Melting pistons.
- Thread starter McRat
- Start date
The only way I buy rail pressure adding timing is: If atomized better and heat created from pressure, fuel will flash faster.
No matter the rail pressure, timing can't start until injection. The higher the pressure the more accurate the tuned injection at " crank angle ".
Lower Rail pressure only delays the start of injection from commanded. So if referenced from a low pressure.....
Wade you say fuel burns from cylinder psi, true but only if temps are present and high enough to flash the fuel.
So with this, do we know the injected quantity and if burned, with different O2 content, its btu numbers?
IMOP if heat was the cause i could see how the high pressure spray could affect the piston material , because the material would be in a "softer state" ..... maybe . The one plausible thought to me is with the high RPM the piston has just less than half the time to cool down between cycles ,so even though the pyro says 1450 the piston temp could be far greater than that for all we know , especially when considering the time Pat has stayed WOT compared to a standard 1/4 mile run . Which brings us back to what Pat found on Casper making perfect sence ................ to me anyway .
Isn't the piston distance from TDC more a function of stroke rather than rod length? Therefore at the most it's 4.4" (or whatever stock stroke is), right?
If so, divide the stroke by 720* (takes two revs to get from top to bottom and then two more to get from bottom to top I think) will give you how far it moves for a given crank angle degree.
As an example:
Stroke = 4.4" (might be wrong but it's just a demo anyways)
4.4"/720* = 0.006111 inches/CAD
30* BTDC only requires 690* (720-30 = 690) of rotation to get to so
0.00611" x 690* = 4.2167"
the difference is the distance from TDC in inches at 30* BTDC = 4.4"-4.2167" = 0.1833"
Is that right or am I out in left field? I've seen this posted by someone else somewhere on here but can't find it to see if I'm remembering it right or not. Thought maybe it was Johnboy who posted it??
I still don't know what causes pistons to melt though.
If so, divide the stroke by 720* (takes two revs to get from top to bottom and then two more to get from bottom to top I think) will give you how far it moves for a given crank angle degree.
As an example:
Stroke = 4.4" (might be wrong but it's just a demo anyways)
4.4"/720* = 0.006111 inches/CAD
30* BTDC only requires 690* (720-30 = 690) of rotation to get to so
0.00611" x 690* = 4.2167"
the difference is the distance from TDC in inches at 30* BTDC = 4.4"-4.2167" = 0.1833"
Is that right or am I out in left field? I've seen this posted by someone else somewhere on here but can't find it to see if I'm remembering it right or not. Thought maybe it was Johnboy who posted it??
I still don't know what causes pistons to melt though.
Why 720* Mitch? Wouldn't that be 2 revolutions? This would be true for Cam to crank relationship?
I went to finish clay baring the pulling truck in the garage. I am trying to get stuff done between posts.:rofl:
Totally agree with above.:cool2:
Good question. Not sure where we can get the answer but still a good question.
Isn't the piston distance from TDC more a function of stroke rather than rod length? Therefore at the most it's 4.4" (or whatever stock stroke is), right?
If so, divide the stroke by 720* (takes two revs to get from top to bottom and then two more to get from bottom to top I think) will give you how far it moves for a given crank angle degree.
As an example:
Stroke = 4.4" (might be wrong but it's just a demo anyways)
4.4"/720* = 0.006111 inches/CAD
30* BTDC only requires 690* (720-30 = 690) of rotation to get to so
0.00611" x 690* = 4.2167"
the difference is the distance from TDC in inches at 30* BTDC = 4.4"-4.2167" = 0.1833"
Is that right or am I out in left field? I've seen this posted by someone else somewhere on here but can't find it to see if I'm remembering it right or not. Thought maybe it was Johnboy who posted it??
I still don't know what causes pistons to melt though.
JoshH has it correct. The Stock stroke is 3.9". To figure your triangle you need to know the hyponetus(sp?) and the angle. The "H" is equal to half the stroke or 1.95" the angle is BTDC. You also need to figure the side opposite. Then you can use rod lenght and side opposite length in a the pathagrean(sp?) formula to figure out the length of the missing side.
I can't remember if the 720 is right or not... but doesn't it take 2 revolutions (720*) to go from TDC to BDC then two more to come back up??? That's where the the 720* comes from.
If it's only 360* then the distance would be double what I posted previously.
