Motion: Motors, Gears and Wheels

Robotics Fundamentals

Course 1 · Chapter 3 · Motion: Motors, Gears and Wheels

Robotics Fundamentals 2 designed the structure and moved the battery until the robot stayed on its wheels. This chapter makes it move. A robot's motion system has a chain of parts: a motor that turns electrical energy into rotation, a gearbox (or belt) that trades speed for torque, and a wheel that turns torque into a push against the ground. Each link has a number attached to it, and the design task is to choose them so that the robot reaches the speed you want, climbs the slope you want, and does not overheat the motor while doing it. Basic Electronics 9 covered the motor from the electrical side (the model, PWM, drivers). Here we meet the same motor from the mechanical side, and then size one for a small robot, with all the arithmetic shown.

The efficiencies, friction coefficients, rolling-resistance coefficients and motor constants in this chapter are typical, approximate values from memory. Real gearboxes, tyres and motors vary by make, wear, temperature and load, so use the figures to compare options and to check the order of size, and then read the datasheet (or measure) before relying on one. The formulas themselves are standard.

1. Torque, speed and power

Torque (τ) is a turning force: a force multiplied by the perpendicular distance from the axis (Robotics Fundamentals 2 called it a moment). Angular speed (ω) is how fast something turns. Power is the rate at which energy is delivered, and for a rotating shaft it is the product of the two. Everything else in this chapter follows from three relations and the fact that a gearbox cannot create power.

 Torque from a force at a radius:       tau = F r                 (N m = newton x metre)
 Angular speed from rpm:               omega = 2 pi n / 60       (rad/s; n in revolutions per minute)
 Rotational power:                     P = tau omega             (W = N m x rad/s)
 Linear power (the same power):        P = F v                   with  v = omega r      (so F v = (tau / r)(omega r) = tau omega)

 Units you will meet on datasheets:
        1 kg cm  = 0.0981 N m          1 N m = 10.2 kg cm
        1 oz in  = 0.00706 N m         1 N m = 141.6 oz in
        Datasheets often quote "torque" in kg cm, which means "kilogram-force x centimetre": a weight of 1 kg hanging 1 cm from the shaft.

 Worked examples:
        6000 rpm   -> omega = 2 pi x 6000 / 60 = 628.3 rad/s
        120 rpm    -> omega = 2 pi x 120 / 60  = 12.57 rad/s
        A motor delivering 0.01 N m at 6000 rpm:   P = 0.01 x 628.3 = 6.28 W
        A wheel of radius 0.04 m pushing the robot with F = 5 N:   tau = 5 x 0.04 = 0.20 N m   (2.04 kg cm)
        The same wheel at 120 rpm:   v = omega r = 12.57 x 0.04 = 0.503 m/s,   P = F v = 5 x 0.503 = 2.51 W   (and tau omega = 0.20 x 12.57 = 2.51 W, the same)
        A servo rated 6 kg cm at its horn:   6 x 0.0981 = 0.589 N m

Speed and torque trade; power does not appear from nowhere

A motor of a given power can give high speed and low torque, or low speed and high torque, depending on the gearing. A gearbox only exchanges one for the other (and loses a little as heat). This is why a small fast motor can drive a robot: gear it down. It is also why a robot that "has plenty of motor" but the wrong gearing still cannot climb a ramp or reach the speed you want.

2. DC motor characteristics (the mechanical view)

The torque-speed line

A brushed DC motor (Basic Electronics 9) has two extreme operating points. With nothing on the shaft it runs at its no-load speed ω_nl and gives no useful torque. Held still (stalled) it gives its stall torque τ_s and draws its stall current I_s = V / R. In between, to a good approximation, torque falls in a straight line as speed rises. The motor's behaviour is described by two constants, k_t (torque per ampere, in N m/A) and k_e (back-EMF per rad/s, in V s/rad), which have the same value in SI units.

