# How does the diameter of a worm gear affect the amount of force needed to turn it

Can anyone help inform a discussion I'm having with fellow modellers on a model railway forum?

I have a locomotive powered by an electric motor turning a 20mm diameter worm gear meshing with a spur gear. The loco performs badly, it lacks power. I am planning to exchange the worm gear for one which is 6.5mm diameter as I'm convinced this will increase the amount of force the motor can exert on the spur gear.

Some on the forum say it will make no difference - one rotation of the worm will move the spur gear forward by one tooth and the effort needed is the same regardless of the diameter of the worm gear.

I've tried supporting my argument with rudimentary physics but I'm confused as to whether i use formulae associated with rotational inertia, torque or centrifugal force. None seems to be directly applicable.

Assuming all other factors are constant, how do I calculate the force needed to turn the two sizes of worm gear?

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.

• So, if we assume the force on the spur gear and its radius are constant, am I correct in saying that the force required to turn the worm is directly proportional to the radius of the worm. In my case, the radius of the smaller worm is more or less a third of the radius of the larger worm and so will require around a third of the torque? – Richard Bennett May 7 at 7:44
• you are right, but don't expect to be exactly one third. There is a large chunk related to friction losses, so my guess is that it will be between 3/5 and 1/2 of the large one. – NMech May 7 at 8:11
• @RichardBennett, since you are new to engineering SE, if you feel that your question has been answered you can accept one of the answers or upvote. – NMech May 7 at 9:20
• I think I've clicked the right symbol. Many thanks for your detailed response. I'm especially pleased because it supports my intuitive view of the situation which, until now, I've not been able to support with maths. – Richard Bennett May 7 at 12:21
• I think your last definition, of $R_W$ is supposed to be "radius of worm gear" . I edited it that way. – Carl Witthoft May 7 at 13:55