Smokindodge

Have you made any head way on this project? A smaller chainsaw engine would probably do what you want to do and have plenty of flight time. I don't know how the rpm would work out though, they run about 14,000 rpm so a gear box may be in order to preserve the integrity of the props

Over built is my middle name. Here's a sorry pic of the 30 ton 39" splitter I built last fall. 19 horse engine with a 28 gpm pump with a 5" ram. Insert Tool man Tim Taylor laugh/grunt here. So a six" two pounder it is. It'll be a bit before I start the fab but I will keep you updated. Thanks, Neal

So your thinking that larger and slower will give it a longer dwell time and more usable force? Here's a machine in the UK, might give us some ideas. http://www.autoguide.co.uk/postmasterfeatures.html Here's one in the US that states it's a 5,000 RPM model http://www.triplexgroup.com/pv400.htm And yet one more that has everything listed in kilo's http://www.randrattachments.co.uk/RVibrPost2.pdf Interesting design, the rotor housing doesn't look to be over 18" dia.

Material flex shouldn't be much of an issue, I'm not going to use any steel under 1" thick on this project. The rotor will be hung on a 2.5" shaft with pillow block bearings on each side of the weight as close as can be. The motor will fit in the end of the shaft to drive it. I've got a Parker hyd motor, I'll check the data plate but I'm pretty sure it'll turn it 3,000. The entire thing will encased in 1/2" thick pipe just to error on the side of caution. 6,000 #'s is a little more than I was shooting for. Will four pounds with a radius of 6" be under 6,000? I'd like to end up at about 3,000 and the more compact I can make the rotor the better for the wallet. John I don't think the vibration will affect the top of the post near as much as the impact style. All the post will feel is the downward swing of the weight, the mast and tractor will absorb the fore and aft "swing" and the upwards rotation is moot.

Thanks for the formula, are you saying that I need to use metric for the calculation? I've tried figuring a 2lb weight with a radius of one foot at 3,000 rpm and I come up with 20,000 Is this the correct pounds of force? I'm also dense when it comes to algebra. I'm good with the hands, not so much with the grey matter. I do appreciate everyone's help.

Spot on IA, the force on the top of the post is all that I'm concerned with. That is what will "Get 'er done"

You got it John! I don't think the fore and aft forces will affect much as long as I don't use too big of a cam, rotor or what ever it should be called. This will be mounted on a 11,000 pound tractor, it's going to take quite a bit to upset that fella. There are several different styles. The vibrating head style only vibrates and provides downward pressure with the mast. There are other types that use a friction wheel to raise a large ram and when the weight is fully raised it trips and falls. A baby Pile driver. And yet another that uses springs and a small weight with a one way hyd. cylinder to tension the springs. When the springs are fully tensioned it releases the hyd pressure and impacts the post. I don't care for the impact style of drivers, too much can go wrong with them. They will also booger up the top of a wood post in a hurry.

Yes that is it exactly. How much weight would it take x many of inches from center at what rpm to generate 3,000 pounds of force. If there is something I haven't explained to where you understand it please ask! I always have trouble trying to articulate my ideas. I'll try my best to describe it, I can't make a diagram worth a hoot. The vibrating head never comes into contact with the post, it simply shakes the bejesus out of everything. The loader will exert downward force and between the shaking and the pushing the post is inserted into the ground. As for the vibrator, think of a 5" round track that completely incases a 1" steel ball. Introduce air to the ball to make it move around the track at 200 rpm. It's gonna be shaking more than J-lo's hind end. Same idea as with a rotor, just attach an arm from the ball to a motor and remove the track. Thanks for the move and the replies!

Two days and thirty veiws and not one suggestion? Can anyone tell me what kind of problem this is so I can find the math for myself?

Howdy all, first post here. An unusual project brings me here, I'm planning to build a vibrating post driver for installing fence posts around my farm. They are availible commercially but they are outside the budget we have and we need to get the fence built. I'm good with my hands and a welder but the physics part not so much. What I need help with is the rotor design, or offset weight that is spun by a hydraulic motor to create the harmonics. The motor speed will be adjustable up to 3,000 rpm. None of the commercial products actually have ratings of force and if they did I'm sure that they all would be measured in a different way so that a comparison would be impossible. Commercial vibrators (no jokes please) for the food industry are capable of producing 3,000 pounds of force, I would like to up end with about that much. The unit will be mounted on a front end loader that will also be able to apply down pressure. Can some one provide me with a formula for figuring this? One pound one foot off center at 100 rpm = X sort of thing. I really don't want to spend four days cutting iron and "testing" it. I'm sure I would make one WAY to large and end up hurling a good size chunk of iron skywards. That's what catapults are for! Would the shape of the "rotor" (is there a proper term?) affect the force much? Say a bell shape versus a round end such as a throw of a crank shaft. Also, is what is the name for this type of calculation? I failed my last physics class because of a cute red head setting behind me. It wasn't all in vein, we are getting married! Thanks, Dodge