Comparing Rod Ratios: Pressure on Pistons & Cylinder Walls

[5.0stang]

Well I am getting into how engines work and am trying to learn.

I was wondering if there is a formula I can follow to find out how much more pressure is applied on the piston end towards the cylinder walls if the rod ratio is steeper.

I want to compare a 1.59 Rod Ratio (347 cubic inch) to a 1.66 Rod Ratio (331 cubic inch).

What information would be needed.

I have the piston/connecting rod weights and measurements.

347 – 3.40 in. stroke with a 5.4 in. rod.
Piston weight is 474 grams. Rod weight is 620 grams.
1.59 Rod Ratio

331 – 3.25 in. stroke with a 5.4 in. rod.
Piston weight is 472 grams. Rod weight is 620 grams.
1.66 Rod Ratio

Do you need to know the crankshaft weight?

Any formulas out there to help predicate or give a good idea how much more thrust/pressure is applied towards the outer cylinder wall.

Thanks for any ideas

 

[Danger]

The crankshaft weight is irrelevant. I can't tell you about any formulae, but I can see right away that you haven't included the wristpin offset in your data (unless you've incorporated it as part of your rod ratio).

 

[5.0stang]

I used Probe Pistons website for the info. They show no info on their specs page about wristpin offset. It does show the weight of the pin. It is 113 grams on each piston.

Any ideas on helping me get an idea on how much more pressure is applied to the cylinder walls?

Thanks for the comment Danger.

 

[brewnog]

I don't follow.

If you're trying to work out the difference in compression ratios between two different piston/conrod setups, the piston, rod and crank weights are all irrelevant.

Chapter Two of Internal Combustion Fundamentals by John Heywood has all the derivation of the geometric formulae you need.

 

[5.0stang]

Sorry brewnog, I'm not trying to figure out the compression calculator. Also how does weight affect the compression ratio...whether static or dynamic?

I thought compression ratio was based mainly on quench area, head cc's, piston cc's, headgasket thickness?

Anyways, what I was trying to figure out was a more general or specific effect on the sidewalls with a steeper rod ratio. Like how much more force does it put "sideways" on the block?

Just seeing if there was any place to get an idea on how much a 1.59 ratio would put instead of a 1.66 rod ratio on the sidewalls of the cylinders. It approaches at a steeper angle, so I'm sure technically there is more friction.

I hope that cleared it up a bit, sorry for the confusion.

 

[5.0stang]

Any takers or suggestions on what I might do to find out some plausable answers?

 

[Q_Goest]

Hi Stang. I think what you're looking for is to determine what side load there is on the cylinder wall.

If I assume forces due to momentum are small, which may or may not be completely reasonable especially at higher RPM, but if we look at this only from a static balance standpoint, we see the side load is linearly proportional to what you're calling the "rod ratio".

The calculation is pretty simple really, take a look at the attached. I've put in a 4" piston and 400 psi of cylinder pressure. Of course, cylinder pressure varies during the stroke, so you'd need to know what cylinder pressure actually is to determine the actual side load. Regardless of this, it seems clear that cylinder wall load is proportional to rod ratio, and thus a higher rr gives lower side wall loading.

 

[5.0stang]

That is sweet man, that is pretty much exactly what I was looking for!

Keep in touch on here, I'll update later with further questions to get a better answer. I'm busy right now.

So where do I need to determine cylinder pressure? BTDC? At maximum compression height? or what? Can I use a compression ratio for that measurement?

 

[Q_Goest]

Stang. Note that the equation I gave you only works for the point where the con rod axis (line between wrist pin and rod bearing) is perpendicular to the crankshaft throw (line between crankshaft center and rod bearing). This is roughly the point where the peak side load will occur due to peak cylinder pressure and geometry of the con rod. Similarly, you can work out the angle the rod makes at different points during rotation and calculate side load from that.

