Conference rusure::math

    It looks something like a window shade being pulled down and raised.
    At the time it first appears in the window it is, for all practical
    purposes, horizontal and it stays that way until it disappears.
    Its rate of descent/ascent is (also, for all practical purposes)
    linear.
    
    You can get somewhat the same effect by turning a VT241 upside-down
    and turning it on; watch the blue region.

    Yes, when it's close enough, it'll look that way.  If you're in
    your front yard (with a better view of the sky), will you be able
    to notice the curvature?

If it is night time, and you could hop into your car with a telescope,
you could probably see that it was a disk, and not an infinite wall
at some short distance away from your front yard by looking for
stars which the disk occluded.

Another thing becomes apparent about your front yard.  No other
hills or mountains can be higher than your yard in the plane with the
disk.  (Unless of course by a very miniscule amount, or that there
is a corresponding ditch or depression in the other direction.)

If this disk is anything but imagination, you will probably feel
lighter on your feet because og the effects of gravity.

Also it is not necessary that you see the window shade effect.  If
the disk is rolling by at an very steep angle*, then you would see (in daytime)
a sliver come accross your window, which would be fuzzy at the top
where the disk left the atmosphere.  As the disk rolled closer to
your home, the dark band would expand until the darkness reached the
top and bottom of your window.

* It just occurred to me that this is a **very** steep angle, in fact
for all practical purposes, the the plane of the disk is tangent with
the earth at the point of contact.


Whew!
Dave F.

    Yes, the HORIZONTAL window shade is the correct answer.  Popular
    wrong answer is to suggest you'd see a slantwise occlusion at upper
    left.  Disk is just too darn big to see such a thing.
    
    Perhaps a more difficult question is:  Suppose we want to
    see this window shade go down at some typical rate, such
    as 1 meter per second.  How fast must the disk's center move
    for such an effect ?
    
    /Eric

Well now that I got my taxes done I can get back to the
fun math!

I think I figured out the speed that the disk would be traveling
at to achieve a 1 M/Sec rate of movement of the 'blind' in the
window.  (This is assuming that the disk is 90 degrees upright.)
The rate of movement of the disk must continuously vary to enable the
'blind' to travel up or down the window at a constant rate.

The rate varies from an average of 3.5 KM/sec for the occlusion to
cover the space between 0 and 1 meters to an average of .8 KM/sec to
cover the distance between the 4th and 5th meters.

The way I arrived at these approximations was by using the pythagrean (sp?)
theorem.  Assuming the radius of the disk at 6375 KM., 

	    2	    2		 2
	6375  + rate  =  6375.001

This gave me the average rate to cover the first meter, then successive
calculations would give you the average rates for different parts of
the window.

Its been a long time since I have used trig functions, and I couldn't
see where they would help, was there a simpler way to do this?

Dave F.

  re .5
  
  I don't understand your model,  except that it appears to be
  assuming that the Earth is flat relative to the disk,  whereas
  it has the same curvature.
  
  Here is my attempt:
  
  Draw *two* touching circles and draw a line connecting their centres.
  This represents the position of the disk just before the one
  metre drop occurs.
  
  Draw a line parallel to this connecting line,  external to and
  between the edges of the circles and mark this line as one metre
  in length.  This length represents the one metre of drop.
      
  Now let R be the radius of the disk (*and* the Earth),
  and Theta be the angle in radians at the centre of the earth
  between where the one metre line touches the Earth and the
  two-centres line at the initial situation.  Think of the one
  metre line as being a piece of elastic connecting the circles.
  
  So in order for the disk to roll round until this piece of elastic
  shrinks from one metre to zero,  its centre must travel through an arc
  of length 2R x Theta.  (length of an arc = radius times angle) Since
  it does so in one second,  the average speed is also 2R x Theta.  So
  since we know R we must find Theta:
  
  Draw a line between where the one metre line touches the earth
  to the connecting line at the initial situation so that it is
  perpendicular to this same connecting line.  Now by similar
  triangles,  this meets the connecting line 0.5 metres below
  the surface of the earth and we have a right angled triangle
  with Theta at the centre of the earth,  a hypotenuse of R
  and a base of R-0.5.  Hence:
  
  	Theta = arccos [ (R-0.5)/R ]

  and so
  
  	v = 2R arccos [ (R-0.5)/R ]
  
  setting R as 6.375 E+06,  we get v = approx 7.921 E-04 metres
  per second.  

  S D Clinch.

  P.S. (.6) This is for a one metre window with no distance between
  the sill and the ground.