So does that mean if I drove a Dodge with a Cummins on a race track with 45 degree banking, the rods would see the same side loading as a V8 would on flat ground?

I'm still waiting to see this dimensional analysis, because you are absolutley wrong.

 

No, it would be different because a V8 crankshaft powers two sets of pistons in different directions instead of one set in the same direction. I don't think it matters enough to even have an opinion on this, though, so I am out of this particular argument.

 

This is funny stuff!!!

 

You yourself said the rods/crank does not care what position they are in or what way they move. You proved yourself wrong.

When I made reference to degrees, that means on flat ground. Not on a banking, not on a slope but even still, the forces STILL would NOT change regardless of orientation in relation to the ground.

 

"the forces STILL would NOT change regardless of orientation in relation to the ground."

Right!! So explain to me how a rod in an inline 6 knows it is an inline 6??? All it sees is one end going up and down in a cylinder attached to a piston and the other end swinging in a circle attached to a crank. It has no idea what the other cylinders are doing or even if other cylinders are present, let alone if they are in a inline or a V configuration. The crank knows and in the design you have to account for the torsional inertia as well as the harmonics generated by multiple cylinders, but an individual rod experiences the same forces, regardless.

 

You act like it has a brain. It doesn't. What I'm saying is that an I-6 will ALWAYS be directly perpendicular to the ground (always 90 degrees), and a V-style will ALWAYS be less than 90 degrees in relation to the ground (135-90=45 and 90-45=45). The V-style stroke will come from about 135 degrees on one side and about 45 degrees from the other, therefore introducing more downward force towards 2 points other than straight down (180 degrees).

Anyway, I'm going to bed, no use arguing <with mules>.

 

Here's the deal...my step dad had a 2001 2500 cummins 4x4 extended cab longbox...that truck had some bahlz...with all-terrains you could lock that sucker in and climb hills where oil would run out of the hood onto the front of the windsheild...also we have a 32ft. fifth wheel trailer and cruising along the interstate you barely noticed that anything's on the back of the truck...plus you see cummins engines in a lot of things...i know guys that that more than 1/2 a million miles on their dodges and they still run new...that's because they've kept things maintanced...

Now to the FORD F-250...my uncle has a 1996 4x4 4door shortbox powerstroke...that too also has plenty of torque for whatever job is needed...every summer he pulls fellow farmers out of the mud as if it were nothing hex he even pulled a new CASE MX280 tractor out that was buried half way up it's tires...both engines have an incredable amount of power...i've even heard that the early strokers had so much torque guys were snapping drivelines...and believe me that takes a lot of force....

As for the all around truck not the deisel battle they are both fairly even...the Dodge is more front heavy it seems but the ride can be rough sometimes...but dodge puts some of the strongest coils in their rigs...the FORD however seems to be a smoother ride and has a little more mid-range power when you punch it...both turbos are excellent on each truck so i don't see a problem there...
Everyone can think what they want but that's all i know from experience with both

 

The part I highlighted in bold has nothing to do with anything at all. The cylinders in the V8 still have the same relation to the crankshaft that those in the I6 do. The force put on the rods during the compression stroke still push the piston and rod straight down the bore toward the crankshaft. It doesn't matter is the cylinder is sideways or upside down, the loads will be the same since thay are always in the same relation to the crank. Where the ground is has nothing to do with the side loading on a rod. The only thing that adds to the downward side loading on the rods in the V8 would be gravity itself. The forces enacted on the rod by that gravity while in the cylinder would be so miniscule that they probably aren't even measureable.

 

"You act like it has a brain. It doesn't"

You are talking semantics now. I was simply using those terms to describe the various parts in relation to the other parts and the forces that are applied to them (or in other words the forces they "feel"). I can repost the discussion from my previous post and throw out all the familiar terms and replace them with more technical terms. I did that as a more friendly way of explaining the priniciples we were discussing.

<"Anyway, I'm going to bed, no use arguing with mules."

I think I've ridden mules that had more smarts than what you are displaying. Think about what you are trying to argue before you post. I think I can see how you are confused and the path you are going down. If you would think about it a little, you would see the flaw in your logic.>

 

Check out this website for a "brief" discussion of the forces applied to connecting rods during a 4 stroke engine cycle. Notice that there is no mention of engine configuration. Why? Because it is IRRELEVANT!!!

http://tinyurl.com/2jywf

 

hmmm you've got me wondering about a Dodge slant 6 engine now and how strong the rods need to be.

 

They only need to be strong on one side!!!!!! LMAO!!

 

<It takes one to know one.>

<You have the most ignorance of anyone on here. You're the one who is confused.>

I have taken this class and we did problems related to what I'm explaining. There IS a DIFFERENCE in the forces applied, I've thought about it, and worked it out in problems using the proper formulas. However, the differences are very miniscule and I only meant this to be a small point, not a rehash of my physics class at school.

Now its time to see if you can back up what you say.

F sub c = mv(squared)/r x a

F= Force

r= radius

M= Mass

V= Velocity

a= angle of force application

So, let's say the mass of the rod is 5 lbs, (nice even number), and the radius of the crank is let's say 4", and the mass is rotating at 2000rpm.

a) What is the force on the crank of an I-6 oriented straight at 90 degrees?

b) What is the force on the crank of a V-8 that has a line of symmetry at 90 degrees? This means you will have to take the average of both sides (one at 135 degrees and the other at 45 degrees).

I'll be waiting for your answers and then we'll see who's right and who's wrong.

 

b) What is the force on the crank of a V-8 that has a line of symmetry at 90 degrees? This means you will have to take the average of both sides (one at 135 degrees and the other at 45 degrees).

This is where you are getting confused. The force on the crank is entirely different form the forces on the rods The line of symetry in the V8 has nothing to do with the force application on the rods. No matter what the line of symetry is, the force on the rods is still applied at a 90 degree angle to the crankshaft, regardless of whhether or not there is another bank of cylinders.

Your formula F sub c = mv(squared)/r x a is correct only for finding force on the crankshaft, not the rods.

 

I've got one of each ('01 PSD and '02 Cummins) and so far the Dodge is holding up the best even though it is the primary towing vehicle. Far too many irritating electrical problems on the Ford on top of just being plain hard to work on. As far as Dana axles go, the Dodge did have a 60 in the front but the rears are 70's or 80's depending upon configuration. The Danas are a pain because they require frequent lube changes. Now Dodge uses axles by American Gear I believe.