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Expected sight distance on earth

Consider the earth as the perfect spheroid

\(\frac{x^2}{400^2}+\frac{y^2}{400^2}+\frac{z^2}{100^2}=1\)

and assume it's surface to be completely land. A person whose eyes are 2 meters above the ground is randomly placed on the earth facing a random direction. What is the expected sight distance of this person?

Sight distance is defined as follows.

A ray is drawn from the person's eyes in the direction he/she is facing such that it is tangent to the spheroid at point B. The linear distance from the person's eyes to point B is the sight distance.

I just thought of this problem today and I have no clue how difficult it is. Approximations are welcomed (for example, calculating the expected distance from the person's feet to point B.

Note by Trevor Arashiro
5 months, 2 weeks ago

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Well, isn't it similar to the method by which we determine the height of any transmission tower (to be used for communication purposes)? There, we draw a tangent to the surface of the earth and then calculate the height of the tower using basic trigonometry and approximations. Here, we have to go in reverse order. Correct me if I'm wrong. Vishesh Arora · 2 months, 3 weeks ago

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@Vishesh Arora Yes, we only have to draw tangents, and that's easy if you assume that the earth is a sphere because all tangents will be of the same length. However, on a spheroid, it's much more difficult. Trevor Arashiro · 1 month, 4 weeks ago

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@Trevor Arashiro It might be good to build this one up one piece at a time. For example, we could start with the following problem:

Given a certain spheroid and a particular point in space, what is the shortest / longest tangent line segment from the point to the spheroid?

I think that would be a fun one to post as a problem. Steven Chase · 1 month, 1 week ago

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@Steven Chase Yes, this sounds promising.

How do we solve the first problem? Agnishom Chattopadhyay · 1 month, 1 week ago

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@Steven Chase I understand most of your comment. However, I am unfamiliar with Hessian matrices.

I think your method will work with one slight change. Rather than trying to solve for a max/min sight distance, we should look for the average sight distance at each test point. Using your first and fifth equations, we can generate a 2D ellipse (of course, rotated in 3D) on the spheroid for each test point.

edit: I'm currently working on some really long equations to find the average distance for each point. Do you have any tips for finding the center, minor axis length, and major axis length of a tilted ellipse (in just 2D) as quick as possible? Trevor Arashiro · 1 month ago

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@Steven Chase Go for it! Sounds like it would be quite fun.

Also, VSauce just posted a video on this... I hope he was inspired by this note hahaha Trevor Arashiro · 1 month ago

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@Trevor Arashiro It's up now. I referenced this note as the inspiration, and deleted some of my comments here which would have been too revealing.

https://brilliant.org/problems/extreme-tangents/ Steven Chase · 4 weeks ago

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@Trevor Arashiro Sounds good. This link is fairly helpful. Under "Ellipse as quadric". https://en.wikipedia.org/wiki/Ellipse Steven Chase · 1 month ago

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@Steven Chase My friend just pointed out a critical observation. If we consider the earth under the equation \(\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}=1\) because it is rotationally symmetric about the z-axis, we only need to consider points along the yz plane because sight distance is the same for every point on a plane parallel to the xy plane. Trevor Arashiro · 1 week, 5 days ago

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@Trevor Arashiro Yep, got it. Thanks for pointing out. Vishesh Arora · 1 month ago

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How about we try a Monte-Carlo Simulation? Agnishom Chattopadhyay · 4 months, 3 weeks ago

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