Here you go, Piston area and force.
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Sometimes, in fact most of the time, a word will be used that is accepted and understood which is not technically accurate.
For example, 12 Volt is accepted as an approximation for system voltage even though in reality the system runs between 14 and 14.8 Volts 99+% of the time. Likewise the connections to the power supply is often marked B, Bat, or Batt for 'battery', even though the system power doesn't usually come from the battery.
If you are someone like Paul Harrell with the skill, patience, and situation to explain something like that as part of the discourse, wonderful. That heads off the "Actually" comments.
So while your observation is more or less technically correct, all your post really does is reinforce Harrison's main point.
For example, 12 Volt is accepted as an approximation for system voltage even though in reality the system runs between 14 and 14.8 Volts 99+% of the time. Likewise the connections to the power supply is often marked B, Bat, or Batt for 'battery', even though the system power doesn't usually come from the battery.
If you are someone like Paul Harrell with the skill, patience, and situation to explain something like that as part of the discourse, wonderful. That heads off the "Actually" comments.
So while your observation is more or less technically correct, all your post really does is reinforce Harrison's main point.
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Mattax,
I have had no interaction with 'harrisonm' in this thread or any other thread that I can think of. In post #96 he chose a personal attack on me. That is fine, I am a big boy. But when you do that, expect a return serve......which is what I did.
What post #96 showed is his lack of knowledge as to the combustion chamber event. You can dress it up but it is still lipstick on a pig.......Of course, not only did he get that wrong, he got the 'force on the piston' wrong. At least he is not alone there........
I have had no interaction with 'harrisonm' in this thread or any other thread that I can think of. In post #96 he chose a personal attack on me. That is fine, I am a big boy. But when you do that, expect a return serve......which is what I did.
What post #96 showed is his lack of knowledge as to the combustion chamber event. You can dress it up but it is still lipstick on a pig.......Of course, not only did he get that wrong, he got the 'force on the piston' wrong. At least he is not alone there........
Dale Davies
Well-Known Member
Piston area × PSI in the cyl = pounds force on the piston, which is almost directly transferred to the con rod.That's like your opinion man
Maybe were looking from the same from you.
It's not instance hp cause your not factoring in stroke or displacement depending on the eg..
This formula has nothing really to do with what were talking about lol.
Now in that, any one cyl block can only be poked so far. Also, is this question limited to boring, or comparing the torque developed between two engines of equal displacement? Are the cylinders in question being supplied through the same size ports and valves, or larger valves that the larger piston allows before cyl wall shrouding?
Dale Davies
Well-Known Member
Sounds like a Paul Harvey Sr moment, "And now for the rest of the story".
Correct, which buddy doesn't get.
Not necessarily talking overboring, generally like how mopar mainly changes displacement by bore.
But what this is really about say a 318 bored and stroked 390 and just overbore a 383 to 390, the general wisdom would be the 318/390 4" would produce more torque cause of the longer stroke ignoring the fact that the larger bore of the 383/390 can produce more force to the piston ideally allowing both these engines to produce the same torque.
Obviously engines are complex devices there lots of variables so it obviously depends on the combos of the different engines of same displacement engines were comparing. For this wrong there would have to be a good reason the larger bore would be generally down in cylinder pressure especially by the same ratio of difference as the difference in piston area, need a little more proof than a garden hose on a turd or a coin in playdough etc..
Just like any NA engines ability to be efficient eg.. Lbs/ft per cid, bmep, 1000-1900 psi however you want to measure it. Since engines of all sizes and bore and stroke is able to reach most of these efficiencies ($$$$$) there would have to be a good reason a larger bore with a shorter stroke couldn't especially when the highest efficient race engines generally use them.
Mmmm. There is some similarity but not quite what I was talking about.
Paul Harrell
Well I think I found the answer even though a larger piston has more piston area 4.03" vs 4.25", area =19.8671"sq vs 20.9517"sq.
A larger cylinder bore has less surface area for a give displacement, 408 LA (4.03" x 4") vs 408 B (4.25" x 3.596") 50.6168"sq vs 47.9886"sq
408 LA. 4.03" x 4.00", (2 x 19.8671) + 50.6168 = 90.351"sq total square inch of surface area.
408 B 4.25" x 3.596", (2 x 20.9517) + 47.9886 = 89.892"sq total square inch of surface area.
Practically the same. So should be no reason the 408 B can't have same cylinder pressure as 408 la they basically have the same total surface just the larger bore has more surface area on the part that actually moves (piston).
