The most common current signal standard in modern use is the 4 to 20
milliamp (4-20 mA) loop, with 4 milliamps representing 0 percent of
measurement, 20 milliamps representing 100 percent, 12 milliamps
representing 50 percent, and so on. A convenient feature of the 4-20
mA standard is its ease of signal conversion to 1-5 volt indicating
instruments. A simple 250 ohm precision resistor connected in series
with the circuit will produce 1 volt of drop at 4 milliamps, 5 volts
of drop at 20 milliamps, etc:

Instrumentation is a field of study and work centering on measurement
and control of physical processes. These physical processes include
pressure, temperature, flow rate, and chemical consistency. An
instrument is a device that measures and/or acts to control any kind
of physical process. Due to the fact that electrical quantities of
voltage and current are easy to measure, manipulate, and transmit over
long distances, they are widely used to represent such physical
variables and transmit the information to remote locations.



With this in mind can you show me how to make some measurments?  And
how to make a math formula using the 4-20 milliamp scale with
pressure, temperature, flow rate, and chemical consistency such as ph
level?
 




----------------------------------------
| Percent of  |   4-20 mA  |   1-5 V   |
| measurement |   signal   |   signal  |
----------------------------------------
|      0      |   4.0 mA   |   1.0 V   |
----------------------------------------
|     10      |   5.6 mA   |   1.4 V   |
----------------------------------------
|     20      |   7.2 mA   |   1.8 V   |
----------------------------------------
|     25      |   8.0 mA   |   2.0 V   |
----------------------------------------
|     30      |   8.8 mA   |   2.2 V   |
----------------------------------------
|     40      |  10.4 mA   |   2.6 V   |
----------------------------------------
|     50      |  12.0 mA   |   3.0 V   |
----------------------------------------
|     60      |  13.6 mA   |   3.4 V   |
----------------------------------------
|     70      |  15.2 mA   |   3.8 V   |
----------------------------------------
|     75      |  16.0 mA   |   4.0 V   |
---------------------------------------
|     80      |  16.8 mA   |   4.2 V   |
----------------------------------------
|     90      |  18.4 mA   |   4.6 V   |
----------------------------------------
|    100      |  20.0 mA   |   5.0 V   |
----------------------------------------

You have a very good start here. I am not sure of how you are going to
apply this so, these are just few approaches you could work with.

If you have minimal electronics abilities and have money to put into
this, then check into National Instruments.  They produce a line of
computer interface boards that are mounted in your PC and simple
graphically programmed software to operate the system.  With this you
can calibrate your sensors so the output data is in the form of values
that make sense with what you are measuring.  Pressure can be in PSI,
Pascals, Bars and so on.  Also, consider what you want to do with the
data and the frequency at which you need to collect values.  The
National Instruments system would be able to storage your data to a
file that can be opened to a spread sheet and collect data points at a
frequency range of less than once a day to more than hundreds of times
a second.

If you do not have several thousand dollars to spend on this project
and the frequency at which you are collecting values is slow, say 10
times a minute or less then the following system may work for you. 
Connect a voltmeter across the precision resistor you mentioned in
your question. A digital multimeter can also be used to measure the
sensor current directly, thus eliminating the resistor.  In either
case you will need to calibrate the sensor.

To calibrate the sensor you will need to make at the least two
measurements, one for the maximum measurement you expect and the
second for the lowest measurement you expect.  The measurements must
be confirmed with known values, this is the critical part of
calibration. If you are trying to calibrate a temperature sensor then
use a thermometer that is in close contact with the sensor so, both
are at the same temperature. The classic method for temperature
calibration, places the sensor in melting ice water for 0deg C and
boiling water for 100deg C (this assumes you are at sea level).

When you take the calibration data you will have a voltage or current
value for a defined condition (a given temperature, pressure, pH,
ect.). You will use the difference to define a ratio of change in the
condition for a change in the measured value plus an offset.

The math:
Vmin = voltage or current measurement for the minimum condition
(temperature, pressure...)
Vmax = voltage or current measurement for the maximum condition
(temperature, pressure...)
Cmin = minimum condition (temperature, pressure...)
Cmax = maximum condition (temperature, pressure...)

R = (Cmax - Cmin)/(Vmax - Vmin)    example units deg C/volt, PSI/amps

To convert a measured voltage (Vmeas) (or current) to condition value
(Cmeas):

Cmeas = Cmin + (R x (Vmeas- Vmin) )


For example:

If we are calibrating a pressure sensor and a reliable pressure gauge
is connected to the same pressure source as the sensor.  The minimum
reading is at 0PSI the digital multimeter reads 1.854volts.  The
pressure is raised to 14.7PSI
and the meter reads 4.150 volts.

Vmin = 1.854     Vmax = 4.150    Cmin = 0    Cmax =  14.7

R =  (14.7 - 0) /(4.15 - 1.854)  =  6.402 PSI/volt

We then measure the pressure of another source with a sensor voltage
of 2.450volts.

Cmeas = 0 + (6.402 x (2.450 - 1.854))  =  3.816 PSI



Because sensors can be "nonlinear" make as many calibration
measurements between the minumum and maximum as feasible. You can use
this data to create a lookup table or plot a chart relating the
calibration measurements.  The more nonlinear you sensor is the more
the calibration points will differ from a straight line between the
minimum and maximum calibration points.  Nonlinearity adds error to
the measurement.

The comments on calibration applies to almost all methods of
collecting data from sensors.

A chart recorder is an old tech method and many were set up to
interface with 4-20mA sensor.  The Hobo data logger can also be used
to collect sensor signals. Data loggers collect sensor data at set
rate and cost from less than $100 to thousands.

I hope this is of help.

You certianly have done your homework on the circuit.  There are many
different ways to measure and many different choices of
instrumentation to perform the measurements.  A good one stop source
for many measurement needs from the sensor up to the software on the
computer to collect the data and everything esle assocaited can be
found at www.omega.com.  The folks there are very good at helping you
to identify a complete system to suit your needs.  From thier products
obviously but they have an excellent assortment.

I would reccomend you visit thier site - check out some of the
excellent documentation and illustrations on measurement systems -
then call them and they can make sure you purchase the correct
equipment which is compatible with your intentions and any existing
hardware.

My experinece with this company has been very good over the last five
years of using them for lab research and actual equipment design and
fabrication.

If you care to provide more detail on the actual situation you are
trying to measure I may be able to provide some advice on the type of
insturmentation to pursue.  The folks at Omega are quite capable of
doing the same.