Hey!

I've been doing a bit more reading...

I've got a question about the degrees of movement on the bottom end of the engine. I know that the camshaft is setup to open at a certain time based on the location of the crank... but what I'd like to know is how to figure out how far down the hole the piston is at a certain degree!

I'm a bit befuddled by this, because I can see in my mind what I'm trying to do on paper, but I cant figure out what the math is to do this.

My engine is apart, or I think I'd just take a degree wheel and a rule to see where everything is!

I hope thats as clear as mud!
Basically what I'm trying to do is see in %'s the differences between camshafts from the perspective of the length of time the valve is open, contrasted against where the piston is in the cylinder.

In my mind if the intake valve opens up right at TDC and closes right ad BDC the cylinder is filled as full as its going to get. Is this correct?

Thanks guys...

 

No. you want to open the intake before TDC. Still air in the intake track takes time to start moving.

Here is my cam card from Crane. It's a solid cam. for my Lt1 motor.

Part Number: 113841 Grind Number: F-278-2 (REPLACES CC-278-2)
Engine Identification:
Start Yr. End Yr. Make Cyl Description
1957 1987 CHEVROLET 8 FAIR IDLE, MODERATE PERFORMANCE USAGE, GOOD MID-RANGE HP, BRACKET RACING, 3400-3800 CRUISE RPM, 10.0 TO 11.5 COMPRESSION RATIO ADVISED. BASIC RPM 3000-6500
Engine Size Configuration
262-400 C.I. V

Valve Setting: Intake .022 Exhaust .022 HOT

Lift: Intake @Cam 320 @Valve 480 All Lifts are based
on zero lash and theoretical rocker arm ratios.
Exhaust @ Cam 3334 @Valve 500
Rocker Arm Ratio 1.50

Cam Timing: TAPPET @.018
Lift: Opens Closes ADV Duration
Intake 29.0 BTDC 69.0 ABDC 278 °
Exhaust 82.0 BBDC 26.0 ATDC 288 °

Spring Requirements: Triple Dual Outer Inner
Part Number 99893
Loads Closed 120 LBS @ 1.875 or 1 7/8"
Open 296 LBS @ 1.415
Recommended RPM range with matching components
Minimum RPM 3000
Maximum RPM 6500
Valve Float 6800

Cam Timing: TAPPET @.050
Lift: Opens Closes Max Lift Duration
Intake 10.0 BTDC 48.0 ABDC 109 238 °
Exhaust 63.0 BBDC 5.0 ATDC 119 248 °

**************************************** ************

As you can see it's already open .050 at 10 degrees. Then mass dynamics- Once in motion it stays in motion. The cylinder continues to fill After Bottom dc. That's how race motors can exceed 100% fill over a narrow rpm band. tuned intake ports and headers all add to the equation.

OH..... I have bigger springs than the cam card and it's a no problem 7000 rpm every day of the week that I have it out. I also have 1.6 roller rocker arms to get the lift up.

 

I don't know of a formula for that.
I'm sure it would vary depending on motor size (even rod length would affect the distance traveled near the top and bottom DC).
It would be one mean calculation to figure out mathmatically, a point on a circle connected to a changing point on a line by a "X" long rod.
easiest way to do it would be with expensive design software, draw everything full size in the computer and it can spit out the data you want. still alot of work.
doing it by hand I'd guess you'd have to figure out a point on the crank (at a rotation angle) from the center line, then use that point to find the distance on the centerline where the connecting rod intersects.
actually it could all be done with "right angle triangle formulas" going to be alot of calculating!

oh and while it would seem logical to open and close the valves @ TDC and BDC it doesn't work as well in the real world as it does on paper.

 

i bet someone know's

 

Makes sense...

Cool stuff guys!

Wonder if any of Duke or CFI knows the formula...

My solid works skills arent up to par for engine creation! I can draw squares and circles...

I like the cam LT1er, a little big on the duration for my motor (I think anyways...) and sounds like not quite enough cylinder preasure... I'll be around 9:1 static!

 

well it seams like it sould be half way simple "but" if you have a 4" stroke and move it 90deg. that should be 1"

 

Clearly I'm going to have to get off my lazy behind and get a motor together to test this!

