I'm trying to work out the wheel force and torque required for a TWIP robot, so that I can size a motor.
I've calculated a maximum traction force of $\small6.51\mathrm N$. My understanding is that a torque force at the wheels of up to and including $\small6.51\mathrm N$ can be applied, to drive the robot without the wheels slipping. This would give the robot a maximum acceleration of $\small3.92\mathrm{ms}^{-2}$
So, assuming I wanted to achieve the maximum force to drive the robot, and hence the maximum acceleration (assuming pendulum is balanced), I would need a wheel force of $\small6.51\mathrm N$.
There is also resistance against the direction of motion/driving force of the robot, in the form of rolling resistance and aerodynamic drag. From what I've read rolling resistance (a type of static friction) is a resistive moment to wheel rotation, which needs to be overcome by the wheel torque force in order to produce acceleration.
I've calculated a rolling resistance value of $\small0.16\mathrm N$. The robot is intended for indoor use but in case I take it outside I calculated an aerodynamic drag value of $\small0.14\mathrm N$, using an average wind flow velocity of $\small3\frac{\mathrm m}{\mathrm s}$ for my location. Taking these resistive forces into account I calculated a wheel force of $\small6.81\mathrm N$ and axle torque of $\small0.20\mathrm{Nm}$, for maximum acceleration of the robot.
I've considered the maximum torque exerted by the pendulum i.e. when it's pitch angle/angle of inclination is at +- 90° from the stable vertical position at 0°.
This torque needs to be matched (or exceeded) by the torque/moment exerted about the pivot by the wheel force, accelerating the robot horizontally. The wheel force and axle torque required to stabilise the pendulum I've calculated as $\small13.7340\mathrm N$ and $\small0.4120\mathrm{Nm}$ respectively, and an axle torque of $\small\approx0.2\mathrm{Nm}$ for one motor. I ignored rolling resistance and aerodynamic drag for these calculations.
The motor will be a brushed DC motor, so I think $\small0.2\mathrm{Nm}$ should be 25% or less of the motor's stall torque.
Can you please tell me if this is correct?
Here are my calculations and FBD:$$$$
Maximum tractive force
$\begin{align} F_{t(max)}&=μN\qquad\qquad\qquad\qquad\qquad\qquad Mass\,of\,robot: 1.66\mathrm{kg}\\ &=(0.4)*(16.28\mathrm N)\qquad\qquad\qquad\,\,\,\,Weight\,of\,robot:16.28\mathrm N\\ &=6.51\mathrm N\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,Number\,of\,wheels:2\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,Wheel\,radius: 0.03\mathrm m\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,Mass\,of\,pendulum:1.4\mathrm{kg}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,Distance\,from\,axle\,to\,pendulum\,COM:0.2575\mathrm m \end{align}$
$F_{t(max)}$: Maximum tractive force
$μ$: Coefficient of friction
$N$: Normal force at wheel$$$$
Maximum acceleration of robot
$\begin{align} a_{r(max)}&=\frac{F_{t(max)}}{m}\\ &=\frac{6.51\mathrm N}{1.66\mathrm{kg}}\\ &=3.92\mathrm{ms}^{-2}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$a_{r(max)}$: Maximum acceleration of robot
$F_{t(max)}$: Maximum tractive force
$m$: Mass of robot$$$$
Rolling resistance force
$\begin{align} F_{rr} &= C_{rr}N\\ &=(0.01)*(16.28\mathrm N)\\ &=0.16\mathrm N\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$F_{rr}$: Rolling resistance force
$C_{rr}$: Rolling resistance coefficient
$N$: Normal force at wheel$$$$
Drag resistance force
$\begin{align} F_{d} &= C_{d}\left(\frac{ρ*v^2}{2}\right)A\\ &=1.28\left(\frac{1.2\frac{\mathrm {kg}}{\mathrm m^3}*(3\frac{\mathrm m}{\mathrm s})^2}{2}\right)0.06\mathrm m^2\\ &=0.14\frac{\mathrm{kg}\cdot\mathrm m}{\mathrm s^2}=0.14\mathrm N\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$F_{d}$: Drag resistance force
$C_{d}$: Drag coefficient
$ρ$: Mass density of fluid
$v$: Flow velocity of fluid relative to object
$A$: Reference area/projected frontal area of object$$$$
Wheel force/tractive force for maximum acceleration of robot
$\begin{align} F_t-F_{rr}-F_d&=ma_{r(max)}\\ F_t-0.16\mathrm N -0.14\mathrm N &=(1.66\mathrm{kg})*(3.92\mathrm{ms}^{-2})\\ F_t&=(1.66\mathrm{kg})*(3.92\mathrm{ms}^{-2})+0.16\mathrm N +0.14\mathrm N\\ &=6.81\mathrm N \end{align}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ OR $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
$\begin{align} F_w&=F_{t(max)}+F_{rr}+F_d\\ &=6.51\mathrm N +0.16\mathrm N +0.14\mathrm N\\ &=6.81\mathrm N\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$F_t$: Tractive force
$F_{rr}$: Rolling resistance force
$F_{d}$: Drag resistance force
$m$: Mass of robot
$a_{r(max)}$: Maximum acceleration of robot
$F_w$: Wheel force
$F_{t(max)}$: Maximum tractive force$$$$
Axle/wheel torque for maximum acceleration of robot
$\begin{align} T_a&=F_w r\\ &=(6.81\mathrm N)*(0.03\mathrm m)\\ &=0.20\mathrm{Nm}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$T_a$: Axle/wheel torque
$F_w$: Wheel force
$r$: Wheel radius (lever arm length)$$$$
Maximum torque exerted by pendulum
$\begin{align} T_{p(max)}&=F_p r\\ &=(1.4\mathrm{kg}*9.81)*(0.2575\mathrm m)\\ &=3.5365\mathrm{kg}\cdot \mathrm m\\ &=3.5365\mathrm{Nm}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$T_{p(max)}$: Maximum torque exerted by pendulum
$F_p$: Force applied to pendulum
$r$: Distance from axle to pendulum COM (lever arm length at +/- 90° )$$$$
Wheel force to stabilise pendulum
$\begin{align} T_{p(max)}&=F_w r\\ 3.5365\mathrm{Nm}&=F_w*(0.2575\mathrm m)\\ F_w&=\frac{3.5365\mathrm{Nm}}{0.2575\mathrm m}\\ &=13.7340\mathrm N\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$T_{p(max)}$: Maximum torque exerted by pendulum
$F_w$: Wheel force
$r$: Distance from axle to pendulum COM$$$$
Axle/wheel torque to stabilise pendulum
$\begin{align} T_a&=F_w r\\ &=13.7340\mathrm N*(0.03\mathrm m)\\ &=0.4120\mathrm{Nm}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$ $\therefore$
$\begin{align} T_{a(one\,motor)}&=\frac{0.4120\mathrm{Nm}}{2}\\ &=0.2060\mathrm{Nm}\\ &\approx0.2\mathrm{Nm}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}$
$T_a$: Axle/wheel torque
$F_w$: Wheel force
$r$: Wheel radius (lever arm length)
$T_{a(one\,motor)}$: Axle/wheel torque for one motor$$$$
FBD
