Re: help me do the math?

On Jul 1, 7:53 pm, "shockwaveriderz" <shockwaverid...@xxxxxxxxxxx>
wrote:
Ned:
explain to me how you got the 275 lbs? For me to measure this amount I
would need some form of pressure gauge correct?

terry dean
nar 1658

--
"Old Rocketeer's don't die; they just go OOP"

"ned" <nedgor...@xxxxxxxx> wrote in message

news:1183331463.181741.42070@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



Yep, Terry, that's the weight of the puck that you're shooting for.
When pressing thin BP pucks, I use 12 tons of force on a 3.5" diameter
puck, which is 2500 lbs/square inch. On your 3/8" diameter puck, you'd
be shooting for 275 lbs of force,,,,heck, if you're a bit chubby like
me you don't even need hydraulics, just rig up a way to stand on your
ram.... ;-)
ned- Hide quoted text -

- Show quoted text -

Terry, 24000 pounds of force on a 3.5" puck, which is what I use is
24000 lbs per 9.6 square inches (3.5 divide by two to get radius is
1.75, square that 1.75 x 1.75 = 3.06 times pi is 3.06 x 3.1416 = 9.6
square inches) 24000 divided by 9.6 = 2500 lbs per square inch,,,the
psi pressure I'm using on the face of the puck.

Now, your 3/8" diameter puck has a face area of .11 square inches
(3/8" = .375" diameter divided by 2 = .1875" radius, squared is .1875
x .1875 = .035, times pi is .035 x 3.1416 = .11 square inches of
area. .11 square inches times 2500 lbs per square inch is .11 x
2500 = 275 lbs of force required,,,which is what you'll weigh after
that Fourth of July picnic,,,, (disclaimer,,,I don't know Terry so I
have no idea what I'm talking about when it comes to his weight...)

And, yes, normally we read that force with a P2F force conversion
gauge, or by a conversion of a psi gauge reading on the bottle jack,
times the area of the bottle jack ram.....

(Now,,,,I'm gonna get more questions,,, ;-)


ned

.

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