Thought Experiments and π

    Suppose in a thought experiment I imagine a perfect rod that is 1 inch long, and then another one that I attach to it. Now I have a rod that is exactly 2 inches long — not approximately 2 inches but exactly 2 inches. I can create a rod of any rational length that is perfectly known.
    Now suppose I choose any integer. I can represent an arbitrary length. For example, suppose I choose 47 inches as the circumference, C, of Humpty Dumpty at the center of his belly.
    Now I bend this object and join the ends in such a way that it forms a circle of length 47 inches. In this case the diameter would be 47 / π , but I’m not allowed to imagine it because a picture has a rational length.
    π cannot be defined as an exact length but only the limit of a series such as: 4 (1- 1/3 + 1/5 – 1/7…). But I can see the length looking across the circle.
    Next, with Humpty Dumpty’s cousin, Harrumph, I take a different approach. I find an idealized machine to make an X-ray image using computerized axial tomography (Cat Scan). The slice through the center of the belly shows a diameter of 15 inches. The Cat Scan machine is connected to an idealized 3D printer. The printer is assigned to make a belt for Harrumph’s waist size. It would be 15 π inches. But even though it is an idealized and perfect machine with no tolerance for error, it cannot do it because it has no image for π. It is not allowed to use an approximation.
    If I can see a circle, I cannot see a diameter. If I see a straight line length, I’m assuming it’s rational because I can see all of it at once without error. Would it be legitimate to see a length and assume that it is π inches long? If I bend it into a circle, the the diameter is 1 inch. The question then is: what am I looking at when I see the π length? Whatever it is, it’s stable. On the other hand, I can look at a 1 inch length and spin it at its center to draw a circle. Then, I can break it open and lay it flat. In this case, the point where I cut it is undefined, isn’t it? Unless, I define the length of the circle as (π – k) + k and I break the circle within the length k.
    I’m going in circles again. Oh well.

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