newbie - 27 February 2009 12:13 AM
So, where have we got too? Nowhere?
As I stated at the beginning of this thread there is something horribly wrong with the way we work out these pressures, it defies basic physics.
A military saying goes ‘bullshit baffles brains’ so lets concentrate on the basics. Lets forget all the fancy talk of Psi, pressure, lbs per,and all the other emotive stuff which gets bandied about when this subject comes up and consider the following.
If you apply force in a downward fashion to say a disc of 4” in diameter, then apply the same force over a lager area, lets say 5 feet in diameter. The way things stand at the moment means you would have to step the pressure up by thousands of percent to achieve the same result? Totally ridiculous. Downward pressure on a disc is the same at the center as it is at the edge, there is no need to increase the pressure to account for a wider disc.
Now take a second scenario. We have a tube with 4” of material inside, applying pressure to the top will need X pressure to compress it. Now increase the contents by 2, this will lead to a necessary pressure increase of 2X to achieve the same amount of compression.
With this in mind, the only time we need to increase the pressure would be by increasing the depth of the mould.
In practical terms. Using the same mould say 6” with 2 gallons it would be one pressure, increase that to 4 gallons and then you would need to increase the pressure.
So, I believe my hypothesis to be correct, “there is something wrong” with the present position on pressure.
Newbie, you are scientifically wrong…
Pressure is force on surface area, so it is weight on surface
If you increased the disk size (diameter) this means u increased the surface, so the weight has distributed on bigger surface and the pressure decreases.
amount of cheese inside the cylinder is not related to pressure, so in 4” mold with 20pounds wight on top is same pressure if u put inside it 1 pound cheese or 4 pound cheese, all it counts is the surface area under the press disk.