1
$\begingroup$

I'm working with a padfoot(spiked) drum compactor. If you roll one time over an area the feet will compact only 13.8% of the area you covered. I'm trying to calculate how many times you need to roll back and forth to have (averaged) 90% coverage.

The point of this is to than increase the surface area of the feet so that 90% can be reached in 8 passes, without unneccessarily increasing the size as that reduces the ground pressure of each pass, giving a diminished return.

  • 0
    So one pass of the drum only touches 13.8% of the fixed area, or does one pass of the drum affect the entire area, but only compacts it by 13.8%?2012-12-14

1 Answers 1

0

Assuming you roll over the area in an independently random way each time: since the compactor leaves $86.2$% of the area untouched after each pass, after $16$ passes the total area untouched will be $.862^{16}\approx.093 < 10$%, so the total compacted area will be just over $90$%.

In general if the compactor affects $p$ fraction of the area each time, after $n$ passes it will have covered $1-(1-p)^n$ of the area. So to cover $90$% in $8$ passes, you'll need just about $25$% coverage on each pass, as $1-(1-.25)^8\approx90$%.

  • 0
    Thanks, It didn't seem to be a very complex question but I couldn't find a technical way to describe it to get the answer. I forgot to mention the completely random part. I spent some time trying to parse out a way to describe how the return pass has a higher percent chance to overlap but after looking at some images from testing it's basically just averages out to random.2012-12-14