$HP = \frac{(Engine Displacement in cubic inches) x (RPM) x (Torque in pound-feet)}{5,252}$
To convert cubic centimeters (cc) to cubic inches, divide by 16.387. To convert RPM to minutes, divide by 60. To convert pound-feet (lb-ft) to pound-inches (lb-in), multiply by 12.
First, convert 347cc to cubic inches:
$347 \ cc \div 16.387 = 21.17 cubic inches$
Next, assume a typical operating speed of 3600 RPM for a snowblower engine:
$3600 \ RPM \div 60 = 60 minutes$
Now, assume a typical torque of 10 pound-feet for a 347cc snowblower engine:
$10 \ lb-ft \times 12 = 120 \ lb-in$
Finally, substitute these values into the formula:
$HP = \frac{(21.17 cubic\ inches) x (60 \ min) x (120 \ lb-in)}{5,252}$
$HP = 2.54$
Therefore, the conversion of 347cc to horsepower for a snowblower is approximately 2.54 horsepower.
