X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Fri, 12 May 2006 17:12:08 -0400 Message-ID: X-Original-Return-Path: Received: from smtpauth07.mail.atl.earthlink.net ([209.86.89.67] verified) by logan.com (CommuniGate Pro SMTP 5.0.9) with ESMTP id 1110614 for lml@lancaironline.net; Fri, 12 May 2006 11:09:26 -0400 Received-SPF: none receiver=logan.com; client-ip=209.86.89.67; envelope-from=rtitsworth@mindspring.com DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=dk20050327; d=mindspring.com; b=rI0ApNZb8XhhGWxQOqVYx+0y3BSkO+iw1w9A1NMQsAQvkagS+BO2MSYQj6P6+vM1; h=Received:Reply-To:From:To:Subject:Date:MIME-Version:Content-Type:X-Mailer:Thread-Index:In-Reply-To:X-MimeOLE:Message-ID:X-ELNK-Trace:X-Originating-IP; Received: from [68.40.94.44] (helo=RDTVAIO) by smtpauth07.mail.atl.earthlink.net with asmtp (Exim 4.34) id 1FeZFx-0004oG-Fd for lml@lancaironline.net; Fri, 12 May 2006 11:08:42 -0400 Reply-To: From: "richard titsworth" X-Original-To: "'Lancair Mailing List'" Subject: RE: [LML] Re: Poor Man's Dyno X-Original-Date: Fri, 12 May 2006 11:08:28 -0400 MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_000_0015_01C675B4.65A800F0" X-Mailer: Microsoft Office Outlook, Build 11.0.5510 Thread-Index: AcZ1vxlILQD3fr5BT86Wz2lt0jS79gABrkaA In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2869 X-Original-Message-ID: X-ELNK-Trace: b17f11247b2ac8f0a79dc4b33984cbaa0a9da525759e265482c197b01c9b1449858f5e4b0ca0f7cf676ae045857c2f36350badd9bab72f9c350badd9bab72f9c X-Originating-IP: 68.40.94.44 This is a multi-part message in MIME format. ------=_NextPart_000_0015_01C675B4.65A800F0 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0016_01C675B4.65A800F0" ------=_NextPart_001_0016_01C675B4.65A800F0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit I believe both viewpoints of this discussion/thread have elements of correctness. If I understand it correctly, then the pictures and discussion below are an attempt to add some clarity. Imagine two identical airplanes except they have different landing gear geometries - but they are the same and symmetrical at the ground (and identical weights). For definition let's call the left side the slant gear and let's call the right side the vertical gear By definition: Torque = Moment Arm X Force (perpendicular to the direction of rotation). For all mention of "forces on the scales" below, I am assuming the scales have been zero'd with the aircraft not running (i.e. subtracting the tare weight of the aircraft). So, lets get to work. If you were attempting to measure torque using the slant gear aircraft, and were using the actual gear geometry as the basis for measuring the relevant forces and moment arm, then the scale would only indicate a portion (the vertical component) of the torque force. (ref: slant gear picture below). Note, the Moment Arm is the slant distance from the axis of rotation to the contact spot on the scale In this case, a geometric (trigonometry) correction factor must be applied to the force indicated on the scale. The correction factor is the 1/cos of the gear angle (measured from horizontal). (Ref: diagram below) Also note: the correct gear angle is to the contact spot on the scale - not necessarily the leg itself. Thus (in this case), the Torque = Moment Arm X Force (on the scale) X 1/Cos(gear angle) However, if you were to use the vertical gear aircraft, the gear geometry is already perpendicular to the moment arm. Thus, the scale would measure the complete (correct) force vector (no correction factor needed). Thus (in this case), the Torque = Moment Arm X Force (on the scale). In both cases the scales will read the same. How can this be??? Note: the moment arm (as measured) is shorter for the vertical gear aircraft. The "magic" is that the slant gear correction factor (trigonometry for decomposing the vertical and horizontal force components) is the same as the trigonometry that affects the length of the respective moment arms. Thus, correction factor is "already included" in the Moment Arm of the Vertical Gear scenario. Of course the actual gear configuration doesn't really matter since "the scale doesn't know". So, if you want to use the force measurement directly off the scale, you need to use the horizontal distance from the centerline as the moment arm (similar to the vertical gear example above). You can do this on any gear configuration. Please comment on any corrections and/or improvements to this attempted explanation. Rick Titsworth rtitsworth@mindspring.com ------=_NextPart_001_0016_01C675B4.65A800F0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

I believe both viewpoints of this discussion/thread have elements of correctness.  =

If I understand it correctly, then = the pictures and discussion below are an attempt to add some = clarity.

 

Imagine two identical airplanes = except they have different landing gear geometries – but they are the = same and symmetrical at the ground (and identical weights).

 

For definition let’s call the = left side the slant gear and let’s call the right side the = vertical gear

 

 

 

 

By definition: Torque =3D Moment = Arm X Force (perpendicular to the direction of = rotation).

 

For all mention of “forces on = the scales” below, I am assuming the scales have been zero’d with the aircraft = not running (i.e. subtracting the tare weight of the aircraft). =

 

So, lets get to = work…

 

If you were attempting to measure = torque using the slant gear aircraft, and were using the actual gear geometry = as the basis for measuring the relevant forces and moment arm, then the scale = would only indicate a portion (the vertical component) of the torque force. = (ref: slant gear picture below).  Note, the Moment Arm is the slant distance = from the axis of rotation to the contact spot on the = scale

 

 

In this case, a geometric = (trigonometry) correction factor must be applied to the force indicated on the scale.  The correction factor is the 1/cos of the gear angle (measured from = horizontal).  (Ref: diagram below)

Also note: the correct gear angle = is to the contact spot on the scale – not necessarily the leg = itself.

 

 

Thus (in this case), the Torque =3D = Moment Arm X Force (on the scale) X 1/Cos(gear = angle)

 

 

However, if you were to use the = vertical gear aircraft, the gear geometry is already perpendicular to the moment = arm.  Thus, the scale would measure the complete (correct) force vector (no = correction factor needed). 

 

 

Thus (in this case), the Torque =3D = Moment Arm X Force (on the scale).

 

 

In both cases the scales will read = the same.  How can this be???

 

Note: the moment arm (as measured) = is shorter for the vertical gear aircraft.  The “magic” is = that the slant gear correction factor (trigonometry for decomposing the = vertical and horizontal force components) is the same as the trigonometry that = affects the length of the respective moment arms.  Thus, correction factor is = “already included” in the Moment Arm of the Vertical Gear = scenario.

 

 

Of course the actual gear = configuration doesn’t really matter since “the scale doesn’t know”. =  So, if you want to use the force measurement directly off the scale, you need = to use the horizontal distance from the centerline as the moment arm (similar = to the vertical gear example above).  You can do this on any gear = configuration.

 

Please comment on any corrections = and/or improvements to this attempted explanation.

 

Rick = Titsworth

rtitsworth@mindspring.com

 

 

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