 Electrical model (Basic Electronics 9):     V = k(e) omega + I R               tau = k(t) I           (k(t) = k(e) = k in SI units)

 Stall (omega = 0):      I(s) = V / R            tau(s) = k V / R
 No load (tau = 0):      omega(nl) = V / k       (in a real motor a little current I(0) is still needed to overcome friction and iron losses)
 Torque-speed line:      tau = tau(s) (1 - omega / omega(nl))
 Mechanical power:       P = tau omega = tau(s) omega (1 - omega / omega(nl))        a parabola, zero at both ends
 Peak power:             at omega = omega(nl) / 2 and tau = tau(s) / 2:              P(max) = tau(s) omega(nl) / 4
 Efficiency:             eta = P(mech) / P(elec) = tau omega / (V I)

 Torque (N m)
 0.018 |*                                       tau(s) = k V / R = 0.006 x 6 / 2.0
       | *
       |  *
       |   *
       |    *
 0.009 |     *          peak power here: P(max) = 0.018 x 1000 / 4 = 4.5 W
       |      *
       |       *
       |        *
       |         *
     0 +----------*---------------------- Speed (rad/s)
       0        500                1000   omega(nl) = V / k = 6 / 0.006 = 1000 rad/s  (9549 rpm)

Here is an illustrative small hobby motor, with the kind of constants a 6 V gear-motor's bare motor typically has (typical values from memory): V = 6 V, R = 2.0 Ω, k = 0.006 N m/A, and a no-load current I₀ = 0.15 A that stands for friction. The table computes, for each current, the speed from ω = (V − I R) / k, the shaft torque from τ = k (I − I₀) and the efficiency.

 Current I  Speed (rpm)   Shaft torque (N m)  P(mech) (W)  P(elec) = V I (W)  Efficiency   Where it is
150 A    9072          0.0000              0.00         0.90               0 %          no load: all the current is friction
300 A    8594          0.0009              0.81         1.80               45.0 %
500 A    7958          0.0021              1.75         3.00               58.3 %
671 A    7413          0.0031              2.43         4.03               60.3 %       peak EFFICIENCY, at about 17 % of stall torque
000 A    6366          0.0051              3.40         6.00               56.7 %
500 A    4775          0.0081              4.05         9.00               45.0 %       about peak POWER (here 4.06 W at 1.575 A)
000 A    3183          0.0111              3.70         12.00              30.8 %
000 A    0             0.0171              0.00         18.00              0 %          stalled: 18 W of heat in the windings, no output

 Peak efficiency:       I = sqrt(I(0) V / R) = sqrt(0.15 x 6 / 2) = 0.671 A;      eta(max) = (1 - sqrt(I(0) R / V))^2 = (1 - 0.2236)^2 = 0.603 (60 %)
 Peak power:            I = (I(0) + V / R) / 2 = (0.15 + 3.0) / 2 = 1.575 A;        P = (I - I(0))(V - I R) = 1.425 x 2.85 = 4.06 W

Three lessons come out of that table, and they decide how a motor is used:

Peak efficiency and peak power are at different places. A small brushed motor is most efficient at a fairly light load, about 10 to 20 % of its stall torque, and at a fairly high speed. At peak power the efficiency has dropped to about 45 %, and half the electrical power is heating the windings.
Run it in the efficient region, not at the peak-power point. A robot that spends its life at low load and high speed wastes the least battery and keeps the motor cool. The usual design rule is to choose the gearing so that the motor's torque at the worst expected load is no more than about a quarter to a third of its stall torque (a typical guide, and the datasheet's rated continuous torque takes priority).
Stall is dangerous. At stall the motor produces no power but dissipates I²R = 3² × 2.0 = 18 W, four times the 4.5 W it could ever deliver. Small motors overheat within seconds. A robot jammed against a wall with the motors driving is a stalled robot: the driver should limit current (Basic Electronics 9).

The datasheet's "stall torque" is not a working torque

Stall torque is the number the motor gives when it cannot turn; it can be held for only a short time. It is useful as a scale (the line above starts there) and for checking that the motor can start the load, but design to the continuous rating. Also, stall torque and no-load speed both scale with the supply voltage: at half the voltage, both halve, which is why a sagging battery makes a robot both slower and weaker.

3. Gear ratios and gearboxes

A hobby motor is typically fast and weak, 5000 to 15000 rpm and a few hundredths of a newton metre, while a wheel wants about 50 to 300 rpm and much more torque. A gearbox bridges the gap. If the driven gear has N times as many teeth as the driving gear (a reduction of N:1), the output turns N times slower and, ignoring losses, delivers N times the torque. Gears lose a little power in friction, so the real output torque is multiplied by an efficiency η.

