⚙️ Nm to Joules Calculator
Compute mechanical work from torque and angular displacement.
Essential for mechanical engineers, robotics designers, and physics students calculating rotational work.
50 N·m over 90 Degrees produces
78.539816 Joules
Angle in radians: 1.570796
Reverse Calculations
Required angle for 50 N·m to deliver 78.539816 J: 90° or 0.25 rev
If the angle stays Degrees, torque needed for 78.539816 J: 50 N·m
How to Use This Calculator
Enter torque
Provide the applied torque in newton-meters from your wrench or motor specification.
Specify angular displacement
Use degrees, radians, or revolutions—the calculator converts to radians internally.
Read the work output
The result shows joules of mechanical work plus reverse calculations for design checks.
Formula
Work (J) = Torque (N·m) × Angle (radians)
Radians = Degrees × π / 180, or Revolutions × 2π
Torque = Work ÷ Angle and Angle = Work ÷ Torque
Use the formula breakdown to confirm the calculation logic or perform the conversion manually if needed.
Full Description
A newton-meter is dimensionally equivalent to a joule when the torque actually causes rotation. That makes converting between them straightforward once you know the angle traveled in radians.
This calculator helps with tasks like estimating the energy stored in torsion springs, evaluating motor performance, or checking the manual work required for valve operations.
Remember that if torque is applied without rotation (for example, a stuck bolt), no mechanical work is done even though force was exerted.
Frequently Asked Questions
Why do I need the angle?
Torque alone does not produce work. Only when torque causes rotation can you compute energy, which equals torque times angular displacement.
What if my angle is in turns?
Select “Revolutions” in the angle unit dropdown. The calculator automatically converts turns to radians.
Can I use negative torque?
Yes. Negative torque or angle simply indicates direction. The resulting work will be negative, representing energy returned by the system.
Does this account for efficiency?
No. It assumes ideal mechanical systems. Apply efficiency factors separately if friction or other losses are significant.