X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from n67.bullet.mail.sp1.yahoo.com ([98.136.44.47] verified) by logan.com (CommuniGate Pro SMTP 5.2.13) with SMTP id 3557455 for flyrotary@lancaironline.net; Tue, 24 Mar 2009 07:53:57 -0400 Received-SPF: none receiver=logan.com; client-ip=98.136.44.47; envelope-from=casey.gary@yahoo.com Received: from [216.252.122.216] by n67.bullet.mail.sp1.yahoo.com with NNFMP; 24 Mar 2009 11:53:20 -0000 Received: from [68.142.237.90] by t1.bullet.sp1.yahoo.com with NNFMP; 24 Mar 2009 11:53:20 -0000 Received: from [69.147.75.182] by t6.bullet.re3.yahoo.com with NNFMP; 24 Mar 2009 11:53:20 -0000 Received: from [127.0.0.1] by omp103.mail.re1.yahoo.com with NNFMP; 24 Mar 2009 11:53:20 -0000 X-Yahoo-Newman-Property: ymail-3 X-Yahoo-Newman-Id: 404427.66155.bm@omp103.mail.re1.yahoo.com Received: (qmail 55589 invoked by uid 60001); 24 Mar 2009 11:53:19 -0000 DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=yahoo.com; s=s1024; t=1237895599; bh=fwh1zz6WvDkW05NMJx+CiioqH8hKtfLJ5UuquD7Ajdc=; h=Message-ID:X-YMail-OSG:Received:X-Mailer:References:Date:From:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=FCrBz2pbTTAVFAqt0iPUva9/sTXv/uMg3I5B83fVtXmLX7DQE6MfDZcowm7BzvgGwuMZR2xfomhT9ZqJ7zPmmfVlGIudbElnC2AclDG+ZEXe8jCpWJpbnPkIJr5lJIVEMFT5O9C4s6WQQalJYPlvN6w2NlOxTiHw4F/1B9PznKE= DomainKey-Signature:a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.com; h=Message-ID:X-YMail-OSG:Received:X-Mailer:References:Date:From:Subject:To:In-Reply-To:MIME-Version:Content-Type; b=pMmENrgkQP1tr//VozXezkPPsnO2fZ8hVqPPHoX+OS7KnqlHb1dKwxw1d3QJ5eOfO5QP6sF9+uAS2hJLtq7cxhm2NH7VXlZyoVcjO6EmZ7l6fkCIJFGkP+t7cVm0lSgodIVOiH5VOgmVrW0/4vZTnD6RzRrcnfo6OKYG7gdQ5do=; Message-ID: <834126.50243.qm@web57504.mail.re1.yahoo.com> X-YMail-OSG: 8RJERzwVM1mP37ooZaAzFvwiMZ6hfKL03OuWlKstmTwRfuQTuY6yl2X5w4UCidDffTW71zMf_djIKlzOnCx2wwQQJiNbWyuJ0loiJyqTSFWeNcFfma2JnkydZq.o3eHypB.mTrI9PE2pS2niHu9rNWobVxhnXOqoDIlKmOqmTL3a.ghVHo0pF7q1r37W Received: from [97.122.186.190] by web57504.mail.re1.yahoo.com via HTTP; Tue, 24 Mar 2009 04:53:19 PDT X-Mailer: YahooMailRC/1277.32 YahooMailWebService/0.7.289.1 References: Date: Tue, 24 Mar 2009 04:53:19 -0700 (PDT) From: Gary Casey Subject: Re: Alternator (Off topic To: Rotary motors in aircraft In-Reply-To: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="0-234894228-1237895599=:50243" --0-234894228-1237895599=:50243 Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: quoted-printable Bill,=0AOkay, I'll admit I was oversimplifying the description a little. T= he rotor will always "cog", meaning that it will tend to line up with the i= ron poles in the stator and stay in one position. with the field fully ene= rgized the cogging torque will certainly be significant, as you state below= .. But if you force it past one pole it will the pull itself to the next an= d attempt to stop there. What I'm trying to say is that once it is spinnin= g even slowly the cogging effect evens out and the net (average) torque wil= l be very low. The voltage capability of the alternator is proportional to= rpm and with full field excitation it will be able to produce battery volt= age well below idle rpm, but probably not at cranking speed, which might be= something like 100 to 200 rpm. I've not tested this, so I'm not sure of t= he numbers. If it can't reach battery voltage it can't produce power and t= herefore won't absorb power. If it did absorb as much power as you suggest and it can't put out any power, where does all that power go? It would ha= ve to go into heat and it certainly doesn't overheat at low rpms. Another = point - with a fixed field current and at a fixed output voltage (like when= the battery is absorbing the current and limiting the output voltage) the = alternator will produce current proportional to rpm until it gets to its in= ternally-limited maximum current (which is independent of rpm). It usually= will reach that condition at about 4,000 rpm, or something less than 2,000= engine rpm. A rule of thumb: Assuming a fixed field current at a fixed v= oltage the current is proportional to rpm (up to a limit) and at a fixed cu= rrent the voltage is proportional to rpm. The internal inductance of the s= tator will limit the output current - regardless of rpm and field current t= he alternator cannot put out more than a given current which is approximate= ly the rating (a "60-amp" alternator cannot put out more than about 60 amps no matter what). =0A=0AHopefully that clarified things a little.= =0AGary=0A=0ABill previously posted:=0A =0AGary,=0AHere is something you c= an try. Take the belt off the alternator and using your hand, spin the alt= ernator pulley, Then turn on the master and spin the pulley. Then turn on= the alternator and spin the pulley. Tell us again what you think happens = when that =E2=80=9C3 amps=E2=80=9D of field current hits that alternator.= =0AIf you can turn that pulley with your hand with the alternator on, I apo= logize profusely! Not because I am wrong, but because you will have to be = one Gorilla!=0A =0ABill B=0A=0A=0A --0-234894228-1237895599=:50243 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: quoted-printable

