Are ft·lb and lb·ft the same unit?

Yes. Both ft·lb and lb·ft mean pound-force times foot. The notation varies by industry. SAE official uses lb·ft while most workshop manuals use ft·lb. They are identical quantities.

What is a torque-to-yield bolt?

TTY bolts are intentionally permanently stretched during installation for consistent clamping force. Installation uses two phases: initial torque (N·m), then additional rotation by angle (degrees). They cannot be re-used. Common for cylinder head bolts since the 1990s.

🔧 Torque Conversion Calculator

Convert between Newton metres, foot-pounds, inch-pounds, kgf·m, kN·m, dyne·cm and 15+ units — with torque definition, power-torque-speed relation, clamping force, preload & bolt stress formulas rendered in MathJax

15+ Units SI · Imperial · CGS N·m ↔ ft·lb · in·lb Free & Instant

🔄 Torque Unit Converter

Enter Value From Unit

Result To Unit

1 N·m = 0.737562 lb·ft

Formula: value × 0.737562

🔩 All Units at Once

💡 Definition: Torque = force × perpendicular distance from pivot: \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \) — all units convert through the newton metre (N·m) as the SI pivot unit.

📖 How to Use This Torque Conversion Calculator

1

Filter by Unit Category (Optional)
Click SI/Metric (N·m, kN·m, kgf·m, kgf·cm, N·cm), Imperial/US (lb·ft, lb·in, oz·ft, oz·in), or CGS (dyn·cm) to narrow the unit dropdowns. "All Units" shows all 17 units together.
2

Enter Your Torque Value
Type the value into "Enter Value." Any numeric input is accepted — from micronewton metres (µN·m) for electronic components to kilonewton metres (kN·m) for bridge engineering. Scientific notation auto-applies for extreme values.
3

Select From and To Units
Choose source in "From Unit" and target in "To Unit." The result and exact conversion factor appear instantly in the green result box.
4

Use Quick-Convert Buttons
Click N·m→ft·lb, ft·lb→N·m, N·m→in·lb, in·lb→N·m, kN·m→ft·lb, N·m→kgf·m, or kgf·cm→N·m for the most common torque conversions. Both dropdowns set automatically.
5

View All Units & Copy
"All Units at Once" shows your torque in every supported unit simultaneously. Click "📋 Copy Result" to copy the primary conversion to clipboard for technical documents, workshop manuals, or engineering reports.

📐 Torque Unit Conversion Factors Reference

From	To	Multiply By	Math Expression
1 N·m	lb·ft	0.737562	\( 1\,\text{N·m} = 0.737562\,\text{lb·ft} \)
1 lb·ft	N·m	1.355818	\( 1\,\text{lb·ft} = 1.355818\,\text{N·m} \)
1 N·m	lb·in	8.850745	\( 1\,\text{N·m} = 8.850745\,\text{lb·in} \)
1 lb·in	N·m	0.112985	\( 1\,\text{lb·in} = 0.112985\,\text{N·m} \)
1 lb·ft	lb·in	12	\( 1\,\text{lb·ft} = 12\,\text{lb·in} \)
1 kgf·m	N·m	9.80665	\( 1\,\text{kgf·m} = g_n \times 1\,\text{m} = 9.80665\,\text{N·m} \)
1 kgf·cm	N·m	0.0980665	\( 1\,\text{kgf·cm} = 0.0980665\,\text{N·m} \)
1 kN·m	N·m	1,000	\( 1\,\text{kN·m} = 10^3\,\text{N·m} \)
1 N·cm	N·m	0.01	\( 1\,\text{N·cm} = 10^{-2}\,\text{N·m} \)
1 dyn·cm	N·m	1×10⁻⁷	\( 1\,\text{dyn·cm} = 10^{-7}\,\text{N·m} \)

🔧 Understanding Torque — A Complete Engineering Guide

Torque is the rotational equivalent of force — the tendency of a force to cause or change the rotation of an object around an axis. While force pushes or pulls an object in a linear direction, torque twists or rotates. Every time you open a jar lid, tighten a bolt with a wrench, turn a steering wheel, or start a motor, torque is the physical quantity doing the mechanical work.

