crank slider mechanism. velocity of piston & angular velocity of link HP needed

[deano_1010]

The instantaneous configuration of a slider crank mechanism has a crank GH 10cm long, the connecting rod HP is 50cm. The crank makes an angle of 60 degree with the inner dead centre position and is rotating at 110 rev/min. Determine the velocity of the piston P and the angular velocity of the link HP.

 

cannot find any simliar examples in text books or online which will help me with this question.

 

 

Right, first off i've converted the 110rev/min=w into 11.5Rads/s

so w=11.5Rads/s

i then assumed you had to find the velocity at H in order to find the velocity at P

so Velocity at H = (Wgh) X gh

Vh = 11.5rads/s X 0.1m

= 1.15ms^-1

dont know whether that is correct, really struggling on this question, step by step help would be more than fantastic. thanks guys

[Janus]

Hint:

Consider the horizontal and vertical components of the velocity of H and their effects on the motion of P and the rotation of HP.

[michel123456]

Sorry taking back my post.

Edited April 11, 2012 by michel123456

[imatfaal]

I have made it to answers - the one the velocity of p is nice and simple. But the angular velocity of HP (not HG as the OP seems to have used) is pretty horrible.

[michel123456]

The crank makes an angle of 60 degree with the inner dead centre position

 

What is this information used for?

[Joatmon]

What is this information used for?

Because the piston speed varies with the angle.

 

 

 

[Janus]

I have made it to answers - the one the velocity of p is nice and simple. But the angular velocity of HP (not HG as the OP seems to have used) is pretty horrible.

 

Tell me about it. I had to solve this type of problem when I was creating a animation of the power and valve train of a engine. I had to come up the equations to describe the movement of the piston rod.

[michel123456]

Because the piston speed varies with the angle.

 

So what?

 

(...)The crank makes an angle of 60 degree with the inner dead centre position (...)

 

I don't understand the statement. In the graph the 60 degree angle is with the PG axis. In this position, P is not the "dead centre" position.

Or am I missing something?

Edited April 12, 2012 by michel123456

[Joatmon]

So what?

 

 

 

I don't understand the statement. In the graph the 60 degree angle is with the PG axis. In this position, P is not the "dead centre" position.

Or am I missing something?

 

The piston is a complete lump of material that slides to and fro in the cylinder under continually changing speed. As I understand the question I think you are asked to calculate the speed when the crank is at 60 degrees. The fact that P is not shown on the centre line does not affect the answer - its just a marker on the cylinder wall.

I notice P is under the piston small end bearing (although not essential to the question, but probably meant to be helpful). I think the triangle G, H, Small end bearing is valid in the calculation.

Edited April 12, 2012 by Joatmon

[imatfaal]

edit

x-posted with Joatmon. Yes - Agree. P being the point of contact between the long rod and the piston - I see what M123456 meant about the diagram now.

 

So what?

I don't understand the statement. In the graph the 60 degree angle is with the PG axis. In this position, P is not the "dead centre" position.

Or am I missing something?

 

Not sure the OP is coming back - but I will keep it vague even so. A crank like this one converts rotary motion into linear - to find out how much it imparts we look at the motion of H around G, we can define offset in x and in y in terms of the distance HG and the angle alpha between the the crank and the horizontal (ie y=HG.Sin(alpha). The position of P is related to x and/or y - the velocity is thus related to the time derivative of those. This will clearly depend on the angle alpha . It is not as simple as I first thought - but not too bad.

 

The angular velocity of HP is the rate of change of angle that of the line HP around the point P - this can also be worked out, I think, but the answers are pretty vile. The angle alpha (60 degrees) in the OP is the angle between the crank and the line joining P and G.

Edited April 12, 2012 by imatfaal

[michel123456]

The piston is a complete lump of material that slides to and fro in the cylinder under continually changing speed. As I understand the question I think you are asked to calculate the speed when the crank is at 60 degrees. The fact that P is not shown on the centre line does not affect the answer - its just a marker on the cylinder wall.

I notice P is under the piston small end bearing (although not essential to the question, but probably meant to be helpful). I think the triangle G, H, Small end bearing is valid in the calculation.

