Diameter of sphere - using only compass and ruler

Diameter of sphere - using only compass and ruler

Postby Guest » Sat Sep 17, 2016 3:25 am

Using only a compass and a ruler, how can we measure the diameter of a hollow sphere?

Re: Diameter of a sphere

Postby Guest » Sat Sep 17, 2016 12:04 pm

Set the compass to some distance less than [tex]\frac{\sqrt{3}}{2}[/tex] times the length of the diameter of the sphere (any value will do as long as it is not too big). Let the distance between the tips of the compass, as measured by the ruler, be [tex]r[/tex].

Draw a circle using the compass (set at distance [tex]r[/tex]) somewhere on the surface the sphere.
Draw another circle (without changing the compass) with its centre somewhere on the circumference of the first circle drawn.
The two circles will intersect at two points, set the compass ends so that they lie on the intersections, then use the ruler to
measure how far apart the ends are, call this distance [tex]d[/tex].

The diameter of the sphere will be

As an example consider a sphere of radius 1. Let us suppose we set [tex]r=\sqrt{2}[/tex] (carefully chosen to make the explanation easier), and choose the centre of the first circle to be the "south pole". The circle drawn will be a line of latitude, specifically the equator (due to the carefully chosen choice of [tex]r[/tex]). Next choose a point on the equator and draw another circle, it will form a "line of longtitude". The intersection of these two circles will be diametrically opposite points, so the distance between the intersections is [tex]d=2[/tex]. Using the formula the diameter is
= [tex]\sqrt{2}\sqrt{\frac{8-4}{6-4}}[/tex]
= [tex]\sqrt{2}\sqrt{2}[/tex]
= 2
As expected.

Hope this helped,

R. Baber.

Re: Diameter of a sphere

Postby Guest » Mon Sep 19, 2016 6:16 am

Obviously it helped!

However, can you please explain how you came to this formula? I understand the second part, which proves it is correct, but I don't understand how you came to this in the first place!

Many thanks!

Re: Diameter of a sphere

Postby Guest » Mon Sep 19, 2016 5:02 pm

There are four points of interest in my construction: the two intersections of the circles, call them [tex]A,B[/tex], and the centres of the two circles, call them [tex]C,D[/tex]. Because of the fact that the compass was fixed at distance [tex]r[/tex], we know that [tex]r=|AC|=|BC|=|AD|=|BD|=|CD|[/tex], so [tex]ACD[/tex] and [tex]BCD[/tex] are two equilateral triangles both with side lengths [tex]r[/tex].

Also these two triangles share an edge [tex]CD[/tex], which is a bit like a hinge, the two triangles don't lie in the same plane as each other. If we could work out the angle between the planes they lie in that would give us information on how "curved" the surface the triangles are lying in is. If the angle between the planes is close to 180, the surface is close to being flat and so the radius of the sphere is very large, if instead the angle is very small, then the surface is very curved and the radius is very small. Instead of explicitly calculating the angle we can equivalently use the distance [tex]d=|AB|[/tex] which is more convenient.

So now you have 4 points that lie on a sphere of radius say [tex]R[/tex] and that these 4 points form two equilateral triangles of side length [tex]r[/tex], with two of the points being a distance of [tex]d[/tex] apart, and all you need to do is work out an equation that relates [tex]R, r,[/tex] and [tex]d[/tex]. You can do this using basic coordinate geometry. We can assume without loss of generality that the sphere is centred on the origin, so any point [tex](x,y,z)[/tex] that lies on the sphere has to satisfy [tex]x^2+y^2+z^2=R^2[/tex] (this is the equation of a sphere, which itself comes from Pythagoras' theorem in 3 dimensions).

To make our lives easier we may as well assume that [tex]C[/tex] and [tex]D[/tex] lie on the equator i.e. their [tex]z[/tex] coordinate is [tex]0[/tex] (we can always rotate to make this true). Furthermore we can assume that they lie either side of the [tex]x[/tex] axis (again we can rotate them along the equator so that this holds true). Consequently the coordinates of [tex]C[/tex] and [tex]D[/tex] take the form [tex](X_1,\pm r/2, 0)[/tex] for some unknown [tex]X_1[/tex] (we forced the [tex]z[/tex] coordinate to be [tex]0[/tex], and that they should have the same [tex]x[/tex] coordinate, the [tex]y[/tex] coordinate is determined by the fact that [tex]|CD|=r[/tex]). You can determine [tex]X_1[/tex] by remembering that [tex]C[/tex] and [tex]D[/tex] lie on the sphere, so their coordinates must satisfy [tex]x^2+y^2+z^2=R^2[/tex], I leave that as exercise for you to do.

Next we consider [tex]A[/tex] and [tex]B[/tex], by symmetry we know that their [tex]y[/tex] coordinate is [tex]0[/tex], and they lie either size of the [tex]x-y[/tex] plane, so their coordinates take the form [tex](X_2,0,\pm d/2)[/tex] (the [tex]z[/tex] coordinate is determined by [tex]|AB|=d[/tex]). Again you can use the fact that [tex]A[/tex] and [tex]B[/tex] lie on a sphere, to determine [tex]X_2[/tex].

Finally you can use the fact that [tex]|AC|=r[/tex], and that the formula for the distance between two points [tex](x_1,y_1,z_1)[/tex] and [tex](x_2,y_2,z_2)[/tex] is [tex]\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}[/tex] by Pythagoras theorem. This will give you a formula involving [tex]R, r,[/tex] and [tex]d[/tex], rearranging gives the formula
[tex]2R = r\sqrt{\frac{4r^2-d^2}{3r^2-d^2}}[/tex].

Hope this helped,

R. Baber.

Re: Diameter of a sphere

Postby Guest » Tue Sep 20, 2016 9:43 am

Dear friend,
Not only you helped me with your solution, but you explained everything in a crystal clear way, using a very correct and understandable English language, as opposed to many others, who sometimes reply giving a correct solution but without any explanation whatsoever!!

Congratulations and thank you so much!

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