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Old Feb 12, 2026 | 07:20 PM
  #1  
grove's Avatar
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From: jacksonville, fl
Car: 1988 firebird
Engine: 305 tpi
Transmission: 95 T56
Axle/Gears: stock 10 built, yukon diff, 4.10
pushrod length

2nd time building a 383. So I'm still learning. I thought I measured correctly with the pushrod checker tool and ordered 7.325" pushrods for my build. they came in, me being a beginner, I decided to go back to UofY (youtube university) because I was 2nd guessing myself. like story short, the 7.325" sweep pattern is not perfectly centered. its more towards the outside of the valve. The width of the patter is about .070". I reinstalled the pushrod checker but set the length to 7.3" and it still showed more to the outside of the valve. dropped the it down to 7.275 and the pattern is directly in the center the width is .060" on the intake and .050 on the exhaust. piston to valve clearance on the exhaust is .191. they're dished pistons with 2 valve reliefs. the intake valve didnt imprint on the clay so i didnt take a measurement. So the question is, could I run the 7.325" pushrods and everything will be fine? or should I buy the 7.275" pushrods. oh btw, when i checked the PtoV clearance I used the 7.325" pushrod. not sure if that matters. while checking for pushrod length I did use checker springs. checked it multiple times with the 7.325" with/ and without 1/2 turn preload, results were the same. the pattern was slightly to the outside of the valve tip. checked multiple times at 7.275" with and with out 1/2 turn preload, and the results were dead center. how did i determine if it was centered. used the valve locks as a reference. this is a hydraulic roller motor. The method I used to find pushrod length the first which gave me 7.325" was the mid lift method has shown on straub technologies youtube channel.
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Old Feb 12, 2026 | 07:41 PM
  #2  
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From: Ocean Springs, MS
Car: '92 RS Heritage Edition, '91 Z28
Engine: '90 5.7, Vortec Heads
Transmission: 700R4
Re: pushrod length

I think you answered your own question. If the 7.275" centers the pattern better and produces a narrower swipe pattern than the 7.325" there is your answer. Could you run the longer pushrods? Sure, but you run the risk of premature guide wear due to side loading.
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Old Feb 12, 2026 | 08:30 PM
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From: jacksonville, fl
Car: 1988 firebird
Engine: 305 tpi
Transmission: 95 T56
Axle/Gears: stock 10 built, yukon diff, 4.10
Re: pushrod length

Originally Posted by kirk35
I think you answered your own question. If the 7.275" centers the pattern better and produces a narrower swipe pattern than the 7.325" there is your answer. Could you run the longer pushrods? Sure, but you run the risk of premature guide wear due to side loading.

Kirk, youre probably right on me answering my own question. Just seeking validation since I'm a beginner and keep 2nd guessing. Thank you for your reply.
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Old Feb 13, 2026 | 08:26 AM
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Re: pushrod length

I have posted this so many times...

"Centering" DOES NOT MATTER, beyond a certain "reasonable" range. Let's say, within the center 1/3 of the valve stem. What DOES MATTER is the width of the rocker arm tip's sweep across the valve stem.

A wide pattern means that the rocker is "scrubbing" across the stem. This is the thing you want to avoid; it means that there's side loading on the valve, which destroys the guide. The narrower the pattern, the less of this there is, which is what you're trying to attain.

Do a thought experiment, involving 2 pencils. Imagine that you have one laying on your desk, and you're gonna push on its eraser with the point of the other one, and move it in a straight line. This duplicates the idea of pushing as straight as possible on a valve stem without creating any force tending to move it to the side. Now: does it matter WHERE on the eraser you put the point of the pushing pencil? NO. What matters is, that you push as perfectly as possible, IN LINE with the pencil you want to move. IN LINE WITH corresponds to MINIMUM SWEEP in the rocker arm situation, which indicates that the rocker is pushing as straight IN LINE as possible with the valve stem, causing as little SIDEWAYS force as possible.

