KiwiBro
Mill 'em, nails be damned.
We've all seen the BS claims about how the 3" cylinder with pump of questionable pressure splitter is capable of 40t of ram force. It's fairly easy to calculate the force of a hydraulic ram, but how about a ram on a kinetic/flywheel splitter?
From here:
http://www.engineeringtoolbox.com/flywheel-energy-d_945.html
we can work out the kinetic energy a flywheel has, but I'm not sure how to compare that to a hydraulic ram force or tonnage rating.
Let's take a DR pro XL model and assume:
weight of each flywheel=74 lbs
max rpm=400
inertial constant=.75 (I'm guess with this)
diameter=18 1/4"
So, moment of inertia = .75 x 74 x .76 x .76 = 32
and kinetic energy = .5 x 32 x (400/9.55) x (400/9.55) = about 28000 ft lb per flywheel
Is that correct or where have I gone wrong with the calcs please?
Then, how to compare that kinetic energy to a traditional hydraulic ram tonnage rating?
From here:
http://www.engineeringtoolbox.com/flywheel-energy-d_945.html
we can work out the kinetic energy a flywheel has, but I'm not sure how to compare that to a hydraulic ram force or tonnage rating.
Let's take a DR pro XL model and assume:
weight of each flywheel=74 lbs
max rpm=400
inertial constant=.75 (I'm guess with this)
diameter=18 1/4"
So, moment of inertia = .75 x 74 x .76 x .76 = 32
and kinetic energy = .5 x 32 x (400/9.55) x (400/9.55) = about 28000 ft lb per flywheel
Is that correct or where have I gone wrong with the calcs please?
Then, how to compare that kinetic energy to a traditional hydraulic ram tonnage rating?