Internals HP/torque vs. stress theory
Originally Posted by Enron Exec
Your right, i foo'ked up there. They should be proportional. 
Im glad there are actually ppl reading this to make sure no one is just spewing out BS.
Id use lbs or Newtons. To get the force exerted on each piston top, id take the torque at the crank and divide it by 6 ( because we have 6 cylinders ), then divide the result by 2 ( because the crank turns 2 times or 720 degrees to complete the 4 strokes ). The result would be the torque each cylinder delivers to the crank. To convert torque to linear force, id have to find the rod length. That should result in the force exerted on top of each piston during the combustion stroke. I think there are a few other things to consider, like the size of the lobs on the crank but im not sure. Maybe there is an equation floating around where we can plug-n-chug.
Im glad there are actually ppl reading this to make sure no one is just spewing out BS.
Id use lbs or Newtons. To get the force exerted on each piston top, id take the torque at the crank and divide it by 6 ( because we have 6 cylinders ), then divide the result by 2 ( because the crank turns 2 times or 720 degrees to complete the 4 strokes ). The result would be the torque each cylinder delivers to the crank. To convert torque to linear force, id have to find the rod length. That should result in the force exerted on top of each piston during the combustion stroke. I think there are a few other things to consider, like the size of the lobs on the crank but im not sure. Maybe there is an equation floating around where we can plug-n-chug.
Will
Originally Posted by overZealous1
i would of thought the lever arm would change effective length as the mixture doesn't fire all in one instant, but for a duration of the stroke, and changes the leverage as it travels down. so that 74 degrees of crank angle is where you just get the highest point of power and leverage, with everything past that trailing off?
these are very good write ups and i am learning a few things from them. most all my experience comes from many books and building many motors and hanging out in machine shops though, and not from a classroom invironment, where some of these theories were not covered, but had to figure out on my own, lol. sounds like you have alot of cool tools to play with there though.
these are very good write ups and i am learning a few things from them. most all my experience comes from many books and building many motors and hanging out in machine shops though, and not from a classroom invironment, where some of these theories were not covered, but had to figure out on my own, lol. sounds like you have alot of cool tools to play with there though.
Will
EDIT: to be perfectly fair, the force acting on the piston does act all the way through the power stroke, but you have to realize that so friggin much is lost through heat and friction, that not long afte the rod is at 90 to the crank throw, like several more degrees of rotation, the applied torque just drops off so fast its not really contributing anyhting. And on a high revving NA engine like the Yamaha, the valves open pretty damn fast to clear things out, but technically, the force acting on the piston is applied throuh the entire power stroke. For the most part, cause a funny thing happens at BDC due to the accelerative forces and crank/rod angles at BDC, but that's WAY off topic.
Last edited by Resolute; Jan 2, 2006 at 10:47 AM.
i understand completely what you are saying. and this brings up a good question. so in a V style engine, the cylinder bank angles have to be specifically set to take advantage of this?
how does this crank angle/leverage point apply to inline or "boxer" style engines?
how does this crank angle/leverage point apply to inline or "boxer" style engines?
Okay... I am on lunch and have had a chance to read over all my little rants through the day so far, and am surprised at some of the mistakes I have made. My apologies, for what it's worth, I am at work doing a job, and I'm just not able to focus enough on this to be COMPLETELY accurate, but for the most part, the theories are right. For those who have read this far, damn, I'm sorry, but you'll have to re-align somethings. Here is what is right and really the important part to answer the question of which engine has the most stress. When the engine produces torque, it is not a function of rpm, but engine speed. the difference lies in the fact that at certain speeds, a certain amount of torque is made, and the fact that the engine is physically turning at x times per minute has nothing else to do with torque production, just the piston speed at that x rpm. What is affected by the number of physical times the crank revolves is the amount of work the engine has done, or the hp. The error I have posted earlier is that the torque placed on the crank is the result of each individual power stroke from a single cylinder in and of itself. This is wrong. It is the amount of torque placed on the crank by all the cylinders that rotation. Even at 4000 rpm, the crank has the same torque applied for one rotation as it does at 2000 rpm if the measured torque is the same. the rpm itself has nothing to do with torque, the piston speeds at 2000 rpm are just better for producing that same amount of torque on one engine and the piston speeds are better at 4000rpm for producing the same amount of torque on the other engine. The physical strain on the engine from the torque is the same. In other words, if an engine making 400lb/ft has one piston, and another making 400lb/ft has ten, then the engine with one piston is taking a much higher load than the one with ten, and again, the rpm at which the peak torque occurs doesn't mean anything for load but when and how much hp will be produced. Since our original question was with two identical engines, each with the same piston area, stroke, etc... then the loads from compression are still the same since the torque is the same. If it were different engines, like which is under more load a V8 or my lawnmower when both make 400 lb/ft, then the stress on the engines would be different, even if the discplacemnt was the same. Ok, even this took about an hour to do inbetween working, so I'll keep checking in, and will do a better job after work when I can better explain things.
