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92 omc 4.3 valve adjustment

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Spepple08

New member
I've got a 92 4.3l I'm working on. I got hosed on it a bit and had to replace head gaskets. Everything im finding on the 4.3 is adjust to no lash then 1 1/2 turns. Looking for some verification on this.

Thanks for any and all input.
 
Although we will see and read "adjusting valves" in many manuals and instructions, it is actually a misnomer.
Once a valve has been installed, spring fitted and the retainer in place....... there is no adjusting it.

You will actually be setting the depth of the hydraulic cam follower's plunger.
(There is approximately .100" + internal plunger travel available. You DO NOT want to exceed this, or the plunger may bottom out!)

Once ZERO lash is achieved, you will rotate (as in tighten) the rocker arm nut until the push rod has properly set the depth of the hydraulic plunger.
The OEM uses a formula for this.
Rocker arm ratio X's rocker arm stud thread pitch X's rocker arm nut revolution. This sets the plunger depth!

1-1/2 turns past zero lash seems excessive to me. I would look further.


With the V-6 engine, I'll suggest using what we call the 6 stop static procedure.
(I would NOT suggest using the 2 or 3 stop!)

Please watch this video....... this guy knows what he's doing and how to explain it.


https://www.youtube.com/watch?v=89o5rLpbCgI

NOTE: the GM SBC V-8 is being used in this video. Your 4.3L SBC adjustment will be very similar.

Note that he's doing this prior to the intake manifold being installed while he can view the cam followers (aka lifters).
I would suggest doing the same.

Note what he says @ :43 to 1:00 min/sec into the video.
This represents the #1 piston @ TDC on the Compression Stroke.
It also represents both valves being closed, and when both cam followers are at the bottom of base circle.

At the 1:10 min/sec mark, he shows what to do when the rocker arms are adjustable.

Note his comment @ 1:49 min/sec where he explains how this sets the depth of the hydraulic plunger, of which is the actual goal!

At 2:27 min/sec, he explains which cylinder will be adjusted next.

The V8 firing order is 1-8-4-3-6-5-7-2, so the next cylinder in line on this V8 engine would be #8.
The #8 cylinder will be at TDC C/S after rotating the crankshaft only 90*.

With your 4.3L V6 engine, your firing order is 1-6-5-4-3-2, so the next cylinder in line would be #6
NOTE: the #6 cylinder will be at TDC C/S after rotating the crankshaft 120*.

After making the #6 adjustment, you will rotate the crankshaft another 120*, and adjust for #5.
Keep going right down the firing order until you come back to #1.


It's actually very simple once we understand what the actual goal is.






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In every GM shop manual I have seen they say to go one full turn after zero lash; which is determined by trying to lift the pushrod up and down not spinning it. Some high performance tuners use 3/4 turn but I used 1 full turn when I did mine.
 
Bt Doctur said:
Depending on the Engine, it may just get torque`d down to 20Ft/Lbs
Click to expand...
That would only apply to the non-adjustable system, as shown @ 1:00 min/sec into the video that I linked us to above.
Judging from the thread title, I believe that the OP is asking about the adjustable system.


louc said:
In every GM shop manual I have seen they say to go one full turn after zero lash; which is determined by trying to lift the pushrod up and down not spinning it. Some high performance tuners use 3/4 turn but I used 1 full turn when I did mine.
Click to expand...

If we are working on a fresh SBC engine, and with the intake manifold not yet installed, we can view the top of the cam follower plungers.
If the plungers are not yet sitting on a full cushion of oil, we can move them against their plunger springs, making it easier and more accurate to find ZERO rocker arm/valve stem lash.
In other words, we can see the moment at which the plunger begins to move downwards into the body.


(excuse the bold font..... for some reason I can't get rid of it)

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


Perhaps this will help explain what occurs when we use 1/2 turn, 3/4 turn, full turn or 1-1/2 turn after having found zero lash:

As we know, rocker arm ratio refers to the amount of movement on the valve side of the rocker arm compared to that on the push rod side.
In other words..... X.0 push rod movement upwards (on the rocker arm) equals X.5 movement downwards against the valve stem side.

For example:
A 1:1.5 rocker arm will move a valve 1.5 times that of the actual cam lift.

Let's say that the cam lift is .300".
Multiply that by 1.5 and the valve travel amount is .450".

Now consider the the thread pitch of the rocker arm adjusting nut.

First think in terms of:
.... the valve stem being stationary.
.... the rocker arm adjustment setting the depth of the cam follower's plunger.

If the rocker arm stud/nut is 7/16" X 20 TPI, one full revolution moves the center of the rocker arm .050".
Now factor in the amount that the ratio and the thread pitch affects push rod side, and the number becomes .025" (at the push rod side).

That means that one full nut turn past ZERO lash, sets the plunger into the follower body by .025".
This leaves approximately .075" plunger travel that is un-used.

Add another 1/2 turn to that, do the math, and you will see where this would set the plunger.

Not deep enough, and eventual component ware may cause the plunger to top out, causing what we call rocker arm clatter.





By the way, the OP has not been back. Maybe he figured it out.


.

 
based on the stated assumptions, the plunger will move a lot more than 0.025" with a full turn applied to the stud's nut....
 
makomark said:
based on the stated assumptions, the plunger will move a lot more than 0.025" with a full turn applied to the stud's nut....
Click to expand...

Mark, you are right.... my apologies...... I did my math incorrectly.

The affect of the rocker arm ratio (on the push rod side), would actually cause one full turn past zero lash (using a 7/16" X 20 TPI stud thread pitch) to set the plunger more like .062" or .063".

Somewhere I have an old drawing that I'll try to find.
If I can find it, I'll post it.
 
So does that mean that the setting should be from 3/4 to one full turn?
when I did mine I used one turn, and the engine sounded the same as it always did...
 
more like 0.083"....

one turn is what GM has recommended for decades...as long as you are running fine, I'd leave it alone...
 
yep this is one of those times where the more you read the more confused you get, there are too many experts out there today, but this discussion was helpful because it makes it easier to understand what you're actually adjusting with hydraulics vs when you do mechanical valve lifter adjustments....
 
................
louc said:
yep this is one of those times where the more you read the more confused you get, there are too many experts out there today,
I'm sorry that you became confused.

but this discussion was helpful because it makes it easier to understand what you're actually adjusting with hydraulics
That is why some of us go into so much detail.

vs when you do mechanical valve lifter adjustments....
Click to expand...
 
makomark said:
more like 0.083"....
Click to expand...
Mark, can you explain the math and how that works out to be at/near .083" after one full turn.
I must not be doing my math correctly.

During the static adjustment, consider the valve stem to be stationary and the fulcrum point.
The push rod will be the area that will realize the result of the rocker arm nut adjustment..... one full turn!




cam follower 11 .jpg
 
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probably your geometry vs the math....

Assuming rigid bodies and ignoring the minor issues with non-perpendicular interfaces, here's my view:

with the valve stem tip being fixed, rotating the rocker nut one turn moves it 0.050" downward. the pushrod end will move down even more, in proportion to its 'extra' length from the fixed point. so pushrod moves down by Y = 0.050"/1.5 *2.5 ~ 0.083".

For an alternative view, assuming the rocker starts in a 'horizontal orientation', spinning the rocker nut down one turn caused the rocker arm to sweep an arc defined as tan Ф = 0.050/1.5. The pushrod's displacement is then 2.5 * tan Ф which yields the same value...
 
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