Dump wagon build

[CDN Farm Boy]

Thanks Aaron, that would be an option I hadn't considered.

Spent the afternoon looking at scissor designs and playing with CAD and came up with the following. Please feel free to poke holes in it. This would reuse my existing 4x30 cyl and the existing lower cyl mount.

Point Legend:
A - Existing cyl base mount & scissor base mount
B - Scissor hinge pivot (lowered position)
C - Cyl/scissor mount (lowered position)
D - Scissor/bed mount (lowered position)
E - Scissor hinge pivot (raised position)
F - Cyl/scissor mount (raised position)
G - Scissor/bed mount (raised position)
H - Bed hinge

When the cyl A-C extends, it will be applying force on the B-C axis at 45 deg. This translates into a force of 19550 lbs at point C. This force will then try to rotate the BCD link around point B. As B-D = 2xB-C then force at point D would be 9775. This force is rotational around point B which is, in fact, a floating point moving in an arc from point B to E around point A. As point D raises, due to the floating pivot of point B and the fixed pivot of point H, D will move in a straight line to the fully raised point G. Line D-G is at 90 deg to Line DH therefore there would be 9775 lbs of lift at point D which would diminish as the box raises.

I don't know how to calculate lift force at anywhere higher than the fully down position but in reality, I guess it doesn't matter as the most lift is required at the bottom.

With 19550 of effective force at point C, where does the other 8150 of the 27,700 maximum push from the cyl go?

 

[tow653]

I have said for years that if I had it all to do over I would get a degree in engineering,now I know why I have one in economics instead.

 

[CDN Farm Boy]

I have said for years that if I had it all to do over I would get a degree in engineering,now I know why I have one in economics instead.

Lol. I'm guessing you didn't follow then?

 

[tow653]

Understatement!

 

[rswyan]

... Spent the afternoon looking at scissor designs and playing with CAD and came up with the following. Please feel free to poke holes in it.

Ok ... your A-E line looks longer than your A-B line in the drawing ... is that in fact the case ?

Because that won't fly in the real world.

I don't know how to calculate lift force at anywhere higher than the fully down position but in reality, I guess it doesn't matter as the most lift is required at the bottom.

You need something like Solidworks ... which has a simulation module:

SOLIDWORKS Simulation | SOLIDWORKS

But yes, it's probably true that as long as you're good at the bottom of the lift cycle you'll be good the rest of the way - since the load should be decreasing and you will (or should) be gaining mechanical advantage.

With 19550 of effective force at point C, where does the other 8150 of the 27,700 maximum push from the cyl go?

Available (effective) force is dependent on the geometrical (is that a word ?) relationships of the various components in a system.

IOW, mechanical advantage (or disadvantage)

The cylinder isn't always capable of exerting 27,770 lbs of (effective) force - it is dependent on existing conditions - specifically only when certain relationships exist between the components being acted upon.

 

[CDN Farm Boy]

I checked my drawing, line A-E = line A-B. Must be just an illusion. The only line that length changes is A-C to represent the cyl extension.

I thought a hyd cyl would always have the same amount of force available based on the surface area of the piston and the input pressure. Where and how that force is distributed is dependent on the relationship of the components. Am I wrong?

 

[rswyan]

I checked my drawing, line A-E = line A-B. Must be just an illusion. The only line that length changes is A-C to represent the cyl extension.

Good.

I thought a hyd cyl would always have the same amount of force available based on the surface area of the piston and the input pressure.

True - that's why I said "effective" previously.

Where and how that force is distributed is dependent on the relationship of the components. Am I wrong?

No ... but you're (kinda) answering your own previous question ...

 

[CDN Farm Boy]

Effective, in this case would mean "doing what I'm trying to do"? IOW, how much of the force is actually doing the lift. The remainder of the force still has to go somewhere.

In my current (poorly designed) setup, I've got about 5000 lbs of lift and 22700 lbs trying to push the bed horizontally off the back. As we all agree, not an ideal situation.

So would that not mean that in this scissor lift with 19500 of lift at point C, the other 8200 goes ???? Just guessing here but into the hinge of the scissor and then down to the lower pivot to point A?

If so, would this be a tension or compressive load on the lower link? Depending on which way I think about it I get either or both. Seems to me that during lift the hinge would be the fulcrum of the upper link puting the lower link in compression but in a stationary raised position, point C would be the fulcrum which would put the lower link in tension. This doesn't make sense as for this to be true, at some point the force on the lower link would have to be zero.



Just trying to understand and learn how all this works so that when I rebuild this I size components accordingly and only have to do it once (more)

I appreciate all the help.

 

[bcp]

In my current (poorly designed) setup, I've got about 5000 lbs of lift and 22700 lbs trying to push the bed horizontally off the back.

And the cylinder is pushing equally as hard on the base mount trying to move it down and forward.

Bruce

 

[dragoneggs]

Effective, in this case would mean "doing what I'm trying to do"? IOW, how much of the force is actually doing the lift. The remainder of the force still has to go somewhere.

In my current (poorly designed) setup, I've got about 5000 lbs of lift and 22700 lbs trying to push the bed horizontally off the back. As we all agree, not an ideal situation.

So would that not mean that in this scissor lift with 19500 of lift at point C, the other 8200 goes ???? Just guessing here but into the hinge of the scissor and then down to the lower pivot to point A?

If so, would this be a tension or compressive load on the lower link? Depending on which way I think about it I get either or both. Seems to me that during lift the hinge would be the fulcrum of the upper link puting the lower link in compression but in a stationary raised position, point C would be the fulcrum which would put the lower link in tension. This doesn't make sense as for this to be true, at some point the force on the lower link would have to be zero.



