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Required force to rotate the Engine piston 
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#1
Dec1413, 06:43 AM

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Dears, please help me:
How to Calculate the Required Force to Push the Car Engine Piston Down if I Want to Rotate my shaft in 1500rpm, My Piston Dia Is 50mm, Piston Hight Is 100mm, Piston Mass Is 1kg, My Engine Is 4cylender and 4stroke? regards ibrahim 


#2
Dec1413, 11:40 PM

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P: 12,447

Welcome to PF;
Unless the piston is right at the top  it should fall all by itself. You need to find out what is stopping this. In a 4cylinder engine, it is probably attached to three other pistons, via the crank, so pushing one down means lifting another one and variously moving the other two ... this usually means that the pistons won't move by themselves as they are arranged to be balanced. You will have to overcome the friction (and drag) between the components  it's the sort of thing you measure rather than calculate. If the car is in gear  then the engine is attached to a load ... the additional force depends on the load. What do you need to know this force for? 


#3
Jan714, 11:22 PM

P: 16

The easier way to do this is to use these formulas:
BMEP (FourStroke Piston Engine) in PSI = [(150.8 x FtLbs Torque) / (CID)] CID = Cubic Inch Displacement of entire piston engine BMEP (TwoStroke Piston Engine) in PSI = [(75.4 x FtLbs Torque) / (CID)] CID = Cubic Inch Displacement of entire piston engine Brake Horsepower (BHP) = [(BMEP x L x A x N x K) / (33,000)] BMEP = Average Gas Pressure Exerted On Piston Head Area in PSI per powerstroke L = Piston Stroke Length in Feet. A = Piston Head Area in Square Inches = [(Piston Diameter / 2)² x (Π)] N = Number of Powerstrokes Per Minute ==> FourStroke = RPM / 2 ==> Two Stroke = RPM K = Number of Cylinders FtLbs Torque at Crankshaft (4Stroke Piston Engine) = [(BMEP x CID) / (150.8)] FtLbs Torque at Crankshaft (2Stroke Piston Engine) = [(BMEP x CID) / (75.4)] Force (pa) = [(Pressure) x (Area)] Force in Lbs Pressure in PSI (Lbs/Sq In) Area in Square Inches Force (ma) = [(Mass) x (Acceleration)] Force in Newtons Mass in kilograms Acceleration in meter/sec² 1 kilogram = 2.2 Lbs 1 meter = 3.28 feet 1 Lbs = 4.45 Newtons Piston Acceleration (GMax) = [(N² x L) / (2,189) x [(1) + (1 / (2A))] N = crankshaft speed (RPM) L = stroke in inches A = the ratio of the connecting rod length, (length between the center of crankpin to the center of the wristpin) to the piston stroke length. Total Force = [Force (pa)] + [Force (ma)] Use Lbs values for both Force types and add together for Total Force Work = Force x Distance Work in FtLbs Force in Lbs Distance in Feet Power = [(Force x Distance) / (Time)] Power in FtLbs/Sec ==> 550 FtLbs/Sec = 1 Horsepower = 746 Watts = 746 Joule/Sec. Force in Lbs Distance in Feet Time in Seconds Instead of figuring out what linear force on piston head is required to develop a certain torque and RPM at the crankshaft the better way to do this is to use the formulas up above. They translate the linear force on the piston head (based on BMEP & Piston Head Area) into Torque at the crankshaft. Regards,  MisterDynamics  January 08, 2014 


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