Understanding torque — and accurately converting between its many measurement units — is indispensable in virtually every branch of mechanical engineering, structural engineering, automotive design, aerospace, robotics, and precision manufacturing. A motor specification in a Japanese service manual uses Newton metres (N·m); an American workshop manual specifies the same bolt in foot-pounds (ft·lb); a RC servo datasheets quotes kilogram-centimetres (kgf·cm). Converting accurately between these is not optional — it directly determines whether a fastener holds or fails, whether a motor is correctly specified, and whether a structure is safely loaded.

Torque — Vector Definition

\[ \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \qquad \tau = r F \sin\theta \]

\(\boldsymbol{\tau}\) = torque vector (N·m) · \(\mathbf{r}\) = position vector from pivot to force application point (m) · \(\mathbf{F}\) = force vector (N) · \(\theta\) = angle between \(\mathbf{r}\) and \(\mathbf{F}\) · Maximum torque when \(\theta = 90°\): \(\tau = rF\) · SI unit: newton metre (N·m)

⚖️ The Newton Metre — SI Standard for Torque

The newton metre (N·m) is the SI derived unit of torque. It equals the torque produced by a force of one newton applied perpendicularly at a distance of one metre from the pivot point. The newton metre is also the SI unit of energy (the joule, J), and while 1 N·m = 1 J mathematically, torque and energy are physically distinct quantities — torque involves a rotational axis, while energy is a scalar. Context always disambiguates: torque is expressed as N·m, energy as J.

The practical range of torque spans an extraordinary scale: from a few µN·m (micronewton metres) for MEMS gyroscopes and precision watch mechanisms, through tens of N·m for automotive fasteners, hundreds of N·m for car engines, thousands of N·m for heavy diesel trucks and electric motors, to millions of N·m for ship propellers and wind turbines.

N·m Scale — Practical Reference Points

\( 1\,\text{N·m} = 0.737562\,\text{ft·lb} = 8.850745\,\text{in·lb} = 0.101972\,\text{kgf·m} \)

\( 1\,\text{kN·m} = 1{,}000\,\text{N·m} = 737.562\,\text{ft·lb} \)

\( 1\,\text{kgf·m} = g_n \times 1\,\text{m} = 9.80665\,\text{N·m} \approx 7.233\,\text{ft·lb} \)

\( 1\,\text{kgf·cm} = 0.0980665\,\text{N·m} = 0.8679\,\text{in·lb} \)

\(g_n = 9.80665\,\text{m/s}^2\) (standard gravity, exact) — this exact value converts all gravitational torque units (kgf·m, gf·cm, etc.) to SI (N·m)

🔩 N·m to ft·lb — The Most Common Torque Conversion

The foot-pound (ft·lb or lb·ft) is the imperial unit of torque dominant in US and UK automotive workshops, aviation maintenance (FAA-approved procedures), and American civil engineering. One foot-pound is the torque produced by a one-pound force applied perpendicularly at a distance of one foot from the pivot. Because the international yard and pound were fixed in 1959, the conversion between N·m and ft·lb is exact:

N·m ↔ ft·lb — Exact Conversion Derivation

\( 1\,\text{ft} = 0.3048\,\text{m} \quad \text{(exact, 1959 agreement)} \)

\( 1\,\text{lbf} = 0.45359237\,\text{kg} \times 9.80665\,\text{m/s}^2 = 4.448222\,\text{N} \quad \text{(exact)} \)

\( 1\,\text{ft·lb} = 0.3048\,\text{m} \times 4.448222\,\text{N} = 1.355818\,\text{N·m} \quad \text{(exact)} \)

\( 1\,\text{N·m} = \frac{1}{1.355818} = 0.737562\,\text{ft·lb} \)

Quick memory aid: 1 N·m ≈ ¾ ft·lb (accurate to ~1.7%). More precisely: multiply N·m × 0.7376 to get ft·lb, multiply ft·lb × 1.3558 to get N·m.

📌 Worked Example — Vehicle Wheel Lug Nut Torque

Problem: A European car manufacturer specifies the wheel lug nut torque as 120 N·m. A US workshop technician needs this in ft·lb to set their torque wrench.

\[ 120\,\text{N·m} \times 0.737562 = \mathbf{88.5\,\text{ft·lb}} \]

Reverse check: \( 88.5\,\text{ft·lb} \times 1.355818 = 120.0\,\text{N·m} \) ✓

Answer: Set the torque wrench to 88.5 ft·lb. Under-torquing risks the wheel loosening at speed; over-torquing risks stud fracture or warping the brake rotor disc. Neither outcome is acceptable — precise conversion is a safety-critical calculation.