Will
Will
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From: Houston
Originally Posted by Resolute
LOL, that's cool. I know the feeling. I'm at work, and would love to just put everything aside and hack this out.
Dorm buddy! I have a wife and a house man.
Now here's the kicker, I'm finishing up my ME, but am planning on going to CSU for my masters asap in Motorsports Engineering. Right now, I get to hang out and help with their prototype racer and facilities on the weekends. Besides being hella fun, I get to schmooze the three profs into liking me enough to enter the program. Now, the facilities they have are pretty awesome. Besides a typical chassis dyno, they have a brake friction dyno, chassis flex dyno, shock dyno, all kinds of geeky metallurgy stuff, and most important for our discussion, an engine brake dyno.
Now, I would love to use some of this stuff on the VQ, and maybe I'll have a chance one day, like hooking up a pressure indicator to the head would be of enormous value, but for now I have to settle for playing with a one liter Yamaha. Now, this is important because of some things you said.
First, why isn't the torque applied over the course of all the revolutions at a certain rpm? In our case, lets imgain the engine with a power peak at 6100rpm makes its peak torque at 4000rpm. Now, if I understood you correctly, you are saying that at 4000rpm, the 400lb/ft of torque is pread over 2000 power strokes. If the max torque occured at say 2000rpm, it would be spread over 1000 power strokes, and therefore more stress is on the engine making torque at the lower rpm. Here is where an engine brake dyno comes in handy, and you pretty much alluded to the same in your last post. The dyno could give a crap about rpm, in fact it doesn't even measure rpm. We only need to moniter rpm to calculate hp. What it does measure is the amount of torque applied on EACH INDIVIDUAL POWER STROKE. Ours actually measures this twofold, by measureing the amount of pressure placed on the mounts and with a brake on the flywheel. Think about it like this, Even with our V6, no two pistons are firing at the same time, and if max torque is applied on every piston when the crank angle is at 74 degrees (it is on the VQ, just trust me on that one cause I couldn't even begin to type out the math) then every time the piston on any cylinder is on its power stroke and the crank angle for that piston is 74.23 degrees, then the max torque is being applied to the crankshaft centerline. Even with 6 cylinders, there are seperate, individual power strokes and seperate individual "surges" of max torque. As you stated, you can feel this at low rpm without that dual-mass flywheel to dampen them out. The dyno records the max amount of torque applied at any one time, regardless of rpm. It just happens that at 4000rpm in our engine, the max torque developed from any one of our six "surges" is 400lb/ft of torque. Now, as far as we are concerned, the rpm is useless, we might as well just say that the engine develops its peak torque, or even better, the individual cylinders each develop their peak torque when the piston speed is 17.7 meters per second. Because that's how fast our piston is moving when the engine is at 4000rpm. The only reason we care about 4000rpm is to calculate the hp. The reality is, if we were measuring the torque over a one minute period, then the max torque developed over the whole minute would be divided out among all the seperate strokes, but we aren't because that's not how torque is measured, that's how horsepower (work) is calculated, the engine at any one instant of time while spinning at 4000rpm has a max torque applied to the crank of 400 lb/ft. If we plot this out with torque as a function of engine speed and time, and kept our engine speed constant at 4000rpm, you would see that the engine has a "push" developing 400lb/ft of torque every .0075 seconds. Every .0075 seconds, another power stroke would occur on the piston and produce 400lb/ft of torque, and this max torque occurs with a piston speed of 17.7 m/s, or when the engine is at 4000rpm. The amount of work the engine does is a function of this torque occuring on each power stroke, and the number of times it occurs in that one minute time period.
I think this is confusing because people see that hp and torque are the same, with a few constants changed around. They are not. They are so different its crazy. As Dave Coleman once said, asking what's the difference between hp and torque is like asking what's the difference between orange and soft, or so he said something to that extent. If we deal with this in terms of Kilowatt output as a function of Newton/meters of torque applied over time, then I think this would be way easier to grasp, especially since the term horsepower is a bit decieving as a term for power.