 

i'm probably off but if you divide the stroke by 360 it should work

 

All it takes is a little trigonometry. Basically what you have is a triangle. Two sides of the triangle are constant. One is the length of the crank throw (half the stroke), the second is the rod length. The third, the distance from the center of the main bearing centerline to the center of the piston pin, is the variable you want to solve for. Based on the angle between the crank throw and the rod, you can solve for the leg of the triangle from the main bearing centerline to the pin, and then add the compression height of the piston to get the distance from the main centerline to the top of the piston. If you subtract that from the deck height, you will get the distance the piston is down the hole. Simple, EH?

RACE ON!!!

 

You notice that they recommend a minumum of 10:1 compression.

This Crane solid is really nice when compared to that 36 year old LT1 solid cam for the 375 hp This new cam has 114 LC and steeper ramps. The wide lobe center traps pressure for such a big 238/248 if it was 110 and advanced like a Comp Cams it would sound like a race motor. I just have race ported old double hump heads. They might flow like the Sportman II 200 cc iron heads.

 

I'm with CFI-EFI

The relationship of the piston position relative to the crankshaft angle of rotation is a function of your stroke and rod length. You can visualize this by assuming you have a very long rod relative to the stroke. In this case, when the crank is at 0 degrees, the piston is at TDC, at 90 degrees it is halfway down, at 180 degrees it is down the hole by whatever the stroke is.

Now take the case where the rod length is half the stroke (a very unfriendly rod ratio of .5). In this case, when the crank is at 0 degrees, the piston is again at TDC. However, when the crank is at 90 degrees, the piston is down the full stroke. It remains in that position until the crank gets to 270 degrees and then it starts moving up the bore again. This is an impractical case, but illustrates that for shorter rods the piston gets pulled down “faster” than for longer rods.

If your trigonometry is good, you can figure the formula. I have been fooling with piston velocity and acceleration for the various rod/stroke combinations, but it all starts with the basic equation for piston position vs crank angle. Here’s what I use. (You may see this equation in different forms, they should ultimately match up, unless I’ve goofed in my trig somewhere.)

P=(sqrt(R*R-S*S*sin(A)*sin(A))+S*cos(A))-(R+S)

Where P is the piston position relative to TDC
R is the rod length
S is one half the stroke
A is the crank angle
* means multiply

 

Not what I call simple... but makes sense.

I'll try to run some equations... <Frantically waving hand above head to signify "WAY OVER MY HEAD!">

 

It's ONLY high school math.

I can see, in rereading my post that I misspoke. I said:I should have said the angle between the crank throw and the line from the centerline of the crank to the piston pin. THAT angle will be the same as the number of degrees the crankshaft has been turned off of TDC until it reaches 180* (BDC).

Happy ciphering.

RACE ON!!!

 

ZD75blue:

For what you are doing, you might just use an approximate expression. If you assume the rod length is much greater than the stroke, the above equation reduces to:

P = R + S*cos(A) - (R+S)

For the typical chevy this will give you a basic idea of the piston position. It shows the piston follows a sinusoidal pattern.

I have written computer programs to do all my calculating of piston postion and piston velocity. (I'm trying to find out what the all the fuss is about with "longer" rods.) If you need real accurate piston positions, I could run the program with your rod and stroke and give you data every 10 degrees or so of crank angle starting at TDC.

 

I dont mean to be a dunce... but what is a *cos ?

You guys must have paid attention in math class...

 

Cosine. It is a trigonometry function.

RACE ON!!!

 

Here's how that equation would read:

P (the distance of the piston from TDC) equals R (the rod length) plus S (half the stroke) times the cosine of A (the crank angle) minus the sum of R plus S.

You should be able to find the cosine function on any scientific calculator. Just as a check for you, the cosine of a 45 degree angle is .7071

 

Ok...

I have a rod length of 5.7, my stroke is 3.48... Standard small block 350.

I have a TI84 Texas instruments calculator... I clicked the COS button and typed 45, and I got .525321

I have a feeling that a 45 degree angle in a decimal is not just "45" but I dont remember...

Thanks for the help guys....

 

So...

I should have 5.7 + 1.74 * .7071 - 7.44 I get -2.1791

Is that the correct way to use the cosine of a 45 degree angle?

 

Real quick:

How can I measure the crankshafts throws? If I know the center point of both the main and rod journals I can start drawing it up in solid works...

Or if measuring isnt needed... Its a 1975 L48 Small block 350...

Is a 5.7 rod the center to center distance?