Erg to Foot-pound
erg
ft-lb
Conversion History
| Conversion | Reuse | Delete |
|---|---|---|
1 erg (Erg) → 7.375621492773e-8 ft-lb (Foot-pound) Just now |
Quick Reference Table (Erg to Foot-pound)
| Erg (erg) | Foot-pound (ft-lb) |
|---|---|
| 1 | 0.00000007375621492773 |
| 10 | 0.00000073756214927727 |
| 100 | 0.00000737562149277265 |
| 10,000 | 0.00073756214927726536 |
| 1,000,000 | 0.07375621492772653639 |
| 10,000,000 | 0.73756214927726536388 |
About Erg (erg)
The erg is a unit of energy in the CGS (centimeter-gram-second) system, equal to the work done by a force of one dyne over one centimeter — which equals exactly 10⁻⁷ joules. Once standard in physics and astronomy, ergs are now largely superseded by joules in most scientific work, though astrophysicists still use them for the luminosity of stars and the energy of astrophysical events. The total solar energy output is about 3.8 × 10³³ erg/s.
The kinetic energy of a mosquito in flight is about 1 erg. A supernova explosion releases roughly 10⁵¹ ergs of energy in total.
Etymology: From the Greek word ἔργον (ergon), meaning "work". Adopted as part of the CGS system formalised in the 1870s by the British Association for the Advancement of Science.
About Foot-pound (ft-lb)
The foot-pound (ft·lb) is the standard unit of torque and mechanical energy in the US customary system, equal to approximately 1.35582 joules. It represents the work done by a force of one pound-force through a displacement of one foot. Engine torque in American automotive engineering is quoted exclusively in foot-pounds (e.g., a V8 pickup truck producing 400 ft·lb). Bolt torque specifications in the US use foot-pounds for larger fasteners. One foot-pound equals 12 inch-pounds.
A typical car engine produces 150–400 ft·lb of torque. A cylinder head bolt on an engine is typically torqued to 60–90 ft·lb.
Erg – Frequently Asked Questions
Why do astrophysicists still use ergs instead of joules?
Astrophysics literature built decades of reference data in CGS units before SI became dominant. Key constants like solar luminosity (3.828 × 10³³ erg/s) and supernova energy (10⁵¹ erg, called a "foe") are baked into textbooks and databases. Switching to SI would require rewriting thousands of reference values, so the field maintains CGS by convention.
How many ergs of energy does a supernova release?
A core-collapse supernova releases roughly 3 × 10⁵³ ergs total, of which about 99% escapes as neutrinos. The visible light and kinetic energy of the ejected shell account for about 10⁵¹ ergs — a unit so common in astrophysics it has its own name: one "foe" (ten to the Fifty-One Ergs). In joules, that is 10⁴⁴ J, or the Sun's total output over 10 billion years.
What is an erg per second and why does it appear in stellar luminosity tables?
An erg per second is the CGS unit of power, equivalent to 10⁻⁷ watts. Astronomers quote stellar luminosities in ergs per second because the numbers align well with astrophysical scales: the Sun emits 3.846 × 10³³ erg/s, and supernovae peak at ~10⁴³ erg/s. Using watts would give the same exponents minus seven — less tidy for a field that already juggles 40-digit numbers daily.
What is the CGS system and why does it use the erg?
CGS (centimeter-gram-second) is a metric system that predates SI, formalised in the 1870s. It derives mechanical units from cm, g, and s: force in dynes (g·cm/s²) and energy in ergs (dyne·cm). CGS was standard in physics until the mid-20th century, and its Gaussian variant remains preferred in electromagnetism and astrophysics because Maxwell's equations take a simpler form.
How small is an erg in everyday terms?
One erg is 10⁻⁷ joules — roughly the kinetic energy of a mosquito in flight or the energy of a single grain of sand falling one centimeter. You would need about 10 million ergs to equal one joule, or 42 billion ergs to match the energy in a single dietary Calorie. The erg is useful precisely because atomic and astronomical quantities span so many orders of magnitude.
Foot-pound – Frequently Asked Questions
Why is engine torque measured in foot-pounds in the US?
American automotive engineering adopted foot-pounds because it was the natural imperial torque unit — one pound-force at one foot from the crankshaft center. The convention became entrenched through SAE standards, shop manuals, and dyno testing. Converting to newton-meters (1 ft·lb ≈ 1.3558 N·m) is straightforward, but the entire US aftermarket ecosystem — torque wrenches, spec sheets, and mechanics' training — runs on foot-pounds.
Why do diesel engines produce more torque but less horsepower than petrol engines at the same displacement?
Diesel engines compress air to much higher ratios (15–22:1 vs 8–12:1 for petrol), creating higher cylinder pressures that push harder on the piston — more force per stroke means more torque. But diesels rev lower (typically 4,000–4,500 RPM max vs 6,000–8,000 RPM) because the heavier rotating assembly and slower combustion limit speed. Since horsepower = torque × RPM / 5,252, the lower RPM ceiling caps peak horsepower despite the torque advantage.
What is the torque-to-horsepower relationship and why does it cross at 5,252 RPM?
Horsepower = torque (ft·lb) × RPM / 5,252. The constant 5,252 comes from unit conversion: 1 HP = 33,000 ft·lb/min, and 33,000 / (2π) ≈ 5,252. This means torque and horsepower curves on a dyno chart always intersect at exactly 5,252 RPM. Below that speed, torque is numerically higher; above it, horsepower is. This is why trucks optimize for low-RPM torque (pulling force) while sportscars chase high-RPM horsepower (speed).
What foot-pound torque values are typical for car lug nuts?
Most passenger cars specify 80–100 ft·lb for wheel lug nuts; light trucks and SUVs call for 100–140 ft·lb; and heavy-duty trucks may require 450–500 ft·lb. Under-torquing risks the wheel coming loose, while over-torquing can warp brake rotors or snap studs. A calibrated torque wrench — not an impact gun alone — is the safe approach.
How does foot-pound muzzle energy relate to firearm stopping power?
Muzzle energy in foot-pounds measures the kinetic energy of a bullet leaving the barrel. A 9 mm pistol produces about 350–400 ft·lb, a .45 ACP about 350–500 ft·lb, and a .308 rifle about 2,600–2,800 ft·lb. While muzzle energy is one factor in terminal performance, bullet construction, sectional density, and shot placement matter at least as much in real-world ballistics.