Cecil, W5DXP wrote:
"Such an antenna could be fed with 1/4 WL of 600 ohm open wire line and
achieve an impedance of 60 ohms looking into the matching section."

Exactly.

The actual impedance of the 1/4-wave folded dipole is an inverse
function of the wire diameter. It can be chosen to present 6000 ohms.
Arnold B. Bailey has studied the behavior, developed the formulas, and
published graphs for the user to choose wire sizes for folded dipoles of
either 1/4-wave or 1/2-wave overall length. He does a sample calculation
and showes his work for one particular 1/4-wave folded dipole.

On page 414 of "TV and Other Receiving Antennas" Bailey writes:
"At first resonance (where the total wire length is 1/2-wave and overall
length is 1/4-wave), the current in the continuous rod (entire wire
length) approaches a constant value, varying from a maximum in the
center (farthest from the drivepoint) to about 0.7 of maximum at the
ends. For this reason the radiation resistance RR approaches the value
for a constant-current rod. The practical value is approximately
RR=570(l squared), where l is the nominal over-all length and equals
0,25 wavelength, thus resulting in a value of 35.6 ohms."

The 1/4-wave folded dipole shares some of the characteristics of a
short-circuited stub. It is a resonant circuit element which converts
the low impedance at its short-circuit into a high impedance at its
feedpoint.

If the 1/4-wave folded dipole is a little short of 1/2-wave around its
perimeter, it presents an inductive reactance. If a little longer
electrically than 1/2-wave around, it presents a capacitive reactance.

Best regards, Richard Harrison, KB5WZI