RCVTR

I did it that way because I wanted a way for it to oscillate without damping out. The bottom of the rod journal was fairly close to the centerline of the main journal - closer than the radius of the main journal.

I wonder about the accuracy of translating the axes, because of the part asymmetry. My thought being that the closer to the actual rotational axis, I could make it oscillate, the more accurate my result would be.

Another way to do it would be to suspend the crankshaft and flywheel vertically on 3 wires. The flywheel would keep the wires equidistant from the centerline, so that it would oscillate about the centerline. Knowing the wire length, you could then calculate MOI of the crank and flywheel and of the flywheel itself and do a subtraction.

gboezio

I do like self inflicted headaches and find your project very interesting, I'm glad to find some more engine nuts over here.
You may have a look here, there enough engineering **** here to read the whole winter.

http://forums.autosport.com/showthre...threadid=82571

RCVTR

We've gotten a bit ahead of ourselves here. I will be providing descriptions for non-engineers for what all of this MOI stuff is about and how it is used.

I can't really get to it right now.

RCVTR

Ok. I finally got the motion equations solved. It took a while, and a considerable amount of head-scratching. I rewrote the previous descriptions. Hopefully they are a little more clear, and added the solutions to the equations of motion.

Now we have the basis for doing the dynamics. More on that later.

tapatio

Very interesting. Please post the Matlab code. Oh, and please do not encourage people to become engineers. We don't need more competition :]

Smileyhawk

For position, velocity and acceleration, you may want to adopt a sign convention that would indicate the direction of travel, thus simplifying some of the descriptions in future. If TDC is 1, and BDC is 0 (or -1, there are arguments for either), then simply stating the sign (+/-) of both the velocity and acceleration will indicate the direction the piston is moving and whether its velocity is increasing or decreasing because V & A are 90 degrees out of phase.

Just my $0.02, but nice work so far.

j shizzy wizzy

suddenly i find myself crying and drooling in the corner after reading this thread......my head hurts!

RCVTR

I chose the reference as the midstroke, because the acceleration changes sign at that point. It seemed more intuitive to me. When we get into forces, and moments (torques), I think it will be a good reference, especially when I get to the balancing problem.

Thanks for the input.

RCVTR

It's not for everyone. But I think a lot of your problem is that it's a lot of completely new concepts.

I think it's pretty cool that you can describe what seems like a complicated mechanism with four simple equations and their first and second time derivatives.

It takes a lot of dreary math classes before you get to do something useful with it. That's when the fun starts.

JamieDaugherty

I think you need to add an equation to your set. The rod vector is ok for transmitting piston forces, but once you start to calculate loads you'll need something else: the equation of motion for the rod center of gravity. Since all body loads act on the Cg you'll need that to apply the rod mass to. Let's not forget to inculde the moment of inertia too, the rod not only translates but also rotates. The equations for the crank and pistons are easy, but the rod is equally important. I think you'll find that it plays a significant part of the overall loads.

Noodle that one for a while if you want your head to hurt. Actually, it's probably not that bad. Solve it for one cylinder and offset it by 90° phase angle for the other and you are done.

RCVTR

The motion is actually completely described by this set of equations. The only thing I need to do is convert the rod angle to the the angle relative to the wrist pins, before doing the force and moment (torque) computations.

The rest will come from the free body diagrams of the 5 moving parts. I have figured out how to make it work, but have not have time to develop the presentation. It's a beautiful thing, as you will see...

JamieDaugherty

Explain to me again how the rod angle alone will generate the resultant force of the rod Cg? The rod angle is necessary to resolve the piston loads back to the crank jounrnal, but the rod body loads act in the direction of the acceleration vector of the rod Cg which is not the same vector.

I'm not trying to be a stick in the mud, I've just been down this road with analysis of this type before. Trust me, it's better to get all of your ducks in a row now rather than later!

RCVTR

Rod angle does not generate a force. I don't think that's the question you meant to ask. Angular and linear acceleration generate forces because of the movement of the mass. F=ma, and T=I(aa), where F is force, m is mas, a is acceleration, T is torque, I is mass moment of inertia, and aa is angular acceleration.

When I do the free body diagrams, I will sum forces and moments for static equillibrium. I have big end and small end weights. When I take the mement about the wrist pin, I will be able to locate the center of mass. I think this is where your question is answered. There will be a vector from the wrist pin centerline to the Cg of the rod. It's direction varies with the rod angle. I will need another one for the Cg location on the crank, and another for the imbalance vector (I'll explain that when I get to balancing).

I don't have enough information on the crank to find the Cg, but I think I can iterate through the solution of Cg until the engine is balanced, then change the piston weights and remove weight at the end of the imbalance vector until it is balanced again. It should be interesting.

I did oscillation studies of the crank and rods, so I will have a reasonable estimate of the moments of inertia. I will explain that techinique in detail.

The equations of motion will resolve the sinusoidal variations in force and moment, because of the time-varying accelerations.

JamieDaugherty

Yes, but the direction of motion of the Cg, or better yet the vector direction of the acceleration of the Cg, is the same as the vector direction of the resultant force for that mass. This is not along the line between the wristpin and the crank journal. In fact, there two instances, at TDC and BDC, when the force due to the mass of the rod is perpendicular to this line.

You've done a really great job so far, keep it up.

RCVTR

I'm finally getting some time to get back to this.

Jamie, I finally see what you are saying. It has been a long time since I did one of these types of analysis. I need to describe the accelerations of the centers of mass of the rods.

I was trying to make it work in my mind and had this feeling something was missing, but couldn't put my finger on it. I finally found - or paid attention to - a reference to it.

thanks for the heads up!

JamieDaugherty

No problem, I'm glad I could help. I'm looking forward to following this as it develops!