Even though the spur gear tooth is moved by one rotation, there is a force associated with it which is related to the friction (or in general the load on the spur gear).
To simplify things, let's assume that the system is in steady state (i.e. no angular acceleration). In that case you don't need to worry about inertial effects.
So, lets assume that the torque needed on the spur gear axis is $M_s$. Then the tangential force on the spur gear (which is causing rotation) will be equal to:
$$F_{s,t} = \frac{M_s}{R_s}$$
where:
- $F_{s,t}$ is the Tangential component on the spur gear
- $M_{s}$ is the required torque
- $R_{s}$ is radius of the spur gear
The tangential component of the force on the worm gear $F_{w,t}$ will be equal to the spur gear $F_{s,t}$.
The following image shows the forces that are developed on the worm gear.
Figure 1: Forces of worm gear (source: khdgears
It is impossible to have a worm gear pair mesh without radial and axial forces. However,
the other forces (radial $r$ and axial $x$) will be proportional and will be depended on the angles of the worm gear.
Now the force that dominates the torque on the worm gear is transverse to the plane of the spur gear. Notice that $F_{t,1}$ and $F_{x_2}$ are collinear, and $F_{t,2}$ and $F_{x_1}$ are collinear. To put things into perspective, in this situation, the worm gear is gear 1, and the spur gear is no 2.
So, although, the tangential force of the spur gear $F_{s,t}$ becomes axial on the worm gear, the worm gears tangential force is still proportional to $F_{s,t}$. Therefore:
$$F_{w,t}\propto F_{s,t}$$
However, now the torque required to keep the system moving would need to be equal to (actually its higher due to losses but to simplify things let's assume no losses):
$$M_w = F_{w,t}\cdot R_w$$
where:
- $R_w$ is the radius of the worm gear.
So in the end:
$$M_w \propto \frac{M_s}{R_s}\cdot R_w$$
So as you can see, increasing the diameter will result in higher torque requirements.