 Gear ratio (reduction):     N = teeth(driven) / teeth(driving)
 Output speed:               omega(out) = omega(in) / N
 Output torque:              tau(out) = tau(in) N eta                    eta = efficiency of the train (1 = perfect)
 Output power:               P(out) = tau(out) omega(out) = tau(in) omega(in) eta = P(in) eta        (the loss is the heat: P(in) (1 - eta))
 Several stages in series:   N = N1 x N2 x N3 ...        eta = eta1 x eta2 x eta3 ...     (every stage costs a little efficiency)

 A two-stage spur-gear train (each pair of gears turns the next shaft the opposite way):

     motor ---[ 12T ]=====[ 48T ]--+                         Stage 1:  48 / 12 = 4:1
                                   | (same shaft)
                                   +--[ 15T ]=====[ 45T ]--- output  Stage 2:  45 / 15 = 3:1

 Worked example:  motor at 6000 rpm and 0.010 N m through that train, eta = 0.92 per stage.
        Ratio:         N = 4 x 3 = 12:1               Efficiency:  eta = 0.92 x 0.92 = 0.846
        Output speed:  6000 / 12 = 500 rpm            Output torque:  0.010 x 12 x 0.846 = 0.1016 N m
        Power in:      0.010 x 628.3 = 6.28 W          Power out:  0.1016 x 52.36 = 5.32 W          Lost as heat: 0.97 W (15.4 %)

Note the direction of the trade. Using a bigger ratio multiplies torque and divides speed by the same factor, but every extra stage costs efficiency and adds weight, noise and backlash. Common hobby gear motors come as ready-made packages with a fixed ratio (for example 30:1, 50:1, 100:1 or 298:1, among many others), which is how most small robots obtain their drivetrain.

Types of gearbox

Type	How it works	Typical efficiency	Strengths	Watch for
Spur gears	Parallel shafts, straight teeth, stages stacked side by side	about 90 to 98 % per stage (hobby plastic and sintered gear motors nearer the low end)	Simple, cheap, easy to build or print	Noisy at speed; the output is offset from the motor shaft; backlash adds up through the stages
Planetary	A central sun gear, several planet gears on a carrier, an outer ring gear; the output is the carrier	about 90 to 97 % per stage	Compact and in line with the motor; the load is shared by several teeth, so it carries more torque for its size; ratios of about 3:1 to 10:1 per stage	More parts and cost; stages are stacked for big ratios; cheap ones have noticeable backlash
Worm	A screw-like worm turns a worm wheel at a right angle	about 40 to 90 %, depending on the lead angle (see below)	Large single-stage ratios (10:1 to 100:1 or more); right-angle drive; can be self-locking	Often inefficient and hot; a self-locking worm cannot be back-driven, so a pushed robot cannot coast
Helical and bevel	Slanted teeth (quieter); bevel gears turn the axis through an angle	about 90 to 98 %	Quiet running; direction change	Helical gears push sideways along the shaft and need thrust bearings

 Planetary stage (sun = input, ring gear held fixed, carrier = output):        N = 1 + R / S         (R = ring teeth, S = sun teeth)
        Geometry:  R = S + 2 P  (P = planet teeth).   Example: S = 12, P = 12, so R = 36:      N = 1 + 36 / 12 = 4:1.   Two such stages in series give 16:1.

 Worm drive:   ratio N = (teeth on the worm wheel) / (number of starts on the worm).      Example: 40-tooth wheel and a single-start worm: N = 40:1.
        Efficiency (approximately):   eta = tan(lambda) / tan(lambda + phi)         lambda = lead angle of the worm,   phi = atan(mu) = the friction angle
        Self-locking when:             lambda < phi     (the wheel cannot drive the worm).
        lambda = 5 deg,  mu = 0.10 (phi = 5.71 deg):    eta = tan 5 / tan 10.71 = 0.0875 / 0.1892 = 0.46        (5 < 5.71, so it IS self-locking)
        lambda = 15 deg, mu = 0.05 (phi = 2.86 deg):    eta = tan 15 / tan 17.86 = 0.2679 / 0.3222 = 0.83         (efficient, but NOT self-locking: it can be back-driven)
        The pattern holds for every worm: the more efficient it is, the less it holds a load by itself, and the more it holds, the more power it wastes.