It's been three hours since I started typing, I keep getting interrupted to actually do some work, and then come back and type a little more while I can. so i hope this is somewhat coherent.
Will
Dorm buddy! I have a wife and a house man.
Now here's the kicker, I'm finishing up my ME, but am planning on going to CSU for my masters asap in Motorsports Engineering. Right now, I get to hang out and help with their prototype racer and facilities on the weekends. Besides being hella fun, I get to schmooze the three profs into liking me enough to enter the program. Now, the facilities they have are pretty awesome. Besides a typical chassis dyno, they have a brake friction dyno, chassis flex dyno, shock dyno, all kinds of geeky metallurgy stuff, and most important for our discussion, an engine brake dyno.
Now, I would love to use some of this stuff on the VQ, and maybe I'll have a chance one day, like hooking up a pressure indicator to the head would be of enormous value, but for now I have to settle for playing with a one liter Yamaha. Now, this is important because of some things you said.
First, why isn't the torque applied over the course of all the revolutions at a certain rpm? In our case, lets imgain the engine with a power peak at 6100rpm makes its peak torque at 4000rpm. Now, if I understood you correctly, you are saying that at 4000rpm, the 400lb/ft of torque is pread over 2000 power strokes. If the max torque occured at say 2000rpm, it would be spread over 1000 power strokes, and therefore more stress is on the engine making torque at the lower rpm. Here is where an engine brake dyno comes in handy, and you pretty much alluded to the same in your last post. The dyno could give a crap about rpm, in fact it doesn't even measure rpm. We only need to moniter rpm to calculate hp. What it does measure is the amount of torque applied on EACH INDIVIDUAL POWER STROKE. Ours actually measures this twofold, by measureing the amount of pressure placed on the mounts and with a brake on the flywheel. Think about it like this, Even with our V6, no two pistons are firing at the same time, and if max torque is applied on every piston when the crank angle is at 74 degrees (it is on the VQ, just trust me on that one cause I couldn't even begin to type out the math) then every time the piston on any cylinder is on its power stroke and the crank angle for that piston is 74.23 degrees, then the max torque is being applied to the crankshaft centerline. Even with 6 cylinders, there are seperate, individual power strokes and seperate individual "surges" of max torque. As you stated, you can feel this at low rpm without that dual-mass flywheel to dampen them out. The dyno records the max amount of torque applied at any one time, regardless of rpm. It just happens that at 4000rpm in our engine, the max torque developed from any one of our six "surges" is 400lb/ft of torque. Now, as far as we are concerned, the rpm is useless, we might as well just say that the engine develops its peak torque, or even better, the individual cylinders each develop their peak torque when the piston speed is 17.7 meters per second. Because that's how fast our piston is moving when the engine is at 4000rpm. The only reason we care about 4000rpm is to calculate the hp. The reality is, if we were measuring the torque over a one minute period, then the max torque developed over the whole minute would be divided out among all the seperate strokes, but we aren't because that's not how torque is measured, that's how horsepower (work) is calculated, the engine at any one instant of time while spinning at 4000rpm has a max torque applied to the crank of 400 lb/ft. If we plot this out with torque as a function of engine speed and time, and kept our engine speed constant at 4000rpm, you would see that the engine has a "push" developing 400lb/ft of torque every .0075 seconds. Every .0075 seconds, another power stroke would occur on the piston and produce 400lb/ft of torque, and this max torque occurs with a piston speed of 17.7 m/s, or when the engine is at 4000rpm. The amount of work the engine does is a function of this torque occuring on each power stroke, and the number of times it occurs in that one minute time period.
I think this is confusing because people see that hp and torque are the same, with a few constants changed around. They are not. They are so different its crazy. As Dave Coleman once said, asking what's the difference between hp and torque is like asking what's the difference between orange and soft, or so he said something to that extent. If we deal with this in terms of Kilowatt output as a function of Newton/meters of torque applied over time, then I think this would be way easier to grasp, especially since the term horsepower is a bit decieving as a term for power.
It's been three hours since I started typing, I keep getting interrupted to actually do some work, and then come back and type a little more while I can. so i hope this is somewhat coherent.
Will
So i guess we all agree on this now or where you just repeating what i was saying.
"the max torque developed from any one of our six "surges" is 400lb/ft of torque."
Your absolutely right. I dont know how i came to that in the first place.
Originally Posted by Enron Exec
" and therefore more stress is on the engine making torque at the lower rpm."
