rev limiter

[California Dreamin]

No that doesn't sound as impressive as:

Try holding a coffee cup and thrusting it back and forth 180 times in a second. (full of hot coffee lol)

Still....it's the mass of the piston and connecting rod going from 27mph to zero and back up to 27mph again 180 times in that second......
The Recipricating movement...thing would be a whole lot simpler travelling continuosly in one direction like a rotory enging.

Martin

[California Dreamin]

Actually thats not correct is it.....as the 27mph is the average or mean speed and so includes slowing down and speeding up phases. Therefore, the terminal velocity is going to be much higher....maybe double or three times as fast in the middle section of the bore.

I couldn't find a formula for this but as it's a progressive expedential movement I'm sure there must be one.

Martin

[ghost123uk]

All true Martin, but the difference in MTV v Mean is, whilst obvious, does not affect the "surprise" MPH value above to a great degree.

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

[ajsimmo]

Actually thats not correct is it.....as the 27mph is the average or mean speed and so includes slowing down and speeding up phases. Therefore, the terminal velocity is going to be much higher....maybe double or three times as fast in the middle section of the bore.

I couldn't find a formula for this but as it's a progressive expedential movement I'm sure there must be one.

Martin

Just been wrestling my brain against this one whilst I change a water pump...how to work it out???
Then I realised it's quite simple. (I think...) Just a quick reckon up as I'm busy, but it's all down to the peak speed of the piston at half way down the bore, ie at the moment the point on a circle prescribed by centre of crank journal is at top/bottom vertically down through engine as it sits, roughly perpendicular to conrod travel (ignoring the slight angle formed at this point coz that's too complicated). The diameter of the circle is equal to the stroke, so:
68.9mm x 3.14 = 216.35mm circumference. This circle is spinning at 5400rpm so the linear velocity at the tangential points described is 1,168,290mm per min or 19,471.5 mm/s or approx 19.5m/s.
So the peak velocity of the piston is only about 70kmh or just under 44mph!!!

[California Dreamin]

Well done that man...that sounds about right...

Martin

[The Bishop]

Actually thats not correct is it.....as the 27mph is the average or mean speed and so includes slowing down and speeding up phases. Therefore, the terminal velocity is going to be much higher....maybe double or three times as fast in the middle section of the bore.

I couldn't find a formula for this but as it's a progressive expedential movement I'm sure there must be one.

Martin

Just been wrestling my brain against this one whilst I change a water pump...how to work it out???
Then I realised it's quite simple. (I think...) Just a quick reckon up as I'm busy, but it's all down to the peak speed of the piston at half way down the bore, ie at the moment the point on a circle prescribed by centre of crank journal is at top/bottom vertically down through engine as it sits, roughly perpendicular to conrod travel (ignoring the slight angle formed at this point coz that's too complicated). The diameter of the circle is equal to the stroke, so:
68.9mm x 3.14 = 216.35mm circumference. This circle is spinning at 5400rpm so the linear velocity at the tangential points described is 1,168,290mm per min or 19,471.5 mm/s or approx 19.5m/s.
So the peak velocity of the piston is only about 70kmh or just under 44mph!!!

Yeek - Andy - glad you didn't have your head in my engine bay whilst you were doing that calculation!
Hope all is good with you.

[ajsimmo]

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?

[Ian Hulley]

Fit a rev counter and a standard non-restricted arm, the standard Velle pod with rev counter has a green area to show the optimal rpm to be doing in each gear. You should be paying attention to the engine, transmission and warning lights at all times anyway. I usually have my 3 ladies in the van when we're away so believe me I know all about not being able to hear the engine.