Sizing and Gearing Calculation Audit

Sizing and Gearing Calculation Audit

Ada re-derives this chapter’s own numbers step by step, at full precision

ADA · CALCULATION AUDIT

Sizing and Gearing Calculation Audit

A smart-vent linkage pushes 1.8 N at a 25 mm radius, so the actuator must beat 0.045 N·m of load torque — doubled to a 0.090 N·m target for margin. A candidate motor rated 0.06 N·m near 300 rpm behind a 4:1 gearbox at 80% efficiency delivers 0.192 N·m at 75 rpm, while the valve draws 2.88 W moving but only 0.48 W averaged over 20 s of every 120 s. This audit re-derives each figure and asks whether the geared chain clears the load with room to spare, and which number — the duty-averaged heat or the worst-case stall current — should size the supply.

Companion to the chapter Actuator Assessment and Reference — every number here comes from that chapter.

— load torque, gear reduction, reflected inertia, and duty power, ~4 minutes

Naming "a servo" or "a stepper" is a guess until the torque, speed, and thermal numbers agree. Every value below is one this chapter already stated — I am only carrying the arithmetic to full precision and rounding at the end.

See the relationship before changing it

The figure reads from left to right. The blue card is gear ratio. The middle card applies this page's rule. The green card is delivered motor torque. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only gear ratio, so the numeric fixture does not switch without explanation.

Walk the arrows. A larger ratio multiplies torque while reducing output speed.

Derive the baseline in four named moves

1. 1

   Name the input. The chapter baseline is 4 to one.

2. 2

   Name the relationship. output torque = 0.060 N m x ratio x 0.80

3. 3

   Substitute with units. 0.060 x 4 x 0.80 = 0.192 N m

4. 4

   Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change gear ratio

Try Predict the direction of output torque = 0.060 N m x ratio x 0.80. Test another gear ratio, then compare delivered motor torque.

Gear ratio

4 to one

Chapter baseline

Delivered motor torque

Observe A larger ratio multiplies torque while reducing output speed. Reset gear ratio to 4 and compare delivered motor torque.

Explain A larger ratio multiplies torque while reducing output speed.

Check yourself

What should you do before trusting a moved-control result?

Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.

What does this small model leave out?

Answer: Only gear ratio moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only gear ratio. It is not a field prediction.

1. Load torque is force times lever arm.

The smart-vent linkage pushes 1.8 N at a 25 mm (0.025 m) radius, and good practice doubles it for friction and acceleration headroom:

τ_load = F × r  →  1.8 × 0.025 = 0.045 N·m  →  ×2 margin = 0.090 N·m target

2. A gear reduction N trades speed for torque.

It multiplies output torque by N (times gearbox efficiency η), divides output speed by N, and divides the reflected load inertia the motor feels by N². The chapter's candidate motor gives 0.06 N·m near 300 rpm; a 4:1 box at 80% efficiency is:

τ_out = 0.06 × 4 × 0.80 = 0.192 N·m  |  ω_out = 300 / 4 = 75 rpm  |  J_reflected ÷ N² = ÷16

3. Duty cycle separates instantaneous power from heat load.

The valve actuator runs 12 V at 0.4 A on 60% PWM, and moves for 20 s out of every 120 s cycle:

Design question	Arithmetic shown	Audit result
Torque the actuator must beat	1.8 × 0.025	0.045 N·m
Target with 2× design margin	0.045 × 2	0.090 N·m
Geared output torque (4:1, η = 0.80)	0.06 × 4 × 0.80	0.192 N·m — clears 0.090
Geared output speed	300 / 4	75 rpm — above the 30 rpm need
Reflected inertia the motor feels	÷ 4²	16× lighter
Valve moving power (V × I × D)	12 × 0.4 × 0.60	2.88 W instantaneous
Duty-averaged heat (20 s per 120 s)	2.88 × (20 / 120)	0.48 W average

What this means for your design: the geared motor clears the 0.09 N·m target with room to spare and still turns faster than the vent needs, so the controller trims speed with PWM rather than fighting a shortfall. And because the valve moves only one-sixth of the time, its thermal load is the 0.48 W duty-average — size the enclosure heat path to that number, but size the supply and driver to the worst-case stall current, never to the 0.48 W average.

Every number above is taken from the chapter’s own examples and re-derived step by step.

TryRecompute load torque, the 2x target, geared output torque, and duty-averaged power.

ObserveThe 0.192 N m geared output clears the 0.090 N m target; moving power averages to 0.48 W.

ExplainAverage heat and peak stall current answer different sizing questions.

Technical boundaries. The audit assumes fixed load radius, gearbox ratio and efficiency, motor rating, voltage, current, and duty. It does not model acceleration, friction, gear backlash, efficiency versus load, stall duration, thermal transients, supply sag, or mechanism wear.

Audit result

The torque margin is 0.192 / 0.090 = 2.13x, but the supply still must be checked against worst-case startup or stall current.