Quote:
Originally Posted by trueviking
i was about to make the same joke in reverse....
assuming the ground is flat....here is how to figure it out mathematically:
The vertical line drawn from the top say point T of the building of height h towards the centre, say O of earth measures (R+h), where R is radius of earth. Neglecting viewer's height, let the tangent drawn from viewer's location, say point L to the surface of earth in such a way that it touches the top of the building measure X. Then the triangle LTO is aright triangle with right angle at L and hypotenuse OT and one side LT = X. Let the angle LOT be theta. It is obvious that if the viewers goes beyond L the building will not be visible. Let the location point of the building be B. We know that TO meets earth at B. We are supposed to find out the the distance BL measured on teh surface of earth. We know that
BL = R*theta and tan (theta) is given by
tan (theta) = LT/OL = sq rt[OT^2 - OL^2] / OL where OL = R and OT = R+h, So
tan (theta) = sq rt[(R+h)^2 - R^2] / R or = sq rt[{1 + (h/R)}^2 - 1] = sq rt [2h/R] neglecting higher powers of (h/R).
h for a 30 story building can be estimated as (10 feet)*30 =my b 300 ft = (300*12*2.54)/100 m = 91.44 m or say 91 m
R = 6.37*10^6 m
tan (theta) = [sq rt(2*91/6.37)]*10 ^-3 = 5.345*10^-3 =
or theta = 5.345*10^-3 rad
BL = (5.345*10^-3)*(6.37*10^6) = 34049 m = 34 km
|
whats the " ^ "suppost to represent? ... i can't beleave how confused i have been left with some basic trig blah i need a refresher