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Dit speed to WPM?

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  • Earl Needham
    Is there a general rule of thumb, that so many dits per second equals so many WPM? It would make it a lot easier to figure how fast somebody was sending, how
    Message 1 of 11 , Dec 1, 2006
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      Is there a general rule of thumb, that so many dits per
      second equals so many WPM? It would make it a lot easier to figure
      how fast somebody was sending, how fast a bug is, etc. -- a LOT
      easier than sending PARIS repeatedly! :-)

      Thanks,
      Earl

      KD5XB
      Clovis, New Mexico DM84jk
      http://groups.yahoo.com/group/cw_bugs
    • Richard Meiss
      Hi, Earl - The formula that you are looking for is: WPM = (dits per second) x 2.4 This assumes a 1 to 3 dit-to-dah ratio and a dit space between elements.
      Message 2 of 11 , Dec 1, 2006
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        Hi, Earl -

        The formula that you are looking for is: WPM = (dits per second) x 2.4 This assumes a 1 to 3 dit-to-dah ratio and a "dit" space between elements.

        This is a lot easier than sending PARIS repeatedly.

        73 de Rich, WB9LPU


        -----Original Message-----
        >From: Earl Needham <needhame1@...>
        >Sent: Dec 1, 2006 4:33 PM
        >To: cw_bugs@yahoogroups.com
        >Subject: [cw_bugs] Dit speed to WPM?
        >
        >
        > Is there a general rule of thumb, that so many dits per
        >second equals so many WPM? It would make it a lot easier to figure
        >how fast somebody was sending, how fast a bug is, etc. -- a LOT
        >easier than sending PARIS repeatedly! :-)
        >
        > Thanks,
        > Earl
        >
        >KD5XB
        >Clovis, New Mexico DM84jk
        >http://groups.yahoo.com/group/cw_bugs
        >
        >
      • Donald Kemp
        Earl, Let me try to figure this out. If I send PARIS in one minute that s 1 WPM. P = 2 dits and 2 dahs. 1 dah = 3 dits x 2 = 6 + 2 = 8 + 3 spaces between
        Message 3 of 11 , Dec 1, 2006
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          Earl,

          Let me try to figure this out.
          If I send PARIS in one minute that's 1 WPM.
          P = 2 dits and 2 dahs. 1 dah = 3 dits x 2 = 6 + 2 = 8 + 3 spaces between
          elements & + 1 space between character = 12 dits
          A = 1 dit and 1 dah. Thats 1 + 3 + 1 element space + 1 character space = 6
          dits
          R = 2 dits and 1 dah. 5 = 2 element spaces + 1 character space = 8 dits
          I = 2 dits + 1 element space + 1 character space = 4 dits
          S = 3 dits + 2 element spaces = 5 dits.

          Total dits = 12 + 6 + 8 + 4 + 5 = 35 dits in one minute or 60/35 =
          1.7142857seconds/dit
          Space between words = 1 dah = 3 dits.
          20 WPM would be 60/(20 x 35) + (19 x 3) = 757. 60/757 = .07926 seconds/dit.
          1/.07926 = 12.617 dits/second = 20 WPM.

          Maybe.

          I may be missing something in the spaces between words. Do I actually count
          them for the dit count?
          If not it would be:
          20 WPM = 60/(20 x 35) = .0857 sec/dit.
          1/.0857 = 11.668 dits/sec = 20 WPM

          Not too much difference.

          So for any other WPM calculation (using the second formula) it would be:
          X WPM = 60/(X x 35) = Y sec/dit
          1/Y = Z dits/sec.

          IE for 15 WPM.

          15 WPM = 60/(15 x 35) = .1142857 sec/dit
          1/.1142857 = 8.75 dits/sec.

          Hopefully there's already a table somewhere that will let me know if I'm all
          wet.

          It was fun anyway Earl.

          73,

          Don, NN8B
          SKCC 36




          On 12/1/06, Earl Needham <needhame1@...> wrote:
          >
          >
          > Is there a general rule of thumb, that so many dits per
          > second equals so many WPM? It would make it a lot easier to figure
          > how fast somebody was sending, how fast a bug is, etc. -- a LOT
          > easier than sending PARIS repeatedly! :-)
          >
          > Thanks,
          > Earl
          >
          > KD5XB
          > Clovis, New Mexico DM84jk
          > http://groups.yahoo.com/group/cw_bugs
          >
          >
          >


