TECHNICAL SUPPORT
Published 2026-04-22
A 25kgservoisnotcapable of lifting 25 kilograms (50 jin) under normal operating conditions. The “25kg” rating refers totorque(25 kg·cm), not direct lifting capacity. In practical use with a typical 2–3 cmservohorn, a 25kgservocan safely handle5–12.5kg (10–25kg). This guide explains exactly how to calculate the real-world lifting ability, using common examples and providing actionable steps.
The “25kg” specification is thestall torquemeasured at 1 cm radius from the output shaft.
Torque = Force × Distance
A 25 kg·cm torque means: with a 1 cm long horn, the servo can pull or hold 25 kg (50 jin) at stall (just before stopping).
Crucially, if you use a longer horn, the available force decreases proportionally.
Key formula:
Lifting force (kg) = Torque (kg·cm) ÷ Horn length (cm)
To convert to jin:1kg = 2kg.
These are typical setups you’ll find in hobby robotics, robot arms, and RC mechanisms.
Example from a real robot arm:
A 25kg servo with a 3 cm horn lifted a 750g (1.5 jin) object easily for 10,000 cycles. When the same servo tried to lift 8 kg (16 jin) with a 4 cm horn, it overheated and failed within 2 minutes. This matches the formula: 6.25 kg theoretical limit → actual safe working load is 60–70% of that due to friction,voltage drop, and dynamic forces.
Three factors drastically reduce the real-world weight a 25kg servo can handle:
Horn length (explained above) – the #1 variable.
Operating voltage – most 25kg servos are rated at 6.0V–7.4V. At 5.0V, torque drops by 20–30%.
Dynamic vs. static load – holding a weight is easier than accelerating it. Lifting requires extra torque to start moving.
Safety margin – never operate at stall torque for more than a few seconds. Continuous operation should be ≤50% of rated torque.
Common mistake: Beginners attach a 5 cm horn and expect to lift 10 kg (20 jin) – but the math shows only 5 kg (10 jin). Then the servo stalls, draws high current, and burns out.
Follow this method to know precisely how many jin your 25kg servo can lift in your project.
1. Measure your servo horn length from the center of the output shaft to the hole where the load is attached (in cm).
2. Apply the formula: Max static force (kg) = 25 ÷ horn length (cm).
3. Convert to jin: multiply by 2.
4. Apply a safety factor:
For intermittent lifting (few seconds): use 70% of max.
For continuous or repetitive lifting: use 50% of max.
5. Test at 30% of calculated load first, then increase gradually.
Example:
Horn length = 2.5 cm
Max force = 25 ÷ 2.5 = 10 kg (20 jin)
Safe continuous load = 5 kg (10 jin)
Safe intermittent load = 7 kg (14 jin)
Repeat the core truth: A 25kg servo does not lift 25kg (50 jin) except with a 1 cm horn – which is impractical. For 95% of real builds using 2–4 cm horns, your usable weight is 5–12.5kg (10–25kg).
Always prioritize horn length – if you need more lifting weight, use a shorter horn, not a higher voltage.
Never exceed 70% of calculated static force for moving loads.
Install a current-limiting fuse (e.g., 3A–5A for a 25kg servo) to prevent burnout.
Test your actual setup with a luggage scale or spring scale – pull from the horn tip while powering the servo. Compare the reading to the formula.
Action steps:
1. Measure your horn length now.
2. Calculate: 25 ÷ horn length (cm) × 2 × 0.6 (safe factor) = safe jin capacity.
3. Reduce load by 20% if using battery voltage below 6V.
4. Monitor servo temperature – if hot to touch (>60°C), reduce load by half.
By applying this torque-arm principle, you will never overload your 25kg servo and will get reliable, repeatable performance. Remember: the number on the servo is kg·cm, not kg of lift. Always do the math before you build.
Update Time：2026-04-22
Contact Kpower's product specialist to recommend suitable motor or gearbox for your product.
