I'm trying to make a robot with two legs, the robot will have four
wheels on each foot.  It takes a step or stride by rolling.  A company
wants to sell me a stepper that generates 550 ounces-inches of torque
the half of the robot weighs 150 pounds. What is the formula for
calculating the poundage that the motor can push using ounces-inches
of torque.  The motors must push 150 pounds each. Stepper motors use
ounces-inches of torque to describe them I need a formula to convert
that to pounds of force or push. Each foot has one motor.

Andrew

 In his comment mirrormn-ga provided good analysis of your issue.
 Since you did not expired the question, you perhaps want some links
 to help you to design the locomotion of your robot.

 The rolling friction 
 Depends on surface, tires, as well as surface 

http://www.school-for-champions.com/science/frictionrolling2.htm
and more like this
://www.google.com/search?hl=en&ie=ISO-8859-1&q=rolling+friction&btnG=Google+Search

Here is a good description of Torque / force realationship
http://science.howstuffworks.com/fpte4.htm

It is essentially radius of your wheels (unles you use gears ot belts)
which will convert torque into a force by:

 Force * radius = Torque

The forward force will accelerate the mass of your robot forward --
and so design details, such as 
a) do separate steps (stopping in between) 
or
 b) ride as a car

will afffect the power vs speed trade-off of your robot.

May be a good idea to do a bit more reading:

http://www.omega.com/literature/transactions/volume3/force.html
or
try SEARCH TERM:robot drive trains
that is, look at links at 
://www.google.com/search?hl=en&ie=ISO-8859-1&q=robot+drive+trains&btnG=Google+Search
e.g

 http://www.omega.com/literature/transactions/volume3/force.html
for more info. 

Hedgie

There is no direct conversion factor, because torque and force are two
different types of quantities. Basically, torque = force * distance
(which is evident from the units "ounce-inches"). A 550 ounce-inch
motor will be able to develop a force of 550 ounces at a distance 1
inch away from its axle (550 ounces * 1 inch) or 225 ounces at a
distance 2 inches away from its axle (225 ounces * 2 inches = 550
ounce-inches). Converting ounces to pounds, the motor you talked about
could also be measured to be able to develop 34.275 pound-inches of
torque.
However, this might not be as big of a problem as it sounds. It is
hard for me to envision the design of your robot, but from what I can
tell, it is a rolling machine. If the wheels are only needed to roll
the robot along, and not actually lift the whole robot up, 34.275
pound-iches of torque could be plenty. In the case of a rolling
machine, the motor only needs to overcome the force of friction
between the motor's axle and its bearings, which should be fairly
insignificant. After that, more torque would only allow the robot to
accelerate faster, have more success going over bumps and up hills,
etc.

close but tourque is defined in a cylindrical coordiante system, it is
Force*arclength or force*radial_distance*angular_displacement