          [Non-text portions of this message have been removed]
        • k4oso@aol.com
          In a message dated 12/1/2006 6:44:49 PM Eastern Standard Time, ... Neither of these methods seem particularly easy. I use my keyer (CMOS4) to pace the
          Message 4 of 11 , Dec 1, 2006
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            In a message dated 12/1/2006 6:44:49 PM Eastern Standard Time,
            wb9lpu@... writes:

            > Hi, Earl -
            >
            > The formula that you are looking for is: WPM = (dits per second) x 2.4 This
            > assumes a 1 to 3 dit-to-dah ratio and a "dit" space between elements.
            >
            > This is a lot easier than sending PARIS repeatedly.
            >
            > 73 de Rich, WB9LPU
            >

            Neither of these methods seem particularly easy. I use my keyer (CMOS4) to
            "pace" the sender, be it myself or someone else. It seems to work pretty
            well,...... without the math:-)
            73, Milt k4oso


            [Non-text portions of this message have been removed]
          • Donald Kemp
            Milt and the group, I guess I never thought about it till Earl asked. Didn t know there was already an easier way till Richard told us. Either way I still need
            Message 5 of 11 , Dec 1, 2006
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              Milt and the group,

              I guess I never thought about it till Earl asked. Didn't know there was
              already an easier way till Richard told us.

              Either way I still need to count the dits, that gets tough to do at higher
              speeds.

              I ususlly don't have a clue about what speed I'm running. I just try to keep
              up with the other guy.

              Great Question Earl; got us thinking.

              73,
              Don, NN8B


              On 12/1/06, k4oso@... <k4oso@...> wrote:
              >
              > In a message dated 12/1/2006 6:44:49 PM Eastern Standard Time,
              > wb9lpu@... <wb9lpu%40earthlink.net> writes:
              >
              > > Hi, Earl -
              > >
              > > The formula that you are looking for is: WPM = (dits per second) x 2.4This
              > > assumes a 1 to 3 dit-to-dah ratio and a "dit" space between elements.
              > >
              > > This is a lot easier than sending PARIS repeatedly.
              > >
              > > 73 de Rich, WB9LPU
              > >
              >
              > Neither of these methods seem particularly easy. I use my keyer (CMOS4) to
              >
              > "pace" the sender, be it myself or someone else. It seems to work pretty
              > well,...... without the math:-)
              > 73, Milt k4oso
              >
              > [Non-text portions of this message have been removed]
              >
              >
              >


              [Non-text portions of this message have been removed]
            • Earl Needham
              ... Gee, you got THAT right! ... Thanks all for the info. You see, I send on my bug at what seems pretty fast to me, but one of the locals says it s only about
              Message 6 of 11 , Dec 1, 2006
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                At 03:43 PM 12/1/2006, Richard Meiss wrote:
                >The formula that you are looking for is: WPM = (dits per second) x
                >2.4 This assumes a 1 to 3 dit-to-dah ratio and a "dit" space between elements.
                >
                >This is a lot easier than sending PARIS repeatedly.

                Gee, you got THAT right!

                At 04:31 PM 12/1/2006, Donald Kemp wrote:

                >Earl,
                >
                >Let me try to figure this out.
                >If I send PARIS in one minute that's 1 WPM.

                At 06:10 PM 12/1/2006, k4oso@... wrote:
                ><snip>
                >Neither of these methods seem particularly easy. I use my keyer (CMOS4) to
                >"pace" the sender, be it myself or someone else. It seems to work pretty
                >well,...... without the math:-)

                At 06:27 PM 12/1/2006, Donald Kemp wrote:
                ><snip>
                >Great Question Earl; got us thinking.

                Thanks all for the info.

                You see, I send on my bug at what seems pretty fast to me,
                but one of the locals says it's only about 8-10 WPM. With the above
                figures, I can see that I'm sending (contest exchanges) at just over
                20 WPM -- about 9 dits/second.

                7 3
                Earl

                KD5XB
                Clovis, New Mexico DM84jk
                http://groups.yahoo.com/group/cw_bugs
              • David J. Ring, Jr.
                By defination a word in Morse Code is the 50 baud PARIS. PARIS consists of 50 basic units of time (dots) so it is 50 baud. 60 wpm would be PARIS sent 60
                Message 7 of 11 , Dec 6, 2006
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                  By defination a "word" in Morse Code is the 50 baud PARIS. PARIS consists of 50 basic units of time (dots) so it is 50 baud.

                  60 wpm would be PARIS sent 60 times in one minute.

                  Thus one PARIS (50 baud) in one second would be sent.