 Backlash:  the small free movement between gear teeth when the direction reverses. At the OUTPUT it is amplified by the radius of whatever is attached.
        1.5 degrees of backlash at the output of a gearbox driving a wheel of radius 0.05 m:    0.05 x 0.02618 rad = 1.3 mm of free motion at the rim.
        Backlash does not matter for a wheel that mostly turns one way, but it matters for an arm that must be positioned, and it can make a control loop wobble (Robotics Fundamentals 7).

Back-driving and holding

Whether a gearbox can be turned backwards from the output also decides what happens when power is off. A high-ratio, low-efficiency gearbox (a worm, or a 298:1 spur train) tends to hold its position and resists being pushed, which suits an arm but means the robot will not freewheel when you push it by hand. A high-efficiency gearbox lets the load drive the motor, which can be useful (the robot rolls freely, and the motor can brake it and even recover energy) and can be a nuisance (an arm falls when the power drops).

4. Belts and chains

Where the motor cannot sit directly in line with the axle, a belt or a chain carries the motion over a distance, still with a ratio set by size: for toothed pulleys or sprockets the ratio is the ratio of tooth counts. Neither needs the precision of a gear mesh, so they tolerate small misalignment.

	Toothed belt (GT2, HTD)	Roller chain	Round or V belt
Efficiency (typical)	about 95 to 98 %	about 95 to 98 %	about 90 to 95 %, and depends on tension
Slip	None, if it fits the teeth correctly	None	Can slip, so it is not for timing
Noise and cost	Quiet, light, cheap; 3D-printable pulleys	Noisier, heavier, needs lubrication	Quiet, cheapest
Strength	Good for small robots; rated by belt width	Carries high loads; used in larger robots	Light loads only
Upkeep	Tension it correctly, with an idler or by sliding the motor mount	Chain stretches; needs adjustment and oil	Re-tension from time to time

 Belt or chain ratio:        N = teeth(driven) / teeth(driving)          (the same as for gears; the two pulleys turn the SAME way)
 Pitch diameter of a toothed pulley:      d = (teeth x pitch) / pi
 Open-belt length (approximate):          L = 2 C + (pi / 2)(d1 + d2) + (d2 - d1)^2 / (4 C)          C = centre distance

 Worked example: GT2 belt (pitch 2 mm), a 16-tooth pulley on the motor and a 48-tooth pulley on the axle, centres 60 mm apart.
        Ratio:          N = 48 / 16 = 3:1
        d1 = 16 x 2 / pi = 10.19 mm          d2 = 48 x 2 / pi = 30.56 mm
        L  = 2 x 60 + (pi / 2)(10.19 + 30.56) + (30.56 - 10.19)^2 / (4 x 60) = 120 + 64.01 + 1.73 = 185.7 mm
        Belt teeth:     185.7 / 2 = 92.9 teeth, so choose the nearest stocked belt (about 93 teeth) and adjust the centre distance to fit.
 Efficiency of this stage: about 0.97 (typical), so it is a cheap place to take a ratio of 2:1 or 3:1 without much loss.

A belt stage is a handy way to add a little ratio and move the motor away from the wheel, while a chain (or a belt) joining two wheels on the same side makes one motor drive both of them, which is a form of four-wheel drive from two motors. Direct drive (a wheel bolted to a geared motor's output shaft) is simplest, cheapest and the right answer for most first robots.

5. Wheels and traction

A wheel does two things: it converts rotation into forward speed, and it converts torque into a push on the ground. The first depends on the diameter, the second on grip.

 Speed:            v = omega r = pi D n / 60                       (D = wheel diameter in metres, n = wheel rpm)
 Drive force:      F = tau(wheel) / r                              so a larger wheel gives the same speed at a lower rpm, but a smaller push for the same torque
 Traction limit:   F(max) = mu N                                   N = load carried by the DRIVEN wheels; mu = friction coefficient (tyre and floor)
 Rolling drag:     F(roll) = Crr m g cos(theta)                    Crr = rolling-resistance coefficient (dimensionless; small)
 A robot can climb a slope without the wheels slipping only if:     tan(theta) <= s mu - Crr          s = fraction of the weight carried by the driven wheels (1 if all driven)

 Typical values (approximate; surfaces and tyres vary a great deal):