🪛 N·m to in·lb — Precision Fastener Engineering

The inch-pound (lb·in or in·lb) is the small-scale imperial torque unit, used where foot-pounds would produce values too small to measure accurately with a standard workshop wrench. Typical applications include: cylinder head valve cover bolts, motorcycle engine side covers, bicycle stem bolts, electronics enclosure fasteners, medical device assemblies, and aerospace avionics racks.

The relationship between inch-pounds and foot-pounds is exact and geometrically obvious — there are exactly 12 inches in one foot, so:

in·lb ↔ ft·lb ↔ N·m — Three-Way Conversion

\( 1\,\text{ft·lb} = 12\,\text{in·lb} \quad \text{(exact — 1 foot = 12 inches)} \)

\( 1\,\text{in·lb} = \frac{1}{12}\,\text{ft·lb} = 0.0833\overline{3}\,\text{ft·lb} \)

\( 1\,\text{N·m} = 8.850745\,\text{in·lb} \qquad 1\,\text{in·lb} = 0.112985\,\text{N·m} \)

\( \tau\,[\text{in·lb}] = \tau\,[\text{N·m}] \times 8.850745 \qquad \tau\,[\text{N·m}] = \tau\,[\text{in·lb}] \times 0.112985 \)

Conversion chain: N·m → ft·lb → in·lb (multiply ×0.737562, then ×12). Or directly N·m → in·lb: multiply ×8.850745 = 0.737562 × 12.

📌 Worked Example — Valve Cover Bolt (Motorcycle Engine)

Problem: A Yamaha workshop manual specifies valve cover bolts at 10 N·m. Convert to in·lb and ft·lb.

N·m → in·lb: \( 10 \times 8.850745 = \mathbf{88.5\,\text{in·lb}} \)

N·m → ft·lb: \( 10 \times 0.737562 = \mathbf{7.38\,\text{ft·lb}} \)

Insight: At 7.38 ft·lb, most workshop torque wrenches struggle with fine resolution. But an in·lb torque wrench calibrated to 88.5 in·lb gives accurate, repeatable results. This is exactly why in·lb torque specifications exist for small fasteners — optimal measurement range for the tool.

⚡ Torque, Power & RPM — The Fundamental Relationship

Torque and power are fundamentally related through rotational speed. An engine that produces high torque at low RPM is described as having "good low-end pull" — ideal for trucks, tractors, and electric vehicles. An engine with high power achieved through high RPM is described as "revvy" — characteristic of naturally-aspirated sports car engines. This relationship is the reason electric vehicles feel so instant: they deliver maximum torque from 0 RPM.

Power–Torque–Speed Relationship

\[ P = \tau \cdot \omega = \frac{2\pi \cdot N \cdot \tau}{60} \]

\(P\) = power (W) · \(\tau\) = torque (N·m) · \(\omega\) = angular velocity (rad/s) · \(N\) = rotational speed (RPM) · Divide \(P\) by 1,000 for kW, by 745.7 for hp · In US customary: \(\text{hp} = \tau\,[\text{ft·lb}] \times N / 5{,}252\)

📌 Worked Example — Engine Power from Torque Spec

Problem: An engine specification sheet states: 380 N·m peak torque at 3,000 RPM. Calculate power in kW and horsepower.

\[ P = \frac{2\pi \times 3{,}000 \times 380}{60} = \frac{6.2832 \times 3{,}000 \times 380}{60} = \frac{7{,}162{,}832}{60} = 119{,}381\,\text{W} \approx \mathbf{119.4\,\text{kW}} \]

In horsepower: \( 119{,}381 \div 745.7 = \mathbf{160.1\,\text{hp}} \)

Cross-check in US customary: \( \frac{380\,\text{N·m} \times 0.737562\,[\text{to ft·lb}] \times 3{,}000}{5{,}252} = \frac{280.3 \times 3{,}000}{5{,}252} = \mathbf{160.1\,\text{hp}} \) ✓

📊 Torque and power crossover: In the US customary system, torque (in ft·lb) and power (in hp) always have equal numerical values at exactly 5,252 RPM. Below 5,252 RPM, the torque curve is numerically higher (torque-dominant). Above 5,252 RPM, the power curve is higher (power-dominant). This crossover point is a direct result of the 5,252 constant = 33,000 ft·lb/min per hp ÷ 2π ÷ (1 min/60s).

