When I understand you correctly, the series component of the dipole
input impedance is 200 Ohm? I would expect value lower then 70
Ohms.

the series component is probably just 25 ohm or so (due to the effect of the
parasitic elements), but it is brought up to 200 ohm by shortening the dipole
length and putting an hairpin in parallel (very well known technique to increase
impedance). As a matter fact it matches very well with a 4-to-1 balun

When 200 Ohms real component of Zi, you used a series hairpin to
cancel the capacitive component.

No, the hairpin is in parallel to the dipole

You can guess the loss by calculating
the AC resistance of the pin.

Assuming homogeneous current distribution around the circumference,
you can calculate the AC resistance with (assuming length of hairpin
0.15 lambda):

Rac = 80*10^-9*sqrt(f)*le/d , Rac in Ohm, f in Hz, le (wire length)
in m, d (diameter of wire) in m, valid for copper. For other
materials (non magnetic), when conductance is factor 4 less (so 4
times higher spec. resistance), Rac doubles.

When the result for your hairpin 200 Ohm, the loss is negligible.

In case of a parallel pin, you can use same technique, but in that
case you have to calculate the inductance of the pin. With the
inductance you know the reactance and the current that runs through it
(for example with 100Vrms across it). With Rac you can calculate the
power loss.

Having all data available, it is clearly possible to calculate losses using the
Ohm law. Mine was a question of more general nature. Does anyone have direct
experience on whether the extra loss of a 200-ohm hairpin match (with respect to
the more conventional 50-ohm hairpin match) is in pratice noticeable?

Thanks & 73

Tony I0JX