The geometry of all of the parts involved is more complex than most people give it credit for. First, the rocker moves in an arc, not a straight line. Then, the valve, rocker stud, and push rod are all at different angles; none are parallel to each other. Specifically, if you use the stud as your "reference", both the valve and the push rod are angled TOWARD it. This means that as the valve moves downward, the tip of the stem moves AWAY FROM the rocker stud, which causes "scrubbing" all by itself. Even if it was possible to make the rocker slide up and down the stud in a straight line rather than swinging in an arc, there would STILL be a certain amount of "scrubbing" going on.

Now consider the sideways force applied to the valve stem. It's a very simple relationship: it's the sine of the angle at which the force is applied at each point during the system's motion, times the valve spring pressure. (i.e. a fraction of the valve spring pressure that increases as both the spring pressure itself AND the angle away from straight-line application increase) Since it's impossible to completely eliminate all sideways forces due to the rocker tip itself not moving in a straight line, it makes the most sense to sort of "offset" the unavoidable error, toward the condition where the angle is more nearly straight when the spring pressure is highest; IOW, when the valve is open.

Thus, the best way to set it up is, to find the push rod length that results in the narrowest sweep; then buy push rods that are slightly shorter than this - emphasis on slightly - since this lets the rocker be more nearly IN LINE WITH the valve stem at higher lifts when spring pressure is greatest, and puts the error that you can't avoid at lower lifts when the spring pressure is also lower.

So: "solidify" a lifter THAT YOU"RE GOING TO BE USING (different lifters may locate the push rod seat at slightly different heights, so use one IDENTICAL TO the ones you're installing) at its full height, or, use such light springs on the valve that they can't overcome the spring in the lifter; find the push rod length that results in MINIMUM SWEEP using your adjustable push rod, COMPLETELY IGNORING the location on the stem that the sweep occurs; add to that length the lifter preload you intend to run, which is about .065 - .070" per turn on the nut, depending on your exact rocker ratio; and buy push rods in the next shorter increment of availability than the result. For example, just plucking numbers out of the air, if you find that with a solidified lifter you get the minimum sweep with a PR length of 7.410", and you intend to run ¼ turn of preload (call it .018" at the lifter), then you'll have 7.410" + .018" which is 7.428", so you'd buy 7.400" or even 7.350" PRs. Measure both intake and exhaust at all 4 corners of the engine (#1, #2, #7, & #8) so you can be sure you're taking into account the factory's sloppy machine work, which will usually show up the most by that comparison. Be aware also, that this is affected by how far the valves and seats have been cut over the lifetime of the heads, so that's yet an additional source of potential variance among them.

Last edited by sofakingdom; Feb 13, 2026 at 12:56 PM.
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Old Feb 14, 2026 | 11:51 AM
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From: jacksonville, fl
Car: 1988 firebird
Engine: 305 tpi
Transmission: 95 T56
Axle/Gears: stock 10 built, yukon diff, 4.10
Re: pushrod length

Originally Posted by sofakingdom
I have posted this so many times...

"Centering" DOES NOT MATTER, beyond a certain "reasonable" range. Let's say, within the center 1/3 of the valve stem. What DOES MATTER is the width of the rocker arm tip's sweep across the valve stem.

A wide pattern means that the rocker is "scrubbing" across the stem. This is the thing you want to avoid; it means that there's side loading on the valve, which destroys the guide. The narrower the pattern, the less of this there is, which is what you're trying to attain.

Do a thought experiment, involving 2 pencils. Imagine that you have one laying on your desk, and you're gonna push on its eraser with the point of the other one, and move it in a straight line. This duplicates the idea of pushing as straight as possible on a valve stem without creating any force tending to move it to the side. Now: does it matter WHERE on the eraser you put the point of the pushing pencil? NO. What matters is, that you push as perfectly as possible, IN LINE with the pencil you want to move. IN LINE WITH corresponds to MINIMUM SWEEP in the rocker arm situation, which indicates that the rocker is pushing as straight IN LINE as possible with the valve stem, causing as little SIDEWAYS force as possible.

The geometry of all of the parts involved is more complex than most people give it credit for. First, the rocker moves in an arc, not a straight line. Then, the valve, rocker stud, and push rod are all at different angles; none are parallel to each other. Specifically, if you use the stud as your "reference", both the valve and the push rod are angled TOWARD it. This means that as the valve moves downward, the tip of the stem moves AWAY FROM the rocker stud, which causes "scrubbing" all by itself. Even if it was possible to make the rocker slide up and down the stud in a straight line rather than swinging in an arc, there would STILL be a certain amount of "scrubbing" going on.

