Cart

Wheel No Slip Force [F_NS] when springs/suspension is used.

Specify the Spring Normal Force [F_SpN]. This [F_NS] formula only applies when Springs/Suspension selected and No. Driven Wheels < No. Wheels.

Radial Force [F_R] applied to the Driven Wheel bearing is the sum of two vectors - Accel Force [F_Acc] and Spring Normal Force [F_SpN].

Wheel No Slip Force [F_NS] when fixed wheels/axes.

In this case, gravity force as seen by each driven wheel and Applied Traction Force [F_Tr] are used.

Radial Force [F_R] applied to the Driven Wheel bearing is the sum of the three vectors - Accel Force [F_Acc], force due to gravity [F_g] and Applied Traction Force [F_Tr].

Wheel No Slip Force Margin [W_FNS%] is the margin/reserve of the application Wheel Output Force of One Driven Wheel [F_DW] wrt the Wheel No Slip Force [F_NS] per the following formula:

The most common question about the Cart mechanism is how to configure it for an AGV, which typically has two gear motors each driving a wheel, along with 2 caster wheels.

Therefore, it has 4 wheels in total with 2 driven wheels, as configured below.

A drive train with two gear motors driving the Cart as shown on the Main form, with number in parallel, n=2, for the gear motor, which aligns with the number of driven wheels, nDW=2.

If a gearbox is driving the driven wheels, then often its output bearing is supporting all forces including the acceleration and gravity forces. The force due to gravity includes the total mass of the mechanism including the cart, payload and wheels.

Three component vectors combine to form the total radial load on the gearbox output bearing:

1. Gravity
2. Acceleration
3. Traction force (if there is a clamping force say for a wheel on a track to improve traction)

For example, for simple math, assume no friction losses, and the driven wheels support 100% of the moving mass.

With an incline of 0deg (horizontal), a Cart mass of 100kg, a Payload of 400kg, and 4 wheels each 3.1886kg, the total mass is 512.75kg, which results in a gravity force (dwonward) of 5,030N.

With an acceleration of 1m/s2, the acceleration force is 512.75N (horizontal).

Therefore, these two force vectors combine for SQRT(5030^2 + 512.75^2) = 5,056N total. And since we have 2 driven wheels, then the max radial force seen at each gearbox output bearing is 5,056N / 2 = 2,528N

The resulting calculated radial load is shown in the Gearbox Shaft Loads form.

Rolling friction can be defined as an overall efficiency value under the [Drive] tab, or as a friction coeffcient under the [Traction] tab.