ghost123uk wrote:
Also true, especially in the case of racing engines, is the forces involved in starting and stopping the mass of the piston xxx times per second. It's amazing that con rods handle it.
Now, go away and work out the force experted in Newtons on the con rod of a WBX at 5,400 rpm. I will wait here for the answer
OK, here goes....
The Inertia Force of a rotating mass M is
This force is provided by the push/pull of the connecting-rod and with reference to the following diagram:-
The connecting-rod is in compression and tension and the force Q applied by the rod to the crank pin C is equivalent to an equal and parallel force through O together with a couple Q.x.
The Couple Q.x tends to retard the rotation of the crankshaft and its effect is taken into account when finding the net turning moment on the crankshaft.
The force at O is transmitted from the crankshaft through the main bearings and onto the engine frame.
Both the force at O and that at P may be resolved parallel and perpendicular to the line of stroke. The horizontal components are equal and opposite.
The one acting through P accelerates the reciprocating parts
The other through O is an unbalanced force applied to the frame and this causes the frame to slide backwards and forwards on its mountings as the crank rotates
The two vertical components are equal and opposite and constitute a couple applied to the frame which attempts to rotate the frame in a clockwise sense.
As the triangles Oba and POM are similar:-
The full effect on the engine frame of the inertia of the reciprocating mass is equivalent to a force F along the line of stroke at O and to the clockwise couple of magnitude S.OP
The Inertia Force can be separated into two parts:-
It is clear that the primary force is equivalent to the component along the line of stroke of the centrifugal force due to an equal mass M rotating with the crank and at crank radius. Consequently, in the case of a single-cylinder engine, the primary reciprocating force could be balanced by a rotating mass on the other side of the crank pin. However this would introduce an unbalanced component of the centrifugal force of magnitude Mr\omega^2\sin\thetaperpendicular to the line of stroke. A compromise solution (partial balance) is usually applied, the inertia force being reduced to a minimum when 50% of the reciprocating mass is balanced.
The secondary force is similarly equivalent to the component of the centrifugal force of mass M at radius r/4n rotating at 2\omega being coincident with the crank at inner dead-centre.
So, taking account of the above and making allowance for all losses of efficiency (friction, power losses due to exhaust gas expulsion and charge gas draw on other cylinders etc) I came to the following result:
Total forces acting on conrod = A LOT of Newtons (approx).
Now just need someone to insert the right numbers........any volunteers?