So i guess we all agree on this now or where you just repeating what i was saying.
"the max torque developed from any one of our six "surges" is 400lb/ft of torque."
Your absolutely right. I dont know how i came to that in the first place.
So i guess we all agree on this now or where you just repeating what i was saying.
"the max torque developed from any one of our six "surges" is 400lb/ft of torque."
Your absolutely right. I dont know how i came to that in the first place.
Will
ok, here is another point. we have surely drifted from the original topic, but what the hell. ok, the vq35 has a larger bore compared to stroke. 95.5 boreX 81.4 stroke. now larger bore motors are more rpm friendly than a longer stroke version due to piston speed being slower. the loss of rod length and distance of throws on the crank centerline though, result in a loss of torque compared to a square bore/stroke motor. now i have also heard the larger bore, compared to stroke, is more conducive to higher hp #'s. do you feel this reason being the ability to rev higher thus having more work cycles to end up with ultimately higher hp numbers, or is there an advantage to having the larger piston top area for the increase in hp. i an leaning towards the first answer.
now most of my experience is with the small block chevy. last one i did was a 383. instead of the 400 crank in the 350, they now make a specific piston and crank combo that allows using the stock 350 5.7 rod length and taking up the difference in wrist pin placement. so you still gain the c.i. from the piston traveling farther thus more vacuum to draw in the mixture at the sacrifice of higher piston speeds. do you feel this method takes full advantage of the torque increases compared to using the standard pin height and the original 400 6.0 rod length?
ok, doing some searching, i found another force i didn't think about earlier. which is cylinder wall side loading. the piston will generate more piston side loading pushing the rod at more of an angle in relationship to the bore using the shorter rod length as i describer earlier.
now most of my experience is with the small block chevy. last one i did was a 383. instead of the 400 crank in the 350, they now make a specific piston and crank combo that allows using the stock 350 5.7 rod length and taking up the difference in wrist pin placement. so you still gain the c.i. from the piston traveling farther thus more vacuum to draw in the mixture at the sacrifice of higher piston speeds. do you feel this method takes full advantage of the torque increases compared to using the standard pin height and the original 400 6.0 rod length?
ok, doing some searching, i found another force i didn't think about earlier. which is cylinder wall side loading. the piston will generate more piston side loading pushing the rod at more of an angle in relationship to the bore using the shorter rod length as i describer earlier.
Last edited by overZealous1; Jan 2, 2006 at 12:57 PM.
Originally Posted by overZealous1
ok, here is another point. we have surely drifted from the original topic, but what the hell. ok, the vq35 has a larger bore compared to stroke. 95.5 boreX 81.4 stroke. now larger bore motors are more rpm friendly than a longer stroke version due to piston speed being slower. the loss of rod length and distance of throws on the crank centerline though, result in a loss of torque compared to a square bore/stroke motor. now i have also heard the larger bore, compared to stroke, is more conducive to higher hp #'s. do you feel this reason being the ability to rev higher thus having more work cycles to end up with ultimately higher hp numbers, or is there an advantage to having the larger piston top area for the increase in hp. i an leaning towards the first answer.
now most of my experience is with the small block chevy. last one i did was a 383. instead of the 400 crank in the 350, they now make a specific piston and crank combo that allows using the stock 350 5.7 rod length and taking up the difference in wrist pin placement. so you still gain the c.i. from the piston traveling farther thus more vacuum to draw in the mixture at the sacrifice of higher piston speeds. do you feel this method takes full advantage of the torque increases compared to using the standard pin height and the original 400 6.0 rod length?
now most of my experience is with the small block chevy. last one i did was a 383. instead of the 400 crank in the 350, they now make a specific piston and crank combo that allows using the stock 350 5.7 rod length and taking up the difference in wrist pin placement. so you still gain the c.i. from the piston traveling farther thus more vacuum to draw in the mixture at the sacrifice of higher piston speeds. do you feel this method takes full advantage of the torque increases compared to using the standard pin height and the original 400 6.0 rod length?
For the SB, are you saying the 400 crank has a longer stroke and a longer rod than a 350? Is the deck height higher then on the 400? So they have a crank that keeps the shorter 350 rod with the longer stroke of the 400 crank? I'm sorry, I don't think I have all the info to answer that question.
Will
ok, heres more info.
