Circles on a sphere

I just intend this puzzle as exploratory. I believe there's an answer, but I'm not sure.

Given a sphere radius x, draw a circle on it with a diameter d and circumference c. This circle has two center points, one closer to the circle, and one farther from it. Thus, it has two diameters, one to the closer center, and one to the farther one. Is an equation relating the two ratios of c/d to a constant?

If you use the closest point, then your ratio is between 2 and pi. If you use the farther one, then your ratio is between 0 and 2.

Originally posted by ark13
I just intend this puzzle as exploratory. I believe there's an answer, but I'm not sure.

Given a sphere radius x, draw a circle on it with a diameter d and circumference c. This circle has two center points, one closer to the circle, and one farther from it. Thus, it has two diameters, one to the closer center, and one to the farther one. Is an equation ...[text shortened]... ur ratio is between 2 and pi. If you use the farther one, then your ratio is between 0 and 2.

I don't understand. Do you mean to draw a cirlce on the skin of the sphere, with the diameters being measured: (a) along the great circle encompassing the sphere, on the skin of the sphere; and (b) along the plane on which the circle sits, slicing the sphere? Or do you mean something else?

Originally posted by PBE6
I don't understand. Do you mean to draw a cirlce on the skin of the sphere, with the diameters being measured: (a) along the great circle encompassing the sphere, on the skin of the sphere; and (b) along the plane on which the circle sits, slicing the sphere? Or do you mean something else?

Draw a circle on the surface of the sphere. The diameter should be measured along the great circle that passes through both center points, and two opposite points on the circle. The diameter is either the short arc or the long arc.

Everything is done on the surface of the sphere.

Originally posted by ark13
Draw a circle on the surface of the sphere. The diameter should be measured along the great circle that passes through both center points, and two opposite points on the circle. The diameter is either the short arc or the long arc.

Everything is done on the surface of the sphere.

Are you drunk? Or am I? The "two opposite points on the circle" part makes sense, as those are the points I envision the great circle passing through. But what are these two centre points you're talking about?

Originally posted by PBE6
Are you drunk? Or am I? The "two opposite points on the circle" part makes sense, as those are the points I envision the great circle passing through. But what are these two centre points you're talking about?

If the circle is earth's equator, then the two "center" points are the poles--this would also be true for instance if we were talking about the arctic circle, and in that case you could define two different "diameters" on the surface of the sphere. I think that's what ark's getting at....

Originally posted by PBE6
Are you drunk? Or am I? The "two opposite points on the circle" part makes sense, as those are the points I envision the great circle passing through. But what are these two centre points you're talking about?

A center point of a circle is a point equidistant from all points on the circle. On a plane, there is one center point. One a sphere, there are two on the surface of the sphere.

If you were to draw both diameters, you'd draw a circle around the orange.

Pi is usually defined as the quota of the circumference and its diameter of a circle. 3.14159... Right?
Put this definition is only when the circle is on a plane.

What about when we have a circle on a sphere? Let’s say the equator on the Earth? Its a circle (the largest one possible) has a diameter of twice the circumference of the globe itself – hence pi = 2.

If the diameter of the globe is unknown, then we can deduce its diameter by measuring a circels diameter and circumference, divide and get a value of pi.

But what about the value of pi on a surface which is hyperbolic? Answer: Pi becomes larger than 3.14

Her we have a method of measuring the size of universe itself. If we draw a circle, millions, even billions of light-years in circumference, divide with its diameter, then we will have a measure of the universe’s size. (!) If Universe is flat we will yield pi as 3.14159... We can also with this method deduce the topology of the universe.

But it is very difficult to draw a circle, large enough, not to mention to deduce the value of pi, in this experiment...

Alright, I guess I didn't explain this well. I'll start it off. If the circle is drawn right around the equator, the ratio c/d=2 for either center (as they are the same distance away). But move the circle in one direction. The circumference decreases, while one diamter increases and the other one decreases. What is the relationship in the rate of these changes. The sum of the two diamters will always stay the same, but the circumference doesn't decrease in a linear fashion.

Originally posted by ark13
Alright, I guess I didn't explain this well. I'll start it off. If the circle is drawn right around the equator, the ratio c/d=2 for either center (as they are the same distance away). But move the circle in one direction. The circumference decreases, while one diamter increases and the other one decreases. What is the relationship in the rate of these ch ...[text shortened]... amters will always stay the same, but the circumference doesn't decrease in a linear fashion.

Ah, now we're cooking with gas. This 3-D problem can be reduced to 2-D, with the great circle passing through all the important bits becoming the new base of operations. In this model, the drawn circle will be represented by a chord of length "d" slicing through the great circle of radius "x". For simplicity's sake, the chord will be drawn higher than and parallel to the equator.

With this model in mind, the angle between the vertical and the point where the chord touches the circle is given by:

sin(theta) = (d/2x)

theta = Asin(d/2x) ...where "Asin" is the inverse sine function

The short arc length from the mentioned point to the top of the circle is:

L1 = (1/2)*theta*x^2

The long arc length from the mentioned point to the bottom of the circle is:

L2 = (1/2)*(pi-theta)*x^2

The "diameters" mentioned in the question are just twice the arc lengths, 2*L1 and 2*L2 respectively.

So, the ratio of the drawn circle's circumference pi*d to the diameters 2*L1 and 2*L2 are:

pi*d/(theta*x^2) and pi*d/((pi-theta)*x^2)

To find the rate of change, you can differentiate with respect to theta.

Originally posted by PBE6
To find the rate of change, you can differentiate with respect to theta.

That's generally where the text book would say: the process of differentiation is trivial and the answer is clearly ___________

Originally posted by The Plumber
That's generally where the text book would say: the process of differentiation is trivial and the answer is clearly ___________

That's where the text book review ends, too. 😛

Originally posted by The Plumber
That's generally where the text book would say: the process of differentiation is trivial and the answer is clearly ___________

With most of my math textbooks, they would say:
"the rest of the solution has been left as an exercise to the reader."

Originally posted by PBE6
Ah, now we're cooking with gas. This 3-D problem can be reduced to 2-D, with the great circle passing through all the important bits becoming the new base of operations. In this model, the drawn circle will be represented by a chord of length "d" slicing through the great circle of radius "x". For simplicity's sake, the chord will be drawn higher than and pa ...[text shortened]... eta)*x^2)

To find the rate of change, you can differentiate with respect to theta.

Looks pretty good. Much more difficult than I expected though. Nice work.