Education Lab / Robotics Theory Basics
MIY STUDIO · COBOT PROJECT 05
Robot Motor Sizing:
Peak Torque, RMS Torque & Current
Building a 6-DOF Collaborative Robot from Scratch
Robot motor sizing starts with the joint’s load and motion. Day 5 turns a simple one-axis model into preliminary torque, speed and current requirements—and shows why peak capability and continuous capability must both be checked.
01 / DEFINE THE AXIS
A small load model with clear assumptions
Start with the mechanical conditions. A payload rating alone does not define a motor requirement: reach, link mass and move time also change the result. This example is a single joint rotating a link in a vertical plane.
Compared with Day 4’s gravity-compensation example, this model adds the link’s own mass and inertia.
| Quantity | Value | Model definition |
|---|---|---|
| Payload mass | 1.00 kg | Point mass at the link tip |
| Joint-to-payload distance | 0.300 m | Measured from the joint axis |
| Link mass | 0.30 kg | Slender, uniform rigid rod |
| Link length / center of mass | 0.300 m / 0.150 m | Rod pivoted at one end |
| Move angle / total time | 90° / 2.0 s | Starts and ends at zero speed |
| Speed profile | Triangular | 1 s acceleration + 1 s deceleration |
| Gravity | 9.81 m/s² | Constant for this calculation |
Scope: These are analytical calculations for learning. The first pass excludes friction, transmission inertia, rotor inertia and additional external loads. Gear ratio, efficiency and torque constant introduced later are also teaching assumptions.
02 / GRAVITY AND INERTIA
Count both the payload and the link
In the horizontal position, gravity acts with the largest moment arm. Add the contribution of each mass at its own center-of-mass distance.
p = payload; l = link. Mass is in kg, distance in m, and torque in N·m.
| Contribution | Calculation | Torque |
|---|---|---|
| Payload | 1.00 × 9.81 × 0.300 | 2.94300 N·m |
| Link | 0.30 × 9.81 × 0.150 | 0.44145 N·m |
| Total | 2.94300 + 0.44145 | 3.38445 N·m |
Unit check: Multiplying mass by distance gives kg·m. Multiplying by gravity turns that into N·m. A powered joint can require torque even while it is stationary.
Mass distribution sets the acceleration torque
For the payload, use point-mass inertia. For the uniform link pivoted at one end, use rod inertia. The link’s full length belongs in the rod formula.
Point payload
Jp = mprp2
= 1.00 × 0.300²
0.090 kg·m²
Uniform link
Jl = ⅓mlL²
= ⅓ × 0.30 × 0.300²
0.009 kg·m²
This is load inertia about the joint output axis. Gravity does not appear in an inertia calculation.
03 / DEFINE THE MOVE
Move 90° in 2 seconds
The joint accelerates from rest for 1 second, then decelerates for 1 second. There is no constant-speed segment. The area under the speed–time triangle equals the move angle.
Use radians in the acceleration calculation: 90° = π/2 ≈ 1.5708 rad.
t = 2.0 s is the total move time. θ = π/2 rad. The acceleration interval is t/2 = 1.0 s.
| Quantity | Calculation | Result |
|---|---|---|
| Joint peak speed | ωmax × 60 / (2π) | 15.0 rpm |
| Acceleration torque | Jα = 0.099 × 1.5708 | 0.15551 N·m |
| Acceleration / gravity ratio | 0.15551 / 3.38445 | Approximately 4.6% |
At this move time, supporting gravity dominates the acceleration requirement. Halving the move time would double peak speed and quadruple acceleration torque for the same triangular profile.
04 / INTERPRET THE PEAK
A conservative bound has a specific meaning
Add the maximum gravity-compensation torque and the maximum inertial torque to obtain a conservative upper bound. This step does not add a separate design safety factor.