   Contact                                         mu (friction, dry)       Crr (rolling resistance)
   Soft rubber or silicone tyre on clean tile      0.7 to 1.0               0.02 to 0.04
   Rubber tyre on carpet                           0.6 to 0.9               0.04 to 0.10 (soft carpet is worse)
   Hard plastic wheel on a smooth floor            0.2 to 0.4               0.01 to 0.02
   Rubber on a wet or dusty floor                  0.3 to 0.5 (and falling) about the same

 Worked example: wheel diameter 0.080 m (r = 0.040 m).
        To drive at 0.5 m/s:     omega = v / r = 0.5 / 0.04 = 12.5 rad/s      n = 12.5 x 60 / (2 pi) = 119.4 rpm
        Pushing with F = 5 N:     tau = 5 x 0.04 = 0.20 N m   (shared between the driven wheels)
        Wheel diameter 0.060 m:   119.4 x (0.080 / 0.060) = 159.2 rpm for the same speed, tau = F r = 0.030 per newton instead of 0.040.
        Wheel diameter 0.100 m:   95.5 rpm and 0.050 N m per newton.

Choosing a wheel

Diameter. A bigger wheel gives more speed from the same motor and gearbox, rolls over bumps and gaps more easily (a rough guide is that a wheel will climb a step of up to a fraction of its radius, depending on grip) and raises the robot higher. It also needs more torque for the same push (τ = F r) and may push the centre of gravity up. Small robots often use 40 to 100 mm wheels.
Grip. Soft rubber or silicone gives high μ and wears and picks up dust; hard plastic is fast and slides. Rubber bands over a printed wheel are a cheap way to add grip. A rubber tyre that works on the lab bench may skid on a dusty floor, so test where it will be used.
Width. A wider tyre does not increase the friction coefficient but spreads wear and improves stability; a narrow tyre turns more easily on the spot.
Where the weight sits. Traction depends on the load on the driven wheels. Put the weight over them (Robotics Fundamentals 2 gave 22 % on the caster and the rest on the drive wheels for the example robot).

Wheel type	What it is	Good for	Drawbacks
Standard fixed wheel	A round wheel on a fixed axle	Most robots; efficient and simple	Cannot move sideways; needs steering by differential speed or a steering mechanism
Caster (ball or swivel)	A free-turning support	Third point of contact on a two-wheel robot	Drag and bounce; swivel casters can flip the robot's heading when it reverses
Omni wheel	A wheel ringed with small rollers that let it slide sideways	Robots that must move in any direction (Robotics Fundamentals 4)	Less grip and a bumpier ride
Mecanum wheel	Angled rollers; four of them let the robot drive in any direction	Sideways motion in tight spaces (Robotics Fundamentals 4)	Low efficiency and grip, harder control; in practice mecanum drivetrains tend to be heavier and more costly
Foam or pneumatic wheel	A soft, deformable tyre	Rough ground and shock absorption	Squashes (the radius changes with load) and has higher rolling resistance on hard floors
Tracks	A belt of tread round two or more wheels	Soft ground and obstacles (Robotics Fundamentals 4)	Skid steering, heavy, lossy, and friction in turning

6. Sizing a motor and gearbox for a robot

Now the pieces fit together. We work from the wheels back to the motor, in this order: the force the robot needs, the torque that force needs at each wheel, the motor torque after the gearbox, the speed (which fixes the gear ratio) and the current (which checks the electrical side).

 The motion requirement, in six steps (air drag is ignored; at robot speeds it is small):

 1. Force needed at the ground:     F = m (a + g sin(theta) + Crr g cos(theta))
        a = acceleration wanted (m/s^2),   theta = slope,   Crr = rolling-resistance coefficient,   g = 9.81 m/s^2
        three terms: to speed up, to climb, and to overcome rolling drag
 2. Torque at each driven wheel:    tau(wheel) = F r / n(wheels)               r = wheel radius, n = number of driven wheels
 3. Motor torque after the gearbox: tau(motor) = margin x tau(wheel) / (N eta)         N = gear ratio, eta = gearbox efficiency, margin about 1.5 to 2
 4. Speed:                          omega(wheel) = v / r        omega(motor) = N omega(wheel) = N v / r
 5. Current:                        I = tau(motor) / k + I(0)        (and the motor's terminal voltage needed:  V(need) = k omega(motor) + I R,  PWM duty = V(need) / V(supply))
 6. Checks:                         tau(motor) is below about 25 to 30 % of stall torque;   duty is below about 90 %;   F is less than the traction limit s mu m g cos(theta)

