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  #11  
Old 07-30-2012, 12:34 AM
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alldayrj alldayrj is online now
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Do you have a dump truck or somewhere on the property to toss the old stone with the cat?
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  #12  
Old 07-30-2012, 09:49 AM
Cat_246B Cat_246B is offline
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alldayjr: Theyhave a spot in their back yard where they want the old gravel, if they decide they don't want it, the Greenhouses down the road said they would take it.
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  #13  
Old 07-30-2012, 11:09 PM
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Make sure you give the driveway pitch so the water can run off and compaction is key rent a large plate compactor or a vibratory roller. The soft spot can be taken care of by removing the soft soil (most Likely clay ) and then replacing with structural fill I.e. 3/4" process and compacting. And to figure out the amount of material you need the formula is length* width* depth (in feet )/27 to get cubic yards and remember that because your compacting the material you need to get more I usually use a 10% or 15% compaction factor.
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Old 07-30-2012, 11:22 PM
Cat_246B Cat_246B is offline
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jdo150: thanks for all of the information, I will remeber to compact it. (which I never though about.) Thanks for the formula too.
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  #15  
Old 07-30-2012, 11:46 PM
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AztlanLC AztlanLC is offline
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The problem with the previous formula is we usually install inches not feet of material this is my formula accounts for compaction

Lengh x width = area x (inches) / 240
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  #16  
Old 07-31-2012, 05:56 AM
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Convert inches to feet # of (inches/12) example 4" /12=.33 feet
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  #17  
Old 07-31-2012, 06:11 AM
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Also how much of a factor is that formula it is way off aztlanlc. The problem is each soil compacts more or less and you need to figure that out your formula may work for common fill but not for a process or crusher run (whatever it's called in your area). There is too much of a factor. Also to the op if they sell in tons ask for the weight of the material it's probably close to 3000lbs a cubic yard.
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  #18  
Old 07-31-2012, 09:10 AM
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AztlanLC AztlanLC is offline
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Not to start a discussion but lets say we have 500sqf. Area we need to fill in with 5" aggregate (my formula is for aggregate not soils) so if I use your formula this is all the math I have to do
5 / 12 = 0.416666. Then 500 x 0.416666 = 208 / 7.71 x 1.15 = 8.87 cys or 13.305 tons

I can assure you you will run short

My way. 500 x 5= 2500 / 240 = 10.41 cyd or 15.62

I'm confident and know from experience this will be exactly what I need, I have done like this for the past 8 years.
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  #19  
Old 07-31-2012, 05:17 PM
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Where did you come up with the 240 just curious. I was tought my way at college and it hasn't failed me yet
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