X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 64 [XX] (100%) BODY: contains text similar to "low payment" Return-Path: Received: from ms-smtp-04.southeast.rr.com ([24.25.9.103] verified) by logan.com (CommuniGate Pro SMTP 5.1.7) with ESMTP id 1914880 for flyrotary@lancaironline.net; Sun, 11 Mar 2007 11:16:48 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.103; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-103-061.carolina.res.rr.com [24.74.103.61]) by ms-smtp-04.southeast.rr.com (8.13.6/8.13.6) with SMTP id l2BFFvFV029079 for ; Sun, 11 Mar 2007 11:15:58 -0400 (EDT) Message-ID: <000701c763f0$2f39b380$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: A solution? was : The truth??? / Injector flow rate mystery solved Date: Sun, 11 Mar 2007 11:16:03 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0004_01C763CE.A7D25260" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0004_01C763CE.A7D25260 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable MessageSounds like a reasonable approach to me Joe. A pull-down = resistor would be relatively easy for me to install - I have the = resistor pack required for the peak-hold type injectors. So I could = easily place four additional resistors in that box. If I understand you (please correct me if I don't), the pull down = resistor should go between the injector and the EC2 sinking terminals. = That way the current induced when the intended pulse terminates and the = magnetic field collapses will have a path to ground rather than being = opposed by the diode in the Ec2. The value of said resistor could be around 100 ohms. Since the induced = voltage could reach from 50 - 100+ volts an 100 ohm resistor could flow = from=20 0.5 - 1 Amp (for a very short duration). As far as affecting the 12 = Volt signal it would only draw 12/100 =3D 0.12 amp or 120 milliamps. = That would be pulled through the injectors at all times. The injector = resistance is probably (peak and hold case) around 3 ohms. So the = injector would draw 12/3 =3D 4 amps (DC case - its undoubtedly less due = to the A/C impedance of the coil). It make take some experimenting - = but 100 ohm looks like a good place to start. The wattage should probably be around 5 - 10 watts just to be on the = safe side. =20 So certainly looks like a suggestion that would work, Joe. =20 Ed ----- Original Message -----=20 From: Joe Ewen=20 To: Rotary motors in aircraft=20 Sent: Sunday, March 11, 2007 10:54 AM Subject: [FlyRotary] Re: The truth??? / Injector flow rate mystery = solved ED, George, Steve, A strong contributor to this may lie in the fundamental design of the = EC2. It's output control is sinking rather than sourcing. If it were = sourcing the positive EMF would be switched and the other side of the = could would always be grounded, leaving a place for the coil breakdown = current to go. Well the controller is what it is, so the question = become what can we do to work around the problem? Encoders are used = often in industrial applications, these are generally connected in a = sinking fashion just as the EC2. In high frequency (encoder pulses) = applications the impedance of the input electronics is often to slow to = bleed of the leading edge of the encoder on voltage before the next = pulse cycle. End result is that the input does not detect tithe state = change. The solution is very simple for the described situation is very = simple. Installing a pull down resistor between the sensor signal line = and ground. This technique is used with standard input electronics with = pulse trains up to 50kHz, which translates to a cycle period of 20=B5S. How does this translate to our application? If we were to add a pull = down resistor in injector signal line, which may very well be a simple = method to reducing the off delay time. This would of course add a small = increase in the current for the injector circuits, but that increase = would likely be minimal. The value of the resistor would certainly need = to be determined using factors such as Injector turn on voltage, turn = off voltage, device current, etc. IMO this may be a simple solution to = the delay issue. In the end I defer final recommendations to Tracy, who certainly knows = the characteristics of his controller system better than I. Joe ----- Original Message -----=20 From: Steven Boese=20 To: Rotary motors in aircraft=20 Sent: Saturday, March 10, 2007 11:44 PM Subject: [FlyRotary] Re: The truth??? / Injector flow rate mystery = solved Ed and George, =20 In my plane, at least, injector open times need to be less than 2 ms = at idle and just above the staging point. This is not possible with a = minimum open time of 2 ms due to the delay on closing. You can program = the EC2 for less than 2 ms but the hardware is unable to do this. That = means that the problem can't be fixed with software or programming = different values in the map table. A shorter close delay time is = required unless you lower the flow rate of the injectors by changing the = injector itself or lower the fuel pressure. The injector open time is = at least 2 ms or it doesn't open at all. =20 Steve Boese ------=_NextPart_000_0004_01C763CE.A7D25260 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Message
Sounds like a reasonable approach  to = me=20 Joe.   A pull-down resistor would be relatively easy  for = me=20 to install  - I have the resistor pack required for the = peak-hold type=20 injectors.  So I could easily place four = additional resistors in=20 that box.
 