🔩 Bolt Preload & Clamping Force — Why Torque Specs Matter

When you tighten a bolt to a specific torque, the goal is not torque itself — it is the clamping force (preload) that torque creates by stretching the bolt. The torque specification in a service manual is an indirect way to specify the desired bolt elongation and the clamping force it generates. Understanding this distinction explains why torque specifications are so critical: they are ultimately a proxy for a structural force.

Torque–Preload Relationship

\( T = K \cdot d \cdot F \qquad \text{(Torque–Preload equation)} \)

\( F = \frac{T}{K \cdot d} \qquad \text{(solving for clamping force)} \)

\( \sigma = \frac{F}{A_s} \qquad \text{(bolt stress from preload)} \)

\( \delta = \frac{F \cdot L}{E \cdot A} \qquad \text{(bolt elongation / stretch)} \)

\(T\) = applied torque (N·m) · \(K\) = nut factor / torque coefficient (dimensionless, ~0.12–0.20 dry, ~0.08–0.15 lubricated) · \(d\) = nominal bolt diameter (m) · \(F\) = clamping force (N) · \(A_s\) = bolt tensile stress area (m²) · \(\sigma\) = bolt tensile stress (Pa) · \(\delta\) = bolt stretch (m) · \(E\) = Young's modulus (Pa) · \(L\) = bolt grip length (m)

📌 Worked Example — Clamping Force from Torque Spec

Problem: An M12 × 1.75 steel bolt is torqued to 80 N·m on a dry thread. What clamping force is generated? (K = 0.20 for dry, d = 0.012 m)

\[ F = \frac{T}{K \cdot d} = \frac{80}{0.20 \times 0.012} = \frac{80}{0.0024} = \mathbf{33{,}333\,\text{N}} = \mathbf{33.3\,\text{kN}} \]

With engine oil lubrication (K = 0.14): \( F = 80 / (0.14 \times 0.012) = 47{,}619\,\text{N} = 47.6\,\text{kN} \)

Critical finding: Lubricating the thread increases clamping force by 43% for identical torque! This is why workshop manuals always specify "dry torque" or "lubricated torque" — they are different specifications, not interchangeable. Using the dry torque spec on lubricated threads risks bolt fracture.

⚠️ Lubrication warning: Every published torque specification assumes a defined friction state — almost always "clean and dry." Applying engine oil, anti-seize compound (MoS₂), or threadlocker changes the K factor and therefore the clamping force produced at any given torque. Typical reductions: engine oil reduces required torque by ~15–25%; anti-seize paste reduces by ~25–35%. Always check whether the manufacturer's spec is for dry or lubricated threads before torquing any critical fastener.

🤖 kgf·cm — Servo Motors, Robotics & Micro-Engineering

The kilogram-force centimetre (kgf·cm) is the standard torque unit in the RC hobby, robotics, and consumer servo motor industries worldwide. It represents the torque produced by a one-kilogram force acting perpendicularly at a radius of one centimetre. Despite not being an SI unit, it is universally used in servo datasheets because it maps intuitively to the physical test used to rate servos: attach a lever arm and measure how many grams (or kilograms) it can hold at a given radius.

kgf·cm ↔ N·m — Servo and Robotics Conversion

\( 1\,\text{kgf·cm} = 1\,\text{kg} \times 9.80665\,\text{m/s}^2 \times 0.01\,\text{m} = 0.0980665\,\text{N·m} \)

\( 1\,\text{N·m} = \frac{1}{0.0980665} = 10.1972\,\text{kgf·cm} \)

\( \tau\,[\text{kgf·cm}] = \tau\,[\text{N·m}] \times 10.1972 \)

\( \tau\,[\text{N·m}] = \tau\,[\text{kgf·cm}] \times 0.0980665 \)

A servo rated "5 kg·cm" (often written kgf·cm) = 5 × 0.0980665 = 0.490 N·m at 1 cm arm radius. At 2 cm arm: can hold 2.5 kg. At 0.5 cm arm: can hold 10 kg. Torque is always arm-length dependent — the rating only applies at 1 cm from the servo output shaft.

