X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.120] verified) by logan.com (CommuniGate Pro SMTP 5.1.12) with ESMTP id 2394279 for flyrotary@lancaironline.net; Thu, 18 Oct 2007 15:28:52 -0400 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.120; envelope-from=eanderson@carolina.rr.com Received: from edward2 ([24.74.103.61]) by cdptpa-omta02.mail.rr.com with SMTP id <20071018192813.JVCE8967.cdptpa-omta02.mail.rr.com@edward2> for ; Thu, 18 Oct 2007 19:28:13 +0000 Message-ID: <001c01c811bd$20e04940$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Math Help Date: Thu, 18 Oct 2007 15:28:57 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0019_01C8119B.9992FFD0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3138 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 This is a multi-part message in MIME format. ------=_NextPart_000_0019_01C8119B.9992FFD0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Assuming the plate is fixed and perpendicular to the wind what you would = like to know is the force against it. Since force =3D Pressure/Area we = first need to find the pressure. The dynamic pressure energy of the airstream at 200 mph is found by = 1/2pV^2 where p is air density and V is velocity. I am going to assume = sea level air density even though we know it will be less at altitude. = So that is one square foot of air =3D 0.00234 slugs (sorry I'm an old = English units guy). Velocity (ft/sec) =3D 200mph *5280/3600 =3D 293.3 = ft/sec That gives us a dynamic Pressure =3D 1/2(0.00234*(293.3)^2 =3D = 0.5*(0.00234)*(86034) =3D 100.66 lbs/ft^2 or 100 lbs per square foot. = Dividing by 144 gives us lbs per square inch. So pressure =3D = 100.66/144 =3D .699 lbs/sq inch or about 19 inches of H20. But, = sticking with psi we have=20 That pressure is acting on 15 sq inches so the total force on that plate = =3D 0.699*15 =3D 10.485 lbs of force. =20 That is assuming all the energy in the air is converted to dynamic = pressure which would probably only happens at the very center of the = plate, the air near the edges would tend to flow (at least partially) = around the plate, so the velocity/pressure distribution would not be = even across the plate. It would be maximum in the center and taper off = near the edges. So this means the actually force would undoubtedly be = less than 10 lbs because of that and the fact that the air at altitude = will be considerably less dense, depending of course on your altitude. = Then there is the fact the plate is moving through the air means that = the pressure in back of the plate is going to be less than if it were = standing still. But I am not going to even attempt to get into = quantifying those factors. But, the force will still likely be less = than 10 lb You might try searching the web for things like - air resistance, plate = plate, dynamic pressure, etc Hope this helps Ed ----- Original Message -----=20 From: Steve Thomas=20 To: Rotary motors in aircraft=20 Sent: Thursday, October 18, 2007 2:35 PM Subject: [FlyRotary] Math Help Not being much of a math or engineering whiz, can anyone help me = calculate the force the relative wind would put on a device that hangs = in the breeze on my airplane? Assuming a 200 MPH airspeed, and a flat = plate of 15 sq. inches, what would be the amount of force on the plate = in pounds? =20 Best Regards, Steve ____________________________________________________________________ ------=_NextPart_000_0019_01C8119B.9992FFD0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Assuming the plate is fixed and perpendicular to = the wind=20 what you would like to know is the force against it.  Since force = =3D=20 Pressure/Area we first need to find the pressure.
 
The dynamic pressure energy of the airstream at = 200 mph is=20 found by 1/2pV^2 where p is air density and V is velocity.  I am = going to=20 assume sea level air density even though we know it will be less at=20 altitude.  So that is one square foot of air =3D 0.00234 slugs = (sorry I'm an=20 old English units guy).  Velocity (ft/sec) =3D 200mph =  *5280/3600 =3D=20 293.3 ft/sec
 
That gives us  a dynamic Pressure =3D=20 1/2(0.00234*(293.3)^2 =3D 0.5*(0.00234)*(86034) =3D 100.66 lbs/ft^2 or = 100 lbs per=20 square foot.  Dividing by 144 gives us lbs per square inch.  = So=20 pressure =3D 100.66/144 =3D .699 lbs/sq inch or about 19 inches of = H20. =20 But, sticking with psi we have 
 
That pressure is acting on 15 sq inches so the = total force=20 on that plate =3D 0.699*15 =3D 10.485 lbs of force. 
 
That is assuming all the energy in the air is = converted to=20 dynamic pressure which would probably only happens at the very center of = the=20 plate, the air near the edges would tend to flow  (at least = partially)=20 around the plate, so the velocity/pressure  distribution would not = be even=20 across the plate. It would be maximum in the center and taper off = near the=20 edges.   So this means the actually force would undoubtedly be = less=20 than 10 lbs because of that and the fact that the air at altitude will = be=20 considerably less dense, depending of course on your=20 altitude.   
 
Then there is  the fact the plate is moving = through=20 the air means that the pressure in back of the plate is going to be less = than if=20 it were standing still.  But I am not = going to even=20 attempt to get into quantifying those  factors.  But, the = force=20 will still  likely be less than 10 lb
 
You might try searching the web for things like = - air=20 resistance, plate plate, dynamic pressure, etc
 
Hope this helps
 
Ed
 
 
 
 
----- Original Message -----
From:=20 Steve = Thomas=20
Sent: Thursday, October 18, = 2007 2:35=20 PM
Subject: [FlyRotary] Math = Help

Not being much of a math or engineering whiz, can = anyone help=20 me calculate the force the relative wind would put on a device that = hangs in=20 the breeze on my airplane?  Assuming a 200 MPH airspeed, and a = flat plate=20 of 15 sq. inches, what would be the amount of force on the plate in=20 pounds?  
Best Regards,

Steve
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