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Torque To RPM Calculator

RPM Formula:

\[ RPM = \frac{Power \times 5252}{Torque} \]

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lb-ft

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1. What is the Torque to RPM Formula?

The RPM formula calculates rotational speed from power and torque using the relationship: RPM = Power × 5252 / Torque. This formula is commonly used in automotive and mechanical engineering to determine engine speed.

2. How Does the Calculator Work?

The calculator uses the RPM formula:

\[ RPM = \frac{Power \times 5252}{Torque} \]

Where:

Explanation: The constant 5252 comes from the relationship between horsepower, torque, and RPM, where 1 horsepower is defined as 550 foot-pounds per second, and there are 60 seconds in a minute and 2π radians in a revolution.

3. Importance of RPM Calculation

Details: Calculating RPM from torque and power is essential for engine performance analysis, gear ratio selection, and understanding the power characteristics of rotating machinery.

4. Using the Calculator

Tips: Enter power in horsepower (hp) and torque in pound-feet (lb-ft). Both values must be positive numbers greater than zero for accurate calculation.

5. Frequently Asked Questions (FAQ)

Q1: Why is the constant 5252 used in the formula?
A: The constant 5252 comes from the conversion between different units of measurement and ensures the formula produces accurate RPM values when using horsepower and pound-feet units.

Q2: Can this formula be used with metric units?
A: No, this specific formula is designed for imperial units. For metric units (kW and Nm), a different conversion factor would be needed.

Q3: What is the relationship between power, torque, and RPM?
A: Power is the product of torque and angular velocity (RPM). At 5252 RPM, horsepower and torque values are numerically equal in this formula.

Q4: When is this calculation most useful?
A: This calculation is particularly useful for automotive engineers, mechanics, and performance enthusiasts who need to understand engine characteristics and performance curves.

Q5: Are there limitations to this formula?
A: This formula assumes ideal conditions and may not account for mechanical losses, efficiency factors, or variable conditions in real-world applications.

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