Curvature of horizon using trig

Click For Summary

Homework Help Overview

The discussion revolves around the curvature of the horizon and its relationship with height and angles, particularly in the context of a person holding a stick at eye level. The original poster references a formula for calculating the distance to the horizon based on height and seeks clarification on a specific example involving a 1m stick.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the height of an observer and the angle of the horizon. Questions arise about how to calculate the angle using trigonometric principles and the implications of holding a stick at eye level. There is also a consideration of the curvature of the horizon and how it affects the perceived height of the stick's ends.

Discussion Status

Some participants are exploring the geometric relationships involved, while others are questioning the assumptions made about the stick's position relative to the horizon. There is an ongoing dialogue about the calculations and interpretations of the results, with no clear consensus yet reached.

Contextual Notes

Participants note potential misunderstandings regarding the curvature and distance calculations, as well as the need for diagrams to aid in visualizing the problem. The discussion includes references to specific angles and measurements that may require further clarification.

nobahar
Messages
482
Reaction score
2
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!
 
Physics news on Phys.org
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.
 
nobahar said:
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".
 
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
 
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!
 
Last edited:

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K