MIDNIGHT RIDER

TO STEVE ALBERS:  I had only two beers all afternoon, and I’m laying here thinking  Screen shot 2016-08-21 at 7.30.18 AMabout the radius of curvature of my very near-field, and how two weeks ago
your diagram showed shockingly even hammer springs all the way across
!!!!!!!!! THIS IS COOOOOOOOL. [[FOLKS, this plot shows 5/6 of the whole near-field.]]
I SHALL COMPOSE THIS RIGHT NOW. We write the dx/dy, slope of
this field. as: ksin(theta) cos(theta)/ y^2, and “at first” we may say the radius of
curvature is the derivative of this wrt ‘x’. I am only a little attled at the moment, but we get an expression if we SET THE SLOPE EQUAL TO 1. This is the 45-degree point in the sharp bend above the axis, and it would seem to be changing.HOWEVER. YOU HAVE NOW SPENT ENOUGH TIME to know how at the pole radius equal to ‘k’ is also equal to ‘y’. As we look outward, this declines with the ellipse, down to a value of k^2 at the ring edge. WOW, wow, wow, the opposite tendency, yah??? CONGRATULATIONS TO US BOTH, your pic almost all the way across, with max ‘r’ at 2500k, showed odd regularity, and this is why!!!!!!!!!!

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