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(OT) Question for the engineers on the list

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Looks like the 'engineers' are in agreement.
For once.
:)
Larry
  ----- Original Message -----=20
  From: SailingFC@aol.com=20
  To: VeeWdriver@aol.com ; scirocco-l@scirocco.org=20
  Sent: Wednesday, October 30, 2002 11:11 PM
  Subject: Re: (OT) Question for the engineers on the list


  Tom,
  You won't get any more horsepower, you will actually lose some due to =
frictional losses in your reduction box.  What you will get is a =
multiplication of torque.  Horsepower is TORQUE times RPM divided by =
5252.  All things being equal, in a gearbox, when you reduce the RPMs by =
4, the torque available increases by 4.  BTW, you can't make more power =
without putting more energy into the system.
  HTH.

  -Dick-
  78 Scirocco
  Original Owner
  http://members.aol.com/sailingfc/
  (B.S. Mechanical Engineer & farm boy)


  In a message dated 10/30/02 7:39:52 PM Pacific Standard Time, =
VeeWdriver@aol.com writes:

  OK, I need some mechanical engineering advice.  I'm working with a =
small=20
  tractor that has a diesel rated at 14HP.  It has a mid-mounted PTO =
rated at=20
  11HP at the shaft at 2200RPM.  I am going to reduce that through a 1:4 =
gear=20
  ratio chain and gear setup at the rear to run at the standard 540 RPM =
for a=20
  tractor PTO.  If I do this, will I in fact get more HP at the PTO?  =
Like say=20
  44HP???=20

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<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Looks like the 'engineers' are in=20
agreement.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>For once.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>:)</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Larry</FONT></DIV>
<BLOCKQUOTE=20
style=3D"BORDER-LEFT: #000000 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: =
0px; PADDING-LEFT: 5px; PADDING-RIGHT: 0px">
  <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
  <DIV=20
  style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
  <A href=3D"mailto:SailingFC@aol.com"=20
  title=3DSailingFC@aol.com>SailingFC@aol.com</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
href=3D"mailto:VeeWdriver@aol.com"=20
  title=3DVeeWdriver@aol.com>VeeWdriver@aol.com</A> ; <A=20
  href=3D"mailto:scirocco-l@scirocco.org"=20
  title=3Dscirocco-l@scirocco.org>scirocco-l@scirocco.org</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Wednesday, October 30, =
2002 11:11=20
  PM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: (OT) Question for =
the=20
  engineers on the list</DIV>
  <DIV><BR></DIV><FONT face=3Darial,helvetica><FONT face=3DArial =
lang=3D0 size=3D2=20
  FAMILY=3D"SANSSERIF">Tom,<BR>You won't get any more horsepower, you =
will=20
  actually lose some due to frictional losses in your reduction =
box.&nbsp; What=20
  you will get is a multiplication of torque.&nbsp; Horsepower is TORQUE =
times=20
  RPM divided by 5252.&nbsp; All things being equal, in a gearbox, when =
you=20
  reduce the RPMs by 4, the torque available increases by 4.&nbsp; BTW, =
you=20
  can't make more power without putting more energy into the=20
  system.<BR>HTH.<BR><BR>-Dick-<BR>78 Scirocco<BR>Original=20
  Owner<BR>http://members.aol.com/sailingfc/<BR>(B.S. Mechanical =
Engineer &amp;=20
  farm boy)<BR><BR><BR>In a message dated 10/30/02 7:39:52 PM Pacific =
Standard=20
  Time, VeeWdriver@aol.com writes:<BR><BR>OK, I need some mechanical =
engineering=20
  advice.&nbsp; I'm working with a small <BR>tractor that has a diesel =
rated at=20
  14HP.&nbsp; It has a mid-mounted PTO rated at <BR>11HP at the shaft at =

  2200RPM.&nbsp; I am going to reduce that through a 1:4 gear <BR>ratio =
chain=20
  and gear setup at the rear to run at the standard 540 RPM for a =
<BR>tractor=20
  PTO.&nbsp; If I do this, will I in fact get more HP at the PTO?&nbsp; =
Like say=20
  <BR>44HP???</FONT> </FONT></BLOCKQUOTE></BODY></HTML>

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