puts a 72" prop close to  mach 0.9 but=20 I've never ran
mine above=20 6500
Perry=20 Casson
--
Homepage:  http://www.flyrotary.com/
Archive and=20 UnSub:  =20 http://mail.lancaironline.net/lists/flyrotary/
Buly
atlasyts@bellsouth.net
--
Homepage:  http://www.flyrotary.com/
Archive and=20 UnSub:  =20 http://mail.lancaironline.net/lists/flyrotary/
--
Homepage:  http://www.flyrotary.com/
Archive and=20 UnSub:  =20 http://mail.lancaironline.net/lists/flyrotary/

Bulent "Buly" Aliev

http://home.bellsouth.net/p/s/community.dll?ep=3D16&ext=3D= 1&groupid=3D164323&ck=3D


Bulent "Buly" Aliev

http://home.bellsouth.net/p/s/community.dl= l?ep=3D16&ext=3D1&groupid=3D164323&ck=3D

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