383 using 5.7 rods- 4.000 boreX3.80 stroke
383 using 6.0 rod(original 400 crank) 4.030 boreX3.75 stroke
deck height should be identical. this is just one example i am using for a reference. just looking at possible effects of rod length here yet creating the same displacement.
also, the reason i brought up the bore/ stroke relationship and how it can relate to rpm, was based around our own vq's. even with all the forged goodies and the stock main cap girdle and overhead cams, not many people i know about have gotten the thing to be able to rev very high. i was shifting my 383 at 7300rpm and didn't have as good of a crank or rods in it as in my vq. if i am remembering correctly some were floating valves right around that same rpm. now i remember putting in my nismo cams and i was getting pretty close to coil bind with the stock springs (right around 0.012" cold). now there are many ways for the aftermarket to make larger dia. springs or double springs for this motor. i guess just not many people have done it. but waiting for the day i see a street vq35 turning 9000rpm with not much issues.
which brings me to another question, boost effects on valve springs. i have heard from some that it makes no difference, but i have a tough time imagining there is no effect. the higher the boost, the more spring pressure required on the intake side to help keep the intake valve closed at high rpm and boost. correct or false. i believe true.
just trying to keep ya busy here and answering some of my own questions i have, lol.
383 using 5.7 rods- 4.000 boreX3.80 stroke
383 using 6.0 rod(original 400 crank) 4.030 boreX3.75 stroke
deck height should be identical. this is just one example i am using for a reference. just looking at possible effects of rod length here yet creating the same displacement.
also, the reason i brought up the bore/ stroke relationship and how it can relate to rpm, was based around our own vq's. even with all the forged goodies and the stock main cap girdle and overhead cams, not many people i know about have gotten the thing to be able to rev very high. i was shifting my 383 at 7300rpm and didn't have as good of a crank or rods in it as in my vq. if i am remembering correctly some were floating valves right around that same rpm. now i remember putting in my nismo cams and i was getting pretty close to coil bind with the stock springs (right around 0.012" cold). now there are many ways for the aftermarket to make larger dia. springs or double springs for this motor. i guess just not many people have done it. but waiting for the day i see a street vq35 turning 9000rpm with not much issues.
which brings me to another question, boost effects on valve springs. i have heard from some that it makes no difference, but i have a tough time imagining there is no effect. the higher the boost, the more spring pressure required on the intake side to help keep the intake valve closed at high rpm and boost. correct or false. i believe true.
just trying to keep ya busy here and answering some of my own questions i have, lol.
overZealous1, you're killing me here.
For all this who are following this so far, I figure I'll recap everything said so far in the thread on hp,tq, and stress I started in the main Engine/Drivetrain forum. I'll go over, simply, how the cylinder pressures and torque are related but not identical, the vibrations caused byt this phenom, and how the torque we are applying with multiple cylinders is the mean of one cycle, or two revolutions.
For overZealous1, here you go:
Using the shorter rod does several things, one of which is increase the side loading on the cylinder as you have already discovered. Second, it increases the speed the piston must travel because it forces a greater angle to the crank, which in turn increases acceleration and inertial loads on the conrod. Here's the math:
Your first engine with the 5.7in rods at 8500rpm has a max accel of:
167,230 ft/s^2
Your second engine with the 6.0in rods at 8500rpm has a max accel of:
162,451 ft/s^2
I expect both to grenade at this rpm without a LOT of work. By comparison, the VQ at 8500rpm has a max accel of:
97,321 ft/s^2, and can rev much higher without too much worry from the inertial forces at this rpm.
But, as one of our buddies down under is finding out, the VQ needs improvements to the head to breathe and operate this high. Springs, retainers, guides, cams, and valves. Such is the price of high rpm.
As far as boost effect on springs, I assume you mean the pressure acting on the back of the valve would push it open a little? Not gonna happen. Most valve springs have at least 60 lbs of seat pressure, which menas you'll have to boost at least 45lbs of air to budge a valve with a one inch square area on the backside. And if you're running that much boost, i guarantee you'll be running even stronger valves to use that boost at the appropriate rpm.
Will
For all this who are following this so far, I figure I'll recap everything said so far in the thread on hp,tq, and stress I started in the main Engine/Drivetrain forum. I'll go over, simply, how the cylinder pressures and torque are related but not identical, the vibrations caused byt this phenom, and how the torque we are applying with multiple cylinders is the mean of one cycle, or two revolutions.
