# Intersecting secant lines with a fixed angle becoming tangents.

• Aug 4th 2010, 10:36 PM
Winawer
Intersecting secant lines with a fixed angle becoming tangents.
Hello! This is my first posting on this forum, so please forgive me for any errors. :)

My question is motivated by a physical problem: I was wondering today how to calculate how high up you would have to go before the Earth just filled your field of vision, and the following is the best way I could come up with to express the question.

Let there be a pair of secants to a circle with a point P outside the circle where they intersect so that they form an angle A. Now, hold the angle A constant but move the point P away from the center of the circle along a line that goes through the center of the circle (i.e. "straight back"). At some point, the secants will become tangents. I'm trying to figure out how far the intersection of the tangents (P) will be from the circle when this happens, for a given A and knowing the diameter of the circle.

I'm interested in trying to answer the question myself, but I could use a pointer in the right direction because I've run out of ideas on how to set this up. Is there a method using straight geometry, or do I have to do something with calculus? (I'm also open to better ways to express the question, too. This isn't a homework question, it's just my curiosity running amok).
• Aug 5th 2010, 12:19 AM
earboth
Quote:

Originally Posted by Winawer
Hello! This is my first posting on this forum, so please forgive me for any errors. :)

My question is motivated by a physical problem: I was wondering today how to calculate how high up you would have to go before the Earth just filled your field of vision, and the following is the best way I could come up with to express the question.

Let there be a pair of secants to a circle with a point P outside the circle where they intersect so that they form an angle A. Now, hold the angle A constant but move the point P away from the center of the circle along a line that goes through the center of the circle (i.e. "straight back"). At some point, the secants will become tangents. I'm trying to figure out how far the intersection of the tangents (P) will be from the circle when this happens, for a given A and knowing the diameter of the circle.

I'm interested in trying to answer the question myself, but I could use a pointer in the right direction because I've run out of ideas on how to set this up. Is there a method using straight geometry, or do I have to do something with calculus? (I'm also open to better ways to express the question, too. This isn't a homework question, it's just my curiosity running amok).

1. Draw a sketch.

2. If P has reached it's final position you are dealing with 2 right triangles: The hypotenuse has the langth r + x and the leg opposite the angle $\frac A2$ has the length r.

3. Use the Sine function to calculate x:

$\sin\left(\frac A2\right) = \dfrac{r}{r+x}$

Solve for x.
• Aug 5th 2010, 01:02 AM
Winawer
Quote:

Originally Posted by earboth
1. Draw a sketch.

2. If P has reached it's final position you are dealing with 2 right triangles: The hypotenuse has the langth r + x and the leg opposite the angle $\frac A2$ has the length r.

3. Use the Sine function to calculate x:

$\sin\left(\frac A2\right) = \dfrac{r}{r+x}$

Solve for x.

Damn, I've been drawing diagrams like that all day, but I forgot that the radius would be perpendicular to the tangent. Thanks a bunch, earboth, especially for the diagram, that really made it clear.

Let's see, using the radius as 6378 km and a conservative estimate of 160 degrees for a human field of view, that means that the Earth would just fill the visual field at about ... just over 98 km up. That sounds like it's in the ballpark.
• Aug 6th 2010, 05:59 AM
bjhopper
Hello Winawer,
This is an interesting question.My first thought was infinity. I looked thru the google for info and found that NASA has a photo taken from 22,300 miles reporting this to be the fullest image on record.Another source indicated a 98% view from the moon.

bjh