From: Ted Viens <email@example.com> 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 <firstname.lastname@example.org> > From: Mark Cashman <email@example.com> > Subject: re: UFO UpDate: Re: The sky over Roswell > Date: Thu, 9 Oct 1997 00:53:28 -0700 > > From: Ted Viens <firstname.lastname@example.org> > >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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