valve springs or no?
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From: Plano, TX
My theroy that i would going on was Higher temp = More pressure ; Lower temp = Lower Pressure. Due to the cooler dense air, you would not need as much boost to fill the same amt of volume? Some heat soak skewing boost numbers? Somthing to do with Boyles law, or the Ideal Gas Law....or Charles law.....heck i don't know! That was along time ago when i did this for a grade! haha
! 
HOHN!!!
! HOHN!!!
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From: Raleigh, NC
Did Don ever divulge this fix?? He cannot be reached via PM's and I have no idea what his e-mail or phone number is??
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From: Cummins Technical Center, IN
OK fellas, here's some numbers we can put to calculate spring rates.
First, let's calculate effects of boost and backpressure (both of which tend to make it harder for intake and exhaust, respectively, to close the valve. For example purposes, I'm just going to make stuff up.
Let's say you have 50psi of boost max. The applies a force to the valve that tends to keep it open. If the valve is, say, 2 square inches in area, then it is ONE HUNDRED POUNDS of force trying to keep the valve open.
Same thing for exhaust. If you have 50psi of backpressure on that same 2 sq in valve, then you'll have 100# pushing the valve open.
The springs have to overcome these forces.
How to you calculate maximum acceleration of the valve?
First, you need to know the maximum rate of lift of the cam, in inches per degree of rotation (polar coordinates, if you're familiar with that).
Say that the max cam lift RATE is .300" in a space of 60º cam rotation. Assuming it's linear in that section (which is usually the case at peak). So we can calculate the rate as (.300/60º), or .005" per degree cam rotation.
The cam turns at half the speed of the engine, so, then we know the rate relative to ENGINE speed is double this, or .01" per degree of ENGINE rotation
Enter engine RPM. Let's say we're figuring valve pressure for 3000 rpm.
One engine revolution is 360º. At 3000 rpm, that means 360*3000, or 1,080,000 degrees per minute of angular velocity. Call it 18000º per second. At that speed, it will only take 1/600th of a second to move the valve that .300" distance. IN other words 180ft/sec valve velocity.
So now we have to slow this valve down and accelerate it to 180ft/sec in the OPPOSITE DIRECTION, all within a space of about 60º engine rotation.
So the spring has to control an acceleration sufficient to produce 360ft/sec velocity change in only 1/300th of a second. That's an acceleration rate of 108,000 ft/second/second--- or in other words, about 3354Gs!!!!!!
Let's say the valve in question weighs 2 ounces (again, all fictional numbers). Using F=MA, we can see that this 2 oz valve when accelerated by 3354 Gs will act as if it weighed ~420 lbs!
Now, to this 420# figure we add the previously calculated effect of boost pressure or backpressure, which is another 100# of force.
That means that 520# is the MINIMUM spring pressure it would take to control the valve under the circumstances in this hypothetical example.
Still think the Stock springs are up to turning 3K rpm at high boost???
Justin
First, let's calculate effects of boost and backpressure (both of which tend to make it harder for intake and exhaust, respectively, to close the valve. For example purposes, I'm just going to make stuff up.
Let's say you have 50psi of boost max. The applies a force to the valve that tends to keep it open. If the valve is, say, 2 square inches in area, then it is ONE HUNDRED POUNDS of force trying to keep the valve open.
Same thing for exhaust. If you have 50psi of backpressure on that same 2 sq in valve, then you'll have 100# pushing the valve open.
The springs have to overcome these forces.
How to you calculate maximum acceleration of the valve?
First, you need to know the maximum rate of lift of the cam, in inches per degree of rotation (polar coordinates, if you're familiar with that).
Say that the max cam lift RATE is .300" in a space of 60º cam rotation. Assuming it's linear in that section (which is usually the case at peak). So we can calculate the rate as (.300/60º), or .005" per degree cam rotation.
The cam turns at half the speed of the engine, so, then we know the rate relative to ENGINE speed is double this, or .01" per degree of ENGINE rotation
Enter engine RPM. Let's say we're figuring valve pressure for 3000 rpm.
One engine revolution is 360º. At 3000 rpm, that means 360*3000, or 1,080,000 degrees per minute of angular velocity. Call it 18000º per second. At that speed, it will only take 1/600th of a second to move the valve that .300" distance. IN other words 180ft/sec valve velocity.
So now we have to slow this valve down and accelerate it to 180ft/sec in the OPPOSITE DIRECTION, all within a space of about 60º engine rotation.
So the spring has to control an acceleration sufficient to produce 360ft/sec velocity change in only 1/300th of a second. That's an acceleration rate of 108,000 ft/second/second--- or in other words, about 3354Gs!!!!!!
Let's say the valve in question weighs 2 ounces (again, all fictional numbers). Using F=MA, we can see that this 2 oz valve when accelerated by 3354 Gs will act as if it weighed ~420 lbs!
Now, to this 420# figure we add the previously calculated effect of boost pressure or backpressure, which is another 100# of force.
That means that 520# is the MINIMUM spring pressure it would take to control the valve under the circumstances in this hypothetical example.
