Broken Valve Spring

[cycleguy55]

Originally Posted by splenderleith

I was able to remove the intake and all 10 of the head bolts. I can definitely see a bent valve from the intake port.

I haven’t figured out how to get the head loose. Any ideas?

You might try using air. Make sure all the head bolts are loose or removed. Put a fitting in one of the spark plug holes in an end cylinder (not the one with the broken valve) and pressurize it. Area of a 4.36" bore is 4.75 sq inches, so 100 PSI is 475 lbs pressure pushing up on the cylinder head.

2021-10-17 Edit: 460 CID standard bore is 4.36", so (4.36 ÷ 2)² x 3.14159 = 4.7524 x 3.14159 = 14.93 in² x 100 PSI = 1,493 lbs.

My earlier numbers were incorrect.

[ZOERA-SC7XX]

Originally Posted by cycleguy55

You might try using air. Make sure all the head bolts are loose or removed. Put a fitting in one of the spark plug holes in an end cylinder (not the one with the broken valve) and pressurize it. Area of a 4.36" bore is 4.75 sq inches, so 100 PSI is 475 lbs pressure pushing up on the cylinder head.

Sounds like a good plan. However, you should leave at least two head bolts loosely attached. 100 psi is quite a bit of air.

[767Jockey]

Originally Posted by ZOERA-SC7XX

Sounds like a good plan. However, you should leave at least two head bolts loosely attached. 100 psi is quite a bit of air.

Make sure the parking brake is firmly set, 100PSI is also enough to rotate the engine in gear and move the car.

[767Jockey]

That piston is likely going to be ugly. Hopefully your cylinder walls were spared.

[Gaz64]

Originally Posted by cycleguy55

You might try using air. Make sure all the head bolts are loose or removed. Put a fitting in one of the spark plug holes in an end cylinder (not the one with the broken valve) and pressurize it. Area of a 4.36" bore is 4.75 sq inches, so 100 PSI is 475 lbs pressure pushing up on the cylinder head.

Bore size is irrelevant. 100 psi from an air line into the cylinder will still be 100 psi.

[patrickt]

Originally Posted by Gaz64

Bore size is irrelevant.

Uhhh, are you sure about that? Lift force (for blowing that head up through the ceiling) is found by multiplying the pressure times the area. That's why a hovercraft can get off the ground by using only 7 lbs. of pressure. Of course, you should not just take my word for it. See: https://www.discoverhover.org/infoin...ors/guide4.htm

[Gaz64]

No, I don't agree.

Let's assume 5 square inches.

A typical compression test might yield 180 psi, about 12 x 1 atmosphere (14.7psi). Compression ratio and inlet valve closing point are the determining factors.

And so do we get 5 times our typical compression pressure?

[patrickt]

The page I referenced gives a wonderful example, including the math, but an easy way to visualize the difference is to think of a thimble full of air at x pressure and imagine how easy it would be to hold your thumb over it. Now take the same pressure and have it inside a barrel. Do you think the strength necessary to hold the lid of the barrel down is the same as the thimble even though the pressure is the same?

[Gaz64]

Originally Posted by patrickt

The page I referenced gives a wonderful example, including the math, but an easy way to visualize the difference is to think of a thimble full of air at x pressure and imagine how easy it would be to hold your thumb over it. Now take the same pressure and have it inside a barrel. Do you think the strength necessary to hold the lid of the barrel down is the same as the thimble even though the pressure is the same?

Hmm, I have to ponder over this one for a bit.

[cycleguy55]

Originally Posted by Gaz64

Bore size is irrelevant. 100 psi from an air line into the cylinder will still be 100 psi.

Bore size matters. 100 PSI (Pounds per Square Inch) has a greater overall force when applied to a greater area. It would be 200 lbs force when applied to 2 square inches, but 500 lbs force when applied to 5 square inches.

That's also why a narrow bicycle tire (e.g. 700 x 25c) needs more pressure (100-125 PSI) to support a given rider weight than does a wider tire (e.g. 700 x 38c) - which might only need 60-80 PSI.

[patrickt]

Originally Posted by cycleguy55

Bore size matters. 100 PSI (Pounds per Square Inch) has a greater overall force ...

I think Gaz gave up when I hit him with my thimble and barrel example.

[splenderleith]

Thanks for all of the help this far. I am sure I wouldn’t have gotten this far without it. Diagnosis spot-on Gary.

I put the plugs back in, cranked it and saw the head move. With some very light prying pressure, it seems to move a little on the intake side. However, I can’t get much more than that. I am wondering if I might be missing something. I can’t get the exhaust side of the head to budge, even the slightest bit.

Valve covers removed, intake manifold removed, headers removed, rocker arms and push rods removed (rocker arm studs and 6 of the valves still in place), 10 head bolts removed, spark plugs removed.

There are 3 large Allen head screws/plugs in the head, between the valves, which I assume are something to do with the machining process. I haven’t messed with them. Does it seem like I missed some fastening device?

[splenderleith]

Picture of one of the Allen bolts.

[Gaz64]

Originally Posted by splenderleith

Picture of one of the Allen bolts.

I thought you said you had head bolts.

They look like studs, after nuts have been removed.

The heads will not move sideways while studs are fitted. The studs need to be removed if the head won't break free.

The hex plug is a coolant plug. No need to remove it.

[Gaz64]

Originally Posted by cycleguy55

You might try using air. Make sure all the head bolts are loose or removed. Put a fitting in one of the spark plug holes in an end cylinder (not the one with the broken valve) and pressurize it. Area of a 4.36" bore is 4.75 sq inches, so 100 PSI is 475 lbs pressure pushing up on the cylinder head.

I'll say my interpretation of Brian's statement lead to some confusion.

I read it as 100 psi is 475 psi, as in 100 psi becomes 475 psi, which is impossible while you have a shop line connected to the cylinder.

100 psi would create 475 lbs of force, not pressure.

The theoretical force output of a cylinder is the usable piston area multiplied by the applied air pressure (F = PA). For example, a cylinder with a 11⁄2-in. bore and an extend force of 80 psi generates 141 lb of force (1.767 × 80 = 141).