Here you go, Piston area and force.

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Dale Davies said:
When you state 1000# force in the cylinder, I assume you mean PSI.
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No he doesn't, that's the problem, he keeps saying a (1000) lbs i keep tell him it's a (1000) psl.

This what he thinks is going on, the combustion has eg.. 1000 lbs so say a piston with a surface area of 20" sq psi = f/a so 50 psi of pressure applied to the piston that's what Bewy is saying he definitely thinks the 1000 lbs / piston area he states over and over and that we got it backwards.
 
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Newbomb Turk said:
It’s simple math.

P*L*A*N

It’s that basic. And the math doesn’t lie.

When you confuse unit loading with area/psi and you are so bull headed that you refuse at least do the math to see how it works you are just being idiotic.
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I admit before this thread I had an vague idea psi was based on volume, knew how that psi interreact with the surface area of the piston and Plan, Bmep etc.. only cared about overall volume and that fit with everything I seen, But wasn't until I just found ideal gas law I understand how.
 
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Not one thing about surface area just Pressure, Volume, Temperature.

1720452094594.png


Look how this imep and bmep graph would over lay on the ideal gas law's one.

1720452817458.png
 
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Instead of this, you got to use scientific principles related to the subject matter.

1720452373918.png
 
I guess the next time I clay a piston, I'll lay dimes on top & do a cranking compression test, if they're all buried I've been successful....
 
Yes,
I should have used the word pressure in post #118. Sorry y'all still don't get it. I have led the horse to the water...& he doesn't want to drink it.
 
Dale
You are 100% wrong in post #125, with 'PSI spread over a larger area creates more force'. Look at post #130. Pressure = force divided by area. The force or pressure [ or whatever you want to call it ] is created by the expanding gas. I used an example of 1000 what-u-ma-call-its. Call it what you like & whatever quantity. The point is that the loading per sq in is less on the bigger piston even though both small & large pistons had the same initial force/pressure applied. That is what the picture in post #130 shows: the flat coin did not sink into the plasticine as it did on it's end because the loading per area was less even though the same initial force was used.

I have already given this example, but maybe you missed it. The piston transmits it's load to the crank via the pin & conrod. If I make the rod eye wider, according to your theory, because I have increased the area, the rod now transmits more force to the crank & the engine makes more hp!!!
Increasing the area of the piston should increase the weight of air drawn into the cyl. If it does that, then the pressure generated from the burn has increased. It is this increase in gas pressure, not the piston area that has created more force.
 
Bewy said:
Dale
You are 100% wrong in post #125, with 'PSI spread over a larger area creates more force'.
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It's literally in it's name PSI, Pounds Per Square Inches, Force = Pounds Per Square Inches x Amount of Square Inches, Eg.. 1 psi x 100 square inches of surface area = 100 lbs of force.

Bewy said:
Look at post #130. Pressure = force divided by area.
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Were not looking for pressure, were looking for what the pressure does to the piston, were looking for the force the piston is passing on to the crank which is Force = pressure x area.
Bewy said:
The force or pressure [ or whatever you want to call it ]
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It matters a lot what it's called.
Bewy said:
is created by the expanding gas. I used an example of 1000 what-u-ma-call-its. Call it what you like & whatever quantity.
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Again psi, basically pressure = temperature / volume, P = nRT / V
Bewy said:
The point is that the loading per sq in is less on the bigger piston even though both small & large pistons had the same initial force/pressure applied.
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Again they both have the same psi, again pressure = nR Temperature / Volume of combustion chamber opening into total volume of the cylinder. Area doesn't have anything to do with the psi created by the combustion process. I know that don't jive with your coin and playdough or water hose and dog **** analogies, but maybe you should use science that's related to the topic.
Bewy said:
That is what the picture in post #130 shows: the flat coin did not sink into the plasticine as it did on it's end because the loading per area was less even though the same initial force was used.
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This is your issue, your basing your whole argument on this, maybe should be on how combustion actually creates pressure, psi. Do some research on the combustion process.
Bewy said:
I have already given this example, but maybe you missed it. The piston transmits it's load
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Force
Bewy said:
to the crank via the pin & conrod. If I make the rod eye wider, according to your theory, because I have increased the area, the rod now transmits more force to the crank
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No cause the piston is applying force to the pin/con rod , combustion psi is applied to the pistons surface area (f=pa) which now the piston applies force to the pin/con rod, the more area between the contact of pin and con rod would be your favorite formula p = f / a so the load (force) would be spread across the wider rod eye but the same force will be passed on.
Bewy said:
& the engine makes more hp!!!
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Torque, guess you don't understand hp either.
Bewy said:
Increasing the area of the piston should increase the weight of air drawn into the cyl. If it does that, then the pressure generated from the burn has increased. It is this increase in gas pressure, not the piston area that has created more force.
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No cause we've been talking same displacement this whole time just different bore stroke ratios so same volume and..

