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Location: UFOUpDatesList.Com > 1997 > Oct > Oct 12

Re: The sky over Roswell

From: Ted Viens <drtedv@smart1.net>
Date: Sat, 11 Oct 1997 22:27:07 -0700
Fwd Date: Sun, 12 Oct 1997 11:10:01 -0400
Subject: Re: The sky over Roswell


> To: UFO UpDates - Toronto <updates@globalserve.net>
> From: Mark Cashman <mcashman@ix.netcom.com>
> Subject: re: UFO UpDate: Re: The sky over Roswell
> Date: Thu, 9 Oct 1997 00:53:28 -0700

> >  From: Ted Viens <drtedv@smart1.net>

> >It is not a question of familiarity with the actual characteristics of
> >secret military aircraft. First comes the observation. From the observation
> >comes a range of possible flight characteristics. These characteristics
> >determine the level of response. They can range from known flight
> >characteristics through just past the rumored edge of secret technology
> >to "Good Golly Miss Molly!" I am suggesting that an NL that goes from
> >zenith to horizon in a minute or less is in the "Miss Molly" range.

> Ted -

> In the interest of providing some material for discussion, I ran
> your estimate of 90 degrees of travel in 1 min (1.5 degrees /
> sec) through my distance / speed spreadsheet for a range of
> altitudes ranging from 1,300 - 21,000 feet. I used the
> intermediate angle of 45 degrees elevation to calculate the
> resulting speeds.

> I also calculated the speed at 100 and 200 miles.

> At 1300' = 33.33 mph
> At 2600' = 66.66 mph
> At 5280' = 133.32 mph
> At 10,560' = 266.63 mph
> At 15,840' = 399.95 mph
> At 21,120' = 533.27 mph
> At 100 mi =13,331.69 mph or 21,455.33 km/h
> At 200 mi = 26,663.39 mph or 42,910.65 km/h

> Most speeds at aircraft altitudes are not abnormal for aircraft
> within those ranges. The rate at 100 mi is also not abnormal,
> since the space shuttle travels at 28,160 km/h at between 100-600
> mi altitudes.

> The shuttle info is from

> http://tommy.jsc.nasa.gov/~woodfill/SPACEED/SEHHTML/technology.html

> The following are the equations I used (please feel free to
> indicate if you think there are any errors):

> Sight Distance = Ground Distance /cos(Elevation*(@PI/180))
> Altitude = tan(Elevation*(@PI/180))*Ground Distance
> Actual Size =(tan(Angular Size*(@PI/180))*Sight Distance)*5280
> Estimated Speed = (tan(Angular Speed Per Second *(@PI/180))*
>         Sight Distance)*3600
> :Ground Distance In Feet = Ground Distance *5280
> :Sight Distance In Feet = Sight Distance *5280
> :Altitude In Feet = Altitude *5280
> 'Estimated Speed km/h' = Estimated Speed *1.6093472

> Because the base and elevation side of the triangle are of the same
> length, the altitude is the same as the ground distance.


Very good work, Mark.  A few weeks ago I prepared a spreadsheet
of the very same material as a tool for skywatchers but I have
been too lazy to type it in yet.  Let us examine two different
cases of a ninety degree overhead pass.

One will be a pass from 45 degrees before zenith to 45 degrees
past zenith just as you have calculated above.  The other will be
a pass from zenith to horizon which more represents the sighting
that I had.  We will arbitrarily use a constant 100 mile altitude
for the bogey.

First, passing from 45 degrees before zenith to 45 degrees past
zenith.  The object passes just a hair over 200 miles (some 1.409
degrees) which in a minutes time would be some 12,000 miles an
hour.

Second, passing from zenith to horizon.  From observer to horizon
intercept to earth center makes a nice right triangle.  Arccos
(radius of earth/ radius to flight path) equals some 12.734
degrees.  This indicates some 903.63 miles of travel in a minute
or some 54,217.8 miles an hour.

This mostly illustrates how treacherous trig can be if you don't
define the situation well.



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