Math Question

[Smokindodge]

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

[Smokindodge]

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?

[m4rc]

I am not certain which force you are trying to calculate. The Force that the rotating object will have on the post will be equal to the force required to keep that object rotating.

 

F=mrw^2, where F is the force in Newtons, m is the mass in kg, r is the radius of rotation of the object in meters and w is the angular velocity in radians/second. This will apply if the size of the rotating object is small compared to r. If it is not, then give a good description of the object rotating.

This will be the force where the rotating object meets the axis of rotation.

 

If you want to calculate the force that this would have in the ground, you will need to consider the leverage that your design will give you.

[Phi for All]

Unless you object strenuously, let's try moving this thread to Engineering and see if we can stir the pot, as it were.

 

I can always move you back if our engineers find this question too... challenging.

[insane_alien]

so, are you wanting to calculate the force it'll hit things with if its rotating at X RPM? or is it something else. The more information you can give us then the more likely we'll be able to help you.

[CPL.Luke]

I think part of the problem is that nobody here really understands what the device is, maybe a diagram?

[CPL.Luke]

although if I understand right you have a very large rotor which spins some weight around, and eventually hits the top of the fence post driving the post into the ground.

 

this is then an energy problem, you have to calculate the kinetic energy of the roter, and you also need to know the amount of energy required to drive the post into the ground, the first is a straight forward calculation, whereas the latter is a bit more difficult.

[insane_alien]

i was imagining one of those pneumatic rams they use for flattening tarmac(they're basically pneumatic drills with a flat bit on the end instead.)

 

in fact, one of those would probably work if you made some sort of a frame to hold it. otherwise it would be a bit awkward.

[Smokindodge]

so, are you wanting to calculate the force it'll hit things with if its rotating at X RPM? or is it something else. The more information you can give us then the more likely we'll be able to help you.

 

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.

 

 

I think part of the problem is that nobody here really understands what the device is, maybe a diagram?

 

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!

[Realitycheck]

If you use galvanized posts, they make a machine that literally drives the posts into the ground very quickly. There is a company in Dallas that specializes in making temporary fences around construction projects. If you are interested, they are called National Fences.

[Rocket Man]

you want the centripetal force.

[math]Force = Mass . Radius \left( \frac {rpm \pi}{30} \right) ^2 [/math]

i don't know how this works with imperial measurement.

[insane_alien]

no he said he wanted the impact force it would have on the pole. i'm not sure how you would go about calculating that though.

[JohnF]

I think he's talking about an offset rotating cam to provide positive and negative force. The upward centripetal force is supposed to offset the weight of the machine whilst the downward part will add to the weight. This will allow the hammering/driving action to occur.

 

He may also need to consider the forward and backward force of the cam. I think the cam in these devices is normally used to put energy in to a spring, or lift a weight, then quick release it.

[Smokindodge]

I think he's talking about an offset rotating cam to provide positive and negative force. The upward centripetal force is supposed to offset the weight of the machine whilst the downward part will add to the weight. This will allow the hammering/driving action to occur..

 

You got it John!

 

He may also need to consider the forward and backward force of the cam. I think the cam in these devices is normally used to put energy in to a spring, or lift a weight, then quick release it.

 

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.

[Smokindodge]

no he said he wanted the impact force it would have on the pole. i'm not sure how you would go about calculating that though.

 

 

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"

[Rocket Man]

then it's still the same equation, just add to that the force the machine applies without oscillations.

[Smokindodge]

you want the centripetal force.

[math]Force = Mass . Radius \left( \frac {rpm \pi}{30} \right) ^2 [/math]

i don't know how this works with imperial measurement.

 

 

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.

[JohnF]

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.

 

I don't think you could run it at 3000rpm. Don't forget that whatever force you achieve for hammering will also be applied to the axel of the cam; that force will also transfer through to the rest of the running gear. It is only by means of the physical connections of the cam to the main unit that force gets to where you want it.

 

At 3000rpm you would have to use a very small weight which would create a vibration rather than a thump; unless that's what you want. The vibration will transfer to the post and may damage/weaken it.

[Rocket Man]

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

 

 

i just tried those figures in metric, it's about 20 000N which is about 6000 lb i think.

3000rpm is pretty fast. that's 50 times per second (you'll hear the thing buzz), you'll have a few issues with materials flex making it not transmit that force properly.

i think 600rpm is more the bench mark.

by all means, set the maximum to 3000rpm but just don't get lead footed on the first test

 

what sort of motor are you thinking of using?

[Smokindodge]

i just tried those figures in metric, it's about 20 000N which is about 6000 lb i think.

3000rpm is pretty fast. that's 50 times per second (you'll hear the thing buzz), you'll have a few issues with materials flex making it not transmit that force properly.

i think 600rpm is more the bench mark.

by all means, set the maximum to 3000rpm but just don't get lead footed on the first test

 

what sort of motor are you thinking of using?

 

 

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.

[Rocket Man]

a hydraulic motor ought to do it. i was just concerned that the forces might damage the innards.

 

with this equation, if you halve either the radius or the mass, you halve the output force.

if you halve the rpm, you divide the force by 4

 

2lb on six inch at 3000rpm puts almost exactly 3000 pounds on the top of the

post.

 

i still think you ought to slow it down a little. depending on the soil, 3000 revs might just sit there buzzing

even 1inch steel can flex a lot under 3000 revs. you might find it doesn't transfer the force by the time the stroke is on the rest curve. try to get the post as close as possible to the bearing (mounted underneath is ideal)

[Smokindodge]

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.

[CPL.Luke]

20000 newtons is the sam as 2000 kilograms resting on top of the post, which is equivalent to about 4000 lbs resting on top of the post, should be sufficient however its hard to say how fast it would be driven in.

[JohnF]

From looking at the videos I think it needs to be high speed and a low weight cam to provide vibration rather than a thump. The vibration will transfer through the post and cause the soil to become fluid allowing the post to sink.

[Rocket Man]

The vibration will transfer through the post and cause the soil to become fluid allowing the post to sink

 

yeah, that sounds more likely.

in which case the bearing mount needs to be very solid and the post secured firmly to it.

you'll still need a lot of force to do that. a smaller rotor arm won't be put under the same stresses as a long one. i think you should go the six inch 2 pounds if you can get the motor up to speed.

(with this sort of device, theres no such thing as over built)