318 John Deere, 4200 John Deere, 1947 John Deere "M"
If nothing else, your photo-documentery is first class! I concur with the others, that the turn buckles give me some heartburn! I'm planning to mimic your design to a point. Looks like we've got simular toothbars, and I thought that I only had five teeth that would leave one dead center, but not the case, I've got six also. I'm thinking that I'm likely to weld a 3" wide,2'long,3/8" plate to the base of my lift pipe horizontally, and then capure two teeth (one on each side of center) with square holes, so my beam is directly centered. I assume that you don't have welding equipment. That would be an asset, but, a nicely thought out concept! Keep the dialoge & pics comming! We'll all learn something:thumbsup: !! ~Scotty
Philcase.. nice work.... PLEASE be very careful.... PLEASE take into account of the hydraulic pressures that will be developed INSIDE the FEL cylinders as the weight is extended out from the bucket.....
let's see if anyone wants to figure out the INTERNAL pressure at the base of a 4 inch cylinder is if you have the weight of 300 pounds in the center of the bucket? lets say that from the FEL mounting pin to the cylinder pin is 4 feet, and from the cylinder pin to the bucket pin is 4 feet..... (lets do this at 90 degrees for easy calculations.)
NOW let's extend the 300 pounds of weight out another 10 feet for a total of 14 feet from the cylinder pin to the weight center???
what are the internal pressures of the cylinder????
what is the cylinder rated for???
what are the hoses rated for??
what are the valve blocks rated for???
can you see a problem being developed as the weight is extended ???
let's hope you never find out. Jim
I have a rig very similiar i used with no troubles but instead of the cables I used chains. I didnt have a winch. either had to drive in lowest gear to place the trusses.
Ya know I had to try... Lets see...4" cyl area = Pi X r(sq) = 12.56 sq inches.
300lbs at 18 ft = 5400ft lbs (8' from bottom of cyl to end of bucket.)
5400 lbs /12.56 sq in = 430 psi...2 cyls...215 psi???
A 2" cylinder will be 1719 psi assuming my numbers are correct. but the weight will be between two cyls. so... 860 psi???
The valves would get the full pressure...right???
Take into account I struggled in physics.... These numbers are by no means to be assumed to be correct. I was origionally concerned with the geometry as the tractor weight is probably close to that with loaded tires.
I like the idea and especially the determination to find a way to do this by yourself!
My main concerns are tractor stabillity and over-stressing the fel with the winch.
Put as much weight as you can on the 3ph! A longer pole attached to the 3ph instead of the fel would be a little more "sure footed"
Be carefull winching too hard (if the truss gets snagged) cause the winch will lift more than the fel is capable of lifting, which will stress the fel more than it's designed to take.
Also be careful of the truss swinging side to side!
Good luck! Can't wait to see the action pics!