Originally posted by twhiteheadI thought I was making that clear, using inches, sorry, 8 X 4 inches on a perfect vacuum with atmosphere on top there would be 8 X 4 X 14.7 pounds of force holding the object in place, or 352.8 odd pounds of force.
Its not very clear what you are asking, a diagram would help.
My first guess is that you are not taking area into account but should be.
Originally posted by sonhouseYou're giving us lots of information, which seems to be the same thing over again in lots of different units (none of which are S.I.), but leaving out important details, such as what materials are in contact and not stating clearly what the actual question is:
I thought I was making that clear, using inches, sorry, 8 X 4 inches on a perfect vacuum with atmosphere on top there would be 8 X 4 X 14.7 pounds of force holding the object in place, or 352.8 odd pounds of force.
So suppose it was a very strong metal, say 1/2 inch SS and you put a bolt partially threaded in the center and pull upwards with a force mea ...[text shortened]... rge part of it but I would think I would hear that much leakage. Any idea what is going on here?
Originally posted by DeepThoughtI was not pulling on a screw, just the corner which sticks out a bit. The substrate is alumina about 1 mm thick. I used pounds and such and inches because the gauges we use for the rough vacuum is in inches of mercury, not sure how to convert to SI.
You're giving us lots of information, which seems to be the same thing over again in lots of different units (none of which are S.I.), but leaving out important details, such as what materials are in contact and not stating clearly what the actual question is:
1) What is the actual question, is it why the force required is so small when pulling on the ...[text shortened]... ressure internal to this thing based on a measurement or based on the rating of the vacuum pump?
Originally posted by twhiteheadNo, the total area doesn't matter except as a leak source. The actual vacuum pressure is from the little channels cut into the chuck metal. They are about 1 mm wide and roughly 50 mm long in various styles, straight lines and circles all connected to the central vacuum source. So the total force is just the area of the channels times the vacuum level. So if it was half vacuum, say 300 torr, and 50 mm^2 of channel surface area and two completely flat surfaces, the bottom metal chuck surface V the bottom of the substrate being worked, then with a very tight fit between the two surfaces there will be very little leakage to atmosphere and that seals fairly well like that, leaving only the surface area of the channels as the force vector. So using PSI, not sure what it is in metric talk, say 15 PSI is one atmosphere and we are positing half atmosphere or around 350 torr or say 7 PSI for grins, then, 50 mm^2 is about 1 kg of force, which is about what we feel when we try to pull up on the corner of the substrate, the corners of the rectangular substrate sticks out from the chuck a few mm so we can test the force of the vacuum in a crude way by just lifting up on the corners and feel what kind of force it takes to force the substrate away from the chuck.
I am still not 100% certain what your set up is, but I think you are going about your calculations all wrong. If I understand you correctly, you are pulling a flat object off something else by hand. The key factor will be how good a seal the two objects make between them, how much they deform when pulled and where you pull. If you pull on the edge then ai ...[text shortened]... n where you had an object against a flat surface. In sum, the area doesn't really matter at all.
Originally posted by sonhouseVacuums don't exert pressure.
No, the total area doesn't matter except as a leak source. The actual vacuum pressure is from the little channels cut into the chuck metal.
Originally posted by twhiteheadThe pressure I was referring to was the force needed to pull the two parts apart. The total area of the surface under the vacuum area is what tells how much pressure or pulling force you need to pull the pieces apart.
Vacuums don't exert pressure.
I am still not certain about your setup, but I imagine you as having a flat surface with a hole in leading to a vacuum. On top of this hole you have placed a flat sheet of something. We are discussing the force needed to lift that flat sheet.
If we assume a perfect seal around the edge of the sheet (lets say we put wet rubb ...[text shortened]... essure and vacuums are irrelevant. What matters is how good the seal is and how easily it fails.
Originally posted by sonhouseNo, you are mistaken.
The pressure I was referring to was the force needed to pull the two parts apart. The total area of the surface under the vacuum area is what tells how much pressure or pulling force you need to pull the pieces apart.
Originally posted by twhiteheadI deg to biffer with you. The area directly impinging on the vacuum is what determines the force needed to separate the two pieces.
No, you are mistaken.
With a perfect seal, the whole area is as good as vacuum.
With an imperfect seal, what counts is how good the seal is.
Originally posted by joe shmoWhich is exactly what I have been saying, it is the area of the slots times the vacuum that gives the total pressure. 14.7 PSI or 6.6 Kg/square inch or 6630 grams per square inch or 6630 grams per 645.16 square mm or about 10.2 grams/mm^2 with full vacuum and 5.1 grams/mm^2 with half vacuum. So 25 mm^2 of slots would give 128 grams of connecting force. So the total size of the substrate has no bearing on the issue, it could be just slightly larger than the area of the slots or 10 times bigger, the force on the substrate would be the same, 128 grams.
The vacuum pressure will always be between 1 atm and 0 atm. The pressure that can be supplied at any given flow ( in this case the total leakage of the chamber/connections or system ) is unique to the turbo-machine/or positive displacement pump that is doing the work. If there are no leaks; the pump will be able to pull a vacuum pressure corresponding to n ...[text shortened]... the Area of your slots in which a vacuum pressure ( less than 1 atm, greater than 0 atm ) acts.
Originally posted by sonhouseAnd that is where you are going wrong. There is no such thing as a vacuum force. The vacuum does not suck.
That area would not participate in the vacuum force ....
Originally posted by twhiteheadThat is semantics. Of course the resultant is A minus B. It is just convenient to refer to it as a force. The force of the atmosphere on top V the lower pressure on bottom results in a downward force on the substrate. I understand that. But it is ONLY the area exposed to the lower pressure that causes the downward pressure not the total surface area of the substrate.
And that is where you are going wrong. There is no such thing as a vacuum force. The vacuum does not suck.
The flat object has three different forces acting on it:
1. The atmosphere on one side - proportional to the area.
2. The solid support which gives a static non-moving force that keeps the flat object stationary but doesn't take part in our calc ...[text shortened]... exposed to the vacuum.
So, total force required to move the object is force 1. minus force 3.