 Choosing N:  run the motor at about 60 to 70 % of its no-load speed when the robot is at its target speed, so there is voltage headroom to accelerate and to control:
        N  =  (0.6 to 0.7) x omega(nl) / omega(wheel)

Worked example: a 2 kg two-motor robot

A two-motor robot of mass 2 kg (the heavier cousin of Robotics Fundamentals 2's example, with the same layout so that 22 % of its weight is on the caster) must drive at 0.5 m/s, reach that speed in 1 second (a = 0.5 m/s²), and be able to do that while climbing a 10 degree slope. Wheels: 80 mm diameter rubber tyres on a hard floor, C_rr = 0.03 and μ = 0.8 (typical). The motor is the illustrative one from section 2 (6 V, 2.0 Ω, k = 0.006, I₀ = 0.15 A, stall torque 0.018 N m, no-load speed 9549 rpm), in a three-stage spur gearbox with 0.90 per stage.

 Step 1  Force.   sin 10 deg = 0.17365,  cos 10 deg = 0.98481
        accelerate:  m a             = 2 x 0.5                       = 1.000 N
        climb:       m g sin(theta)  = 2 x 9.81 x 0.17365            = 3.407 N
        rolling:     Crr m g cos(theta) = 0.03 x 2 x 9.81 x 0.98481  = 0.580 N
        F = 1.000 + 3.407 + 0.580 = 4.987 N                (steady climb with no acceleration:  3.407 + 0.580 = 3.987 N)

 Step 2  Wheel torque (n = 2, r = 0.040 m):    tau(wheel) = 4.987 x 0.040 / 2 = 0.0997 N m each     (steady climb: 0.0797 N m each)

 Step 4  Speed and the gear ratio first:
        wheel:   omega = 0.5 / 0.04 = 12.5 rad/s = 119.4 rpm
        motor no-load speed 9549 rpm,  so  N = (0.6 to 0.7) x 9549 / 119.4 = 48 to 56:    choose 50:1  (a common stock ratio)
        At 50:1 the motor turns at 12.5 x 50 = 625 rad/s = 5968 rpm at 0.5 m/s, which is 62.5 % of its no-load speed.
        Gearbox efficiency (three spur stages):  eta = 0.9^3 = 0.729

 Step 3  Motor torque with the 1.5 margin:     tau(motor) = 1.5 x 0.0997 / (50 x 0.729) = 0.00410 N m      (without margin: 0.00274 N m)
        Percentage of stall torque:            0.00410 / 0.018 = 22.8 %        (inside the 25 to 30 % guide)

 Step 5  Current:    with margin:  I = 0.00410 / 0.006 + 0.15 = 0.834 A          without margin:  I = 0.00274 / 0.006 + 0.15 = 0.606 A
        Voltage needed at 0.5 m/s (no margin):   V(need) = 0.006 x 625 + 0.606 x 2.0 = 3.75 + 1.21 = 4.96 V    duty = 4.96 / 6 = 83 %
        With the margin load:                    V(need) = 3.75 + 0.834 x 2.0 = 5.42 V                         duty = 90 %      (OK, but headroom is thin)
        Electrical power, two motors (no margin):  2 x 4.96 x 0.606 = 6.01 W
        Mechanical power at the wheels:            F v = 4.987 x 0.5 = 2.49 W      overall drive efficiency = 2.49 / 6.01 = 41 %
        (motor about 57 % at this point x gearbox 73 % = about 41 %. Real robots are inefficient: size the battery on this.)

 Step 6  Checks.
        Stall wheel torque:       0.018 x 50 x 0.729 = 0.656 N m each,  so the pair could push with 2 x 0.656 / 0.04 = 32.8 N (6.6 times what is needed)
        Traction limit:           s mu m g cos(theta) = 0.78 x 0.8 x 2 x 9.81 x 0.98481 = 12.06 N       (greater than 4.99 N, so the wheels will not slip)
        Steepest slope without slipping:   tan(theta) = s mu - Crr = 0.78 x 0.8 - 0.03 = 0.594   theta = 30.7 degrees (the motors would stall or overheat before this)
        Speed at the worst load:   with I = 0.606 A the motor can still turn at (6 - 0.606 x 2.0) / 0.006 = 798 rad/s = 7620 rpm,
                                   so the wheel speed could reach 798 / 50 x 0.04 = 0.638 m/s: above the 0.5 m/s target. The speed controller uses PWM to hold 0.5.

