Conference rusure::math

    
    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

    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.
    

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

    
    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
    

>    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

    
    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? 
    

re .6
    
>but the depth of the hole is the same value of the hight you'd go up to , by
>symmetry?
    
    Algebraically the answer for going up is different from the answer for
    going down. The symmetry isn't there. Indeed it's not completely clear what
    is meant by 'horizon' when you dig down. From inside a (hollow) sphere
    all points of the surface are visible.
    
    If D << R
        (D is the 150 from the original problem - the required line of
         sight distance to the horizon) then
    
        H = D�/(2R) is an approximation - correct for going up or down.

    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

    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

    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

        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
        

    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.

        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

    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