                  Each dot is 1/50 of a second since 50 baud is sent in one second, then the basic element of the baud is the dot, and it takes 1/50th of a second at 60 wpm.

                  Thus at 30 wpm the dot takes 2/50th of a second of 1/25 of a second.

                  10 wpm would be 3/25 of a second dots.

                  The reason that American Morse is faster than International Morse is because of the "dotty" letters allow it to have more and fast characters than International Morse.

                  E
                  I
                  S
                  H
                  T
                  M = all remain the same - fast letters.
                  O is faster ee
                  C is faster ee e
                  R is faster e ee
                  Y is faster ee ee
                  L is faster daaaah

                  You could calculate this by taking the "most common letters in newsprint (messages)" and compare the baud count in both American Morse and International Morse.

                  I believe according to my memory when I did this it figured out to about 25% faster for American.

                  But American Morse would also be faster than International for CODE GROUPS because of the introduction of "dotty" characters.

                  In International Morse, code groups sent at 20 gpm (groups per minute) equal the "transmission speed" of English sent at 25 wpm (words per minute) - words and groups both being counted as "5" characters.

                  Also in the old estimation of code speed by American Morse operators, the speed is calculated as "expanded words" from Phillips Code at the ear, to English at the typewriter.

                  In any event, the limiting factor is the typewriter!

                  73

                  DR
                  ----- Original Message -----
                  From: Donald Kemp
                  To: cw_bugs@yahoogroups.com
                  Sent: Friday, December 01, 2006 6:31 PM
                  Subject: Re: [cw_bugs] Dit speed to WPM?


                  Earl,

                  Let me try to figure this out.
                  If I send PARIS in one minute that's 1 WPM.
                  P = 2 dits and 2 dahs. 1 dah = 3 dits x 2 = 6 + 2 = 8 + 3 spaces between
                  elements & + 1 space between character = 12 dits
                  A = 1 dit and 1 dah. Thats 1 + 3 + 1 element space + 1 character space = 6
                  dits
                  R = 2 dits and 1 dah. 5 = 2 element spaces + 1 character space = 8 dits
                  I = 2 dits + 1 element space + 1 character space = 4 dits
                  S = 3 dits + 2 element spaces = 5 dits.

                  Total dits = 12 + 6 + 8 + 4 + 5 = 35 dits in one minute or 60/35 =
                  1.7142857seconds/dit
                  Space between words = 1 dah = 3 dits.
                  20 WPM would be 60/(20 x 35) + (19 x 3) = 757. 60/757 = .07926 seconds/dit.
                  1/.07926 = 12.617 dits/second = 20 WPM.

                  Maybe.

                  I may be missing something in the spaces between words. Do I actually count
                  them for the dit count?
                  If not it would be:
                  20 WPM = 60/(20 x 35) = .0857 sec/dit.
                  1/.0857 = 11.668 dits/sec = 20 WPM

                  Not too much difference.

                  So for any other WPM calculation (using the second formula) it would be:
                  X WPM = 60/(X x 35) = Y sec/dit
                  1/Y = Z dits/sec.

                  IE for 15 WPM.

                  15 WPM = 60/(15 x 35) = .1142857 sec/dit
                  1/.1142857 = 8.75 dits/sec.

                  Hopefully there's already a table somewhere that will let me know if I'm all
                  wet.

                  It was fun anyway Earl.

                  73,

                  Don, NN8B
                  SKCC 36

                  On 12/1/06, Earl Needham <needhame1@...> wrote:
                  >
                  >
                  > Is there a general rule of thumb, that so many dits per
                  > second equals so many WPM? It would make it a lot easier to figure
                  > how fast somebody was sending, how fast a bug is, etc. -- a LOT
                  > easier than sending PARIS repeatedly! :-)
                  >
                  > Thanks,
                  > Earl
                  >
                  > KD5XB
                  > Clovis, New Mexico DM84jk
                  > http://groups.yahoo.com/group/cw_bugs
                  >
                  >
                  >

                  [Non-text portions of this message have been removed]






                  ------------------------------------------------------------------------------


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                  Checked by AVG Free Edition.
                  Version: 7.5.430 / Virus Database: 268.15.3/562 - Release Date: 12/1/2006 1:12 PM


                  [Non-text portions of this message have been removed]
                • David J. Ring, Jr.
                  Further to my last message - I ve been looking for the little box graph showing the baud of morse code, but I found this at Kent Engineers - the makers of Kent
                  Message 8 of 11 , Dec 6, 2006
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                    Further to my last message - I've been looking for the little box graph showing the baud of morse code, but I found this at Kent Engineers - the makers of Kent keys.

