Richard Harrison wrote:

Tom Donaly, KA6RUH wrote:
": how do you take the gradient of the current at a point on a
transmission line?"

Not sure I understand the question. Gradient is the rate of change and
that`s the derivative of the current at a given point. Over a certain
path it is the difference between the path ends and can be averaged for
the path.

For convenience, Kraus has collected transmission line formulas. I`m not
a typist so I`ll just say they are near the end of the new edition, page
890. In the 1950 edition they can be fornd on pages 506 and 507, also
near the end of the book.

Work out your own example.

Best regards, Richard Harrison, KB5WZI


Thank you, Richard, you just made my point.
73,
Tom Donaly, KA6RUH