Still think the Stock springs are up to turning 3K rpm at high boost???
Justin
Wow! This example needs over 500# of spring force at only 3K rpm.
JLH
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From: Airdrie, Alberta
Sorry fellas-- I was reviewing this and noticed that my cam/engine speed conversion was off. See above quote from proper numbers. Turns out the effective weight of the valve is DOUBLE what I previously calculated!
Wow! This example needs over 500# of spring force at only 3K rpm.
JLH
Wow! This example needs over 500# of spring force at only 3K rpm.
JLH
If we are assuming that everything (base circle to total lift) happens within 60º of cam rotation, that would be 120º of crank rotations. So your lift/degree of crank would be .0025in/ºcrank, instead of .0100in/º.
Likewise, angular velocity of crank is 18000ºcrank/sec, cam velocity would be 9000ºcam/sec. Since its the cam we care about, 9000ºcam/sec divided by 60º would be 1/150 sec. Which would equal out to 45 in/sec. .300inx150/sec=45in/sec. Valve closing would be in another 60 degrees. So in total, you'd have .600" travel in 120 degrees cam rotation, you'd still get 45 in/sec. 9000ºcam/sec divided by 120º is 1/75th sec. Being the metric kid that I am, thats what I'll work in.
45in x 25.4mm/in = 1143mm, or 1.143m. So velocity is 1.143m/s
1.143m/s divide by 1/75 sec is 85.725m/s/s.
G=9.81m/s/s. 85.725/9.81 = 8.73G
This gives a pretty low spring pressure, but you'd still have the mass of the lifters, pushtubes, rockers, and bridges to take into account as well.
Now, its been a while since I took physics, but I'm pretty sure all that rambling works out.
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From: Airdrie, Alberta
If we are assuming that everything (base circle to total lift) happens within 60º of cam rotation, that would be 120º of crank rotations. So your lift/degree of crank would be .0025in/ºcrank, instead of .0100in/º.
Likewise, angular velocity of crank is 18000ºcrank/sec, cam velocity would be 9000ºcam/sec. Since its the cam we care about, 9000ºcam/sec divided by 60º would be 1/150 sec. Which would equal out to 45 in/sec. .300inx150/sec=45in/sec. Valve closing would be in another 60 degrees. So in total, you'd have .600" travel in 120 degrees cam rotation, you'd still get 45 in/sec. 9000ºcam/sec divided by 120º is 1/75th sec. Being the metric kid that I am, thats what I'll work in.
45in x 25.4mm/in = 1143mm, or 1.143m. So velocity is 1.143m/s
1.143m/s divide by 1/75 sec is 85.725m/s/s.
G=9.81m/s/s. 85.725/9.81 = 8.73G
This gives a pretty low spring pressure, but you'd still have the mass of the lifters, pushtubes, rockers, and bridges to take into account as well.
Now, its been a while since I took physics, but I'm pretty sure all that rambling works out.
Likewise, angular velocity of crank is 18000ºcrank/sec, cam velocity would be 9000ºcam/sec. Since its the cam we care about, 9000ºcam/sec divided by 60º would be 1/150 sec. Which would equal out to 45 in/sec. .300inx150/sec=45in/sec. Valve closing would be in another 60 degrees. So in total, you'd have .600" travel in 120 degrees cam rotation, you'd still get 45 in/sec. 9000ºcam/sec divided by 120º is 1/75th sec. Being the metric kid that I am, thats what I'll work in.
45in x 25.4mm/in = 1143mm, or 1.143m. So velocity is 1.143m/s
1.143m/s divide by 1/75 sec is 85.725m/s/s.
G=9.81m/s/s. 85.725/9.81 = 8.73G
This gives a pretty low spring pressure, but you'd still have the mass of the lifters, pushtubes, rockers, and bridges to take into account as well.
Now, its been a while since I took physics, but I'm pretty sure all that rambling works out.
Only real flaw in your math that I can tell is you used 180ft/sec, where it would be 180in/sec. In which case, the acceleration from your numbers come out to 279.5G.
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From: fredericksburg, virginia
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From: Texas/Oklahoma Border
How's this! MASSIVE forces are involved in the speed and directional changes of valves opening & closing! Therefore, over 45-50 psi boost... and/or over 3000 rpm.... stronger valve springs are beneficial and highly recommended!
For detailed explanation, in laymans terms, re-read post #32 in this thread. IMO, Don M. covered it all....way back then.
RJ
For detailed explanation, in laymans terms, re-read post #32 in this thread. IMO, Don M. covered it all....way back then.
RJ
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From: Lake Charles,Louisiana
How's this! MASSIVE forces are involved in the speed and directional changes of valves opening & closing! Therefore, over 45-50 psi boost... and/or over 3000 rpm.... stronger valve springs are beneficial and highly recommended!
For detailed explanation, in laymans terms, re-read post #32 in this thread. IMO, Don M. covered it all....way back then.
RJ
For detailed explanation, in laymans terms, re-read post #32 in this thread. IMO, Don M. covered it all....way back then.
RJ
No,No, i understand what the springs do or allow. i was talking about the math they were doing (i think).