                    Unfortunately - a picture is worth "how many are there?" words pasted herein.

                    But it does answer some of your questions.

                    73

                    David Ring, N1EA
                    Calculating Morse Code Speed

                    The word PARIS is the standard for determing CW code speed. Each dit is one element, each dah is three elements, intra-character spacing is one element, inter-character spacing is three elements and inter-word spacing is seven elements. The word PARIS is exactly 50 elements.
                    Note that after each dit/dah of the letter P -- one element spacing is used except the last one. (Intra-Character).
                    After the last dit of P is sent, 3 elements are added (Inter-Character). After the word PARIS - 7 elements are used.
                    Thus:
                    P = di da da di = 1 1 3 1 3 1 1 (3) = 14 elements
                    A = di da = 1 1 3 (3) = 8 elements
                    R = di da di = 1 1 3 1 1 (3) = 10 elements
                    I = di di = 1 1 1 (3) = 6 elements
                    S = di di di = 1 1 1 1 1 [7] = 12 elements
                    Total = 50 elements
                    () = intercharacter
                    [] = interword
                    If you send PARIS 5 times in a minute (5WPM) you have sent 250 elements (using correct spacing). 250 elements into 60 seconds per minute = 240 milliseconds per element.

                    13 words-per-minute is one element every 92.31 milliseconds.

                    The Farnsworth method sends the dits and dahs and intra-character spacing at a higher speed, then increasing the inter-character and inter-word spacing to slow the sending speed down to the overall speed. For example, to send at 5 wpm with 13 wpm characters in Farnsworth method, the dits and intra-character spacing would be 92.3 milliseconds, the dah would be 276.9 milliseconds, the inter-character spacing would be 1.443 seconds and inter-word spacing would be 3.367 seconds.

                    ----- Original Message -----
                    From: David J. Ring, Jr.
                    To: cw_bugs@yahoogroups.com
                    Sent: Thursday, December 07, 2006 1:02 AM
                    Subject: Re: [cw_bugs] Dit speed to WPM?


                    By defination a "word" in Morse Code is the 50 baud PARIS. PARIS consists of 50 basic units of time (dots) so it is 50 baud.

                    60 wpm would be PARIS sent 60 times in one minute.

                    Thus one PARIS (50 baud) in one second would be sent.

                    Each dot is 1/50 of a second since 50 baud is sent in one second, then the basic element of the baud is the dot, and it takes 1/50th of a second at 60 wpm.

                    Thus at 30 wpm the dot takes 2/50th of a second of 1/25 of a second.

                    10 wpm would be 3/25 of a second dots.

                    The reason that American Morse is faster than International Morse is because of the "dotty" letters allow it to have more and fast characters than International Morse.

                    E
                    I
                    S
                    H
                    T
                    M = all remain the same - fast letters.
                    O is faster ee
                    C is faster ee e
                    R is faster e ee
                    Y is faster ee ee
                    L is faster daaaah

                    You could calculate this by taking the "most common letters in newsprint (messages)" and compare the baud count in both American Morse and International Morse.

                    I believe according to my memory when I did this it figured out to about 25% faster for American.

                    But American Morse would also be faster than International for CODE GROUPS because of the introduction of "dotty" characters.

                    In International Morse, code groups sent at 20 gpm (groups per minute) equal the "transmission speed" of English sent at 25 wpm (words per minute) - words and groups both being counted as "5" characters.

                    Also in the old estimation of code speed by American Morse operators, the speed is calculated as "expanded words" from Phillips Code at the ear, to English at the typewriter.

                    In any event, the limiting factor is the typewriter!

                    73

                    DR
                    ----- Original Message -----
                    From: Donald Kemp
                    To: cw_bugs@yahoogroups.com
                    Sent: Friday, December 01, 2006 6:31 PM
                    Subject: Re: [cw_bugs] Dit speed to WPM?


                    Earl,

                    Let me try to figure this out.
                    If I send PARIS in one minute that's 1 WPM.
                    P = 2 dits and 2 dahs. 1 dah = 3 dits x 2 = 6 + 2 = 8 + 3 spaces between
                    elements & + 1 space between character = 12 dits
                    A = 1 dit and 1 dah. Thats 1 + 3 + 1 element space + 1 character space = 6
                    dits
                    R = 2 dits and 1 dah. 5 = 2 element spaces + 1 character space = 8 dits
                    I = 2 dits + 1 element space + 1 character space = 4 dits
                    S = 3 dits + 2 element spaces = 5 dits.