For overZealous1, here you go:
Using the shorter rod does several things, one of which is increase the side loading on the cylinder as you have already discovered. Second, it increases the speed the piston must travel because it forces a greater angle to the crank, which in turn increases acceleration and inertial loads on the conrod. Here's the math:
Your first engine with the 5.7in rods at 8500rpm has a max accel of:
167,230 ft/s^2
Your second engine with the 6.0in rods at 8500rpm has a max accel of:
162,451 ft/s^2
I expect both to grenade at this rpm without a LOT of work. By comparison, the VQ at 8500rpm has a max accel of:
97,321 ft/s^2, and can rev much higher without too much worry from the inertial forces at this rpm.
But, as one of our buddies down under is finding out, the VQ needs improvements to the head to breathe and operate this high. Springs, retainers, guides, cams, and valves. Such is the price of high rpm.
As far as boost effect on springs, I assume you mean the pressure acting on the back of the valve would push it open a little? Not gonna happen. Most valve springs have at least 60 lbs of seat pressure, which menas you'll have to boost at least 45lbs of air to budge a valve with a one inch square area on the backside. And if you're running that much boost, i guarantee you'll be running even stronger valves to use that boost at the appropriate rpm.
Will
my point being not just floating the valves constantly but you will be reducing effective spring pressure by having the force always on the back side of the valve under boost. the vq springs are pretty weak and i was able to disassemble with my hands, lol. now to make it worse, when making even 16psi you are going to be doing it at a higher rpm greatly increasing the chance of floating the valves.
the only way that possibly it wouldn't be something to worry about is if the cylinder fills fast enough to match the intake psi before closing.
the only way that possibly it wouldn't be something to worry about is if the cylinder fills fast enough to match the intake psi before closing.
Originally Posted by overZealous1
my point being not just floating the valves constantly but you will be reducing effective spring pressure by having the force always on the back side of the valve under boost. the vq springs are pretty weak and i was able to disassemble with my hands, lol. now to make it worse, when making even 16psi you are going to be doing it at a higher rpm greatly increasing the chance of floating the valves.
the only way that possibly it wouldn't be something to worry about is if the cylinder fills fast enough to match the intake psi before closing.
the only way that possibly it wouldn't be something to worry about is if the cylinder fills fast enough to match the intake psi before closing.
Will
Just a minor correction. Stock valve spring seat pressure on the VQ is near 35 lbs of pressure, rather than 65 lbs average you mentioned. JWT upgraded springs have 43lbs of seat pressure. Ferrea valve springs have 85ft/lbs and higher worth of seat pressure, depending on which spring you choose.
Originally Posted by Sharif@Forged
Just a minor correction. Stock valve spring seat pressure on the VQ is near 35 lbs of pressure, rather than 65 lbs average you mentioned. JWT upgraded springs have 43lbs of seat pressure. Ferrea valve springs have 85ft/lbs and higher worth of seat pressure, depending on which spring you choose.
Will
Originally Posted by Quamen
So can anyone confirm if the engine in the dyno chart was recieving to much stress for the stock internals or not?
Originally Posted by Resolute
Wow, only 35lbs, OverZealous1 was right, pretty weak. Good to know. Sharif, you have a lot of experience with FI VQ's, any problems you've seen with boost on stock VQ springs? I guess with only 35lbs of seat pressure, a built engine could have enough boost to unseat a valve with the stock spring, but again, anyone running that much pressure would certainly upgrade the valvetrain... I would think.
Will
Will
Can you explain again, how 15psi of boost pressure effects the valve seat, assuming the valve spring is running 35lbs of seat pressure? I was assuming that the air will find the "open" valves, and reduce the pressure on the closed valves. I am wrong?
my idea behind it is just thatwhatever boost you are running, it will try opening the intake valve off the seat. now lets say we did a test with valve springs that have 15lbs seat pressure, and through 15psi boost at it , for simplicity, say the motor is not running and head off car, the valve should lift off the seat, correct? now throw in 6000 rpm and the fact the valve spring needs to try and close against the rushing in boost, it seems to me you will have lowered the effectiveness of the normal atmospheric spring pressure and run alot higher chance of floating a valve compared to a n/a motor.
it is a theory i read along time ago, and have heard different ideas on it since, saying it has no effect. i could not even possibly imagine it has no effect though at high boost.
it is a theory i read along time ago, and have heard different ideas on it since, saying it has no effect. i could not even possibly imagine it has no effect though at high boost.
the above test is also figuring the inside valve area at 1" sq. area. lol. at 15 psi and a 2" sq head area, you would be able to lift 30lbs with it, taking 30lbs off of the effective spring seat pressure at that boost.