If I understand you (please correct me if I = don't), the=20 pull down resistor should go between the injector and the EC2 sinking=20 terminals.  That way the current induced when the intended pulse = terminates=20 and the magnetic field collapses will have a path to ground rather than = being=20 opposed by the diode in the Ec2.
 
The value of said resistor could be around 100 = ohms. =20 Since the induced voltage could reach from 50 - 100+ volts an 100 ohm = resistor=20 could flow from
0.5 - 1 Amp (for a very short duration).  = As far as=20 affecting the 12 Volt signal it would only draw 12/100 =3D 0.12 amp or = 120=20 milliamps.  That would be pulled through the injectors at all = times. =20 The injector resistance is probably (peak and hold case) around 3=20 ohms.  So the injector would draw 12/3 =3D 4 amps (DC case - its = undoubtedly=20 less due to the A/C impedance of the coil).  It make take some=20 experimenting - but 100 ohm looks like a good place to = start.
 
The wattage should probably be around 5 - 10 = watts just to=20 be on the safe side. 
 
So certainly looks like a suggestion that would = work,=20 Joe. 
 
 
Ed
----- Original Message -----
From:=20 Joe = Ewen=20
Sent: Sunday, March 11, 2007 = 10:54=20 AM
Subject: [FlyRotary] Re: The = truth??? /=20 Injector flow rate mystery solved

ED, George, Steve,
A strong contributor to this may lie = in the=20 fundamental design of the EC2.  It's output control is sinking = rather=20 than sourcing.  If it were sourcing the positive EMF would be = switched=20 and the other  side of the could would always be grounded, = leaving a=20 place for the coil breakdown current to go.  Well the controller = is what=20 it is, so the question become what can we do to work around the = problem? =20 Encoders are used often in industrial applications,  these are = generally=20 connected in a sinking fashion just as the EC2.  In high = frequency=20 (encoder pulses) applications the impedance of the input = electronics is=20 often to slow to bleed of the leading edge of the encoder on voltage = before=20 the next pulse cycle.  End result is that the input does not = detect tithe=20 state change.  The solution is very simple for the described = situation is=20 very simple.  Installing a pull down resistor between the sensor = signal=20 line and ground.  This technique is used with standard input = electronics=20 with pulse trains up to 50kHz, which translates to a cycle period of = 20=B5S.
 
How does this translate to our = application? =20 If we were to add a pull down resistor in injector signal = line, which may=20 very well be a simple method to reducing the off delay time.  = This would=20 of course add a small increase in the current for the injector = circuits, but=20 that increase would likely be minimal.  The value of the resistor = would=20 certainly need to be determined using factors such as Injector turn on = voltage, turn off voltage, device current, etc.  IMO this = may be a=20 simple solution to the delay issue.
 
In the end I defer final = recommendations to=20 Tracy, who certainly knows the characteristics of his controller = system better=20 than I.
 
Joe
 
 
 
----- Original Message -----
From:=20 Steven = Boese=20
To: Rotary motors in = aircraft=20
Sent: Saturday, March 10, = 2007 11:44=20 PM
Subject: [FlyRotary] Re: The = truth??? /=20 Injector flow rate mystery solved

Ed and=20 George,

 

In my = plane, at=20 least, injector open times need to be less than 2 ms at idle and = just above=20 the staging point.  = This is not=20 possible with a minimum open time of 2 ms due to the delay on = closing.  You can program the EC2 = for less=20 than 2 ms but the hardware is unable to do this.  That means that the = problem can=92t be=20 fixed with software or programming different values in the map = table.  A shorter close delay time = is=20 required unless you lower the flow rate of the injectors by changing = the=20 injector itself or lower the fuel pressure.  The injector open time is = at least 2=20 ms or it doesn=92t open at all.

 

Steve = Boese

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