                    Total dits = 12 + 6 + 8 + 4 + 5 = 35 dits in one minute or 60/35 =
                    1.7142857seconds/dit
                    Space between words = 1 dah = 3 dits.
                    20 WPM would be 60/(20 x 35) + (19 x 3) = 757. 60/757 = .07926 seconds/dit.
                    1/.07926 = 12.617 dits/second = 20 WPM.

                    Maybe.

                    I may be missing something in the spaces between words. Do I actually count
                    them for the dit count?
                    If not it would be:
                    20 WPM = 60/(20 x 35) = .0857 sec/dit.
                    1/.0857 = 11.668 dits/sec = 20 WPM

                    Not too much difference.

                    So for any other WPM calculation (using the second formula) it would be:
                    X WPM = 60/(X x 35) = Y sec/dit
                    1/Y = Z dits/sec.

                    IE for 15 WPM.

                    15 WPM = 60/(15 x 35) = .1142857 sec/dit
                    1/.1142857 = 8.75 dits/sec.

                    Hopefully there's already a table somewhere that will let me know if I'm all
                    wet.

                    It was fun anyway Earl.

                    73,

                    Don, NN8B
                    SKCC 36

                    On 12/1/06, Earl Needham <needhame1@...> wrote:
                    >
                    >
                    > Is there a general rule of thumb, that so many dits per
                    > second equals so many WPM? It would make it a lot easier to figure
                    > how fast somebody was sending, how fast a bug is, etc. -- a LOT
                    > easier than sending PARIS repeatedly! :-)
                    >
                    > Thanks,
                    > Earl
                    >
                    > KD5XB
                    > Clovis, New Mexico DM84jk
                    > http://groups.yahoo.com/group/cw_bugs
                    >
                    >
                    >

                    [Non-text portions of this message have been removed]






                    ----------------------------------------------------------------------------


                    No virus found in this incoming message.
                    Checked by AVG Free Edition.
                    Version: 7.5.430 / Virus Database: 268.15.3/562 - Release Date: 12/1/2006 1:12 PM


                    [Non-text portions of this message have been removed]
                  • k4oso
                    While we re on captivating subjects, does anyone know how many pieces of lint makes a bellbutton full? 73, Milt k4oso ... consists of 50 basic units of time
                    Message 9 of 11 , Dec 7, 2006
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                      While we're on captivating subjects, does anyone know how many pieces
                      of lint makes a bellbutton full?
                      73, Milt k4oso