 

Oh yes, you must be right.

So the angular velocity of HP is also asked for this particular instant I presume.

Edited April 12, 2012 by michel123456

[Joatmon]

Oh yes, you must be right.

So the angular velocity of HP is also asked for this particular instant I presume.

Since the angular velocity will change with angle I also presume the same. I am not very good with maths - come in Imatfaal!

 

 

 

[michel123456]

Since the angular velocity will change with angle I also presume the same. I am not very good with maths - come in Imatfaal!

Except that if the piston was at the dead centre position, the speed would be null, and the angular speed also null.

But the sketch disagrees. As it is drawn, the piston is not at the dead position.

So, after reading all this twice, I still don't understand the statement.

 

I had in mind a stupid mechanism like this:

 

Because it is nowhere said that the piston is aligned with the center G (although it may seem evident).

[Joatmon]

Because it is nowhere said that the piston is aligned with the center G (although it may seem evident).

I think the shading under the triangle under the bearing at G is supposed to indicate it is fixed.I think we can assume the cylinder is fixed. In other words I think we can assume the normal relationship between piston and crank assembly (e.g. as in a car engine).

This means in the two positions where HGP is a straight line the angular velocity will be at its maximum and when angle GHP is 90 degrees the angular velocity will be zero. (NB point P is the small end bearing)

 

Therefore when the angle is 60 degrees the angular velocity will be somewhere between its maximum value and zero.

Edited April 12, 2012 by Joatmon

[doG]

Except that if the piston was at the dead centre position, the speed would be null, and the angular speed also null.

But the sketch disagrees. As it is drawn, the piston is not at the dead position.

So, after reading all this twice, I still don't understand the statement.

Nowhere does the OP say the piston is at the dead center position, it says, "The crank makes an angle of 60 degree with the inner dead centre position...". This is simply the instantaneous position the problem is to be evaluated at because the piston velocity is variable from 0, when the crank is in line with the center of the bore, to a maximum, when the crank is at 90° to that line.

 

The sketch itself looks like a scanned image from the textbook the problem is from, not a drawing by the OP, so I would not assume any other arrangements to be representative of what is being asked.

[Joatmon]

Nowhere does the OP say the piston is at the dead center position, it says, "The crank makes an angle of 60 degree with the inner dead centre position...". This is simply the instantaneous position the problem is to be evaluated at because the piston velocity is variable from 0, when the crank is in line with the center of the bore, to a maximum, when the crank is at 90° to that line.

 

The sketch itself looks like a scanned image from the textbook the problem is from, not a drawing by the OP, so I would not assume any other arrangements to be representative of what is being asked.

I think that the maximum piston speed is when angle GHP is 90 degrees? (P being the piston bearing centre)

Edited April 13, 2012 by Joatmon

[imatfaal]

Nowhere does the OP say the piston is at the dead center position, it says, "The crank makes an angle of 60 degree with the inner dead centre position...". This is simply the instantaneous position the problem is to be evaluated at because the piston velocity is variable from 0, when the crank is in line with the center of the bore, to a maximum, when the crank is at 90° to that line.

 

The sketch itself looks like a scanned image from the textbook the problem is from, not a drawing by the OP, so I would not assume any other arrangements to be representative of what is being asked.

The point H has its highest horizontal velocity at 90 degrees - but this does not translate to the highest horizontal veloxity of P.

 

 

 

I think that the maximum piston speed is when angle GHP is 90 degrees? (P being the piston bearing centre)

 

I almost agree. Both using eqautions and numerical estimation I get damn close to this - but not exact. Your answer would give the angle with line PG as 78.69 whereas my answer is 79.1001 - as does my numerical estimation.

 

your answer seems so much more likely that I am redoing me sums

[Joatmon]

I almost agree. Both using eqautions and numerical estimation I get damn close to this - but not exact. Your answer would give the angle with line PG as 78.69 whereas my answer is 79.1001 - as does my numerical estimation.

I based my opinion on the fact that the line PH (extended) makes a tangent with the circle. My first thought was any further rotation would reduce the angle HPG, but on second thoughts this doesn't take into account piston movement. I think your answer may well be correct! I'm no mathematician, just trying to visualise what happens with a technician's eye! lol.

