Suggestions:
1. If you use a control block diagram, then the output signal from the capacitor to be converted must pass through the same value capacitor division, and then be multiplied by Resistance R, i.e. [2(PI)fC/2(PI)fC]xR=R
or

----|2(PI)fC|-----|1/[2(PI)fC]|------|R|------ = ----|R|----

This is good for the AC signal.
2. If you have the DC signal, then the capacitance has its own resistance in the parallel connection, and some resistance in the series connection depending on its leakage current.

If you are looking for capacitive reactance (effectively AC resistance), the formula is 1/2(PI)FC.
PI = 3.14....
F = frequency in Hertz (or Cycles Per Second)
C = capacitance in Farads

A 1uF capacitor would have a reactance of about 2653 ohms at 60Hz.

The answer that Bill gave you is correct, you can not convert either from capacitor to resistor or from resistor to capcitor, but you can know the capacitive reactance which is a sort of resistive capacitance, but only works with ac, and is frequency dependent.
There is quite a lot of theory on inductives and capacitives reactances, which together with any resistance involved,(for ac only), that conform the term "Impedance".

Suggestion to the above statement:
In series of control blocks, one may add multipliers 2x(PI)f, if only the capacitance is considered and not capacitive reactance. I thought that it was evident, but I realize the the more accurate information is needed for its clarity. Specifically,

---|1/[2(PI)fC]|-----|2(PI)f|-----|2(PI)fC|-----|1/[2(PI)f]|-----|R|----=

-----------|1/C|---------------------------|C|----------------------|R|----=-----|R|------
However, the question is what is this actually needed for since the above reasoning may have a very little practical use, if any.

hi cadman,
  The guys above have pretty well covered the calcs (ps jbartos I have copied your calc I,ve been after that one for a long time)
  But just to be controversial, remember if we're using smaller caps in a signal or lower power circuit it's odds on they are +/- 20%. The bloody big ones in banks etc are different (and you pay for it).
  So the rule of thumb is good. For ac circuit do as the guys have suggested, Just don't get too excited about decimal places (anybody got a 1.0234 uf cap on the bench?)
  If its a small signal cct calc the dc current as zit, ac as per above.
IF it's a big cap for a bank then use jbartos's formula. If it's a filter for gp power supply about 1000uf per amp
best of fun
Don

You cannot convert an apple into an orange and you cannot convert a capacitor into a resistor.  There is no conversion.

I am a little mystified by this post - can cadman please expand the original question to clarify what conversion is required?
I certainly agree with what BillJ, algarin & Wirenut have said - there is no "conversion" between resistance and capacitance.  What jbartos appears to be saying is that R=R=R (a rose is a rose is a rose?), which is true but doesn't seem to have any bearing on any "conversion".
It may help to know what the origin of the question was - back to you, cadman.

What about dimensional calculus/conversion? How does one convert # to lb? What about (#/#)xlb=lb, assuming that #=lb? This is in essence lb=lb or a rose is a rose. What about signal processing? It is normal to multiply, divide, square, etc. the signal. This is somewhat what was done above.

To jbartos -
I am still mystified.  You may have processed the signal, but in no way does that processing "convert" C into R.  Your calculation, if I understood it, goes as follows -
- Start with Xc = (1/(2(PI)fxC)
- Multiply (1/2(PI)fC) by 2(PI)f = 1/C
- Multiply (1/C) by (2(PI)fC) = 2(PI)f
- Multiply 2(PI)f by (1/(2(PI)f)= 1
- Multiply 1 by R = R

As regards conversion of # to lb, please explain what you mean by "#" and by "lb".  As far as I am aware, both are common abbreviations for the word "pound", so your conversion is saying take pounds, divide by pounds and then multiply by pounds to get pounds.  Am I missing something here?

1. In some cases # symbol may or may not be acceptable, e.g. some old typewriters were missing various symbols/characters.
2. There is no question regarding equalities that were stated by you and others. Thing is, if you have an input signal and the need to have it converted, the output signal may look differently. Similarly, if you have a capacitance C and it needs to be converted to R, for some unknown and still pending reason from the author of the original posting, how else to do it? Also, notice that "convert" may mean to "alter the physical or chemical nature of properties of, especially in manufacturing." See Webster Dictionary.  The "Conversion" does not necessarily mean the equality between the input and output of the conversion process to me.

can anyone show me the formulas for calculating capacitance for 3 and 4 conductor cables (shielded and unshielded)?
Billtk

If you know the Belden equivalent, they list the capacitance per foot for a number of their cables.  I believe they have a catalog that you can down load

Suggestions: References:
1. Wadhwa C. L., "Electrical Power Systems," Second Edition, John Wiley and Sons, New York, 1991
2. Simmons R. M., "Cable Geometry and the Calculation of Current-Carrying Capacity," AIEE Transactions, Volume 42, pp 600-620, June 1923
3. Insulated Conductors Committee Task Group 7-39, Cost of Losses, "Loss Evaluation for Underground Transmission and Distribution Cable Systems," IEEE Transactions on Power Delivery, Vol. 5, No. 4, November 1990
4. Stevenson, Jr, W. D., "Elements of Power System Analysis," Third Edition, McGraw-Hill Book Co., 1975
Unshielded cables (3-conductor, 4-conductor, etc.) are special cases of shielded cables, i.e. the capacitances between cable conductor and shield do not exist (or each capacitor has zero value causing the capacitive reactance to be equal to the infinity).
Three-conductor shielded cable:
Reference 1 includes "Capacitance of a 3-Core Cable" section on pages 200-203. It depicts 3-conductor shielded cable capacitances as a delta capacitor, C2, connection among three conductors which is circumscribed by a circle representing the cable shield. Each delta end is connected to the shield over the capacitor, C1. The next step transfigures the delta capacitor connection into the three end star connection having C1 capacitor link to the cable shield. The star transfigured capacitances become C0=C1+3C2 between each star end and the fictitious neutral (C0 is the capacitance per phase). Normally, C0 is determined by measurements. Reference 1 gives an approximate value of C0 for circular conductors as:
C0={0.0299 x er / [ln(1+((T+t)/d ) x (3.84 - (1.70 x t/T) + (0.52 x (t/T)**2))]} in microFarads/kilometer
Where er is relative permitivity of the dielectric, d is conductor diameter, t is cable belt thickness, T is conductor insulation thickness.
The C0 equation is from Reference 2 (Equation 18 on page 613) having the error less than 2 % for most cases. Reference 1 refers to the Simon formula and does not provide reference. It is probably meant to be Simons, Reference 2.  Generally, Reference 2 includes the following formulae for the cable geometric factor:
1. Mie's formula
2. Russell's formula No. 1
3. Russell's formula No. 2
and Atkinson experimental curve.
Reference 3 presents the cable (assumed 3-conductor, shielded) capacitance C:
C={7.354 x et x 10**(-12) / log10[(2T + Dc)/Dc]}, in Farads/foot
where T is insulation thickness, et is dielectric constant, Dc is conductor diameter, log10 is base 10 logarithm, ** means the exponent sign.
Four-conductor shielded cable
Instead of capacitors connected in delta, as it is for the three conductor cable, there will be a square connection with two diagonal capacitances between the diametrically placed conductors in the cable (assuming that the fourth conductor is a current-carrying conductor). The four corner capacitor mesh can be transfigured into the four-end star, Reference 4 page 157 equation 7.7.
Please, notice that the cable construction, i.e. details, have a significant impact on the capacitance accurate determinantion, selection of a right formula, etc.