Stomper, there is a formula, and the result is a variable graph, but it won't be necessary to do that for approximate force calculations. What your designing is considered a third class lever, with a velocity ratio. For a somewhat accurate force applied after the mechanical disadvantage, use distance from *grapple pivot pin* to the *hyd. cyl. pin* that pushes on the grapple. (let's say 6" for this example) You state to expect 6908# push at that pin, so, at same distance (6") measured from grapple pivot pin toward end of grapple, that force would be the SAME (excluding frictional losses). Now, (assume grapple is 24"), measuring out 6" = 6908#, 12" out = 3454#, 18" out = 1727#, and lastly at 24" at the tip of grapple = 863#. Your definitely not going to be "pinching" stumps in two, or cracking rocks with 863#'s of force, lol, but sure enough will crush old limbs and branches in two if they are caught nearer the middle of grapple at 3454#'s. Think the best analogy (albeit reversed) to lever vs. force is picture torque wrench, one foot long, would take 100# pull on the handle to get 100# torque. Likewise, a TWO foot torque wrench would only need 50# pull on the handle to get the *same* 100# torque on bolt, and so on and on with 3' wrench etc. Torque wrench GAINS power with length, but grapple LOSES power the further away from pivot pin. I have designed and built several hydraulic excavator thumbs, bucket grapples, and dump gate latches over the years. It can be very disappointing to build something without a solid design, only to find your efforts wasted because it is not strong enough, or worst yet, TOO strong and breaks equipment. I commend your undertaking the task to research it fully, you won't regret it!
Thanks, That is how I thought it was figured out but wasn't 100% positive. Where I was confused was, I wasn't sure if the angle of the cylinder push direction, in relation to the direction of the force I wanted, had anything to do with it. At one point in the cycle when the grapple is being closed (or opened)the cylinder push is 90 degrees to the direction of force I want applied. In other words, I want the force to be straight down but the force applied is horizontal. I hope that makes sense.