whiz kid mathematical types please

   / whiz kid mathematical types please #1  

wroughtn_harv

Super Member
Joined
May 12, 2002
Messages
6,000
Location
Denison, Texas
Tractor
2013 Volvo MC85C
Boy I bet that's got the woodwork shaking./w3tcompact/icons/laugh.gif

Here's the sitchyation.

My tractor's a JCB165HF. I call her Iris cause, well, my grandma on my mama's side was called Iris.

She's a skid steer sorta like a Bobcat if someone capable of thinking out of the box had designed her.

The problem with drilling holes with a skid steer is they are designed for lifting, not pushing down.

For some time I've been thinking along the line of adding weight to the front lifting arm and then ingenuously devising a way to extend it out front compounding the weight, sorta like that Don guy did down there in God's country outside of Austin.

Man that's a lot of complicated thinking. Got myself a headache even trying to figure all the ins and outs of that configuring.

So now I've decided I'm gonna be simple.

The bottom of the tractor is protected by a solid belly pan front to back. There are two scupper kinds of pieces that run front to back on each side that are for draining debris etc.

I'm thinking about putting a couple of pieces of heavy wall rectangular tube, say three by four quarter wall along the scupper thingys.

In these will be other pieces of steel tube that slide in and out.

Here's the thought. When I need to apply down pressure on the auger I manually slide the inner tubes out to the rear. I pin these in place.

On the end of these tubes I have a method of attaching a pad.

What this does is when I apply the down pressure on the auger instead of the tractor raring up like a goosed pony and taking weight off the auger the arms out the back with the pads pick up the weight of the tractor and transfer it back to the auger.

Yup, I thought of that. Bent tubular arms are a definate for sure maybe most likely. It just so happens I have a couple of good points up high on the back of the cab to attach two other arms to angle down and back to the pads to pick up the bulk of the weight transfer.

I'm thinking I'd like this to be a simple pull tubes and pin kind of thing without any need for additional hydraulics and controls etc. I figure if the pads can be four feet off the back of the tractor and four inches or so off the ground I'll just have to be aware in tight spaces.

The pads won't be activated until they're needed when the weight transfer initiates them into place.

Here's the math part. Tractor weighs fifty four hundred pounds. It's about ten feet long. With a skid steer the bulk of the weight is around and behind the rear wheels. As I said, it's for lifting.

Now the shorter the distance from the rear of the tractor the pads are located the easier it's gonna be on everything involved. So if someone can figure out the down pressure with the pads at say two feet back versus three feet and even four feet.

I'm not real good at this word drawing stuff. But I'm even worse at the computer drawing.

A little help will be appreciated.

Keep in mind. This modification will only be needed when I'm in serious rock, which isn't that often. It's just that when you need it bygawd you need it.

I generally know when I'm gonna be in an area where there might be rock. So this thing will be one of those deals where hopefully it'll be just a matter of minutes to set it up. With the pads four inches off the ground I'll still be plenty mobile for moving hole to hole. But if and or when I hit rock it will only be a matter of applying downforce to go from bushel baskets of grins instead of truckloads of bad words.
 
   / whiz kid mathematical types please #2  
OK......how about this: You put a pad on the OPPOSITE end of the tractor as the auger is. By applying downpressure on this pad, you "bridge" the tractor from the pad to the auger, only leaving enough track touching the ground to keep the tractor from flopping over sideways. The weight of the tractor will be evenly divided....half to the auger, half to the pad. .....Dave
 
   / whiz kid mathematical types please #3  
Cj that is what he is talking about doing.
Since the CG (center of gravity) is about the rear wheels on
a SSL.
If u want half the weight on auger, the distance from auger to rear axle would be same as pad to rear axle.
If this distance is different divide the two to get the
% on auger or pad.
Hope this helps.
 
   / whiz kid mathematical types please #4  
EXACTAMENTO, thanks for making that point so clear.** OR.....how about TWO pads so that the auger and the pads act as a "tripod" and that way he can drill on any surface, regardless of the slope.
 
   / whiz kid mathematical types please
  • Thread Starter
#5  
So I guess the further back I go with the pads the more weight will be on the auger?????????

That makes sense. I was trying to find out how far back I'd have to go to get the various weights to the auger.

I've since decided that the heavier tubes will come from the top back of the tractor down. The horizontal tubes will just be locators and actually won't be carrying any weight.

These things will appear like wheely bars on a drag car when I'm driving around the job site.
 
   / whiz kid mathematical types please #6  
<font color=blue>So I guess the further back I go with the pads the more weight will be on the auger?????????</font color=blue>

Yup.

I did up a drawing for you, but can't convert it to a jpg. I have it in powerpoint if you want I can email it to you. My email is in my profile or send me a PM. I'll try to describe it here.

This problem is solved by what we mechanical engineers call a Free Body Diagram (FBD). To simplify we will assume that IRIS isn't accelerating, then we can apply two rules.

1. The sum of the forces = 0
2. The sum of the momements (torque) = 0

To calculate the maximum downforce, lets assume that you are able to lift both wheels off the ground (you may not be able to). Then we calculate moments at your pads sticking out the back of the tractor

There are two moments acting at this point

1. The distance between the pad and the Center of Gravity of IRIS (I will assume that it is slightly ahead (6") of the rear wheel). This is Dimension "A" in my drawing. = 3.5 ft X 5400 lb = 18,900 ft-lbs

2. the distance between the pad and the auger (dim "B" in my drawing) X auger downforce. = 10 ft * ???? We are trying to solve for the auger downforce so we use rule 2. The sum of the moments = 0 this means that moment 1 + moment 2 = 0.

Here's the algebra

A*Weight + B*Downforce = 0
rearrange A*Weight = -B*Downforce
Downforce = (-A*Weight)/B
The negative sign just keeps track of the direction of the force.

In numbers = (-3.5*5400)/10
= 1890 pounds

If you go to 4' behind
= (4.5*5400)/11 (this number gets bigger as you move the pad back)
= 2209 pounds

For reference w/o the pads there use the rear wheel as the pivot point to sum the moments (Dims C & D in my drawing)
downforce = (-C*weight)/D
= (-.5*5400)/7
= 385 pounds

BIG DIFFERENCE.

I guestimated at these dimensions, so pleas take out your tape measure or your highly calibrated eye & insert your numbers.
 
   / whiz kid mathematical types please #7  
Found a converter on the web. Drawing is attached
 

Attachments

  • 53-200582-wharv_jcb_auger_pads.jpg
    53-200582-wharv_jcb_auger_pads.jpg
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   / whiz kid mathematical types please #8  
Now if for the full picture (and to help you size your tubes) you can calculate the reaction force at the pad using rule 1

Pad force + Weight + Auger Downforce Reaction = 0

The weight is pushing down & the reaction forces from the ground at the pad and auger are pushing up this means they will have different signs (+/-)

Weight - pad force - auger downforce = 0
Pad force = Weight - Auger downforce
So somewhere near 3,500#
 
   / whiz kid mathematical types please #9  
Hazmat
Thanks for the great explanation and formulas for calculating the forces involved. I wish had known these for a few past projects. Now the next question, is there a formula or table somewhere to calculate the strength of steel tubing/flat/pipe/etc? So that you know it will be able to withstand 3500lbs without bending.

Tim
 
   / whiz kid mathematical types please #10  
This is off the subject but what converter did you use. A DWG to jpg? I could use somthing like this quite a bit.

Thanks
Tim
 

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