Curvature of horizon using trig

[nobahar]

Hello!

In New Scientist this week (actually next week!), there was a question concerning the curvature at the horizon.

The formula is as follows for the distance to the horizon:
(2*6373*h)^1/2 km; where h is the height of the individual from the ground.
Using the exaple it states a towe 150m high the horizon will be 44km away and displaced down by 0.39 degrees.
That's all fine, but the next bit states that by holding a 1m stick 1m in front of you, the ends of the stick will be 0.8mm above the horizon. Intrested to know how this was arrived at.
Although I'm not sure whether this is in the right place or not, I'm pretty sure it'd be beneficial for anyne else with trig questions!

Thanks in advance!

 

[Dick]

You said the for the tower, the horizon was some angle below horizontal. Do you know how to find that angle. If so then I think you are supposed to assume that the tip of the meter stick is at the same horizontal level as your eye. Since the horizon is at some angle below that, it should be pretty straightforward trig to find the distance on the meter stick that marks the level of the horizon. You have a right triangle and you know the horizonal leg is 1m and the angle at the vertex where your eye is.

 

[HallsofIvy]

Hello!

In New Scientist this week (actually next week!), there was a question concerning the curvature at the horizon.

The formula is as follows for the distance to the horizon:
(2*6373*h)^1/2 km; where h is the height of the individual from the ground.
Using the exaple it states a towe 150m high the horizon will be 44km away and displaced down by 0.39 degrees.
That's all fine, but the next bit states that by holding a 1m stick 1m in front of you, the ends of the stick will be 0.8mm above the horizon. Intrested to know how this was arrived at.
Although I'm not sure whether this is in the right place or not, I'm pretty sure it'd be beneficial for anyne else with trig questions!

Thanks in advance!

You have a right triangle in which the "near" side is 1 m and the angle is 0.39 degrees.
The "opposite" side is given by 1*tan(.39)= 0.006806 m. or about 6 mm, not "0.8".

 

[DaxInvader]

HallsofIvy, What is your actual scholarship? I mean, you help a lot around here and i try to do the same. I feel bad asking questions when i don't even try to anwser others :P

 

[nobahar]

Isn't this telling me the degree's downwards from horizontal the horizon is?
If I was to hold the metre stick along the horizon, there would be a slight downwards curvature at either end. Is this what has been calculated? (If so, I've misunderstood the explanations!) A diagram of some sort, if possible, would be appreciated!
Furthermore, progressively left or right from the stick is THE SAME distance from me, since I am on a sphere. Instead of being an increasing distance. Can anyone elaborate for me?
As always... Thanks in advance!