1 kgf·cm = 0.072 ft·lb/s
1 ft·lb/s = 13.826 kgf·cm
Example:
Convert 15 Kilogram-Force Centimeter to Foot-Pound per Second:
15 kgf·cm = 1.085 ft·lb/s
| Kilogram-Force Centimeter | Foot-Pound per Second |
|---|---|
| 0.01 kgf·cm | 0.001 ft·lb/s |
| 0.1 kgf·cm | 0.007 ft·lb/s |
| 1 kgf·cm | 0.072 ft·lb/s |
| 2 kgf·cm | 0.145 ft·lb/s |
| 3 kgf·cm | 0.217 ft·lb/s |
| 5 kgf·cm | 0.362 ft·lb/s |
| 10 kgf·cm | 0.723 ft·lb/s |
| 20 kgf·cm | 1.447 ft·lb/s |
| 30 kgf·cm | 2.17 ft·lb/s |
| 40 kgf·cm | 2.893 ft·lb/s |
| 50 kgf·cm | 3.617 ft·lb/s |
| 60 kgf·cm | 4.34 ft·lb/s |
| 70 kgf·cm | 5.063 ft·lb/s |
| 80 kgf·cm | 5.786 ft·lb/s |
| 90 kgf·cm | 6.51 ft·lb/s |
| 100 kgf·cm | 7.233 ft·lb/s |
| 250 kgf·cm | 18.083 ft·lb/s |
| 500 kgf·cm | 36.165 ft·lb/s |
| 750 kgf·cm | 54.248 ft·lb/s |
| 1000 kgf·cm | 72.33 ft·lb/s |
| 10000 kgf·cm | 723.3 ft·lb/s |
| 100000 kgf·cm | 7,233.003 ft·lb/s |
The Kilogram Force Centimeter (kgf·cm) is a unit of torque that measures the rotational force applied at a distance. This tool is essential for engineers, mechanics, and anyone involved in fields requiring precise torque calculations. By converting torque values into a standard unit, users can ensure accuracy in their projects, whether they are designing machinery or conducting experiments.
The kilogram force centimeter (kgf·cm) is defined as the torque resulting from a force of one kilogram-force applied at a radius of one centimeter from the axis of rotation. This unit is commonly used in mechanical engineering and physics to quantify rotational forces.
The kilogram force centimeter is part of the metric system, which is widely adopted globally for scientific and engineering applications. It is essential to standardize measurements to ensure consistency and reliability in calculations across various fields.
The concept of torque has been around since the early days of physics, but the specific unit of kilogram force centimeter gained prominence with the development of the metric system in the 18th century. Over the years, as engineering practices evolved, so did the need for precise torque measurements, leading to the widespread use of kgf·cm in various applications.
To illustrate how to use the kilogram force centimeter, consider a scenario where a force of 5 kgf is applied at a distance of 10 cm from the pivot point. The torque can be calculated as follows:
[ \text{Torque (kgf·cm)} = \text{Force (kgf)} \times \text{Distance (cm)} ]
[ \text{Torque} = 5 , \text{kgf} \times 10 , \text{cm} = 50 , \text{kgf·cm} ]
The kilogram force centimeter is particularly useful in mechanical engineering, automotive design, and any field where rotational forces are critical. It helps in determining the effectiveness of engines, motors, and other machinery.
To interact with the Kilogram Force Centimeter tool, follow these simple steps:
What is the kilogram force centimeter (kgf·cm)?
How do I convert kgf·cm to other torque units?
Why is kgf·cm important in engineering?
Can I use this tool for automotive applications?
What is the relationship between kgf·cm and Newton meters?
By utilizing the Kilogram Force Centimeter tool effectively, users can enhance their understanding of torque and improve their engineering calculations, ultimately leading to more successful projects and designs.
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.
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