X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.122] verified) by logan.com (CommuniGate Pro SMTP 5.3.3) with ESMTP id 4151930 for flyrotary@lancaironline.net; Thu, 04 Mar 2010 07:46:04 -0500 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.122; envelope-from=eanderson@carolina.rr.com Return-Path: X-Authority-Analysis: v=1.0 c=1 a=MS9J9ufBWioA:10 a=ayC55rCoAAAA:8 a=arxwEM4EAAAA:8 a=QdXCYpuVAAAA:8 a=7g1VtSJxAAAA:8 a=ekHE3smAAAAA:20 a=UretUmmEAAAA:8 a=Ia-xEzejAAAA:8 a=kviXuzpPAAAA:8 a=nUuTZ29dAAAA:8 a=AzPexmebVJ3lcCoK1lIA:9 a=WVtIuiJLqtXLUslLGG8A:7 a=3agb8X4BFKIlXpWgNGlHu3Bc1fwA:4 a=1vhyWl4Y8LcA:10 a=EzXvWhQp4_cA:10 a=4vB-4DCPJfMA:10 a=67CWjeNaVGUFAdDh:21 a=8w-3VwYYPVOoB0Yf:21 a=SSmOFEACAAAA:8 a=pDDUHcaWVue9YpnjEEIA:7 a=xnhwDho9WehMUgnuyKPaN_-Q_aoA:4 a=G211DGs7y-zO-Hpl:21 a=75l9qiRtbuIouwgp:21 a=tC6YKQLvsyiDG588E-sA:9 a=sNVqgiWbfm6gx21eht5QA4Equm4A:4 a=1Vq_FK4TplAA:10 a=rK7_Snnp8w1mzslLtbcA:9 a=tY9SOkmA9sggeAwJ54AA:9 a=OZJGW_Z213-B3NEgXKdzjVFaZkcA:4 a=q6PCrga3cWlUD47UAaQA:9 a=CzjJBf81rvx4PltPpKAA:9 a=Cw0yJJRF8uhkOVBUJO4Zf-kaV7wA:4 a=e2n4KQ7JTmKbnDR3ovAA:9 a=Yblhoe6ZdYcz7Y2O:18 a=i6GYtFeWOMyvAXi3:18 a=Uvyq_eEvy86UaCuJ:18 X-Cloudmark-Score: 0 X-Originating-IP: 75.191.186.236 Received: from [75.191.186.236] ([75.191.186.236:4877] helo=computername) by cdptpa-oedge03.mail.rr.com (envelope-from ) (ecelerity 2.2.2.39 r()) with ESMTP id D8/89-06903-76BAF8B4; Thu, 04 Mar 2010 12:45:29 +0000 From: "Ed Anderson" Message-ID: To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: Fixed Pitch Prop Thrust was RE: [FlyRotary] Re: single rotor Date: Thu, 4 Mar 2010 07:45:32 -0500 MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_000_0024_01CABB6E.AB5D8530" X-Mailer: Microsoft Office Outlook, Build 11.0.5510 In-Reply-To: Thread-Index: Acq7ef6ThuoJbHUySGO2+ZuysJAMiQAFsplQ X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 This is a multi-part message in MIME format. ------=_NextPart_000_0024_01CABB6E.AB5D8530 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0025_01CABB6E.AB5D8530" ------=_NextPart_001_0025_01CABB6E.AB5D8530 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable I=92ll give it a shot, George =96 even thought I believe I have = previously admitted to not understanding this prop stuff. =20 (1) You are correct V is the velocity of incoming air flow to the prop disc which if the aircraft is not moving is equal to zero. =20 So if the air coming into the prop and the air going out of the prop = disc are equal - then that means the prop did not provide any acceleration or additional velocity to the air mass or in other words Dv (or the change = in air velocity) =3D 0.=20 =20 So if Dv =3D 0 and every term in the equation is multiplied by this = factor, if it is zero then the entire equation =3D 0 and that leaves us with Thrust = T =3D 0. So if no change in air velocity then no change in momentum (pDv)of = the air mass and therefore again no thrust produced. =20 Perhaps another way to look at it is the prop didn=92t push against the = air and therefore didn=92t change (increase) its velocity, therefore no = thrust =96 if it had pushed against the air then the air would have gained in = velocity or Dv <> 0 =20 Strickly looking at the formula for Thrust above you can see that all = terms of the equation or multiplied by the tail end part p*Dv. Well that part = is actually the equation for momentum or rather since its Dv rather than V, = it is the change in momentum imparted to the air mass by the spinning prop. Since the density p is constant, then the change in momentum is entirely = due to the Dv factor. Since Dv =3D v2 =96 v1 change in velocity =96 that is = also the definition of Acceleration =3D Dv/Dt =96 here the Dt (change in time is = assumed to be the same unit time factor for all factors in the equation) and so = is not shown). =20 We know the area for a circle is P R2 or equivalently P (D/2)2 =3D P = D2/4 note that this is the first term in the above formula - the area of the prop disc. So if we now multiply this by p (air density) we get the = air mass that the prop disc is trying to impart a change in momentum to or = P D2/4 * p that leaves (v + Dv/2) *Dv =96 since the aircraft is not = moving then V =3D 0 leaving Dv/2*Dv or Dv2/2 . if we move the density factor p back = to this part of the equation we have p* Dv2/2 . This is the formula for = the kinetic energy increase imparted to the air mass by the additional = velocity imparted to the air by the spinning prop. =20 So probably have not helped your understanding =96 but in summary, if = there is no change in velocity Dv imparted to the air mass flowing through the prop disc, then there is no thrust T. Or at least that is what the = equation appears to say to me. =20 Best Regards =20 Ed =20 Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://www.andersonee.com http://www.dmack.net/mazda/index.html http://www.flyrotary.com/ http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.rotaryaviation.com/Rotorhead%20Truth.htm =20 _____ =20 From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of George Lendich Sent: Thursday, March 04, 2010 3:49 AM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Fixed Pitch Prop Thrust was RE: [FlyRotary] Re: single rotor =20 Ed, Propellers are something else I don't understand and need to learn, so a couple of question if I may. 1. I assume 'Velocity of incoming flow ' means speed of the aircraft. 2. Additional velocity, acceleration by propeller - don't quite = understand that one, Could you explain how one gets 2. value. TIA George ( down under) Arggg! Props! =20 Ok, Bill =96 here is my stab at it. The following equation tells the = tale =96 well some of it. =20 (1) where: T thrust [N] D propeller diameter [m] v velocity of incoming flow [m/s] additional velocity, acceleration by propeller [m/s] density of fluid [kg/m=B3] =20 (air: =3D 1.225 kg/m=B3, water: =3D 1000 kg/m=B3) =20 =20 =20 Thrust is about increasing the momentum of the air mass passing through = the prop disc. Static thrust occurs while sitting still - therefore v (the = air velocity of air in front of the prop disc) =3D 0. So the addition = momentum imparted to the air mass by our spinning prop is p*DV. Since the air = mass p during our run up is essentially constant ), that leaves two variables - = the diameter of the prop D and the amount the spinning prop accelerates the = air (Dv) to affect the thrust (T) generated.=20 =20 The following are extracts from some of the better article (more understandable) material I have read about props and static thrust. As = it concludes and Tracy points out Static Thrust does not really tell the = whole story. =20 =20 For a typical, fixed pitch propeller, the largest induced velocity = occurs under static conditions, where the efficiency is small. It decreases = with increasing flight speed, until it reaches zero: no thrust is generated. For a given power P, it is always desirable to use the largest possible propeller diameter D, which may be limited by mechanical restrictions (landing gear height) or aerodynamic constraints (tip Mach number). = That's why most man or solar powered airplanes use large, slowly turning props. These catch a large volume of air and accelerate it only slightly to = achieve the maximum efficiency. As long as an aircraft does not move, its propeller operates under = static conditions. There is no air moving towards the propeller due to the = flight speed, the propeller creates its own inflow instead. A propeller, with = its chord and twist distribution designed for the operating point under = flight conditions, does not perform very well under static conditions.=20 As opposed to a larger helicopter rotor, the flow around the relatively small propeller is heavily distorted and even may be partially = separated. From the momentum theory of propellers we learn, that the efficiency at lower speeds is strongly dependent on the power loading (power per disk area), and this ratio for a propeller is much higher than that for a helicopter rotor. We are able to achieve about 80-90% of the thrust, as predicted by momentum theory for the design point, but we can reach only = 50% or less of the predicted ideal thrust under static conditions. So much for theory. My personal experience when I went from the faster turning 68x72=94 prop to the slower turning (2.85 gear box) 76x88 prop = =96 my take off acceleration increased significantly indicating (in my opinion) more thrust was being generated. With the 76x88 prop and my old 13B I would generate 5800 rpm static (for what its worth), with it cut down to 74x88 I picked up 200 rpm for a static of 6000 rpm. Plus I got another = inch of ground clearance =96 