Again, P = nRT / V, pressure is based on volume of combustion chamber and total volume and of course the amount of fuel and air (nR) and how well it combust (temperature ). Nothing to do with surface area. Read post #129 it's plane as day.

It's amazing you can be this far off and still be so confident :) It's also amazing you just brush aside anything anyone else says ignore any evidence provided and just keep repeating the same baseless analogies.
 
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Bewy said:
Dale
You are 100% wrong in post #125, with 'PSI spread over a larger area creates more force'. Look at post #130. Pressure = force divided by area. The force or pressure [ or whatever you want to call it ] is created by the expanding gas. I used an example of 1000 what-u-ma-call-its. Call it what you like & whatever quantity. The point is that the loading per sq in is less on the bigger piston even though both small & large pistons had the same initial force/pressure applied. That is what the picture in post #130 shows: the flat coin did not sink into the plasticine as it did on it's end because the loading per area was less even though the same initial force was used.

I have already given this example, but maybe you missed it. The piston transmits it's load to the crank via the pin & conrod. If I make the rod eye wider, according to your theory, because I have increased the area, the rod now transmits more force to the crank & the engine makes more hp!!!
Increasing the area of the piston should increase the weight of air drawn into the cyl. If it does that, then the pressure generated from the burn has increased. It is this increase in gas pressure, not the piston area that has created more force.
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Pressure - is a value exerted equally in all directions.
Force - is a value exerted in one direction.
In a cylinder;
Force is equal to the pressure exerted on an object × the area the pressure is applied to, in our case -
Pounds per square inch × the area of the piston, or more easily referenced as the area of the bore, in square inches.
OR, pounds/square inches × square inches. NOTE, the square inches cancel eachother out, leaving pounds force.
Obviously the same pressure, in PSI, acting on a smaller piston will exert less force pushing that piston down. Conversely a larger piston with the same pressure acting on it will exert more force.
Given a given cylinder pressure, it acts equally on all surfaces of the cylinder. A larger or smaller piston has no effect on the pressure acting on it, the pressure is the same. Now this is accepting we "freeze frame" the piston motion at say TDC.
Increasing the area of the piston "can" increase the mass or weight of air drawn into the cylinder, if the stroke is the same. This "may" increase HP, but that is dependent on a number of other factors. The main one there is camshaft LCA and duration, which affect where in the RPM range the torque peak will occur.
In post #133, you stated you still do not get it, force and pressure. Obviously you are correct in that when dividing you pressure by the area instead of multiplying. Back to science class you go.
 
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Dale Davies said:
Pressure - is a value exerted equally in all directions.
Force - is a value exerted in one direction.
In a cylinder;
Force is equal to the pressure exerted on an object × the area the pressure is applied to, in our case -
Pounds per square inch × the area of the piston, or more easily referenced as the area of the bore, in square inches.
OR, pounds/square inches × square inches. NOTE, the square inches cancel eachother out, leaving pounds force.
Obviously the same pressure, in PSI, acting on a smaller piston will exert less force pushing that piston down. Conversely a larger piston with the same pressure acting on it will exert more force.
Given a given cylinder pressure, it acts equally on all surfaces of the cylinder. A larger or smaller piston has no effect on the pressure acting on it, the pressure is the same. Now this is accepting we "freeze frame" the piston motion at say TDC.
Increasing the area of the piston "can" increase the mass or weight of air drawn into the cylinder, if the stroke is the same. This "may" increase HP, but that is dependent on a number of other factors. The main one there is camshaft LCA and duration, which affect where in the RPM range the torque peak will occur.
In post #133, you stated you still do not get it, force and pressure. Obviously you are correct in that when dividing you pressure by the area instead of multiplying. Back to science class you go.
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He'll just ignore all this, he's built his whole argument on p = f/a which be fine if were talking hydraulics
not combustions of gasses in a sealed chamber that the volume expands over time.

Obviously this paragraph from his grade six text book is all that's needed to explain thermodynamics of an engine. Were just not able to grasp the intricacies of coin being smushed into playdough is obviously the problem :)

1720546782573.png
 
273 said:
He'll just ignore all this, he's built his whole argument on p = f/a which be fine if were talking hydraulics
not combustions of gasses in a sealed chamber that the volume expands over time.

Obviously this paragraph from his grade six text book is all that's needed to explain thermodynamics of an engine. Were just not able to grasp the intricacies of coin being smushed into playdough is obviously the problem :)

View attachment 1716274067
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Yup, hard to teach astrophysics to a flea.
 
This seems to be why p=f/a doesn't simply work for a gas in a sealed volume.


"It's the volume that determines the pressure (for a given amount of gas at a given temperature), so if you consider a rectangular container, increasing the surface area of two opposite sides and/or increasing the distance between those sides will increase the volume and therefor decrease the pressure. But increasing the total area while keeping the volume the same (in other words: changing the shape, not the volume) will have no effect.