🌍 Real-World Torque Reference Table

Application	Typical Torque (N·m)	In ft·lb	In in·lb
⌚ Luxury watch screw	0.001–0.010	0.00074–0.0074	0.009–0.088
🤖 RC micro servo (5 kgf·cm)	0.49	0.36	4.3
💻 Laptop hinge bolt	0.5–1.5	0.37–1.11	4.4–13.3
🔩 Bicycle stem bolt	5–8	3.7–5.9	44–71
🏍️ Motorcycle valve cover	8–12	5.9–8.9	71–106
🚗 Spark plug (automotive)	20–30	14.8–22.1	177–265
🔧 Wheel lug nut (passenger car)	80–140	59–103	708–1,239
🚗 Economy car engine (peak)	150–220	111–162	1,327–1,947
🏎️ Performance car engine (peak)	400–700	295–516	3,540–6,195
⚡ Tesla Model S Plaid (motor)	~1,840	~1,357	~16,283
🚚 Heavy diesel truck (peak)	1,000–3,000	738–2,213	8,851–26,552
🚢 Large ship propeller shaft	50,000–500,000	36,878–368,781	—

📦 Torque Unit System Guide

⚖️

Newton Metre (N·m)

SI unit. 1 N·m = 1 J = force of 1 N at 1 m radius. Global engineering standard. All service manuals outside North America use N·m exclusively. Never confuse with joules (same dimensions, different physical context).

🔩

Foot-Pound (ft·lb)

US/UK imperial unit. 1 ft·lb = 1.355818 N·m. SAE standard for all US automotive, aviation (FAA AC 43.13), and US military specifications. Note: "ft-lb" and "lb-ft" both appear in practice, always meaning the same unit.

🪛

Inch-Pound (in·lb)

Small-scale imperial unit. 1 in·lb = 1/12 ft·lb = 0.112985 N·m. Used for small fasteners: electronics, avionics, medical devices, bicycle components. 1 N·m = 8.85 in·lb — quick rule: ×9 for rough N·m→in·lb.

🤖

kgf·cm (Servo/Robotics)

Non-SI gravitational unit. 1 kgf·cm = 0.0980665 N·m. Universal in RC servos, stepper motors, hobby robotics. "10 kg·cm servo" = 0.981 N·m. Convert to N·m: multiply kgf·cm × 0.0981.

Written & Reviewed by Num8ers Editorial Team — Mechanical Engineering, Automotive Fastener & Structural Analysis Specialists Last updated: April 2026 · Conversion factors verified against NIST SP 811 (2008), BIPM SI Brochure 9th edition (2019), SAE International fastener standards, ISO 898-1 (mechanical properties of fasteners), and Machinery's Handbook 31st edition.

❓ Frequently Asked Questions — Torque Conversion

How do I convert N·m to ft·lb (Newton metres to foot-pounds)?

Multiply by 0.737562. \(1\,\text{N·m} = 0.737562\,\text{ft·lb}\). Example: 120 N·m × 0.737562 = 88.5 ft·lb. Quick approximation: multiply by 0.74 (accurate to ~0.3%). The exact factor derives from the 1959 international definitions: 1 ft = 0.3048 m (exact), 1 lbf = 4.448222 N (exact), so 1 ft·lb = 1.355818 N·m (exact).

How do I convert ft·lb to N·m (foot-pounds to Newton metres)?

Multiply by 1.355818. \(1\,\text{ft·lb} = 1.355818\,\text{N·m}\). Example: 100 ft·lb × 1.355818 = 135.6 N·m. Quick approximation: multiply by 1.36. Example: 60 ft·lb = 81.35 N·m. This is the most common torque conversion in cross-border automotive and machinery work.

How do I convert N·m to in·lb (Newton metres to inch-pounds)?

Multiply by 8.850745. \(1\,\text{N·m} = 8.850745\,\text{in·lb}\). Example: 10 N·m = 88.5 in·lb. Quick approximation: multiply by 9 (≈1.7% overestimate). Reverse (in·lb to N·m): multiply by 0.112985. Example: 120 in·lb = 13.56 N·m.

