The Physics

How fast do you want the robot to go? How quickly do you want your robot to accelerate to its top speed? How much torque do you need?

Let us make our lives easy and assume linear acceleration.

In physics, $v(t)=v_{0}+a\times t$, in which $v(t)$ is the velocity at time $t$, $v_{0}$ is the initial velocity in units of meters per second and $a$ is acceleration in units of meters per second squared. The amount of force to accelerate your robot is $F=m\times a$, in which $F$ is force in units of Newton (N) and $m$ is mass in units of gram (g).

For example, if your robot has a mass of 1kg, you want the top speed to be $0.2\mathrm{m}\mathrm{s}^{-1}$, and you want it accelerate to top speed in 0.5s, the required force is $F=1\mathrm{kg}\times \frac{0.2\mathrm{m}\mathrm{s}^{-1}}{0.5\textrm{s}}= 0.4\textrm{m}\textrm{s}^{-2}=0.4\textrm{N}$. How much is a Newton (N)? A ``pound'' is the force of a 0.4545kg at one G. One G is $9.8\mathrm{m} \mathrm{s}^{-2}$. Therefore, a ``pound'' is approximately 4.45N. In other words, 1N is about 0.22 pounds. That's not much.

How can a robot's wheels exert 0.4N? Assuming the robot is driven by two drive wheels, each wheel only needs to push at 0.2N at the axle. Intuitively, the large the wheel, the more torque it needs to deliver the same amount of force at the axle. Torque is $\tau =Fl$. If we use a wheel has a radius of 30mm, the required torque becomes $\tau = 0.2 \mathrm{N}\times 0.03\mathrm{m}=0.006\mathrm{Nm}$.

Most stepper motors are specified by its detent and holding torque. The holding torque is easy to explain: this is the amount of torque to force the axle to turn one step when the motor is energized. The detent torque is the amount of torque that a motor has to turn the axle. This is what we need to know. Most stepper motors have both the holding torque and detent torque specified. Surplus motors, however, seldom has any specifications. You will need to purchase one and find out by testing.

In the case that torque is specified, it is often specified in oz-in (ounce-inch). How does an ounce-inch relate to newton-meter? $1\mathrm{oz}\mathrm{in}=\frac{1}{16}\mathrm{lb}0.0254\mathrm{m}$. 1 pound is 4.45N, so 1 oz-in is $\frac{4.45}{16}\mathrm{N}0.0254\mathrm{m}$, which computes to 0.00706Nm. In our example, we need 0.006Nm, which translates to less than 1 oz-in.

Some other times, torque is very incorrectly specified in g-cm (gram centimeter). The unit should have been g-G-cm (gram at one G centimeter). How does one 1 g-G-cm relate to N-m? It is $0.001\mathrm{kg}9.8\mathrm{m}\mathrm{s}^{-2}0.01\mathrm{m}$, which computes to $9.8\times 10^{-5}\mathrm{Nm}$. An oz-in is, therefore, 72.04 g-G-cm.

Note that our computation does not include any friction and angular momentum of the wheels. Furthermore, it assumes that the torque remains the same from stationary to full speed. In reality, a design needs a lot of margins. In this case, it is not unreasonable at all to specify stepper motors with at least 3 to 4 oz-in of torque.