How sensitive is the answer to the ratio? Repeating the margin-load calculation (acceleration on the slope, torque margin 1.5) for neighbouring ratios shows why 50:1 is a reasonable pick:

 Ratio N   Motor torque (N m)   Current I   % of stall torque   Voltage needed at 0.5 m/s    Duty     Verdict
 30:1      0.00684              1.290 A     38.0 %              4.83 V                       80.5 %   too much torque for the motor: hot and inefficient
 40:1      0.00513              1.005 A     28.5 %              5.01 V                       83.5 %   acceptable
 50:1      0.00410              0.834 A     22.8 %              5.42 V                       90.3 %   chosen: cool, with enough voltage left
 60:1      0.00342              0.720 A     19.0 %              5.94 V                       99.0 %   no voltage left: it cannot hold 0.5 m/s on the slope

 The trade: a lower ratio asks more of the torque (heat), a higher ratio asks more of the voltage (speed). The window between them is the design.

Steady-state force on a slope scales strongly with the slope angle, which is why a robot that is fine on the flat can fail on a ramp:

 Slope     Steady force F = m g (sin + Crr cos)      Torque per wheel (r = 0.04, 2 wheels)
deg     0.589 N                                   0.0118 N m
deg     2.296 N                                   0.0459 N m
deg    3.987 N                                   0.0797 N m
deg    5.647 N                                   0.1129 N m
deg    7.264 N                                   0.1453 N m

 From flat to 10 degrees the force needed rises almost seven times.

A small sizing calculator

Because the same six steps are repeated every time a design changes, it pays to put them in a few lines. This is hardware-neutral pseudocode in the style of Python (real board code is left to the course Programming for Robotics):

 def size_drivetrain(mass, n_driven, wheel_d, v_target, accel, slope_deg, crr,
                    ratio, gear_eta, volts, resistance, kt, i_noload, margin=1.5):
     g = 9.81
     theta = radians(slope_deg)
     r = wheel_d / 2

     force = mass * (accel + g * sin(theta) + crr * g * cos(theta))     # step 1
     wheel_torque = force * r / n_driven                                # step 2
     motor_torque = margin * wheel_torque / (ratio * gear_eta)          # step 3
     motor_speed = ratio * v_target / r                                 # step 4 (rad/s)
     current = motor_torque / kt + i_noload                             # step 5
     volts_needed = kt * motor_speed + current * resistance
     duty = volts_needed / volts                                        # must stay below about 0.9
     stall_fraction = motor_torque / (kt * volts / resistance)          # should be below about 0.3

     return force, wheel_torque, motor_torque, current, duty, stall_fraction

 # the 2 kg example:
 # size_drivetrain(2, 2, 0.08, 0.5, 0.5, 10, 0.03, 50, 0.729, 6, 2.0, 0.006, 0.15)
 #   -> force 4.99 N, wheel torque 0.0997 N m, motor torque 0.0041 N m, current 0.834 A, duty 0.90, stall fraction 0.228

What this model leaves out

It treats the motor as an ideal line, assumes steady friction and a constant battery voltage, and ignores air drag, gearbox losses that vary with load and temperature, and the extra force when turning (a skid or a heavy caster resists the turn). It also assumes the weight is shared as designed; accelerating shifts weight to the rear wheels. Build the first robot with a motor and ratio that give a margin, then measure the actual current on the bench (Basic Electronics 9) and adjust. The calculation tells you the neighbourhood, and the tests tell you the street.

7. Challenges

Challenge 1: Reading a motor's torque-speed line

A brushed DC motor is rated at 12 V with a stall current of 4 A and a stall torque of 0.12 N m. Use the ideal model (ignore friction). (a) Find the winding resistance R, the torque constant k (which also equals the back-EMF constant), and the no-load speed in rad/s and rpm. (b) Express the stall torque in kg cm. (c) At a load torque of 0.04 N m, find the speed (rad/s and rpm), the mechanical power, the current, the electrical power and the efficiency. (d) Find the peak mechanical power, and the speed, torque, current and efficiency at which it occurs. (e) How much heat is produced in the windings at stall, and how does it compare with the peak output power? (f) At half the supply voltage (6 V), what are the stall torque and the no-load speed?

