| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1439.1 | | PENUTS::DDESMAISONS | Are you married or happy? | Tue May 06 1997 11:39 | 4 |
|
neighbors running and screaming?
|
| 1439.2 | | 39044::16.121.160.233::slab | [email protected] | Tue May 06 1997 18:51 | 4 |
|
You definitely wouldn't see much ... if there's anything on the disc, you'll see a blur of
color [but I guess that depends on the speed of the disc].
|
| 1439.3 | Like a fool, I'll try answering | NETCAD::ROLKE | The FDDI Genome Project | Wed May 07 1997 13:54 | 27 |
| This answer is "really serious" so don't laugh:
If there were any distinguishing marks (such as "P185/HR506880000")
then this mark would appear to fall vertically from the sky, linger
for a moment, and then rise straight up again. To me the disc
would look like a horizontal line with the disc above and my street
below.
This is because we are constrained to looking at something in my
street from my window: my window gives but a small, mostly horizontal
view of the street some 20 meters distant. The disc with a radius
of 4000 miles would have no noticable curvature.
I'd have taken pictures but my lens cap was stuck!
Chuck
Now....
Tell us more about your problem.
How long have you been concerned about these "discs"?
How many have you seen?
How do you know how big they are?
When did you stop seeing discs this big?
Have you seen the rubber bands wrapped around the earth?
Are you tempted to climb the rope which goes to the moon?
Keep your mirrors clean!
|
| 1439.4 | | 4446::OSMAN | Eric Osman, dtn 226-7122 | Thu May 08 1997 10:32 | 11 |
|
Yes, that's getting to the point. Specifically, as the disk
is rolled by your window, you'd see a horizontal line go down and then
up again, like a window shade.
(For a disk the size of the earth, how fast would it need to be rolled
from left to right for the window shade affect to be at some reasonable
speed, i.e. not so fast as to be done in an instant, and not so slow
as to be impatient waiten for ?)
/Eric
|
| 1439.5 | | CSC32::MACGREGOR | Colorado: the TRUE mid-west | Thu May 08 1997 18:32 | 6 |
|
I would venture a guess that the rate would be variable. I mean my
eight year old would be bored in about a nanosecond, whereas the monk
at the top of the mountain in B.C. might be able to wait a lifetime 8^)
Marc
|
| 1439.6 | | 4446::OSMAN | Eric Osman, dtn 226-7122 | Fri May 09 1997 19:00 | 8 |
|
I didn't mean it to be a trick question.
If you need me to pick a value, o.k. suppose we want to see the shade
affect cover a 1 meter high window in 1 second, then have the window
clear again 1 second later. How fast is disk rolling ?
/Eric
|
| 1439.7 | whoosh! | NETCAD::ROLKE | The FDDI Genome Project | Fri May 16 1997 11:30 | 31 |
| This is my reward for typing "open bra" so often!?
-C
Two circles (radius = r = 6367470 m) join at a point (point = P).
The height of the window is one meter (h = 1m).
Their common tangent has a point A where the distance from A to
either circle's center is (r + h/2). This is were the circles are
"h" apart.
So the distance from P to A is
PA = sqrt( (r+(h/2))^2 - r^2 )
PA = sqrt( rh + (h/2)^2 )
For the disc to "roll" this far in any unit of time (time = t = 1S)
then the whole mess is:
disk speed = sqrt( 2rh/2 + (h/2)^2 ) / t
disk speed = 2523.38 m/s
This disc makes one revolution around the earth in 4.4 hours.
Note the loose assumptions that 1) the earth is circular, 2) the window
is at ground level, 3) the earth/disc radius is so big compared to the
window that I can ignore the curvature effects as the disc covers the
window.
This solves for the speed of point of contact of the earth and disc.
Would the "speed" of the disc be better measured as the disc's axle
speed in space around the center of the earth?
|