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Need help with my Wolter gage


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Posted

Okay,I have posted this over at passfire and haven't found the answers there,so I figured I'd try here.

 

I have a Harbor freight 12 ton press with the rod that comes down(1" total diameter)

 

The Wolter pressure gage is just below my spindle.

 

I am having an extreemly hard time figuring out thae gage.

 

My tooling is 4 OZ and the rams are 1/2" total Diameter.

 

I have been trying to use this,but nothing makes since.

 

http://www.passfire.com/archives/issue3_7/design3_7_2.asp

 

I enter the 1" for the press,and enter .5" for the rams. I enter 1750 for the pounds.

 

Thanks

hkmgs@yahoo.com

Posted

If you want to use the Passfire calculator to figure out your loading pressure based on the readout on the gauge, you have to enter 1.1284 for the Piston dia. This is the diameter of the piston inside the Wolter gauge which is exactly 1 sq in. of area.

 

The neat thing about the assembly is that you don't have to do any measurement of your hydraulic cylinder and math to figure out the force applied your tooling (lbs). You simply take the readout on the gauge which *is* the force. To find the loading pressure, divide it by the area of your rammer face.

 

Example: If your Wolter gauge reads ~1725 and your tooling is 1/2":

 

Area of tooling = Pi * R^2 = Pi * (.25)^2 = .196

 

Loading pressure = (pounds of force applied) / (tooling area) = 1725 /.196 = ~8800 PSI

 

Simple, no?

Posted
If you want to use the Passfire calculator to figure out your loading pressure based on the readout on the gauge, you have to enter 1.1284 for the Piston dia. This is the diameter of the piston inside the Wolter gauge which is exactly 1 sq in. of area.

 

The neat thing about the assembly is that you don't have to do any measurement of your hydraulic cylinder and math to figure out the force applied your tooling (lbs). You simply take the readout on the gauge which *is* the force. To find the loading pressure, divide it by the area of your rammer face.

 

Example: If your Wolter gauge reads ~1725 and your tooling is 1/2":

 

Area of tooling = Pi * R^2 = Pi * (.25)^2 = .196

 

Loading pressure = (pounds of force applied) / (tooling area) = 1725 /.196 = ~8800 PSI

 

Simple, no?

Thanks,i think i finally understand.

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