needed on my nose geared Rv-6a. =20 =20 Interestingly enough the larger slower turning prop not only did not = hurt my top speed it actually increase around 4 mph =96 perhaps due to the = increased HP due to high rpm of the lighter loaded engine? =20 Ok, Bill that=92s my take and what I could pull out of references. = Don=92t know if it really tells us a whole lot =96 there are some good NACA studies = on Prop =96 but the math makes my head hurt. =20 Ed Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://www.andersonee.com http://www.dmack.net/mazda/index.html http://www.flyrotary.com/ http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.rotaryaviation.com/Rotorhead%20Truth.htm =20 _____ =20 From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Bill Bradburry Sent: Wednesday, March 03, 2010 5:01 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: single rotor =20 I would like to get some educational (for me) discussions going on this. = =20 A prop of 76 X 88 is pretty common in our usage. Tracy, Ed, and I have = a Performance Prop in this dimension. Dennis and maybe others have a = Catto prop in this dimension. We all seem to be getting static rpm of about 52-5400 rpm (except for Dennis with his new DIE manifold). Tracy and Ed = had their prop cut down to 74 X 88 and are getting increased static to = around 6000 rpm. Higher rpm =3D higher HP for the rotary. We should get = higher thrust with a slightly smaller diameter prop? This has something to do = with the idea of sizing the prop to the engine. I wonder what is the proper size? What is the proper static rpm for best performance with the = rotary? What did Tracy and Ed lose in prop performance and what did they gain in total performance when they cut the prop down? =20 It seems to me that a prop sized for climb would allow around 7500 rpm = at about Vx or Vy? Max speed would require 7500 rpm at WOT sea level? I wonder what rpm our props allow at these speeds? If you had a prop that would do the above, I wonder what the static rpm would be? Then since = most of us have fixed pitch props, I wonder where we should try to be for the best of both worlds (a compromise)? =20 We have some really good engineers in this group and they have made = these selections. I know they know why they made the selection they did. How about sharing? :>) Don=92t worry, you can not ramble on too much for me! =20 Bill B=20 =20 _____ =20 From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Tracy Crook Sent: Wednesday, March 03, 2010 2:32 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: single rotor Al is correct about it taking HP to make static thrust with a prop but = the assumption about the relationship between HP and static thrust is = subject to a lot of variables. There is no fixed relationship between static = thrust and HP. If there were, you could not account for the ability of most helicopters to hover. =20 =20 You could easily increase static thrust by 1.18 by increasing the = diameter of the prop and the reduction ratio of the redrive with NO increase in = HP. =20 =20 But my real point was that static thrust is not a very useful = measurement to us. =20 Tracy On Wed, Mar 3, 2010 at 11:06 AM, Al Gietzen wrote: Looking at the two sizes of the engine, it takes 1.6 times as much horsepower to develop 1.18 times as much static thrust! Somehow this = does not compute for me=85.I always doubt the performance figures in a sales presentation and when they don=92t make sense to me=85..??? =20 Bill B (hoping this generates an educational experience for me :>) =20 =20 We=92re talking about the amount of force exerted by the prop with the = plane (motor) standing still. So, it seems to make sense to me that the power needed to accelerate the = air to generate the thrust would go as the cube root; and the cube root of = 1.6 is very close 1.18. =20 To move the amount of air it takes to generate the thrust certainly does take horsepower. Very much the same as the power it takes to drive the = pump (or generator) on a dyno. So I don=92t know how Tracy was interpreting = the question. =20 Al =20 __________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com __________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com ------=_NextPart_001_0025_01CABB6E.AB5D8530 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