Edited April 13, 2012 by Joatmon

[michel123456]

The point H has its highest horizontal velocity at 90 degrees - but this does not translate to the highest horizontal veloxity of P.

 

 

 

 

 

I almost agree. Both using eqautions and numerical estimation I get damn close to this - but not exact. Your answer would give the angle with line PG as 78.69 whereas my answer is 79.1001 - as does my numerical estimation.

 

your answer seems so much more likely that I am redoing me sums

 

I made this on the original OP sketch:

point A is the Bottom Dead Centre (BDC), point B is the Top Dead Centre (TDC)

Both at points A & B the speed of the piston is null. The distance traveled by the piston is equal to the diameter of the circle 2R= 2 times the distance HG.

 

So without calculating anything I suspect that the top speed of the piston will be when the crank will be exactly in between i.e. at 90 degrees (point F) and not when the angle HPG will be 90 degrees. IOW the horizontal speed of the piston corresponds to the horizontal speed of point H at any time. and since as Imatfaal wrote "The point H has its highest horizontal velocity at 90 degrees", so does the piston.

Edited April 13, 2012 by michel123456

[Joatmon]

I made this on the original OP sketch:

point A is the Bottom Dead Centre (BDC), point B is the Top Dead Centre (TDC)

Both at points A & B the speed of the piston is null. The distance traveled by the piston is equal to the diameter of the circle 2R= 2 times the distance HG.

 

So without calculating anything I suspect that the top speed of the piston will be when the crank will be exactly in between i.e. at 90 degrees (point F) and not when the angle HPG will be 90 degrees. IOW the horizontal speed of the piston corresponds to the horizontal speed of point H at any time. and since as Imatfaal wrote "The point H has its highest horizontal velocity at 90 degrees", so does the piston.

 

Looks like I got my wires crossed!

 

Now considering the angle HPG, do we agree that its maximum angular velocity occurs at Top and bottom dead centre? I think everyone does.

 

Now an unresolved question seems to be at what angle does the angular velocity become null? The angular velocity will be above zero and less than maximum when the angle is as shown (at 60 degrees). My first thought, which I now have doubts about, is when the angle GHP is 90 degrees.

Edited April 13, 2012 by Joatmon

[Xittenn]

step 1) Draw a phase diagram

 

step 2) find [math] \phi_0 [/math]

 

step 3) find the angular frequency

 

step 4) write the standard wave equation

 

step 5) find the equations for v and a

 

step 6) apply the mechanical advantage of a piston rod

 

step 7) learn physics guys! j/k

Edited April 13, 2012 by Xittenn

[Joatmon]

step 1) Draw a phase diagram

 

step 2) find [math] \phi_0 [/math]

 

step 3) find the angular frequency

 

step 4) write the standard wave equation

 

step 5) find the equations for v and a

 

step 6) apply the mechanical advantage of a piston rod

 

step 7) learn physics guys! j/k

 

Edited April 13, 2012 by Joatmon

[michel123456]

I made this on the original OP sketch:

point A is the Bottom Dead Centre (BDC), point B is the Top Dead Centre (TDC)

Both at points A & B the speed of the piston is null. The distance traveled by the piston is equal to the diameter of the circle 2R= 2 times the distance HG.

 

So without calculating anything I suspect that the top speed of the piston will be when the crank will be exactly in between i.e. at 90 degrees (point F) and not when the angle HPG will be 90 degrees. IOW the horizontal speed of the piston corresponds to the horizontal speed of point H at any time. and since as Imatfaal wrote "The point H has its highest horizontal velocity at 90 degrees", so does the piston.

 

This part must be wrong. I am really sorry.

The vertical vector also induces horizontal motion to the piston.

Only proper calculation can tell.

[Xittenn]

Related Rates:

 

step 1) write out the equation for the triangle for which you wish to solve

 

step 2) find the derivative of both sides with respect to time

 

step 3) substitute the known value [math] \frac{d \phi}{dt} [/math]

 

step 4) solve for [math] \frac{ds}{dt} [/math]

 

step 5) profit $$$