I've recently done some NEC-2 (MultiNEC) modeling of folded dipoles
which might help answer some of your questions. Translating the
results to folded monoploes should be fairly straightforward.

The model is a half-wave folded dipole for 14.2 MHz in freespace,
resonant at 33.15 feet using #18 wire with 2 inch spacing. The
center-fed input impedance is 289 - j0.01, which is 4 times the
resonant impedance of 72 ohms for a conventional dipole. A folded
1/4-wavelength monopole would have half that impedance, or about 144
ohms.

Examining the R-X curves for this dipole shows that it has
characteristics very similar to a 3/2-wavelength dipole, operating at
its third harmonic, and on a relatively low-slope part of the curves,
indicating a low Q and good bandwidth, similar to a fat dipole.

Shortening the antenna increases capacitive reactance, as might be
expected. However, input resistance *increases* as the length
decreases, which is contrary to our experience with common
1/2-wavelength dipoles. This is because we're on the high side of
full-wave resonance, where very high resistance values exist at its
peak. As we shorten the antenna, we're climbing the full-wave
resistance curve, which peaks when the antenna length is 22 feet. If
we further shorten the antenna past full-wave resonance, we now begin
experiencing a "normal" decrease in resistance as we "slide" back down
the low side of the full-wave resistance spike. However, capacitive
reactance has now quickly changed to inductive reactance as we crossed
full-wave resonance.

If we continue to shorten the folded antenna length, we come to a
length of about 17 feet where the input impedance is 50 + j2000 ohms.
Notice that the impedance is *inductive*, not capacitive as we are
accustomed to seeing with ordinary short dipoles. The inductive 2000
ohms can be cancelled with a series capacitor (or other suitable
matching network). Q has increased (because we're on a relatively
steep part of the R-X curves) and bandwidth has narrowed considerably
from the resonance at 33.15 feet.

So, by reducing the length of the 1/2-wavelength folded dipole from
33.15 feet to 17 feet, we have a 50 ohm resistive impedance by
matching the inductive reactance with a capacitor (or split capacitor)
instead of the usual lossy, low-Q loading coils. Gain and patterns
appear to be the same as a conventional dipole.

Translated to a monopole, the length would be a little more than half
the dipole's 17 feet, to boost feed point resistance from 25 ohms to
50 ohms. My guess is (I haven't modeled it) that this antenna
functions much like a 3/8-wavelength monopole, although much shorter.
Actually building this antenna and placing it the real world will
obviously change the above values.

Unfortunately, it doesn't appear that any combination of element size
and spacing will offset the need for impedance matching with the
shortened folded dipole or monopole.

I hope this makes sense. I'm sure Roy, Cecil, Tom, and others might
have comments/corrections that will be helpful to me and others who
are relative neophytes in the wonderful world of antennas.

Al WA4GKQ

Even better, is there some choice of the folded section wire diameters and
spacing that will give an inductance that will exactly offset the
capacitance due to shortness? So, then, is there a folded monopole of such
dimensions that the resistance is 50 Ohms (due to being shorter than 1/4
wave) with no terminal reactance (due to the inductive design of the
"transmission line" cancelled by the shortness of the antenna's
capacitance)?