To demonstrate this, assume the original container is a cube of 1 m³ (with each side a square of 1 m * 1 m) , and you change it to a rectangular box with width and length = 2 m and height = 0.25 m (top and bottom side become squares of 2 m * 2 m; left, right, front and back side become rectangles of 2 m * 0.25 m): now the average distance between particles and the top (or bottom) side of the box is reduced by a factor 4, so 4 times more particles hit the surface each second, but the surface area is also four times bigger (4 m²), therefor the pressure stays the same. The same is true for the sides of the box: the average distance has doubled, so the number of particles hitting them each second is halved, but the surface area is also reduced to half the original (0.5 m² instead of 1 m²), resulting in the same pressure."
 
273 said:
This seems to be why p=f/a doesn't simply work for a gas in a sealed volume.


"It's the volume that determines the pressure (for a given amount of gas at a given temperature), so if you consider a rectangular container, increasing the surface area of two opposite sides and/or increasing the distance between those sides will increase the volume and therefor decrease the pressure. But increasing the total area while keeping the volume the same (in other words: changing the shape, not the volume) will have no effect.

To demonstrate this, assume the original container is a cube of 1 m³ (with each side a square of 1 m * 1 m) , and you change it to a rectangular box with width and length = 2 m and height = 0.25 m (top and bottom side become squares of 2 m * 2 m; left, right, front and back side become rectangles of 2 m * 0.25 m): now the average distance between particles and the top (or bottom) side of the box is reduced by a factor 4, so 4 times more particles hit the surface each second, but the surface area is also four times bigger (4 m²), therefor the pressure stays the same. The same is true for the sides of the box: the average distance has doubled, so the number of particles hitting them each second is halved, but the surface area is also reduced to half the original (0.5 m² instead of 1 m²), resulting in the same pressure."
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Volume the same, so pressure remains the same.
 
Dale Davies said:
Volume the same, so pressure remains the same.
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I knew that but didn't exactly know how, Ideal gas law confirmed that 100% for me and this gave a good explanation how, didn't realize the space of the particles changed in relation to the different areas of the chamber at various distance that have different amounts surface area averaging it out when the overall volume is the same.
 
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Look, guys.

The nickel in the clay.... that's not showing how pressure moves a nickel into clay. It's show how the pressure applied by the nickel on the clay is higher if the force is applied edgewise (small area) vs side ways (large area).

Maybe a way of looking at this that is easier for some folks to grasp.... let's talk about braking systems.

Larger caliper pistons, and more caliper pistons, yield more rotor clamping force for a given fluid pressure.

Smaller master cylinder pistons give greater fluid pressure for a given amount of pedal force.

Pressure = Force/Area. This can be expressed as PSI (POUNDS PER SQUARE INCH, or stated another way, pounds (force)/square inch (area).

Let's play with some numbers.

Ex. 1: 10 PSI = 10 lbs/Area
Area=10lbs/10psi = 1 square inch

Ex. 2: 10 PSI = 100 lbs/Area
Area=100lbs/10psi = 10 square inches

Ex. 3: 10psi = force/10 square inches
Force = 10psi*10 square inches =100 pounds

Ex. 4: 10psi = force/100 square inches
Force= 10psi*100sq in = 1000lbs

In these examples....the area is the piston face area. The pressure is fluid pressure. The force is the Force the piston will exert on the rotor. Given constant pressure, if you increase piston size, force will increase. If pressure reduces, force decreases.

This is the most basic, fundamental physics guys. It's absolutely irrefutable.

It's not whatchamacallits or what have yous. Anyone who's talking like that just doesn't have good basic arithmetic skills.
 
Mean416 said:
Look, guys.

The nickel in the clay.... that's not showing how pressure moves a nickel into clay. It's show how the pressure applied by the nickel on the clay is higher if the force is applied edgewise (small area) vs side ways (large area).

Maybe a way of looking at this that is easier for some folks to grasp.... let's talk about braking systems.

Larger caliper pistons, and more caliper pistons, yield more rotor clamping force for a given fluid pressure.

Smaller master cylinder pistons give greater fluid pressure for a given amount of pedal force.

Pressure = Force/Area. This can be expressed as PSI (POUNDS PER SQUARE INCH, or stated another way, pounds (force)/square inch (area).

Let's play with some numbers.

Ex. 1: 10 PSI = 10 lbs/Area
Area=10lbs/10psi = 1 square inch

Ex. 2: 10 PSI = 100 lbs/Area
Area=100lbs/10psi = 10 square inches

Ex. 3: 10psi = force/10 square inches
Force = 10psi*10 square inches =100 pounds

Ex. 4: 10psi = force/100 square inches
Force= 10psi*100sq in = 1000lbs

In these examples....the area is the piston face area. The pressure is fluid pressure. The force is the Force the piston will exert on the rotor. Given constant pressure, if you increase piston size, force will increase. If pressure reduces, force decreases.

This is the most basic, fundamental physics guys. It's absolutely irrefutable.

It's not whatchamacallits or what have yous. Anyone who's talking like that just doesn't have good basic arithmetic skills.
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You have some messed up math and science skills.
 
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