1 ft·lb/s = 13,825.516 gf·cm
1 gf·cm = 7.2330e-5 ft·lb/s
Example:
Convert 15 Foot-Pound per Second to Gram-Force Centimeter:
15 ft·lb/s = 207,382.745 gf·cm
| Foot-Pound per Second | Gram-Force Centimeter |
|---|---|
| 0.01 ft·lb/s | 138.255 gf·cm |
| 0.1 ft·lb/s | 1,382.552 gf·cm |
| 1 ft·lb/s | 13,825.516 gf·cm |
| 2 ft·lb/s | 27,651.033 gf·cm |
| 3 ft·lb/s | 41,476.549 gf·cm |
| 5 ft·lb/s | 69,127.582 gf·cm |
| 10 ft·lb/s | 138,255.164 gf·cm |
| 20 ft·lb/s | 276,510.327 gf·cm |
| 30 ft·lb/s | 414,765.491 gf·cm |
| 40 ft·lb/s | 553,020.654 gf·cm |
| 50 ft·lb/s | 691,275.818 gf·cm |
| 60 ft·lb/s | 829,530.982 gf·cm |
| 70 ft·lb/s | 967,786.145 gf·cm |
| 80 ft·lb/s | 1,106,041.309 gf·cm |
| 90 ft·lb/s | 1,244,296.472 gf·cm |
| 100 ft·lb/s | 1,382,551.636 gf·cm |
| 250 ft·lb/s | 3,456,379.09 gf·cm |
| 500 ft·lb/s | 6,912,758.179 gf·cm |
| 750 ft·lb/s | 10,369,137.269 gf·cm |
| 1000 ft·lb/s | 13,825,516.359 gf·cm |
| 10000 ft·lb/s | 138,255,163.588 gf·cm |
| 100000 ft·lb/s | 1,382,551,635.88 gf·cm |
The foot-pound per second (ft·lb/s) is a unit of power that quantifies the rate at which work is done or energy is transferred. Specifically, it measures the energy expended when one foot-pound of work is performed in one second. This unit is commonly used in engineering and physics to express mechanical power, particularly in the context of torque and rotational systems.
The foot-pound per second is part of the Imperial system of measurements, which is primarily used in the United States. It is defined as the power required to move one pound a distance of one foot in one second. This unit is often compared to other power units such as watts, where 1 ft·lb/s is approximately equal to 1.35582 watts.
The concept of measuring power dates back to the late 18th century when Scottish engineer James Watt introduced the term "horsepower" to describe the output of steam engines. The foot-pound per second emerged as a practical unit for measuring power in mechanical systems, especially in the context of engines and machinery.
To understand how to use the foot-pound per second, consider an example where a motor exerts a torque of 10 ft·lb at a rotational speed of 100 revolutions per minute (RPM). To calculate the power output in ft·lb/s, you can use the formula:
[ \text{Power (ft·lb/s)} = \frac{\text{Torque (ft·lb)} \times \text{RPM} \times 2\pi}{60} ]
In this case:
[ \text{Power} = \frac{10 \times 100 \times 2\pi}{60} \approx 104.72 \text{ ft·lb/s} ]
The foot-pound per second is widely used in various fields, including mechanical engineering, automotive engineering, and physics. It helps engineers and technicians assess the performance of engines, motors, and other machinery by providing a clear understanding of power output.
To interact with the foot-pound per second tool on our website, follow these steps:
What is the foot-pound per second (ft·lb/s)?
How do I convert foot-pounds per second to watts?
What is the significance of using ft·lb/s in engineering?
Can I use this tool for other units of power?
Is there a formula to calculate power using torque and RPM?
By utilizing the foot-pound per second tool effectively, users can enhance their understanding of power measurements and improve their engineering projects. For more information and to access the tool, visit Inayam's Torque Converter.
The gram force centimeter (gf·cm) is a unit of torque that represents the rotational force applied at a distance of one centimeter from a pivot point. It is derived from the force exerted by one gram of mass under standard gravity (approximately 9.81 m/s²) acting at a distance of one centimeter from the axis of rotation. This unit is particularly useful in various engineering and physics applications where precise torque measurements are essential.
The gram force centimeter is part of the centimeter-gram-second (CGS) system of units. While the metric system has evolved, the CGS system remains relevant for certain scientific and engineering contexts. The standardization of this unit allows for consistent measurements across various applications, ensuring accuracy and reliability in calculations.
The concept of torque has been utilized since ancient times, but the formalization of units like the gram force centimeter emerged in the 19th century with the development of the CGS system. As engineering and scientific disciplines advanced, the need for precise torque measurements led to the adoption of various units, including the gram force centimeter, which remains in use today alongside the more commonly known Newton-meter (N·m) in the International System of Units (SI).
To illustrate how to calculate torque in gram force centimeters, consider a scenario where a force of 5 grams is applied at a distance of 10 centimeters from the pivot point. The torque can be calculated using the formula:
[ \text{Torque (gf·cm)} = \text{Force (g)} \times \text{Distance (cm)} ]
In this case:
[ \text{Torque} = 5 , \text{g} \times 10 , \text{cm} = 50 , \text{gf·cm} ]
The gram force centimeter is widely used in fields such as mechanical engineering, physics, and robotics. It is particularly valuable in applications involving small-scale mechanisms, where precise torque measurements are critical for performance and safety.
To effectively use the gram force centimeter tool on our website, follow these steps:
What is the gram force centimeter (gf·cm)?
How do I convert gram force centimeters to other torque units?
What are the applications of gram force centimeters?
How can I ensure accurate torque calculations?
Where can I find more information on torque measurements?
By utilizing the gram force centimeter tool effectively, you can enhance your understanding of torque measurements and apply this knowledge in various scientific and engineering contexts.
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