You need the cylinder pressure to calculate side load though. Cylinder pressure varies considerably over the course of the cycle, from atmospheric to some peak which occurs soon after ignition. To calculate that, you might read through this article:
http://hcs.harvard.edu/~jus/0303/kuo.pdf

The equations given are relatively simple. You could put them into an Excel spread sheet at 1 degree cranshaft angle increments and plot everything you're interested in including cylinder wall side loading & cylinder pressure. You could also put frictional forces in due to cylinder side loading and come out with total power, then graph all that in any way you wanted. If you or anyone is interested in doing all that, feel free to post your spreadsheet and I'll look it over as you go.

 

[5.0stang]

Sorry I'm so slow at responding and pretty much uninformed. This stuff is not my forte.

What I do know is that when a compression check is done on these particular engines I'm looking at. They are usually ranging around 150-160psi. I would assume that is peak psi calculations?

A gauge is installed on the end of a tap that goes into the head. The engine is then turned over and the compression gauge jumps to what it reads.

So instead of adding 400 to the psi reading, for simplicity reasons, I could add 150 in there instead right?

PSI increases as compression increases right?

Also the "1580 lbs" is in actual pounds, not inch pounds.

Thanks for being patient with me.

 

[Q_Goest]

Regarding the compression check you do, that doesn't give you the cylinder pressure when the gas ignites since you've removed the plug and you aren't burning any fuel. When the engine is running and fuel is being burned, cylinder pressure is much higher.

Regarding the force on the wall, the load is a force and thus is measured in pounds. Inch pounds is a measure of torque.

 

[5.0stang]

I got about 12.7 for the area.

How did you get the outcome of 1580lbs, my math isn't coming out right?

 

[Q_Goest]

I checked it again. The example still comes out to 1580.

 

[5.0stang]

I was getting 3,173...looks like I'm missing a division somewhere...

I know your right, and I'm wrong...just trying to see where, thanks:)

 

[5.0stang]

Anybody? Thanks!

Also does the force on the sidewall vary with rpm?

 

[Danger]

Hi Stang;
Again, I can be of no use mathematically. Regarding your last question, though, I believe that the answer has to be 'yes'. You have the same weight of piston/rod/ring assembly at the same angle, but moving faster as RPM's increase. Faster impact means more kinetic energy to be absorbed.

 

[brewnog]

I'd clarify that further to say that engine speed and applied torque are both liable to effect the in-cylinder pressures seen. Torque demand will have an effect by requiring greater BMEPs, and speed will have an effect principally due to breathing, but also as a result of the driver running the engine at whatever speed is required to provide sufficient torque for the demanded load, at that vehicle speed. Bit of a mouthful, sorry.

 

[Q_Goest]

I was getting 3,173...looks like I'm missing a division somewhere...

Probably the /2 in the numerator (ie: S/2).

Also does the force on the sidewall vary with rpm?

You can break down sidewall forces into what is causing this force. One is the pressure on the piston creating a compressive load on the rod. This accounts for the vast majority of the sidewall load. The other would be some minor amount of momentum as the rod rotates about some point. The rod's rotational inertia will create some very minor loading.

If however, you are thinking that RPM affects cylinder pressure such as brewnog mentions:

I'd clarify that further to say that engine speed and applied torque are both liable to effect the in-cylinder pressures seen.

Then yes, any time you have some change in the engine which is affecting pressure inside your cylinder, there will be a change in sidewall forces. I think it's important to learn how to apply basic principals when analyzing things like this. A statics analysis of the rod/piston provides the sidewall loading due to pressure and a dynamics analysis of the moving parts provides the sidewall loading due to inertia and momentum of the individual parts.

I've looked over the reference I'd mentioned before. It looks quite useful, very straightforward. It seems as if one could fairly easily create a spread sheet using Excel to determine cylinder pressure for a 'typical' engine. Very nicely written paper.
http://www.hcs.harvard.edu/~jus/0303/kuo.pdf

 

[5.0stang]

I looked at the formula and multiplied 400*12.6*3.4 then divided the answer by 5.4 and then by 2 and got 1,586.67. You got 1580. Did I do the math correctly?

So for the 331 engine I got (using the same format) I got 1,516.67.

 

[Q_Goest]

Yep, you have the math right now.

 

[5.0stang]

Thank you very much!