Ok, I maybe proving the theory here that some of us need to spend more time chasing women but...

I've been trying understand the workings of valve gear for some time. For me that would be being able to define the position of various components given the position of the main crank.

I'm stuck on the eccentric crank (one last detail before I can move on)

Basically, the equation that governs the position of the eccentric crank in the X (horizontal) axis, can be simplified to an equation of the form:

X= a*sin(è) +b*cos(è)

where X,a, and b are constants. è is the angle of rotation of the the main crank and the independent variable i'm trying to solve for.

I know that the path of the eccentric is circular, and so the function produces a sinusoidal wave. I've plotted it on EXCEL just for fun.

This is easily solvable for the specific case when X=0, but I'd like a general form, if it can be understood by a guy who had three classses of calculus but skedaddled out of engineering when it came time for differential equations.

Any ideas??



superheater@rrmail.com

Try constraining the system with

R^2 = X^2 + Y^2

You also have the equation for the Y position or coordinate.

Hope that this helps



pkurilecz@yahoo.com

> Try constraining the system with

> R^2 = X^2 + Y^2

> You also have the equation for the Y
> position or coordinate.

> Hope that this helps

Yes, have a similar situation in solving the Y position. Will give it a try. Thanks.

superheater@rrmail.com

The eccentric crank is secured to the drive wheel and rotates in unison with it, so the end of the crank connected to the eccentric rod just decribes a small circle out of phase with the crank pin. Find the distance from the end of the eccentric crank to the center of the drive wheel and call this r. Find the angle between the eccentric crank and the line between the crank pin and the center of drive wheel and call this a. Now, if b is the angle of the crank pin (independent variable), X=r cos (b+a).

> The eccentric crank is secured to the drive
> wheel and rotates in unison with it, so the
> end of the crank connected to the eccentric
> rod just decribes a small circle out of
> phase with the crank pin. Find the distance
> from the end of the eccentric crank to the
> center of the drive wheel and call this r.
> Find the angle between the eccentric crank
> and the line between the crank pin and the
> center of drive wheel and call this a. Now,
> if b is the angle of the crank pin
> (independent variable), X=r cos (b+a).

Thanks to one and all for responses but I'm still vexed.

You equation is slightly different than the one I was using but leaves me with same problem:

Cos (a + b)= cos(a)cos(b)-sin(a)sin(b)

I am once again simplifying an equation of the form Constant1*cos(a) + costant2*sin(a)

Or am I getting something wrong?



Superheater@rrmail.com

I think you're trying too hard on this. The end of the eccentric crank describes a circle, the diameter of which corresponds to the length the eccentric rod must be displaced in order to move the valve a complete stroke. Call this dimension 2r. The actual length of the eccentric crank and the angle it makes with the line from the crank pin to the drive center is not critical for your analysis (I'm guessing this is where you're getting the sine term in your equation) All that you need to know is that the end of the crank describes a circle of the prescribed radius, and that the end of the crank is displaced some angle from the main crank pin. For simplicity, assume this displacement is 90 degrees leading. In practice, it won't be exactly 90 due to lead and lap, but for your purposes, its a reasonable assumption. Now, call the angle of drive wheel rotation B. B is your independant variable, so you don't need to solve for it. It assumes all values from 0-360. You need a reference line on the driver to measure rotation from. Use the line from the main crank pin to the driver center. Set B=0 when this line is horizontal and the piston is as far forward as it will go (TDC). Now, since the eccentric crank is fixed to the driver, the angle between your reference line and a line from the end of the eccentric crank to the center of the driver is a constant, call it A. We're ignoring the length/angle of the eccentric crank itself because the end of the crank is the only point of interest. We know the distance the end of the crank is from the driver center (r) and we know its angular displacement from the reference line (A). We can now tie the x position of the eccentric crank to the position of the drive wheel using the equation x=r cos (B-A) (angles subtract because driver rotates CW and because eccentric pin leads main pin. Note: use angles from 0-360, not 0 to 90 to insure proper sign). However, we can simplify this expression even further since we assumed A=90. cos(B-90)=-sin(B), so the equation you need is simply -r*sin(B). Hope I explained better this time.
-Chris

engine_lathe@yahoo.com