How do I convert in·lb to ft·lb (inch-pounds to foot-pounds)?

Divide by 12 (exact). \(1\,\text{ft·lb} = 12\,\text{in·lb}\) because 1 foot = 12 inches exactly. Example: 120 in·lb ÷ 12 = 10 ft·lb. Example: 60 in·lb ÷ 12 = 5 ft·lb. Reverse (ft·lb to in·lb): multiply by 12. Example: 8 ft·lb = 96 in·lb. This conversion is exact and requires no rounding.

How do I convert kgf·cm to N·m for servo motors?

Multiply by 0.0980665. \(1\,\text{kgf·cm} = 0.0980665\,\text{N·m}\). Example: A "10 kg·cm" servo = 10 × 0.0980665 = 0.981 N·m. Quick approximation: multiply by 0.098. Reverse (N·m to kgf·cm): multiply by 10.1972. Example: 2 N·m = 20.39 kgf·cm.

What is the difference between torque and energy if both are measured in N·m?

Torque (N·m) and energy (J = N·m) have identical SI dimensions but are physically distinct quantities. Torque is a rotational force — it acts around an axis and does not require motion to exist (a stationary bolt under tension has torque). Energy is a scalar quantity representing work done. Context always disambiguates: torque specifications use "N·m"; energy uses "J" (joule). Never confuse them in calculations.

How does lubrication affect torque specifications?

Lubrication drastically increases clamping force for a given torque. The torque–preload equation \(T = K \cdot d \cdot F\) shows that reducing the nut factor K (from DRY ~0.20 to LUBRICATED ~0.12) increases clamping force \(F = T/(K \cdot d)\) by ~67% for the same torque. Rule of thumb: reduce the published dry torque spec by 20–30% when using anti-seize paste. Always check whether specifications are for dry or lubricated threads.

How is torque related to power and RPM?

\(P = \tau \cdot \omega = 2\pi N \tau / 60\) (where P is in watts, τ in N·m, N in RPM). In US customary: \(\text{hp} = \tau\,[\text{ft·lb}] \times N / 5{,}252\). At 5,252 RPM, numerically torque (ft·lb) = power (hp). Below 5,252 RPM, torque numerically exceeds hp; above it, hp exceeds torque. Power is the rate of doing torque work; torque is the instantaneous rotational force.

What is the difference between N·m and kN·m?

\(1\,\text{kN·m} = 1{,}000\,\text{N·m} = 737.562\,\text{ft·lb}\). Kilonewton metres are used for large structural engineering (bridge bolts, wind turbine blades, ship shafts) where N·m values would be inconveniently large. A large wind turbine gearbox may specify fasteners at 10–50 kN·m (7,376–36,878 ft·lb) — values utterly beyond the range of any hand tool.

What is a torque-to-yield (TTY) bolt and how does it change torque procedures?

TTY bolts are intentionally yielded (permanently stretched) during installation to provide consistent clamping force. They cannot be re-used. Installation typically involves two phases: (1) initial torque to a specified N·m value, then (2) rotation by a specified angle (e.g., +90°). The angle phase cannot be calculated with a standard torque conversion — it requires a protractor gauge or angle-torque wrench. Common for cylinder head bolts, main bearing caps, and wheel hub nuts since the 1990s.

Are ft·lb and lb·ft the same thing?

Yes — in torque context, ft·lb and lb·ft are the same unit. Both mean "pound-force times foot." The notation varies by industry and country: SAE official standard uses "lb·ft"; most workshop manuals and informal usage writes "ft·lb." Neither is "more correct." By ISO convention, for torque the unit should be explicitly N·m (SI) or lbf·ft (Imperial). The confusion with lb·ft of energy (= ft·lbf of work) is avoided by context.

How accurate is the Num8ers Torque Conversion Calculator?

Uses exact or high-precision conversion factors per NIST SP 811 and BIPM: 1 ft·lb = 1.355818 N·m (derived from exact 1959 yard/pound definitions); 1 kgf·m = 9.80665 N·m (exact standard gravity); 1 lbf = 4.44822162 N (exact); 1 in·lb = 0.1129848 N·m (exact). JavaScript double precision provides ~15 significant digits. Results are mathematically exact within floating-point arithmetic.

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