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Re: Dit speed to WPM?

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  • 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 1 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]
      >
      >
      >
      >
      >
      >
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      >
    • k4oso
      By the way, a bellbutton is the same as a bellybutton :-) Milt k4oso
      Message 2 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 3 of 11 , Dec 7, 2006
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          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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