Had a curve ball thrown to me today about an hour before leaving a road way construction site. The local city inspector asked for a change to a driveway to remove an existing hump that was causing cars to drag. This was to be a field fit that was not included in the plans.
What he wanted to do was lower the sidewalk 3 inch's then from the back of the sidewalk slope the driveway up at 12:1 to meet the existing grade that was falling away to a drop inlet. I shot the drop inlet and the edge of the existing sidewalk and calculated an existing -2% grade. The only thing left was how far back to saw-cut and remove the existing drive so as not to exceed a 8.3% (12:1) grade. Did it with a level by hunt and peck quick enough, but I know there is a formula in a rusted up file drawer in the back of my brain I am forgetting.
Can someone help jog my memory on this.
(0.25/((1/12)+0.02))=2.42' horizontal.
Personally I would go back 6.25' so that you have 2% up to meet the 2% down with a less abrupt transition.
Formula:
(VertD/(slope1-slope2))=horizontalD.
Pay attention signs of the slopes.
Isn't that the formula to find the P.I. of a vertical curve?
Thank you Dave it was driving me nuts.
Yes, that's probably the recess of my mind that it came from, the mental vertical curve file.