[IntegraBad18c]

Tuan, in your article "Engine Package I: Cam Timing, Rod Ratio, Port Size" you go indepth about rod/stroke ratios. I have a question about that:

Lets say, in a 1.58:1 rod ratio, what would be the angle (not crank degrees) of a rod, at 90 crank degrees ABDC or ATDC, to a rod (vertical line if ya wanna use that instead....a little more possible in this hypothetical crap I'm posting) at BDC or TDC. I'm trying to get the angle of the rod, that has a low rod ratio, when it is being pushed up/down the cylinder at 90 crank degrees. I want to see what kinda angles (with one leg being the rod at ABDC OR ATDC rod and the other leg being the rod at BDC or TDC or a vetical line) that are causing this extra friction in the cylinder when a low rod ratio is being used. Would these angles be relative to all rod ratios regardless of rod length and stroke length so long as the ratio was the same?

 

[MichaelDelaney]

ugggghhh! you want to get into geometry and trigonometry? this will involve sines,cosines, and trigonometry?... triangles:sides, hypotenuses, angles, Pythagorus Thereom, yuck!....you sure you want to get into this? I feel a headache coming on already...where's my advil bottle?

Here's a diagram of 2 engines with the same stroke. The top one has a long rod and high rod ratio. The bottom one has a short rod and low rod ratio.

Let's just look at one engine for now...let's pick the top one just randomly: Imagine the circle to the left is the piston. To the right is the crankshaft (square) with the rod big end attached or anchored to the crank.

Back to the circle on the left.We know that a piston travels up and down but the motion of the piston pin where the rod attaches to the piston travels in a circle described by a radius, small letter r, when the piston goes up and down.


A + C is the stroke. A + C+ D is the rod length. In the low rod ratio engine, A+B is the rod length and one side of your "triangle". What is little c?

To calculate the tangential force pressing on the cylinder wall, you need to use triangle trigonometry to solve for little r's length. The rod to crank angle is on the right side in the square, inside the triangle. We know the rod length and stroke of an engine. You can obtain the crank angle by using Pythagorus to calculate the sides of the triangle (knowing the stroke and the rod length as the sides) and then use sine or cosine.

Now, to calculate sideload force in a low vs high rod ratio motor: The tangential force of a rotating object (circular motion or our piston travelling an arc of a circle) requires that you calculate the centripetal acceleration using the radius of the circle (little r which is 1/2 the stroke) times the mass of the piston.

If you plug in the numbers into the math, you will eventually solve for the lateral force created for each engine. This is how they calculated that a longer rod has less sideload.

Use the B18B dimensions for the low rod ratio engine and the B16A dimensions for the high rod ratio engine. You can find the largest crank angle at the point where the piston is maximally loading the wall and you can also calculate that force.

I'm not going to do it since I'm too lazy right now.....

You guys still taking trig and physics in high school, go for it....

let me know what you get....I'll be back later with the cosine and sine equations used to solve for crank angle...

 

[phatintegra]

I would of help you, but math I use it everyday at work. Im sick of it my brain is almost fryed trying to figure out a bunch of numbers and equations at work I think Michael can explain it (he is doing good so far).

 

[MichaelDelaney]

Here's Larry Meaux's way to calculate what you are asking:

o = axis-to-axis perpendicular offset
x = distance along cylinder axis from crank axis to piston pin (vertical position of piston)
L = connecting rod length
t = crankshaft throw

I. At the top of the stroke:
x^2 = (L+t)^2 - o^2


x_top = sqrt[(L+t)^2 - o^2]

II. At the bottom of the stroke:

x^2 = (L-t)^2 - o^2

x_bottom = sqrt[(L-t)^2 - o^2]

The change in piston height from top to bottom is therefore,

x_top - x_bottom =

sqrt[(L+t)^2 - o^2] - sqrt[(L-t)^2 - o^2]




For this case, it is interesting to note that the terms TDC and BDC don't apply in the normal sense. "TDC" is no longer 180 crankshaft degrees from "BDC."

If TDC is defined as the position where the piston is farthest from the crank axis, and positive crank angle is measured in the rotational direction that gives the shorter distance between TDC and BDC, then the crank angle at BDC can be found from the following relation:

angle_BDC = atan(o / x_top) - atan(o / x_bottom) + 180 deg

 

[MichaelDelaney]

Here's the crib note to solving the piston speed based upon the diagram in my thread above:


V(t) = - r sin(t) - { [ r^2 sin(t)cos(t)] / [ sqrt(c^2-r(sin(t)^2))] }


where V is the calculated max. piston speed, t is the crank angle, c is the crankshaft throw, r is 1/2 the stroke.

I told you this would be a pain in the.....

 

[DaBoyNBlu]

since we're talking about it... i found some pistons that say they need a c-c rod length of 5.429" instead of the stock 5.433" of a b18c1. are these pistons ok if i buy some eagle rods to go with them (eagle rods have the stock length of 5.433")
thanks

 

[Gvtec]

I was going to say the same thing.

 

[MichaelDelaney]

I forgot to state that you will need the piston mass to calculate the rotational tangential (sideload) force that will push against the cylinder wall:

B16A pistons 299 grams

B16B pistons 327 grams

B18B pistons 280 grams

B18C1 pistons 305 grams

B18C5 pistons 310 grams

The "surprise" here is that the Type R pistons are the heaviest..