Now consider the sideways force applied to the valve stem. It's a very simple relationship: it's the sine of the angle at which the force is applied at each point during the system's motion, times the valve spring pressure. (i.e. a fraction of the valve spring pressure that increases as both the spring pressure itself AND the angle away from straight-line application increase) Since it's impossible to completely eliminate all sideways forces due to the rocker tip itself not moving in a straight line, it makes the most sense to sort of "offset" the unavoidable error, toward the condition where the angle is more nearly straight when the spring pressure is highest; IOW, when the valve is open.

Thus, the best way to set it up is, to find the push rod length that results in the narrowest sweep; then buy push rods that are slightly shorter than this - emphasis on slightly - since this lets the rocker be more nearly IN LINE WITH the valve stem at higher lifts when spring pressure is greatest, and puts the error that you can't avoid at lower lifts when the spring pressure is also lower.

So: "solidify" a lifter THAT YOU"RE GOING TO BE USING (different lifters may locate the push rod seat at slightly different heights, so use one IDENTICAL TO the ones you're installing) at its full height, or, use such light springs on the valve that they can't overcome the spring in the lifter; find the push rod length that results in MINIMUM SWEEP using your adjustable push rod, COMPLETELY IGNORING the location on the stem that the sweep occurs; add to that length the lifter preload you intend to run, which is about .065 - .070" per turn on the nut, depending on your exact rocker ratio; and buy push rods in the next shorter increment of availability than the result. For example, just plucking numbers out of the air, if you find that with a solidified lifter you get the minimum sweep with a PR length of 7.410", and you intend to run ¼ turn of preload (call it .018" at the lifter), then you'll have 7.410" + .018" which is 7.428", so you'd buy 7.400" or even 7.350" PRs. Measure both intake and exhaust at all 4 corners of the engine (#1, #2, #7, & #8) so you can be sure you're taking into account the factory's sloppy machine work, which will usually show up the most by that comparison. Be aware also, that this is affected by how far the valves and seats have been cut over the lifetime of the heads, so that's yet an additional source of potential variance among them.

lets see if my math checks out. rocker studs are 7/16-20. one revolution equal .050". the hydraulic roller lifter says to set preload between .030"-.060". I'd like to shoot for somewhere in the middle. just plugging in numbers, lets say pushrod length is 7.300" using a solidified hyd roller lifter, and I give the polylock 3/4 turn which would put preload around .043" 7.300"+.043"= 7.343" which I should then by 7.350" pushrods. correct?
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Old Feb 14, 2026 | 12:22 PM
  #6  
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Re: pushrod length

With 1.5 rockers and 20 TPI studs (.050" per turn), one turn on the nut is about .083" at the lifter. The arithmetic (doesn't quite qualify as "math") is, 1.5+1 ÷ 1.5 gives how far the push rod end of the rocker moves when the trunnion moves and the valve end is held still; so, 2.5 ÷ 1.5, which is 1.66666, × .050", is .08333". If you have 1.6 rockers then instead of 1.5+1 ÷ 1.5, it's 1.6+1 ÷ 1.6, which is 1.625, × .050" which would be .081". Nitpicking almost, butt worth at least thinking about.

So yeah, if you want about .045" or so of preload more or less, and came up with 7.300" on your adj PR, then that would give 7.34x" depending on which rocker ratio you have, and you'd be better off buying slightly shorter PRs than that. The higher your valve lift and the higher your spring rate is, the more shorter than the "nominal" you'd want to go. So, I'd recommend 7.325" or even 7.300", so that the part of the rocker motion at full lift is more nearly vertical and "perfect", and the part of the arc with greater "error" occurs near zero lift, where spring pressure (and therefore that factor of side loading) is lowest.

To "solidify" a lifter, take it apart and fill it with trans gel or Vaseline. Either of those things will easily melt right back out later. You could also stack washers or something under the push rod cup butt might not be able to get it exactly perfectly solid that way if there's any gap left.

Last edited by sofakingdom; Feb 14, 2026 at 12:59 PM.
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