Just trying to understand and learn how all this works so that when I rebuild this I size components accordingly and only have to do it once (more)

I appreciate all the help.

You are now into the engineering world of 'statics'. It's actually not that difficult but it is all about 'freebody' diagraming. Static Forces must be equal and opposing. As long as the joint is statically determinate you can calculate the loads. Say you have three members, if you know one of the members, you can calculate the other two based on their relative angles to each other. Break up your force into x and y components and then figure what the other x,y components must be based on their angles. You can do this with a little algebra or maybe look online for a free or cheap statics program for trusses, etc. Been too long for me but it does bring back memories. I wonder if I could still pass a college test? :ashamed:

 

[CDN Farm Boy]

And the cylinder is pushing equally as hard on the base mount trying to move it down and forward.

Bruce

Yup. That's why I put it at the front axle and not out in the middle of the reach.

 

[CDN Farm Boy]

You are now into the engineering world of 'statics'. It's actually not that difficult but it is all about 'freebody' diagraming. Static Forces must be equal and opposing. As long as the joint is statically determinate you can calculate the loads. Say you have three members, if you know one of the members, you can calculate the other two based on their relative angles to each other. Break up your force into x and y components and then figure what the other x,y components must be based on their angles. You can do this with a little algebra or maybe look online for a free or cheap statics program for trusses, etc. Been too long for me but it does bring back memories. I wonder if I could still pass a college test? :ashamed:

Sounds about right. Basically since I would have a WAY better design geometrically with the scissor lift and most of the forces are acting between 45 and 90 deg, don't over think my problem? Just build a 'reasonable' amount of beef into the components

 

[farmer2009]

Nice build. Sounds like your still having trouble getting this figured out. There was a suggestion a couple of pages back I thinks was discredited to fast. The side dump. I know his suggestion was a 3 way dump but.
If you used your current cylinder what kind of angle could you get with it. I think you could move almost vertical as your only trying to dump 6feet not 10 feet. So you might have a long enough cylinder now. Just an idea. Take it for what it's worth.

 

[rswyan]

I believe another suggestion that was made was to add a second cylinder ...

You mentioned that you score the first one reasonable ... any chance of getting a second, identical one at a similar price ?

Would require some reworking of your design to be sure, but that's going to occur anyways it appears ...

 

[rswyan]

Effective, in this case would mean "doing what I'm trying to do"? IOW, how much of the force is actually doing the lift.

Yup, exactly ... you're trying to accomplish some work in a particular direction ... and not all the force you have available is working to that end.

 

[CDN Farm Boy]

Nice build. Sounds like your still having trouble getting this figured out. There was a suggestion a couple of pages back I thinks was discredited to fast. The side dump. I know his suggestion was a 3 way dump but.
If you used your current cylinder what kind of angle could you get with it. I think you could move almost vertical as your only trying to dump 6feet not 10 feet. So you might have a long enough cylinder now. Just an idea. Take it for what it's worth.

I appreciate it but my cyl is 40" retracted. No way can I get that anywhere near vertical under the bed. Also, far too much structural reworking would be required to make it a side dump. While it could be a good option for some, a side dump isn't how I want to be able to use the wagon. That is why it was discredited, not that it is a bad option for some

I'm confident that this scissor lift will solve my problems, and thankful for all the help in understanding it. Since my first design failed as miserably as it did, which is very unlike me, I'm likely over thinking some details so I don't get bit in the a** again.

 

[CDN Farm Boy]

Yup, exactly ... you're trying to accomplish some work in a particular direction ... and not all the force you have available is working to that end.

That part I'm clear on, thanks in a big part to you. I know I'll now have enough lift with the redesign. I'm just trying to make sure the remaining 8000 lbs is accounted for. Or perhaps I simply don't need to worry about it?

Aside from the fact that the math says the scissor lift will work for me, a BIG plus is that to incorporate it, I don't have to undo anything I already have. Expand my cyl base mount to accept the lower link and add a new bed side mount. The scissor itself can be built on my bench

 

[Xfaxman]

That part I'm clear on, thanks in a big part to you. I know I'll now have enough lift with the redesign. I'm just trying to make sure the remaining 8000 lbs is accounted for. Or perhaps I simply don't need to worry about it?

Aside from the fact that the math says the scissor lift will work for me, a BIG plus is that to incorporate it, I don't have to undo anything I already have. Expand my cyl base mount to accept the lower link and add a new bed side mount. The scissor itself can be built on my bench

And it will be the longest scissor lift that I have ever seen! :laughing:

 

[CDN Farm Boy]

And it will be the longest scissor lift that I have ever seen! :laughing:

Lol. Perhaps it will be. It's only 54" or so from lower mount to hinge. I guess that's on the large side?

If I'd have planned it this way from the start, I'd have bought the 4x20 cyl that was besides this one in the clearance bin and $75 cheaper, but hey, obviously I thought my original way would work. Live and learn.

 

[Xfaxman]

Lol. Perhaps it will be. It's only 54" or so from lower mount to hinge. I guess that's on the large side?

If I'd have planned it this way from the start, I'd have bought the 4x20 cyl that was besides this one in the clearance bin and $75 cheaper, but hey, obviously I thought my original way would work. Live and learn.

Yep, all that I have seen use a short stroke large diameter cylinder.

Here is a thread that showed up when I was searching for drawings: http://www.tractorbynet.com/forums/...help-scissor-lift-dumpp-truck.html?highlight=