Wheeler Radius For Tosses The radius of an arc is equal to half of the rise added to the ratio of chord length squared over eight times the rise.

Radius from chord length and rise You may wish to confirm this by filling in the details of this picture on the right:

For the picture on the left: Take the 32" high arc OBG which requires 0.813 seconds for the toss so light has traveled m(OH) = 9,595,719,169 inches. ( m(OH)² + 16² )^0.5 corresponds to the chord length for a rise of m(DB) = 32 inches. Then calculating the radius m(JB) of this arc using the method above gives m(JB) = 3.59679 times ten to the 17th power inches. That is 96.6 % of one light year.

The odd outcome, to be seen by you, is when you calculate the radius m(IA) for arc OAE, the 16" height of the unit toss, you get ..... [ oops - I almost gave it away - You do it and see!!!! ].

The straight lines ( geodesics ) of spacetime near the surface of the earth are the free fall lines of these juggling paths. Curvatures of all these paths (the reciprocal of radius of curvature) are nearly equal to 1/(.97 light year).

• The Talk Outline


"Out Juggling In My Class" by William V. Thayer
Juggleometry History

copyright © 1998 Wm. V. Thayer, All Rights Reserved

Wheeler's Curvature For Spacetime Tosses



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