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From: orb@whuts.UUCP (SEVENER)
Newsgroups: talk.politics.misc,sci.misc
Subject: Re: The Last Word on Friedman, Sevener, and Cuba
Message-ID: <3969@whuts.UUCP>
Date: 22 Mar 88 17:15:18 GMT
References: <3405@bloom-beacon.MIT.EDU> <3895@whuts.UUCP> <3588@bloom-beacon.MIT.EDU> <3904@whuts.UUCP> <3699@bloom-beacon.MIT.EDU> <3925@whuts.UUCP> <1570@sigma.UUCP>
Reply-To: orb@whuts.UUCP (45263-SEVENER,T.J.)
Organization: AT&T Bell Laboratories, Whippany, NJ
Lines: 69

In article <1570@sigma.UUCP> bill@sigma.UUCP (William Swan) writes:
>In article <3925@whuts.UUCP> orb@whuts.UUCP (45263-SEVENER,T.J.) writes:
>>3)Even if John Carr were right to argue that the index of refraction
>>  between an atmospheric density of 26.81 inches and 29.92 inches
>>  is the same as that between 29.92 inches and 0 inches, he is
>>  *still* wrong in contending that a refraction of 0.5 degrees would
>>  make Cuba visible from Key West.
>>  In fact, even with a refraction of 0.5 degrees the highest point
>>  in Cuba of 344 feet is still 500 feet below the tangent line
>>  of sight *with refraction*'s height of 844.8 feet.
>
>Sevener never ceases to astound me.
>
>John Carr *never* argued this point, and would be foolish to. It is entirely
>and absolutely wrong, and is not really too far from the way Sevener tried
>to smear Friedman by implying that Friedman claimed to represent the official
>opinions of MIT (we never saw an apology for this, by the way, even though
>one has been called for). 

Mr. Swan, you never cease to amaze me!

Mr. Carr indubitably stated the following:

1)the total refractive displacement of the Sun on the horizon
  was 0.5 degrees.  Actually in the Encyclopedias where I looked
  this up it was 35 minutes.
  So Mr. Carr was neither *wrong* nor *foolish* to argue this
  point as it is essentially correct.
  You are so #$%^&* that you apparently do not understand that.
  Look it up yourself, if you don't believe me.

2)Mr. Carr never calculated what height would be visible from
  a refractive displacement of 0.5 degrees over 90 miles.

  I did.
  I discovered that with a *TOTAL* displacement of 0.5 degrees
  one would be able to see a point 844.8 feet high from 90
  miles away.  This is still 500 feet higher than the 344 ft
  "peak" in Cuba which is actually 95-100 miles away.

  John Carr instead presented a convoluted series of equations based
  on false assumptions which purported to prove that Cuba
  could be visible from Key West.  Somehow, despite
  the fact that astronomy texts list a *TOTAL* displacement
  of 0.5 degrees for the Sun, a figure which I proved would
  still make Cuba impossible to see from Key West, John Carr's
  convoluted equations showed that Cuba "miracule ala Reaganista"!
  was visible from Key West.

  Mr. Carr then proceeded to justify these astounding results
  by pointing out that light in a continuously changing medium
  will be refracted continuously rather than at a sharp angle
  as in going from air to water or air to vacumn.
  A point I never disputed.
  Regardless of whether the refractive displacement is continuous
  or not, authoritative texts list a *TOTAL* refractive displacement
  for the Sun of 0.5 degrees after going from a refractive index
  of 1.00293 to 1.000000 and after going through 180 miles of
  continuously changing atmospheric density.
  There is no way that the total refractive displacement going
  the 90 miles from Cuba to Key West with a difference of
  0.000030 in refractive indices versus 0.000293 that there
  will be more refractive displacement than that viewing the Sun.

I don't expect you to understand that, Mr. Swan.

I am concluding it is hopeless to expect you to understand
anything.

tim sevener  whuts!orb