Tom S
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Any more thoughts on if the shape of the valve effects how boost works upon the valve spring?
Is he a cock bag like you?
Stop now. You enjoy hearing yourself talk more then other people do. Its getting old. It would be different if you had a better idea of what you were talking about, but you dont.
Like Dave said, 910's for all your valve spring needs.
Big Bore- Your math seems to be flawed. Is there exactly one square inch that the boost pressure has to act on the valve ?
Why would that matter? Spring exerting pressure one way, boost exerting pressure the other way. It could be 1/10th of an inch of area. That would not change the units of measure. If it is in fact flawed, please correct me by showing the correct math.
I'm not misunderstanding the principle at all. I'm fully aware that the point load is either magnified or reduced based on the area of the load.
In this instance, I would venture the area of pressure of the spring and that being affected by the boost is almost equal.
The spring exerts pressure in an area the diameter of the spring. Boost exerts pressure on the backside of the valve, however only part of that pressure is in direct opposition to the spring, since the valve curves and the directional forces of the boost will be reduced as the surface becomes closer to perpendicular to the valve face, and, the area of the diameter of the valve stem is not being acted upon by the boost at all. The area of pressure from boost is not the area of the valve face, it will be less than that. If I were to hazard a guess it would start exerting pressure in opposition to the spring somewhere around 45*, but that will be significantly reduced, so it's probably almost equal to the area of the spring pressure being exerted in opposition, possibly even less, which in turn makes the higher pressure spring even more overkill.
Big Bore-I agree...this statement above is counter to physics 101. Perhaps it's a forum communication thing, and you just posted something confusing. I'm under the impression that determining the area on the valve that boost/drive pressure forces actually work upon is paramount to determining seat pressure requirements....so it would seem to matter a bit.
BigBore-Using the methodology you posted above...What is the valve surface area number you've determined to be exposed to boost/drive pressure forces?
Can you explain the above bolded/underlined text for me? It confuses me.
My issue with the statement I responded to was not that the area didn't matter. Yes it was, see quotes belowIt was whether or not it was an actual inch.
I have not determined the actual area. But I'm confident that any area that is more than 45* to the valve face is not being affected in opposition to the spring.
The underlined part was qualified as being a "ventured guess". It would probably be more accurately described as a wild ass guess based on my visualization of the areas being acted on by the spring and boost...which is really just a visualization of the diameters involved and the percentage of the valve that I believe is actually being acted upon.
Why would that matter? Spring exerting pressure one way, boost exerting pressure the other way. It could be 1/10th of an inch of area. That would not change the units of measure. If it is in fact flawed, please correct me by showing the correct math.
Hmmm......sounds like you are.No, your entire line of reasoning is incorrect.
If you have 80 lbs of weight on a one inch square, and 80 lbs on a 10 inch square, which has more lbs per square inch. 80 lbs does not become more if the area is larger, and it does not become less if it's smaller.
The spring rate is transferred directly to the valve, otherwise it would have no affect on anything. What you are calling "connecting" is known as load transfer. This should be obvious in that Doug has made larger locks to retain his higher rate springsBB-You keep referring to an area that the spring in contact with the valve...
On the 2000 7.3 engine in my truck, the spring doesn't contact the valve anywhere....it contacts a retainer, that "connects" the spring force to the valve stem.
Where do you think the seat pressure is measured?Can you explain to me, why does this (the surface area that the valve spring contacts/applies force to the valve) matter?
I think the word you are looking for is "implying". You would be the one "inferring". And I'm not implying anything, I'm making a clear statement. Boost acts on the valve, which is connected to the spring. Remove the valve springs, turn the head upside down so gravity will hold the valve in place, and put 60 psi to the sealed intake runner. Make sure no one is standing over the valve and that it's not aimed at anything important.Are inferring that boost/drive pressures can act on the spring?
Ok, I'll play. Not only have I disassembled several sets of heads, I have assembled the new heads (I did not do the actual head assembly, not that it's relevant) to the long block and installed the motors. I have had the VC's off numerous 7.3's and R&R'd several sets of injectors and o-rings. Additionally I've disassembled numerous heads both 4 stroke and 2 stroke and done head work including valve seat lapping and porting and polishing. I'm certified to work on large and small diesel generators, both the engines and the electronics, and I'm currently an engineering student. I have extensive experience in other fields that directly involve loads and supporting them properly. Now it would only be good manners to share your experience and qualifications.Not to insult your intelligence or experience, but have you examined or assembled/disassembled at 7.3 cylinder head off the motor?
It sounds to me like you are applying spring pressures incorrectly...possibly making this harder than it has to be....or maybe I just don't get your point....could you just break this down Barney Rubble/Scooby Doo simple for me so I can understand your point?
Thanks.
Hmmm......sounds like you are.
Alright master-tech goodbyeLOL
Depends on what you are trying to calculate. The lift isn't relevant to our discussion of how boost acts on the valve when closed. It is relevant to calculating spring length. At that point you will also want to know open pressure and spring rate.75 psi boost/drive
1/2" of total valve lift
what other factors am I missing?
I'm not sure what you mean by translating. A coil spring exerts pressure on an area the size of it's diameter. The seat pressure and open pressure are measured on the end of the valve, but that load is directly transferred to the area of the spring diameter by the retainer and locks.What relationship does the diameter of the spring and/or it's retainer have with translating "spring rate" to seat pressure and boost/drive psi?
Choupique108-Sorry for the thread sidetrack. I should've known better than to post in a thread with "Gangsta" in the title. ointlaugh:
So in your daily driver - what pressures are you expecting to see?Thread title has nothing to do with it. It's ones that try to out think themselves that kill it.