Any pneumatic experts out there?

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navycop

navycop

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Trying to figure out some pneumatic problems. Can you guys help with the formulas? Need to know how much psi to move a 250 lbs w/2 in bore cylinder.
Also how many lbs a 4" bore cylinder can move w/100 psi.
The book I am using is not real clear. It says: force (lbs)=pressure times lbs divided by inches squared. Then times area (inches squared). I have the lbs but not the psi. it also mentions: pressure=force (lbs) divided by area (inches squared). I divided 250 by 2 and came up with 125 psi. Maybe you can explain it better than the book. Maybe an example problem so I can see where the numbers actually go..
 
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Mike: I'm no expert, but I'll give this a shot. force= pounds per square inch x the surface area of the top of the piston. 4" = 3.1416x4x100 4" piston@100psi will lift appx. 1256 lbs. not counting friction loss. and etc. formula pi x radius squared x pressure in inches. 2'=3.1416x1=3.1416 sq.inches divided into 250=79.5772 psi to lift 250 lbs. give or take a psi or two.

hope this is correct.

Tim
Nova DVR
 
Mike:
I'm certainly NOT an expert! We used to use a "field measurement" rule of thumb:
1 inch at 100 PSI. should easily lift 100 pounds.

By our "rule of thumb" calculations, 2 inches at 100 psi should lift 200 lbs, etc.

It looks like your numbers are close to the thumb calculations.
 
Force(lbs) = Area(sq In) * Pressure(psi). Area = pi * D-squared/4
ie. for 4" dia area is 3.14 * 16/4 or about 12.6 sq in. At 100psi the force will be about 1260 lbs., not accounting for frictional loss in the cylinder.

2" bore has area of about 3.14 sq in. to exert 250lbs force need 250/3.14 or about 80 psi again not accounting for frictional losses in cylinder.
 
mhbeauford said:
Force(lbs) = Area(sq In) * Pressure(psi). Area = pi * D-squared/4
ie. for 4" dia area is 3.14 * 16/4 or about 12.6 sq in. At 100psi the force will be about 1260 lbs., not accounting for frictional loss in the cylinder.

2" bore has area of about 3.14 sq in. to exert 250lbs force need 250/3.14 or about 80 psi again not accounting for frictional losses in cylinder.
Click to expand...

Thanks guys. I must not be smarter than a fifth grader on this. I am still getting confused on the above statement. 2'' bore has the same area of the 4" bore?
 
navycop said:
mhbeauford said:
Force(lbs) = Area(sq In) * Pressure(psi). Area = pi * D-squared/4
ie. for 4" dia area is 3.14 * 16/4 or about 12.6 sq in. At 100psi the force will be about 1260 lbs., not accounting for frictional loss in the cylinder.

2" bore has area of about 3.14 sq in. to exert 250lbs force need 250/3.14 or about 80 psi again not accounting for frictional losses in cylinder.
Click to expand...

Thanks guys. I must not be smarter than a fifth grader on this. I am still getting confused on the above statement. 2'' bore has the same area of the 4" bore?
Click to expand...

No, 4" dia has 12.6 sq in, 2" dia has 3.14 sq in, see above
 
Mike, no. you just happen to be using some numbers that appear that way. The area of circle is Pi times the radius squared. the radius of a 4" cylinder is 2". 2 squared is 4 or 2X2=4. THis then gives you Pi or 3.1416 X R squared or 4 (3.1416 X 4 = 12.5664) That is the area of a 4 inch piston in square inches. You then multiply that area by the pressure in inch pounds. so 100 p.s.i will lift 1256.64 lbs with a 4 inch piston.

a 2 inch piston will lift only 314.16 lbs. with 100 p.s.i.
radius of 2 inch piston is 1 inc. 1X1=1. 1 X 3.1416= 3.1416. 3.1416 X 100 = 314.16 lbs. or 785.4 lbs at 250 p.s.i.

Now I am sure it is clear as mud.
 
navycop said:
Okay. For all the marbles. I tried this one.
What psi to lift 400lbs with 3.5 inch bore (3.5 squared =5.87).
400/5.87=68.14 or 68psi.
Click to expand...

Mike: 3.5 is the Diameter The Radius "R" is half of that or 1.75 so pi 3.14 x R squared (1.75 x 1.75) (3.0625) =9.62 square inches for the area of your piston. 400psi/9.62=41.58 psi not counting friction leakage etc. and ad nausim
 
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