Torque of a small gasoline engine

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SUMMARY

A 20hp gasoline engine produces approximately 30 lb-ft of torque at 3500 RPM, calculated using the formula T = HP × 5252 / RPM. This formula is applicable for any motor where output is measured in horsepower and speed in RPM. It is crucial to note that the torque output can vary significantly based on the engine's performance characteristics and operating conditions. Instantaneous torque values may fluctuate, especially in single-cylinder engines, where torque can even reverse during certain cycles.

PREREQUISITES
  • Understanding of horsepower (HP) and torque (T) relationships
  • Familiarity with RPM (revolutions per minute) as a measure of engine speed
  • Basic algebra for manipulating equations
  • Knowledge of internal combustion engine operation principles
NEXT STEPS
  • Research the impact of RPM on torque output in gasoline engines
  • Learn about torque curves and their significance in engine performance
  • Explore the differences in torque output between single-cylinder and multi-cylinder engines
  • Investigate how engine tuning affects horsepower and torque relationships
USEFUL FOR

Engineers, automotive enthusiasts, and mechanics interested in understanding engine performance metrics, particularly those working with gasoline engines and torque calculations.

codyo83
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Does someone know how much torque a 20hp gasoline engine produce at 3500rpm? An equation that I can use to figure this out would be helpful also.
 
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Found the formula! 5,252×20÷3,800=27.6421052632 I used 3800rpm instead of 3500rpm - (note so not to confuse anyone)
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hp rpm torque
 

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codyo83 said:
Does someone know how much torque a 20hp gasoline engine produce at 3500rpm? An equation that I can use to figure this out would be helpful also.

HP = T * RPM / 5252, where T is torque in lb-ft

For your motor, HP = 20, RPM = 3500, therefore doing a little algebra, T = 20 * 5252 / 3500 = 30 lb-ft of torque produced by the motor at this output condition.

Since 1 HP = 33,000 lb-ft/min, and 1 revolution of the output shaft = 2 pi radians, then the factor 5252 = 33,000 / (2 pi).

The formula works for any motor where the output is in HP and the speed is in RPM.
 
Keep in mind though that this only works if the motor is producing 20hp at 3500rpm. What is colloquially called a "20hp" motor is really a motor which, when operated at its peak performance point, outputs 20hp. If peak performance occurs at some other point than 3500rpm, the torque at 3500rpm will be lower than your calculated number.
 
Bear also in mind the fact that the torque produced by any IC engine fluctuates quite substantially. On a single cylinder engine, the instantaneous torque may actually reverse. So, while the power calculation (based on average torque and average speed) can give you an average torque value, instantaneous values will be all over the place.
 
Dr.D said:
Bear also in mind the fact that the torque produced by any IC engine fluctuates quite substantially. On a single cylinder engine, the instantaneous torque may actually reverse. So, while the power calculation (based on average torque and average speed) can give you an average torque value, instantaneous values will be all over the place.

Indeed it does. Here's a calculated torque output from a single cylinder engine showing negative torque on compression and exhaust and exhaust strokes.

The peak torque is showing around 430lbft, however this is actually representative of a single cylinder in a four cylinder engine producing an average torque value of a mere 138lbft.
 

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Kozy, thanks for posting the pressure - crank angle curve. That looks just about right.

It also show how much higher the instantaneous maximum torque is compared to the average torque, as well as showing the torque reversal.
 

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