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Challenge 2: Gearbox, belt and worm

A motor delivers 0.020 N m at 4000 rpm. It drives a two-stage spur gearbox: stage 1 is a 12-tooth pinion driving a 48-tooth gear, stage 2 is a 15-tooth pinion driving a 45-tooth gear. Each stage is 92 % efficient. (a) Find the total ratio and the total efficiency. (b) Find the output speed (rpm) and torque. (c) Find the input power, the output power and the power lost as heat. (d) The output drives a wheel of diameter 0.10 m: find the speed in m/s and the push at the rim. (e) A belt stage with a 16-tooth and a 32-tooth pulley (97 % efficient) is added after the gearbox: find the new speed, torque, wheel speed and push. (f) Instead of the spur gearbox, the motor drives a 2-start worm and a 30-tooth worm wheel; the worm has a lead angle of 8 degrees and a friction coefficient of 0.08. Find the ratio, the efficiency, the output torque and speed, and say whether it is self-locking. (g) 1.5 degrees of backlash at the output of the spur gearbox drives a wheel of radius 0.05 m: how much free motion at the rim?

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Challenge 3: Size the drive for a four-wheel robot

A 3 kg robot has four driven wheels of diameter 0.10 m, one motor each. It must drive at 0.8 m/s, accelerate at 0.4 m/s², and climb a 15 degree slope with rolling resistance Crr = 0.04. The motors are 12 V, with winding resistance 4 Ω and k = 0.02 N m/A (ignore friction), and the gearboxes are 75 % efficient. (a) Find the force needed and the torque each wheel must supply. (b) Find the motor's stall torque, stall current and no-load speed (rpm), and choose a gear ratio that puts the motor at about 75 % of no-load speed at 0.8 m/s (choose a whole-number ratio near that). (c) With a gear ratio of 30:1 and a torque margin of 1.5, find the motor torque, current and the percentage of stall torque. (d) Using the current without the margin, find the PWM duty needed at 0.8 m/s, and say whether the headroom is adequate. (e) With a friction coefficient of 0.6, check the traction limit when all four wheels are driven. Then repeat it if only two wheels are driven and they carry half the weight. What happens?

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Quick reference

Torque, speed, power: τ = F r; ω = 2 π n / 60; P = τ ω = F v; 1 kg cm = 0.0981 N m; 1 oz in = 0.00706 N m
DC motor: V = k ω + I R; τ = k I; τ(s) = k V / R; ω(nl) = V / k; τ = τ(s)(1 − ω / ω(nl)); P(max) = τ(s) ω(nl) / 4 at half speed
Efficiency: peak efficiency (about 60 % in a small brushed motor) at about 10 to 20 % of stall torque, peak power at about half stall torque with efficiency about 45 %; stall wastes I²R as heat
Gears: output speed = input / N; output torque = input × N × η; ratios multiply and efficiencies multiply; planetary N = 1 + R / S (ring fixed)
Gearbox types: spur about 90 to 98 % per stage; planetary about 90 to 97 %; worm about 40 to 90 % and self-locking if the lead angle is less than the friction angle; backlash at the rim = r × angle (radians)
Belts and chains: ratio = teeth ratio; pulley d = teeth × pitch / π; efficiency about 95 to 98 %
Wheels: v = π D n / 60; F(max) = μ N; climbable if tan(θ) ≤ s μ − Crr; larger wheel means more speed per rpm and more torque per newton
Sizing: F = m (a + g sin θ + Crr g cos θ); wheel torque = F r / n; motor torque = margin × wheel torque / (N η); N about (0.6 to 0.7) ω(nl) / ω(wheel)
Checks: motor torque below about 25 to 30 % of stall; PWM duty below about 90 %; F below the traction limit; I = τ / k + I(0)

Next, Robotics Fundamentals 4 looks at drive systems and kinematics: how two wheels driven at different speeds steer a robot (differential drive), how omni and mecanum wheels move it sideways, how to calculate where the robot has got to from its wheel speeds (odometry), what turning radius it has, how tracks compare with wheels, and a first look at legs.