C-ya
If it's only 360* then the distance would be double what I posted previously.
C-ya
and if what I posted is wrong, please, by all means, delete it. No need having something wrong posted. Just add confusion.
C-ya
C-ya
Isn't the piston distance from TDC more a function of stroke rather than rod length? Therefore at the most it's 4.4" (or whatever stock stroke is), right?
If so, divide the stroke by 720* (takes two revs to get from top to bottom and then two more to get from bottom to top I think) will give you how far it moves for a given crank angle degree.
As an example:
Stroke = 4.4" (might be wrong but it's just a demo anyways)
4.4"/720* = 0.006111 inches/CAD
30* BTDC only requires 690* (720-30 = 690) of rotation to get to so
0.00611" x 690* = 4.2167"
the difference is the distance from TDC in inches at 30* BTDC = 4.4"-4.2167" = 0.1833"
Is that right or am I out in left field? I've seen this posted by someone else somewhere on here but can't find it to see if I'm remembering it right or not. Thought maybe it was Johnboy who posted it??
I still don't know what causes pistons to melt though.
I believe 1 crank revolution will get you up and down or vice versa and 2 revolutions gets u all four strokes and not just 2.
I can't remember if the 720 is right or not... but doesn't it take 2 revolutions (720*) to go from TDC to BDC then two more to come back up??? That's where the the 720* comes from.
If it's only 360* then the distance would be double what I posted previously.
C-ya
From TDC top BDC is 180* half a crank rotation.
A complete 4 stroke engine cycle take two rotations of the crank. 1 crank rotation if a started a TDC will get clear around back to TDC.
Later guys. Time to make the donuts!
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Good questions.
Last first, the O2 does not affect total BTU the fuel will produce unless there isn't enough of it.
The actual injected qty is hard to nail down. So many factors come into play.
We can't nail down the BTU input into the cylinder unless we know the Fuel qty injected. We can get close by looking at the pressure. The math, however, can get ugly. You need to account for the varying surface area of the chamber over time as well as the varying pressure and volume. Don't forget that some of those BTUs are being turned into rotational energy. Then there is the actual temperature of the piston and cylinder walls to factor in.... Ugly.
Best you can reasonably do is look at the temperatures the piston is exposed to from the chamber and work out roughly how many BTU's it is dumping having an idea how well it conducts heat.
But, what do I know....
So best we can do at the moment is calculate crudely from pressure, Temperatures. Still a huge guessing game.
If we can accurately measure cfm/lbs per minute air and we crudely calculate cylinder temps from pressure, we would have a starting point for mm3 injected?
Does an abundance of o2 increase or lower cylinder temps apples to apples?
An abundance of air could be the ticket to reducing cylinder temps. We've seen it with our supercharged/turbocharged engines. Not saying the is right just saying egt's are greatly reduced.
An abundance of air could be the ticket to reducing cylinder temps. We've seen it with our supercharged/turbocharged engines. Not saying the is right just saying egt's are greatly reduced.
I am curious as well , we have had many of discusions about the possibilty of greater motor longevity records with low egt setups . But truthfully its just based off of weighing the statistics of PUBLIC failures .
Public or private engine failure, the question still exists and is valid. Whether or not we get an acceptable answer is the question/problem.
how are you acheiving more oil sprayed on piston? i would of loved to do this while my motor was apart
Obviously 1800+ EGT's are bad but guys have wasted pistons with twins and really low EGT's... it is a role of the dice when discussing how far a stock engine can go. Trent your truck is a freak plain and simple. I would say Casper was a freak as well... talking about stock engines, i doubt we will ever "unlock" the secret to keep a stock engine alive at huge horsepower. I bet at 600hp the LB7's/LLY's would go for a very long time just like you are proving Trent. The LBZ/LMM's are not so lucky as there pistons are weaker(i know we all know this). Anthony is an example... LOTS of low 12 second passes running lots of nitrous... he adds a few extra parts to help with flow BOOM cracked piston... These engines are so young in comparison to Chevy small block 350's. It is probably going to be a long expensive road to figure it all out. Its poeple like Pat, and Max'd Out that push the bar higher and higher that help in us not having to spend the extra time and money testing combination's. From listening to what Guy has been saying forged pistons seem to be the best thing we have today... I guarantee you we will find out how far these Mahle Forged Piston's go...