Bill,

Okay, I'll admit I was = oversimplifying the description a little. =C2=A0The rotor will always "cog"= , meaning that it will tend to line up with the iron poles in the stator and stay in one position. =C2=A0with = the field fully energized the cogging torque will certainly be significant,= as you state below. =C2=A0But if you force it past one pole it will the pu= ll itself to the next and attempt to stop there. =C2=A0What I'm trying to s= ay is that once it is spinning even slowly the cogging effect evens out and= the net (average) torque will be very low. =C2=A0The voltage capability of= the alternator is proportional to rpm and with full field excitation it wi= ll be able to produce battery voltage well below idle rpm, but probably not= at cranking speed, which might be something like 100 to 200 rpm. =C2=A0I'v= e not tested this, so I'm not sure of the numbers. =C2=A0If it can't reach = battery voltage it can't produce power and therefore won't absorb power. = =C2=A0If it did absorb as much power as you suggest and it can't put out an= y power, where does all that power go? =C2=A0It would have to go into heat = and it certainly doesn't overheat at low rpms. =C2=A0Another point - with a fixed field current and= at a fixed output voltage (like when the battery is absorbing the current = and limiting the output voltage) the alternator will produce current propor= tional to rpm until it gets to its internally-limited maximum current (whic= h is independent of rpm). =C2=A0It usually will reach that condition at abo= ut 4,000 rpm, or something less than 2,000 engine rpm. =C2=A0A rule of thum= b: =C2=A0Assuming a fixed field current at a fixed voltage the current is p= roportional to rpm (up to a limit) and at a fixed current the voltage is pr= oportional to rpm. =C2=A0The internal inductance of the stator will limit t= he output current - regardless of rpm and field current the alternator cann= ot put out more than a given current which is approximately the rating (a "= 60-amp" alternator cannot put out more than about 60 amps no matter what).= =C2=A0


Hopefully that clarified things a little.

Gary


Bill previously posted:

=C2=A0=C2=A0

Gary,

= Here is something you can try.=C2=A0= Take the belt off the alternator and using your hand, spin the alternator = pulley,=C2=A0 Then turn on the master and spin the pulley.=C2=A0 Then turn = on the alternator and spin the pulley.=C2=A0 Tell us again what you think h= appens when that =E2=80=9C3 amps=E2=80=9D of field current hits that altern= ator.

If you can turn that pulley with your han= d with the alternator on, I apologize profusely!=C2=A0 Not because I am wro= ng, but because you will have to be one Gorilla!

=C2=A0

Bill B

<= /div>



=0A

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