| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1584.1 | Yes, can be solved | HERON::BLOMBERG | Trapped inside the universe | Wed Mar 25 1992 15:23 | 11 |
|
Assuming the earth is spherical, yes, it's an application of
Pythagora's theorem. Simplification possible if the heights can
be assumed much smaller than the earth's radius.
However, because of humidity and other effects in the air the
real distance you can see is mostly much shorter. About twice per
year (crispy winter mornings) I can see Corsica (200 m away) from
my office.
/Ake
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| 1584.2 | Ok. What about Ozone? | CSCMA::LABAK | | Wed Mar 25 1992 17:05 | 9 |
| Ake,
Do you mean the Pythagorean theorem? You still have to assume
the radius of the earth. This problem came from a basic Trigonometry
book. When I looked at it I thought the author had made a
mistake in not giving enough information in the problem.
Thanks
Rick L.
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| 1584.3 | | ZFC::deramo | Colorado Rocky Mountain High | Wed Mar 25 1992 17:19 | 6 |
| Is the 150 miles measured in a straight line from the point
at height H above the surface, or along the arc on the surface?
In either case the result depends on R. But R is relatively
constant :-) and can be assumed to be known.
Dan
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| 1584.4 | solve quadratic equation | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Wed Mar 25 1992 18:00 | 11 |
|
as was said befor,
let R be radius of earth, so R^2 = (R-H)^2 + 150^2
solve for H. (watch the units offcourse),
iam assuming Radius of earth is > than 150 miles :-)
/naser
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| 1584.5 | | ZFC::deramo | Colorado Rocky Mountain High | Wed Mar 25 1992 22:41 | 14 |
| > let R be radius of earth, so R^2 = (R-H)^2 + 150^2
>
> solve for H. (watch the units offcourse),
R-H would correspond to digging a hole.
If the 150 miles is arc length along the surface, then the
right triangle has one angle of 150/R with adj. side of length
R and hyp. of length R+H.
If the 150 miles is line-of-sight then the right triangle
has legs R and 150 and hyp. R+H.
Dan
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| 1584.6 | the question did not say you cant dig down :-) | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Thu Mar 26 1992 01:10 | 5 |
|
i assumed line of sight, and digging a hole it is, but the depth
of the hole is the same value of the hight you'd go up to , by
symmetry?
|
| 1584.7 | | AUSSIE::GARSON | | Wed Apr 01 1992 01:26 | 15 |
| 1584.8 | i try to explain better | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Wed Apr 01 1992 04:23 | 27 |
| lets me try to explain, it is hard to draw a curve here, but this is
how it looks
a
|
|
b------------------------c
|
|
d
the person at the ground at point b, bc is 150 miles horizantal, the
earth curves from b to d, so there is a curve going from b to d, drop
a stright vertical line from c to the ground to d. the distance cd
is the same distance the person has to go up so as to see point d.
i.e. line ad is a tangent to the earth.
now, draw a line from d to the center of earth O, and from b to O.
draw line horizantal from d to meet line bO at point X.
look at triangle XdO the distance XO is raduis of earth - bX
but bX = ba , and ba is what we are looking for. so apply phytharogse
(sp?) to triangle XdO, solve for XO, that gives you ba .
which is what i said befor.
/nasser
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| 1584.9 | | FORTY2::PALKA | | Wed Apr 01 1992 09:20 | 18 |
| RE .8
Are you assuming that bc = Xd ? It isn't - because cd is not parallel to
bO.
Furthermore there is a question whether 150 miles is the distance cb or
the distance (along the arc) db.
So now we have three interpretations of '150 miles'
- bc (line of sight)
- bd ('map' distance)
- Xd (dont know what this corresponds to in physical terms)
The difference between the different calculations is however small
(compared with the accuracy of measurement of altitude in a typical
plane, and errors due to the contours of the terrain).
Andrew
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| 1584.10 | yes, they are || | STAR::ABBASI | i^(-i) = SQRT(exp(PI)) | Wed Apr 01 1992 13:09 | 22 |
| offcourse they are parallel ?
line bc is a tangent, so angle Obc is 90.
angle cdX is 90 also .
Xbcd is a rectangle !
b --------- c
| | earth goes along arc bd
X --------- d
| /
| /
| /
| /
| /
|/
O
?
/nasser
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| 1584.11 | | GUESS::DERAMO | Dan D'Eramo, zfc::deramo | Wed Apr 01 1992 23:36 | 12 |
| re .-1
I think he interpreted
>.8
> drop a stright vertical line from c to the ground to d.
to mean Odc should be a straight line, whereas you meant
it to mean bX || cd like in your drawing.
Dan
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| 1584.12 | | AUSSIE::GARSON | | Thu Apr 02 1992 02:01 | 11 |
| re .8
> i.e. line ad is a tangent to the earth.
No, it's not.
For an obvious counterexample let bc=R in which case da clearly cuts
the Earth.
In fact it only looks as if ad is a tangent when bc << R. As bc increases
from 0 the error becomes grosser.
|
| 1584.13 | | GUESS::DERAMO | Dan D'Eramo, zfc::deramo | Thu Apr 02 1992 03:58 | 13 |
| re .-1,
>> > i.e. line ad is a tangent to the earth.
>>
>> No, it's not.
That was the point of his drawing. b is the north pole,
d is another point on the earth the given distance away
(say, 150 miles), a is how high you have to be above b
(the north pole) to be able to see d. c was then chosen
so that ab || cd and bc is perpendicular to them both.
Dan
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| 1584.14 | | FORTY2::PALKA | | Thu Apr 02 1992 08:47 | 11 |
| Yes, I interpreted 'vertical' as meaning 'in a direction towards the
centre of the earth'.
However if you find d by drawing a perpendicular to bc then the
distance bd is not the height of c above the ground. (and the height of
c is the answer to one of the interpretations of the question).
Furthermore I dont see how you can say that bX = ba if ad is tangent to
the earth at d.
Andrew
|