Eagleview
Veteran Member
No mate , it's 8.00am Friday morning . This is the best time to have a yarn . You guy's are finished for the day and i'm just starting .
No mate , it's 8.00am Friday morning . This is the best time to have a yarn . You guy's are finished for the day and i'm just starting .
No mate , it's 8.00am Friday morning . This is the best time to have a yarn . You guy's are finished for the day and i'm just starting .
I believe that they build or order a certain bucket, and then figure the cylinder size that will give a reasonable breakout before the bucket is destroyed. (SNIP) A shorter bucket on the same frame, will have a greater break out force, due to mechanical advantage. The same thing applies to the lift cylinders also. You could go to a larger cylinder, but you might place your front axles at risk, by overload. Just enough is what they settle on.
You're right there, JJ. Later when I tried to do a little grading, the left-hand edge of the bucket seemed to hit the driveway about a half inch before the right-hand edge.
I'm sure the first owner must have bent it!![]()
There is no such thing, only answers can be stupid and Picasso discovered that much in spite of his handicaps when he said "Computers are useless, they only give us answers"This may seem like a dumb question, but ....
Dead on... The bucket force is designed to overcome the crowd which can in-turn overpower the boom. That way the operator always has one more trick in his bag and doesn't get the bucket stuck.The breakout force is the amount the bucket can exert in it's crowd back function ie. tearing out a tree root while rolling the bucket back , as opposed to trying to lift it with the loaders lift arms (lift capacity).
I just looked at some more recent Cat specs and noticed a trend to cite breakout in terms of torque rather than force (i.e. ft-lbs). Still my 426's supposedly 50% above lift capacity breakout force means that, if I hit a bump in the road while transporting, my snow blower kept comfortably aloft will untilt the bucket holding it level as well. So I'm confused too, although only academically cause it isn't of any other interest in my case. I could in fact do more with math if Cat had quoted a torque value instead of a 6000 lbs force.I'm trying to understand the figures here.
Breakout Force is quoted for the pivot pin. So does this assume that with the pivot pin stationary, the breakout force is that which is applied by the bucket cylinders?
And if so, by extrapolation, can one then calculate the breakout force at the lip of this particular bucket?
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Actually, if you look at the more recent specs...they are in lbf...which is NOT ft-lbs.I just looked at some more recent Cat specs and noticed a trend to cite breakout in terms of torque rather than force (i.e. ft-lbs). Still my 426's supposedly 50% above lift capacity breakout force means that, if I hit a bump in the road while transporting, my snow blower kept comfortably aloft will untilt the bucket holding it level as well. So I'm confused too, although only academically cause it isn't of any other interest in my case. I could in fact do more with math if Cat had quoted a torque value instead of a 6000 lbs force.