That Dr. Math thing looks confusing to me too. Your analysis looks close enough, although I don’t know why you can assume the bottom leg of your triangle will be r-1.

Here’s another way you might be able to do what you want: Each section of the Cassini globe is an approximation of a “lune.” A lune is the section of the surface of a sphere that looks like an orange wedge. That is, it goes from pole to pole and is basically a wedge of the sphere. Lunes are determined by the angle they make at each pole. So a 90 degree (pi/2 radians — oops, sorry. Didn’t mean to cause convulsions) lune would cover 1/4 of the whole sphere.

Yeah. Anyway, a nice thing is that the area formula for a lune is very nice. A lune whose angle is theta (in radians — oops) drawn on a sphere with radius r will have area:

lune area = 2 * theta * r^2.

Right. So if you can measure the area of one of your Cassini sections (in square inches) and measure the angle of the lune (the angle at the top of the section) AND convert this angle to radians (ooops! radians = (degree angle) * (pi/180) ) then you can use this lune area formula to calculate what the radius of the sphere must be.

I thought I saw a proof of the Pythagorean Thm via folding before. But I never heard of this Bhaskara II fellow before. Do let me know if you learn or know more!