I’ll give it a shot, George = – even thought I believe I have previously admitted to not understanding = this prop stuff.

 

         &nbs= p; (1)

You are correct V is the velocity = of =A0incoming air flow to the prop disc which if the aircraft is not moving is equal = to zero.=A0

So if the air coming into the prop = and the air going out of the prop disc are equal - then that means the prop did = not provide any acceleration or additional velocity to the air mass or in = other words Dv (or the change in air velocity) =3D 0. =

 

So if D<= /b>v =3D 0 and every = term in the equation is multiplied by this factor, if it is zero then the entire = equation =3D 0 and that leaves us with Thrust T =3D 0.=A0 So if no change in air = velocity then no change in momentum (pDv)of the air mass and therefore again no thrust = produced.

 

=A0Perhaps another way to look at = it is the prop didn’t push against the air and therefore didn’t change (increase) its velocity, therefore no thrust – if it had pushed = against the air then the air would have gained in velocity or = Dv <> = 0

 

Strickly looking at the formula for = Thrust above you can see that all terms of the equation or multiplied by the = tail end part p<= /b>*D<= /b>v.=A0 Well that = part is actually the equation for momentum or rather since its = Dv rather than V, = it is the change in momentum = imparted to the air mass by the spinning prop.=A0 Since the density p is constant, then the change in momentum is entirely = due to the Dv factor.=A0 Since Dv =3D = v2 – v1 change in velocity – that is also the definition of = Acceleration =3D = Dv/D<= /b>t – here the = Dt (change in time = is assumed to be the same unit time factor =A0for all factors in the = equation) and so is not shown).

 

We know the area for a circle is = P R2 or equivalently P<= /b> (D/2)2 =3D P<= /b> D2/4 =A0<= font size=3D2 color=3Dnavy face=3DArial>note that this is the first term in the above formula=A0 - = the area of the prop disc.=A0=A0 So if we now multiply this by = p (air density) =A0we get the air mass that the prop disc is trying to = impart a change in momentum to or =A0P<= /b> D2/4 *= p=A0 = that leaves (v + D<= /b>v/2) *D<= /b>v – since the = aircraft is not moving then V =3D = 0 = leaving D<= /b>v/2*Dv or = Dv2/2 = . if we move the density factor p back to this part of the equation we have p* = Dv2/2 .=A0 This is =A0the formula for the kinetic energy increase imparted to the air mass = by the additional velocity imparted to the air by the spinning = prop.

 

So probably have not helped your understanding – but in summary, =A0if there is no change in = velocity Dv imparted to the air mass flowing through the prop disc, then there is no = thrust T.=A0 Or at least that is what the equation appears to say to = me.

 

Best = Regards

 

Ed

 

From: = Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of George Lendich
Sent: Thursday, March 04, = 2010 3:49 AM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: = Fixed Pitch Prop Thrust was RE: [FlyRotary] Re: single = rotor

 

 Ed,

Propellers are something else I don't understand and = need to learn, so a couple of question if I may.

1. I assume 'Velocity of incoming flow ' means speed = of the aircraft.

2. Additional velocity, acceleration by propeller - = don't quite understand that one,

Could you explain how one gets 2. = value.

TIA

George ( down under)

Arggg!  Props!

 

Ok, Bill – here is my stab at it.  The following equation tells = the tale – well some of it.

 

         &nbs= p; (1)

where:

T

thrust

[N]

D

propeller = diameter

[m]

v

velocity of incoming = flow

[m/s]

additional velocity, acceleration by = propeller

[m/s]

density of = fluid

[kg/m=B3]

 

(air:  =3D 1.225 = kg/m=B3, water:  =3D 1000 = kg/m=B3)

 

 

 

Thrust is about increasing the momentum of the air mass passing through the prop disc.  Static thrust occurs while sitting still -  therefore = v (the air velocity = of air in front of the prop disc)  =3D 0.  So the addition momentum imparted = to the air mass by our spinning prop is p*D= V.  = Since the air mass p during our run up is essentially constant ), that = leaves two variables - the diameter of the prop D and the amount the spinning prop accelerates the air = (Dv) to affect the  thrust (T) generated. =

 <= /span>

The following are extracts from some of the better article (more understandable) = material I have read about props and static thrust.  As it concludes and = Tracy points = out Static Thrust does not really tell the whole story.  

 

For a typical, fixed pitch propeller, the largest = induced velocity = occurs = under static conditions, where the efficiency is small. It decreases with increasing = flight speed, until it reaches zero: no thrust is = generated.

For a given power P, it is always desirable to use the largest possible = propeller diameter D, which may be = limited by mechanical restrictions (landing gear height) or aerodynamic = constraints (tip Mach number). That's why most man or solar powered airplanes use = large, slowly turning props. These catch a large volume of air and accelerate = it only slightly to achieve the maximum efficiency.