The maxima may occur at different times. To find the actual peak, evaluate torque along the trajectory. Define q = 0° as horizontal, with upward rotation positive:
Torque is in N·m and angular acceleration is in rad/s². Use the correct angle mode when evaluating cos q.
| Motion condition | Angular acceleration | Joint torque |
|---|---|---|
| Hold position | 0 rad/s² | 3.384 N·m |
| Accelerate upward | +1.571 rad/s² | 3.540 N·m |
| Decelerate while moving upward | −1.571 rad/s² | 3.229 N·m |
When does the bound match the peak? For the ideal move that starts horizontal and accelerates upward toward vertical, both positive terms are maximal at acceleration onset. A different start angle or motion direction can produce a lower peak. Joint peak torque and peak speed also need not occur together.
05 / APPLY THE REDUCER
Convert joint torque and speed to the motor shaft
Assume a 50:1 reduction ratio and 80% forward transmission efficiency. The motor rotates 50 times faster than the joint. Transmission loss raises the motor torque needed to drive a given output load.
Motor speed
nm,max = Nnjoint,max
= 50 × 15
750 rpm
Motor-side load torque
Tm,load = Tjoint / (Nη)
= 3.540 / (50 × 0.80)
0.0885 N·m
| Assumed efficiency | Calculation | Motor-side load torque |
|---|---|---|
| 100% · η = 1.00 | 3.540 / (50 × 1.00) | 0.0708 N·m |
| 80% · η = 0.80 | 3.540 / (50 × 0.80) | 0.0885 N·m |
| 60% · η = 0.60 | 3.540 / (50 × 0.60) | 0.1180 N·m |
Use a decimal efficiency: 80% means 0.80. The 0.0885 N·m result is a preliminary load-transmission requirement. Add the motor’s own acceleration torque and the transmission’s inertia as appropriate. Confirm speed, load and temperature dependence of efficiency; powered holding and backdriving need separate treatment.
A robot joint needs a complete mechanical and thermal model before selecting hardware. Motor packaging, gearing and cooling belong in the same review as torque and speed. [1]
06 / CHECK REPEATED OPERATION
RMS torque: include the whole duty cycle
Peak torque checks the highest instantaneous demand. RMS torque provides a first thermal screening for repeated operation. With a constant torque constant and a consistent current definition, copper loss scales with current squared, so torque is squared before time averaging.
For constant torque in each segment:
Tᵢ is the motor torque in segment i; tᵢ is that segment’s duration. For continuously varying torque, integrate T(t)² over the complete cycle instead.
Independent worked example: The following motor-side torque history is a separate teaching example. It has not been derived from the 300 mm arm trajectory. The actual arm’s RMS torque remains to be calculated after defining its return move and dwell conditions.
| Segment | Motor torque | Duration |
|---|---|---|
| A | +0.090 N·m | 1 s |
| B | −0.090 N·m | 1 s |
| C | +0.060 N·m | 8 s |
Negative torque still contributes to heating. Squaring prevents opposite signs from canceling. Holding periods must also be included when the motor actively supports a load.
At matching operating and rating conditions, an example motor with 0.050 N·m continuous torque and 0.100 N·m permitted peak torque passes the peak magnitude check but fails this RMS screening.
RMS screening is not a complete thermal validation. Long stationary holding can concentrate heating in particular phases; check the manufacturer’s standstill rating, mounting conditions and thermal limits. [2]
07 / SIZE THE CURRENT
Use the torque constant on motor-side torque
For the independent torque-history example, assume Kt = 0.10 N·m/A. This means 1 A produces 0.10 N·m under the chosen current convention.
The supplied torque is already motor-side torque. Use the torque constant and current ratings on the same definition.
| Quantity | Calculation | Result |
|---|---|---|
| Maximum current magnitude | 0.090 / 0.10 | 0.900 A |
| Cycle RMS current | 0.067082 / 0.10 | 0.67082 A |
| Simple unit check | 0.050 / 0.10 | 0.500 A |
The example reaches |I| = 0.900 A for two consecutive seconds per 10-second cycle: +0.900 A for one second, then −0.900 A for one second. Check the drive’s peak-current duration and repeat-duty limits as well as its continuous capability.