                      --- In cw_bugs@yahoogroups.com, "David J. Ring, Jr." <n1ea@...> wrote:
                      >
                      > By defination a "word" in Morse Code is the 50 baud PARIS. PARIS
                      consists of 50 basic units of time (dots) so it is 50 baud.
                      >
                      > 60 wpm would be PARIS sent 60 times in one minute.
                      >
                      > Thus one PARIS (50 baud) in one second would be sent.
                      >
                      > Each dot is 1/50 of a second since 50 baud is sent in one second,
                      then the basic element of the baud is the dot, and it takes 1/50th of
                      a second at 60 wpm.
                      >
                      > Thus at 30 wpm the dot takes 2/50th of a second of 1/25 of a second.
                      >
                      > 10 wpm would be 3/25 of a second dots.
                      >
                      > The reason that American Morse is faster than International Morse
                      is because of the "dotty" letters allow it to have more and fast
                      characters than International Morse.
                      >
                      > E
                      > I
                      > S
                      > H
                      > T
                      > M = all remain the same - fast letters.
                      > O is faster ee
                      > C is faster ee e
                      > R is faster e ee
                      > Y is faster ee ee
                      > L is faster daaaah
                      >
                      > You could calculate this by taking the "most common letters in
                      newsprint (messages)" and compare the baud count in both American
                      Morse and International Morse.
                      >
                      > I believe according to my memory when I did this it figured out to
                      about 25% faster for American.
                      >
                      > But American Morse would also be faster than International for CODE
                      GROUPS because of the introduction of "dotty" characters.
                      >
                      > In International Morse, code groups sent at 20 gpm (groups per
                      minute) equal the "transmission speed" of English sent at 25 wpm
                      (words per minute) - words and groups both being counted as "5"
                      characters.
                      >
                      > Also in the old estimation of code speed by American Morse
                      operators, the speed is calculated as "expanded words" from Phillips
                      Code at the ear, to English at the typewriter.
                      >
                      > In any event, the limiting factor is the typewriter!
                      >
                      > 73
                      >
                      > DR
                      > ----- Original Message -----
                      > From: Donald Kemp
                      > To: cw_bugs@yahoogroups.com
                      > Sent: Friday, December 01, 2006 6:31 PM
                      > Subject: Re: [cw_bugs] Dit speed to WPM?
                      >
                      >
                      > Earl,
                      >
                      > Let me try to figure this out.
                      > If I send PARIS in one minute that's 1 WPM.
                      > P = 2 dits and 2 dahs. 1 dah = 3 dits x 2 = 6 + 2 = 8 + 3 spaces
                      between
                      > elements & + 1 space between character = 12 dits
                      > A = 1 dit and 1 dah. Thats 1 + 3 + 1 element space + 1 character
                      space = 6
                      > dits
                      > R = 2 dits and 1 dah. 5 = 2 element spaces + 1 character space =
                      8 dits
                      > I = 2 dits + 1 element space + 1 character space = 4 dits
                      > S = 3 dits + 2 element spaces = 5 dits.
                      >
                      > Total dits = 12 + 6 + 8 + 4 + 5 = 35 dits in one minute or 60/35 =
                      > 1.7142857seconds/dit
                      > Space between words = 1 dah = 3 dits.
                      > 20 WPM would be 60/(20 x 35) + (19 x 3) = 757. 60/757 = .07926
                      seconds/dit.
                      > 1/.07926 = 12.617 dits/second = 20 WPM.
                      >
                      > Maybe.
                      >
                      > I may be missing something in the spaces between words. Do I
                      actually count
                      > them for the dit count?
                      > If not it would be:
                      > 20 WPM = 60/(20 x 35) = .0857 sec/dit.
                      > 1/.0857 = 11.668 dits/sec = 20 WPM
                      >
                      > Not too much difference.
                      >
                      > So for any other WPM calculation (using the second formula) it
                      would be:
                      > X WPM = 60/(X x 35) = Y sec/dit
                      > 1/Y = Z dits/sec.
                      >
                      > IE for 15 WPM.
                      >
                      > 15 WPM = 60/(15 x 35) = .1142857 sec/dit
                      > 1/.1142857 = 8.75 dits/sec.
                      >
                      > Hopefully there's already a table somewhere that will let me know
                      if I'm all
                      > wet.
                      >
                      > It was fun anyway Earl.
                      >
                      > 73,
                      >
                      > Don, NN8B
                      > SKCC 36
                      >
                      > On 12/1/06, Earl Needham <needhame1@...> wrote:
                      > >
                      > >
                      > > Is there a general rule of thumb, that so many dits per
                      > > second equals so many WPM? It would make it a lot easier to
                      figure
                      > > how fast somebody was sending, how fast a bug is, etc. -- a LOT
                      > > easier than sending PARIS repeatedly! :-)
                      > >
                      > > Thanks,
                      > > Earl
                      > >
                      > > KD5XB
                      > > Clovis, New Mexico DM84jk
                      > > http://groups.yahoo.com/group/cw_bugs
                      > >
                      > >
                      > >
                      >
                      > [Non-text portions of this message have been removed]
                      >
                      >
                      >
                      >
                      >
                      >
                      > --------------------------------------------------------------------
                      ----------
                      >
                      >
                      > No virus found in this incoming message.
                      > Checked by AVG Free Edition.
                      > Version: 7.5.430 / Virus Database: 268.15.3/562 - Release Date:
                      12/1/2006 1:12 PM
                      >
                      >
                      > [Non-text portions of this message have been removed]
                      >
                    • k4oso
                      By the way, a bellbutton is the same as a bellybutton :-) Milt k4oso
                      Message 10 of 11 , Dec 7, 2006
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                        By the way, a "bellbutton" is the same as a "bellybutton":-)
                        Milt k4oso
                      • Earl Needham
                        ... Gee, now you got me wondering if this same comparison translates over to bell bottoms... Earl KD5XB Clovis, New Mexico DM84jk
                        Message 11 of 11 , Dec 7, 2006
                        • 0 Attachment
                          At 11:53 AM 12/7/2006, k4oso wrote:
                          >By the way, a "bellbutton" is the same as a "bellybutton":-)
                          >Milt k4oso

                          Gee, now you got me wondering if this same comparison
                          translates over to bell bottoms...

                          Earl

                          KD5XB
                          Clovis, New Mexico DM84jk
                          http://groups.yahoo.com/group/cw_bugs
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