Given a sphere and a compass, is it possible to draw a great circle on the sphere?

Question

No straighthedges are allowed.

Answer 1

If a circle on the sphere is drawn, and using the same compass setting, it's found that the length of this circle is equal to four steps, then it's a great circle. How to find the length of this step, I have no idea. I'm not sure if it's possible at all. Hey, are straightedges allowed?

Addendum: Yes, it is possible. The Mohr-Mascheroni theorem states that any compass-and-straightedge construction is possible with compass alone. A key step towards constructing the great circle is to be able to find the center of any symmetric arrangement of four points on the sphere. Starting with a random compass span less than the radius of the sphere, draw 2 circles through each other's center, forming 2 intersections, so that we have our first set of 4 points in a symmetrical arrangement. Find the center. Drawing more circles outwards towards the other side of the sphere, we can similiarly locate 4 more points and their center. We will thus have the antipodes of one great circle. A little more such work simliar to this will get us the midpoints of this great circle, and finally we will have our compass span to draw the great circle anywhere we want to. The only thing I'm leaving out for lack of space is the method of finding the center of any 4 points in a symmetrical arrangment by compass alone. I leave it as an exercise to the reader.

Answer 2

Yes, PROVIDED that the (equal length) arms of the compass exceed the radius of the sphere!

If the arms meet the local horizons at the "pole" (the point of the compass) and "equator" (the great circle) at an angle θ deg., by simple geometry the angle between the arms at the compass's fulcrum will be 90 - 2θ deg. The length of the arms will be R = r / [2^(1/2) cos (45 + θ)], where r is the radius of the sphere.

In order to be able to draw the great circle, we would need
θ > 0, and a minimum R > r.

As an example, if θ = 30 deg., R = 2.732... r.

θ must of course be less than 45 deg. As θ --> 45 deg., the arm-lengths R --> ∞.

Live long and prosper.

LATER EDIT: O.K., I agree that I didn't realise that the problem was how to CONSTRUCT a great circle, either given a pole or having to produce that pole if the great circle was to pass through some pre-assigned points. That is in fact a VERY DIFFERENT QUESTION from simply asking "is it possible to draw a great circle on the sphere?" (The first one is a deep mathematical question, the second one merely a question of physical possibility. My initial answer merely addressed the physical possibility.)

I may be wrong, but scthyian1950's Addendum strikes me as unsatisfactory. The Mohr-Mascheroni theorem is a theorem of Euclidean plane geometry. It says that IF you can make a construction with compass and straight edge then you can do it solely with a compass. However, (i) it doesn't prove you can do it for ANYTHING you might imagine constructing, you've first got to show that you CAN do it with the compass and straight edge! And (ii) there is a special meaning associated with anything where a straight line is required as part of the construction --- that is, that determining two points on it is deemed equivalent to "producing the line." (After all, you can't DRAW a straight line with a compass!) This means that despite the existence of this theorem, you can't actually draw the analogue in the plane (a straight line) of the great circle required in the solid geometry case under consideration.

What's more, though scythian says "A key step towards constructing the great circle is to be able to find the center of any symmetric arrangement of four points on the sphere," he doesn't in fact show how this is to be done.

Furthermore, the four "symmetrically arranged points" his initial construction produces AREN'T SYMMETRICALLY PLACED, at least not in the sense that any of these four points is equivalent to any one of the others.

It's simple to see this if you think of doing that construction in a plane (the limit as the radii of his circles become small). The construction then produces a rhombus with diagonals in the ratio of 3^(1/2) : 1, rather than a square (the symmetrical arrangement). On the sphere, the corresponding four points will be at the vertices of a spherical equivalent of a rhombus, with a diagonal ratio along the relevant great circles depending on the ratio of compass point separation to sphere radius, of course.

My only conclusion is that determining two points that are 90 degrees apart on the sphere (the real essence of the problem?), solely with a compass, is very tricky indeed!

Some clarification of the "rules" would also help. Some purists insist that each use of a compass in the plane case is to be "from scratch" in the sense that the compass points should be collapsed to one another between steps. (In other words, one can't use the compass to "copy" a length from one part of the diagram to another). Others might consider that perfectly acceptable. So: what uses of the compass itself are to be considered acceptable in this case?

Answer 3

Sure seems like it if you have a big enough compass. If it is too small then it doesn't seem geometrically possible. If the legs are long enough it should be sufficient to tilt the compass so that one end is on the top of the sphere and the other can touch the line that would make the equator, which would be a great circle. Then just carefully turn the compass and draw the line!

Answer 4

Hmm... the constructions suggested by the two posters above me seem to assume the top of the sphere is known. However, for this compass and straightedge problem (I suppose in this case we have a bendable straightedge) the points 90° away would have to be constructed -- sorry, no estimating allowed -- and I'm not sure that can be done. In other words, if you are given a point on the sphere and asked to construct a great circle through that point, how do you determine a longest circle on the sphere through that point?

Answer 5

You'd have to modify your compass. The straight pencil part of a standard compass would contact the sphere and not allow it to draw the line. The geometry of the sphere prevents this from being easy.