Match current conventions. Phase RMS, phase amplitude, torque-producing current and DC supply current are different quantities. Manufacturers define torque constants for specific current and commutation conventions. The 0.67082 A here is a time RMS over the assumed cycle, using the selected Kt convention. [3]
08 / PRELIMINARY SIZING SHEET
Keep the load model and the RMS example separate
| Requirement | Result | Basis |
|---|---|---|
| Maximum gravity torque | 3.384 N·m | Horizontal arm |
| Joint load inertia | 0.099 kg·m² | Payload + uniform link |
| Maximum angular acceleration | 1.571 rad/s² | 90° in 2 s; triangular speed profile |
| Joint torque upper bound | 3.540 N·m | Gravity maximum + inertial maximum |
| Joint peak speed | 15.0 rpm | Triangular profile |
| Motor-side load torque bound | 0.0885 N·m | N = 50; assumed forward η = 0.80 |
| Motor peak speed | 750 rpm | N = 50 |
| Actual arm cycle RMS torque | To be determined | Return motion and dwell conditions needed |
| Requirement | Result | Basis |
|---|---|---|
| Peak torque magnitude | 0.090 N·m | Assumed 10 s torque history |
| RMS torque | 0.06708 N·m | Entire 10 s cycle |
| Peak current magnitude | 0.900 A | Assumed Kt = 0.10 N·m/A |
| Cycle RMS current | 0.67082 A | Same Kt convention |
The sizing sheet is ready for the next modeling step. It establishes load requirements, while a purchase specification also needs allowances, compatible motor–drive ratings and the real duty cycle.
09 / BEFORE SELECTING HARDWARE
Check what the simple calculation leaves out
| Check | What to establish |
|---|---|
| Motion and duty cycle | Define start/end angles, return move, dwells and repetitions. Evaluate signed torque along the trajectory. |
| Motor and drive | Compare torque–speed capability at the available bus voltage. Add rotor/transmission inertia; confirm current convention and permitted overload duration. |
| Transmission and support | Check reducer peak, continuous and life ratings, plus output radial/axial and moment-load limits. Review bearing spacing, shaft stiffness and fastener load paths. |
| Thermal conditions | Confirm housing heat flow, mounting and ambient temperature. Validate stationary holding as well as moving operation. |
| Assembly and maintenance | Allow bearing and fastener access, cable clearance and service space. Review backlash, repeatability and the load-holding arrangement discussed in Day 4. |
Three calculation habits to keep
- Check units first. Convert degrees to radians and millimeters to meters. Gravity torque requires g.
- Track the shaft. Joint torque and motor torque need different comparisons. Normalize percentage efficiency before calculating.
- Check magnitude and duration. Peak demand, cycle RMS demand and holding demand answer different sizing questions.
Why does a stopped joint still heat?
A powered motor can use current to balance gravity at zero speed. Mechanical output power is then zero, but winding copper loss can remain.
Can a maximum-torque rating determine the motor choice?
It covers only part of the requirement. The motor and drive must support the required torque–speed trajectory, repeated duty and thermal conditions.
Next lesson: define a horizontal-to-vertical round trip and calculate the arm’s torque history and cycle RMS. The next lesson page will be linked after publication.
REFERENCES
Official sources and further reading
Original teaching explanations, figures and worked calculations by MIY Studio. These sources support sizing practice, holding-load thermal considerations and current conventions. No third-party images or diagrams are reproduced.
- Kollmorgen — Demystifying the Use of Frameless Motors in RoboticsLoad inertia, torque, speed, gearing and thermal requirements in robot joints.
- Kollmorgen — The Difference Between Continuous Ratings and Holding Continuous LoadsRMS sizing and additional considerations for applications that hold loads with little movement.
- maxon — BLDC motors with sinusoidal commutationTorque constants, reference-current definitions and commutation conventions.
- MIY Studio — Day 3: Continuous & Peak Torque, RMS SizingEarlier lesson introducing continuous and peak motor torque.