A larger cylinder bore has less surface area for a give displacement, 408 LA (4.03" x 4") vs 408 B (4.25" x 3.596") 50.6168"sq vs 47.9886"sq
408 LA. 4.03" x 4.00", (2 x 19.8671) + 50.6168 = 90.351"sq total square inch of surface area.
408 B 4.25" x 3.596", (2 x 20.9517) + 47.9886 = 89.892"sq total square inch of surface area.
Practically the same. So should be no reason the 408 B can't have same cylinder pressure as 408 la they basically have the same total surface just the larger bore has more surface area on the part that actually moves (piston).
Dale Davies
Well-Known Member
Cylinder surface area does not have a lot of effect on force on the connecting rod. It is well accepted that part of rotary engines poor fuel economy is the large surface area exposed to the combustion. So apparently the small bore, long stroke engine may suffer poor combustion and power extraction efficiency in addition to higher friction.
More piston area does not automatically mean more force on the rod!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
If more air air is drawn into the cyl as a result of an increase in piston area, then more air is compressed & more force is applied to the piston. If more air was drawn in as a result of a longer stroke, then the same force is applied to the crank whether the piston area was small or large. The small piston will have greater loading [ more lbs/ per sq in ].
More piston area only increases 'push' on the conrod if a secondary event such as ingesting more air occurs that increases the 'push' medium, the medium in this case being more force generated by the expanding hot burning mixture.
Here is yet another example: the claim is that increased piston area will create more 'push' on the con rod. The force on the piston is transmitted to the crank via the conrod & through the piston pin to the rod. Using the same 'logic', if I double the pin/rod contact area, the rod will 'push' harder on the crank!!
If more air air is drawn into the cyl as a result of an increase in piston area, then more air is compressed & more force is applied to the piston. If more air was drawn in as a result of a longer stroke, then the same force is applied to the crank whether the piston area was small or large. The small piston will have greater loading [ more lbs/ per sq in ].
More piston area only increases 'push' on the conrod if a secondary event such as ingesting more air occurs that increases the 'push' medium, the medium in this case being more force generated by the expanding hot burning mixture.
Here is yet another example: the claim is that increased piston area will create more 'push' on the con rod. The force on the piston is transmitted to the crank via the conrod & through the piston pin to the rod. Using the same 'logic', if I double the pin/rod contact area, the rod will 'push' harder on the crank!!
Newbomb Turk
Well-Known Member
For one really were talking an average psi through out the power stroke, it's not a static amount it's forever changing evolving. Your assumption is the 383 B will have less overall average psi over a 383 RB or 408 B vs 408 LA etc.., you have offered zero proof for your position.
The overall surface area for both is basically the same at bdc and at tdc the larger bores full chamber, head gasket, cylinder wall, piston area etc.. Does have more surface area but for a small fraction of the power stroke. The larger piston has less bore surface area (cylinder wall) in the exact opposite percentage ratio. The smaller bore has a longer stroke so the piston moves away from tdc faster at same rpm to cover a greater distance which doesn't take far down the bore for the smaller piston to have more total area but even though the smaller bore will have more total area the larger piston will have more percentage of area for the part that the psi actually moves the piston.
1. Were talking average psi.
2. The larger bore has more surface area only at tdc, shortly after the smaller bore has more surface area and for a greater percentage of the power stroke (the rest of the power stroke) .
3. With the larger bores overall total surface area a higher percentage is the piston area compared to the smaller bore, the part the actually applies force to the crank to be multiplied.
You got to prove that the larger bore for some reason can't have more average psi applied to the piston.
Just assuming and implying it doesn't make so.
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273.
Now you are shifting the goal posts, introducing more red herrings.
What started this discussion was your claim that a larger piston area creates more downward force, increases hp. That only happens if the area increase produces the secondary event of drawing more air into the cyl. Unless additional air is drawn into the cyl, then a bigger piston area adds nothing.
Now you are shifting the goal posts, introducing more red herrings.
What started this discussion was your claim that a larger piston area creates more downward force, increases hp. That only happens if the area increase produces the secondary event of drawing more air into the cyl. Unless additional air is drawn into the cyl, then a bigger piston area adds nothing.
I am not shifting the goal post you seem to take issues that there would be similar psi capabilities so I'm trying to answer you.