 

[IntegraBad18c]

Oh, yes....the gut of the engine...the real thing...mwa ha ha ha ha!!! I appreciate it, Tuan, you have no idea. That's about everything I wanted with some lagniappe. Love your style. The whole subject isn't entirely important b/c I'm not an engineer for whatever company desgning an engine. I still appreciate it though! We already pretty much have a set choice of rods ands strokes, and the option to mix them. I doubt any body is gonna weld on 12 inches of deck height for a 16" rod they had custom made....they would have to be a loon...or worse...a looney. I'm not to loon status...yet...

 

[IntegraBad18c]

Oh yeah...I forgot.

Ni!

 

[IntegraBad18c]

Okay, I'm back now. I did some lookin' at the graph and some reading of your posts. I haven't gotten that far but I have done some. I plan on finishing up more tonight.

When looking at the visual I found it easier to view the circle as the crank, little r as the crank journal (with 2r being the stroke), c as the connecting rod, the square as the piston, A+B being the total length of the rod at BDC on the low rod ratio motor...etc,etc.

Would this be an incorrect way of looking at it? Because it seems a hella easier interpreting the visual that way, now that the crank, and not the piston wrist pin, is moving. I'm looking at this a lot like the visual on how a motor works at How Stuff Works.

With my interpretation I answered your first question. "c" in the low rod ratio motor is the rod at x crank angle degrees. I figured this b/c "c" is the same length as A+B, just a different angle b/c the journal is moving the rod. No, I didn't use the Pythagorem Theorem to figure that one out. The only line present at the end of A+B and "c" was a curve. I didn't take the liberty of drawing a line either (I had all of this printed out) to use "a (squared)+b(squared)=c(squared)". I used a measuring tape. Yes, I cheated.


More to come...

 

[MichaelDelaney]

The whole idea is to get the triangle so you can work out by pythagorus and basic trig the crank angle, piston speed at any crank angle, the rotational tangential torque to determine sideload. Yeah actually I had it assed backwards...the circle is the crank and the square is the piston (duh!)....I haven't looked at this diagram for years so please forgive me...

you still need to calculate sines and cosines for the crank angle. you can't cheat around that unfortunately.

 

[IntegraBad18c]

No, I can't. I had to sort thru some mail, but I'm back at it. I'm REALLY rusty, so I gotta do some research! But ya, I gotta use the Law of Cosines to get one angle (cause I have is goddamned lenghts!)then the Law of Sines and I'm set. Then I wanna make a chart and work on a similar rod ratio except with a different length rod and stroke. I got a lotta work...

 

[MichaelDelaney]

keep the stroke constant and vary the rod length first. get a chart for that...then go to changing strokes...are you using the b18c and b16a lengths and the b18c stroke as your comparison as a starting point?

 

[IntegraBad18c]

I am back.

Gall me if I am wrong. I hate saying I'm right when there is that possibility I may be wrong. I hate trig, too. Holding up foolsgold and yelling "Eureka!" is not my thing. Anyhow.....

Using the Pythagorem Theorem and what I already knew of the b18b/b18a rod/stroke lenghts I solved for (solved for length "a") the mystery length that runs down the crank journal at 180 crank degrees. I am referring to the motor with the high rod ratio. Me and the motor with the low rod ratio got into a bad relationship and the biatch ran away with my calculator and the kids. The mystery length was 129.57 mm.

So far:
a=129.57 mm
b=44.5 mm (length of crank journal)
c=137 mm (length of connecting rod)

Awesome...lets use the Law of Cosines:
b squared= a squared+ c squared - 2*a*c Cos B

Now lets skip the written work I did...
I got 35.5 degrees. As I asked in my first post (scroll up), I asked what angle would be made with the rod and a vertical line at x crank degrees. I have the angle (>b= 35.5 degrees) of the rod at x crank degrees to the vertical line. Now I gotta figure out the crank degrees the piston is at when the rod makes 35.5 degrees with the vertical line. Then I gotta figure the side loadin' crap and everything else I said I would.

BTW!!! I did use the higher rod ratio motor on top as my working visual, but I used the B18B motor specs. So you can say that I still got the bottom motor back. Ha ha ha...payback is a *****!!!

 

[MichaelDelaney]

use an Excel file..it's much easier.... and you can plot the angles on a graph....

I'll leave you to your math...I read the last chapter to this novel first and so I already know how it's gonna end. The butler did it and the hero gets the girl.

 

[Conor07]

uh, I wish I was smart like you guys.

 

[MichaelDelaney]

uh, I wish I was smart like you guys.

you are as smart. just takes the effort.

the question is what is useful to you and what is worth the effort?

 

[IntegraBad18c]

My answer to that question is: "Do it for the HELL of it and the LOVE of it." Every effort to possibly educate and explore the b18 and other engines as well is worth the effort. I always give props to the pioneers.

 

[IntegraBad18c]

Alrighty, I sat down and made some of my own "real-life" visuals. The tricky part to getting the angle of the rod to the imaginary vertical line is figuring out how long that imaginary vertical line is.

Using b18b specs with the crank angle at 90 degrees (has to be if I want to use the Pythagorem Theorem), the vertical line is 144mm. This is putting the piston into the wall at about 40 degrees.

I realized sumthin was f**ked up when I experimented with a longer rod and got a larger angle!!!!!

In order to properly obtain the vertical line length,
which is changed when you change the crank angle , you have to choose your crank angle, and then draw a picture. The Paint program in Windows helps a lot b/c you can use the pixels as mm's.

So, yeah...I almost have the girl but I'm not done f**king up the butler...