A Complicated Geometry Question (No, This Is Not Homework)

Imagine a circle with a known radius (N) and known diameter (N+N). Now imagine a straight line (A-C) beginning at the circle’s “north pole” (for lack of a better choice of words) and extending outward. How can you calculate the distance from points on the A-C line to points on the circle (viz, B-D, C-E, or any other imaginary line)?

Also, if the diameter of the circle were known - specifically, 12742 kilometers, how long would segment A-B have to be in order for segment B-D to be one meter?

In your specific example, what are points A, B, and D? Are they all on the same line? Is the line tangent to the circle, or meet some other criterion?

Here you go.

A is both on the circle and the starting point for line A-C. B is on line A-C (or A-B, if you prefer). D is on the circle.

“Extending outward” is pretty vague.

Is line AC a tangent line? (If so, I think MrDibble’s link has you covered.) Or is AC an extension of the radius of the circle through A? Or some other line?

If those are your only constraints, then any value for AB less than or equal to 1 meter will work, for some choice of D.

If I understand this correctly, if we center the circle at (0,0), then A is at (0,N), B is at (0,N+b) for some value b and D is at some (x,y) on the circle so that (x^2+y^2)=N^2

Then the distance between B and D is Sqrt(x^2+(N+b-y)^2).

This is for a general point on the circle D. If you don’t have the coordinates of D but have it specified in some other way (such as saying that the line BD is tangent to the circle, or you know the x coordinate of D but not the y coordinate) let us know and we can specify the coordinates of D to put in the above formula.

That’s what I assumed, given the drawing in the OP. Otherwise we’re in power of a point territory.

Alternative understanding:

B is an arbitrary point on a line with slope c starting at A. So that B has the coordinates (b,N+b*c), and D is the point on the circle that is closest to B.

We can use [Legrange multipliers](Legrange multipliers) to find points (x,y) that minimize (x-b)^2+(y-N+b*c) subject to the constraint that x^2+y^2=N^2.

This results in the solution that x=bN/sqrt((N+cb)^2+b^2) and y=N*(N+cb)/sqrt((N+cb)^2+b^2)

and the distance is sqrt((b-x)^2+(N+c*b-y)^2)

Didn’t see the image link. :smack:

B= (b,N)
D=(b,sqrt(N^2-b^2))

distance = sqrt((b-b)^2+(N-sqrt(N^2-b^2)) = sqrt(N-sqrt(N^2-b^2))

I’m getting about 3570 meters, or 3.57 km. Can somebody check my work. I’m setting it up like this:

x^2+y^2=r^2

The radius in meters is 6 371 000 meters.

x^2+y^2= 6371000^2

At x = 0, y equals 6 371 000. We want to know what x is when y is 6 370 999 (a difference of one meter on the vertical axis)

Solving for x, I get 3569.593674355668 meters.

Looks good to me.

Thank you, pulykamell! I can’t follow the math (my understanding of mathematics ends at basic arithmetic), but I’m going to take your word for it.

So B is a point on the circle radius R, as is A.
Since the line form A is a tangent, obviously the angle AB is 90 degrees and Pythagoras applies.
Similarly, AB is the angle of the
You get 2 equations. Call C the hypotenuse
The triangle sides A, B, C; A^2 + B^2 = C^2 given B=1 then C=SQRT(A^2 + 1)
The angle AB of the circle Radius N=1274200m, let’s call it angle X then
A = N*Sin(X)
B = N(1-cos(X)) or since B=1 1=N(1-cos(X)) so N=1/(1-cos(X)) or cos(X)=1-1/1274200

From there it’s simple, yeah?

Basically, the formula for a circle is a[sup]2[/sup]+b[sup]2[/sup]=r[sup]2[/sup]

Or, even more simply, you can even just think of it in terms of drawing a triangle. You remember the Pythagorean theorem? Well, it’s the same as the equation for the circle, essentially.

How high is point A in your picture? Well, it’s 6371 km, since the radius of the circle is 6371 km. Now, you want to know how far along the x axis you need to go such that point B is one meter lower than it was in point A. Therefore, you’re looking for a place where B is 6370.999 km, right (as that is one meter lower than 6371). So just draw yourself a triangle. You know the hypotenuse is the radius, you know side b is 6370.999, so now all you need to do is solve for side A, using the Pythagorean theorem.

This is beginning to look like a flat earth question

As a matter of fact, it is. This question came from a thought experiment I’ve been conducting in my own brain. Imagine, if you will, a metal rod, a plank of wood, what have you. For this experiment it would have to be so strong that it wouldn’t bend or bow under its own weight. Anyway, you lay it on the ground at Point A. From Point A you walk towards Point B, and as the Earth curves underneath you (because the Earth is round, of course), as you walk further and further away from Point A the distance 'twixt the rod and the ground becomes greater. I just wanted to know how long that distance would be before it was one meter (and I chose one meter because why not?).

*This, of course, assumes that you’re doing the experiment in a place where there are no natural fluctuations in the surface of the ground, like the Bonneville Salt Flats or that salt lake in Bolivia or wherever.

You may find this page useful

http://earthcurvature.com

You can’t get a rod that rigid, but you can get a straight laser beam. This is an issue, for the 10-km-long arms of the LIGO detectors.

Note also that this question is equivalent to asking “how far away is the horizon, from a height of 1 m?”.

You can’t argue with a flat-earth proponent, because they are deliberately misunderstanding. Forget it. Just laugh at them.