I don't how many times I have to tell you I've never said more hp, I've stated from the beginning a larger piston will have more force for a given psi, that's what I claim now and that's what I claimed in the beginning and when it comes same displacements the larger piston greater force should average out the longer stroke multiplication of that force. Aka similar torque.
Proof ? Since both cylinders have the same volume (same fuel and air potential), shouldn't they have the same torque potential ?
You tell me why the larger bore and shorter stroke in a 383 B, 408 B vs 383 RB, 408 LA etc.. Can't have similar average psi applied to the piston for more average force to be multiplied less by a shorter stroke for similar torque as the smaller bore and longer stroke engines of same displacement?
Just cause you say so or don't make sense to you ain't proof.
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Now you are mixing entities. I am not disputing more [ or less ] tq from the engine families you just listed. There is more than just piston size & stroke that determine the actual TQ/HP an engine produces. Efficiency of the ports, induction & exh systems, friction etc will all affect the tq produced.
I am disputing something far more simple: just raw high school physics. I have already given YOU this example, but it seems the message did not get through.....
Two engines, different bore & stroke, but same cubic inches & same compression ratio. One engine has a 4" piston, other engine 4.25" piston. Both engines produce 1000lbs of force in the chamber at the completion of the mixture burn. 1000 lbs to push on the piston. Your argument is that because the 4.25" piston has more area, it will have more force applied to it than the 4" piston. Nope. Started with 1000 lbs, so it cannot be greater than that. What it means is this: the 4" piston has 80 llbs/ sq in of pressure on it & the larger 4.25" piston has 70 lbs/ sq in [ numbers rounded off ]. That is what this link explains....
I am disputing something far more simple: just raw high school physics. I have already given YOU this example, but it seems the message did not get through.....
Two engines, different bore & stroke, but same cubic inches & same compression ratio. One engine has a 4" piston, other engine 4.25" piston. Both engines produce 1000lbs of force in the chamber at the completion of the mixture burn. 1000 lbs to push on the piston. Your argument is that because the 4.25" piston has more area, it will have more force applied to it than the 4" piston. Nope. Started with 1000 lbs, so it cannot be greater than that. What it means is this: the 4" piston has 80 llbs/ sq in of pressure on it & the larger 4.25" piston has 70 lbs/ sq in [ numbers rounded off ]. That is what this link explains....
No ****, but bore and stroke is what were talking about and isolate as best as can be.
Combustion is in psi, you keep saying in lbs, yes if both engines combustion was a 1000 lbs of force then your right but were talking psi, there is no reason that both of these engines wouldn't have similar psi, also it's not just a singular amount of psi it's the average psi of the complete power stroke.
Like I've said and everyone else has said from the beginning if the psi is the same the force of the piston will be more for the larger piston, I imagine you agree with that statement, now if you don't think the same psi is possible please explain why, saying both engines have a 1000 lbs of force isn't an explanation.
Again this ain't a formula for power, it's away to rank an engines power, like hp per cid, hp per L, hp per piston surface area. Really has no bearing on the discussion.
Unless you can show a valid reason why a larger piston, shorter stroke would have less average psi than a smaller piston, longer stroke engine of same displacement on average for similar level of efficiency.
Then there's really nothing more to discuss.
*Edit, And this (below) explains that so unless you can prove below has no bearing on what were discussing think were done.
Ideal gas law is P = nRT / V
The Main ones that concern this discussion are Pressure = Temperature / Volume
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Well from what I can find and kind of expected, that engines psi is based on volume not surface area, since engines of same displacement and cr have same volumes no matter there bore stroke ratios.
Ideal gas law is P = nRT / V
The Main ones that concern this discussion are Pressure = Temperature / Volume
Ideal gas law is P = nRT / V
The Main ones that concern this discussion are Pressure = Temperature / Volume
Help me, Jesus.
For me it was worth it I have a better/clearer understanding than I did before.
So ... Chrysler engineers in the way back even understood dimensional analysis.
What you are saying above is that:
Power per piston area = (corrected)Power/Area
So ... Power/area = Power/Area
Amazing. Such genius. Man whoever could deduce that 1=1
Dale Davies
Well-Known Member
When you state 1000# force in the cylinder, I assume you mean PSI. That PSI spread over a larger piston area creates more force in pounds pushing the piston and thus the con rod down.
More force on a shorter stroke will probably equalize the maximum torque developed, but at higher RPM. This boils down to more HP at higher RPM. The shorter stroke will give lower piston speed at any given RPM, so there is less loss to friction.
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