As long as an aircraft does not move, its propeller operates under static conditions. There is no air moving towards the propeller due to the = flight speed, the propeller creates its own inflow instead. A propeller, with = its chord and twist distribution designed for the operating point under = flight conditions, does not perform very well under static conditions. =

As opposed to a larger helicopter rotor, the flow around the relatively = small propeller is heavily distorted and even may be partially separated. From = the momentum theory of propellers we learn, that the efficiency at lower = speeds is strongly dependent on the power loading (power per disk area), and this = ratio for a propeller is much higher than that for a helicopter rotor. We are = able to achieve about 80-90% of the thrust, as predicted by momentum theory for = the design point, but we can reach only 50% or less of the predicted ideal = thrust under static conditions.

So much for theory.  My = personal experience when I went from the faster turning 68x72” prop to the = slower turning (2.85 gear box) 76x88 prop – my take off acceleration = increased significantly indicating (in my opinion) more thrust was being = generated.  With the  76x88 prop and my old 13B I would generate 5800 rpm = static (for what its worth), with it cut down to 74x88 I picked up 200 rpm for a = static of 6000 rpm.  Plus I got another inch of ground clearance – = needed on my nose geared Rv-6a. 

 

Interestingly enough the larger = slower turning prop not only did not hurt my top speed it actually increase = around 4 mph – perhaps due to the increased HP due to high rpm of the = lighter loaded engine?

 

Ok, Bill that’s my take and = what I could pull out of references.  Don’t know if it really tells = us a whole lot – there are some good NACA studies on Prop – but = the math makes my head hurt.

 

Ed

From: = Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Bill Bradburry
Sent: Wednesday, March = 03, 2010 5:01 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: = single rotor

 

I would like to get some educational (for me) discussions going = on this. 

A prop of 76 X 88 is pretty common in our usage.  Tracy, = Ed, and I have a Performance Prop in this dimension.  Dennis and maybe others = have a Catto prop in this dimension.  We all seem to be getting static rpm = of about 52-5400 rpm (except for Dennis with his new DIE manifold).  = Tracy and Ed had their prop cut down to 74 X 88 and are getting increased = static to around 6000 rpm.  Higher rpm =3D higher HP for the rotary.  We = should get higher thrust with a slightly smaller diameter prop?  This has something to do with the idea of sizing the prop to the engine.  I = wonder what is the proper size?  What is the proper static rpm for best performance with the rotary?   What did Tracy and Ed lose in prop performance and what did they gain in total performance when they cut = the prop down?

 

It seems to me that a prop sized for climb would allow around = 7500 rpm at about Vx or Vy?  Max speed would require 7500 rpm at WOT sea level?  I wonder what rpm our props allow at these speeds?  If = you had a prop that would do the above, I wonder what the static rpm would be?  Then since most of us have fixed pitch props, I wonder where = we should try to be for the best of both worlds (a = compromise)?

 

We have some really good engineers in this group and they have = made these selections.  I know they know why they made the selection = they did.  How about sharing?  :>)

Don’t worry, you can not ramble on too much for = me!

 

Bill B

 

From: Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net] On Behalf Of Tracy = Crook
Sent: Wednesday, March = 03, 2010 2:32 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: = single rotor

Al is correct about it taking HP to make static thrust with a = prop but the assumption about the relationship between HP and static thrust is = subject to a lot of variables.  There is no fixed relationship between = static thrust and HP.   If there were, you could not account for the = ability of most helicopters = to hover.  

 

 You could easily increase static thrust by 1.18 by = increasing the diameter of the prop and the reduction ratio of the redrive with NO increase in HP.  =

 

But my real point was that static thrust is not a very useful measurement to us.

 

Tracy

On Wed, Mar 3, 2010 at 11:06 AM, Al Gietzen <ALVentures@cox.net> = wrote:

Looking at the two sizes of the engine, it takes 1.6 times as = much horsepower to develop 1.18 times as much static thrust!  Somehow = this does not compute for me….I always doubt the performance figures in a = sales presentation and when they don’t make sense to = me…..???

 

Bill B (hoping this generates an educational experience for = me  :>) 

 

We’re talking about the amount of force exerted by the = prop with the plane (motor) standing still.

So, it seems to make sense to me that the power needed to accelerate the air to generate the thrust would go as the cube root; and = the cube root of 1.6 is very close 1.18.

 

To move the amount of air it takes to generate the thrust = certainly does take horsepower.  Very much the same as the power it takes to = drive the pump (or generator) on a dyno.  So I don’t know how = Tracy was = interpreting the question.

 

Al

 



__________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com



__________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com



__________ Information from ESET NOD32 Antivirus, version of = virus signature database 3267 (20080714) __________

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