Any ideas on how to draw this site plan?

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cbe's picture
Any ideas on how to draw this site plan?
SoftPlan Version:
2016 Plus

I'm trying to draw the site lines for lot 616A.  Attached is the info I was given from the homeowner.  The info is incomplete, but I still can't get it close to what is shown on the plat.  My property lines aren't matching up.



- Cory

Bill Wimberley
Bill Wimberley's picture

I can help you get close but I ran into a limitation in my math skills.

First of all you are given the length and bearing of 3 lines. Let's call these lines A-B, C-D, and E-F. I drew these lines in just at arbitrary poistions then moved them so they sort of matched up with the drawing.

Then I used the curve data to draw the curves that connect A to E and B to F. Since I had no way of knowing any sort of starting bearing for these curves I just drew them at an arbitrary location somewhere off to the side.

If you connect the end points of Curves 1 you get the distance from A to E.

If you connect the end points of Curves 2 you get the distance from B to F.

Next I drew a circle with the center at A and a radius of A-E.

Next I drew a circle with the center at B and a radius of B-F.

So we know that E-F would touch both circles.

This is where I ran into my math would you precisely place E-F so that it touches both Circle 1 & 2. If you can figure that out then you can precisely place E-F.

Once you have E-F in place you would draw a line from A-E and another from B-F. Note the bearings for A-E and B-F and rotate Curves 1 and Curves 2 to match.

Then move Curves 1 and 2 into place.

Then move C-D into place.

I am attaching the drawing file for what I did.

Binary Data site_cory.spd35.16 KB

Bill is the owner and maintainer of

Bill Wimberley
Bill Wimberley's picture

I have submitted a question on the Math Help Forum to see if I can find a way to locate the position of E-F. If I find a solution I'll supply it here.

cbe's picture
Bill - Thanks.  That is

Bill - Thanks.  That is pretty impressive.  I did the same as you to get the chord length of the curved lines.  I viewed the 'chords' as 2 of the 4 lines of a quadrilateral with the other two being the outside lines which i have the bearing & length.  The problem is I don't know how to solve for the interior angles.  I know they'll all add up to 360, but the math to get me there is above my head.  I did some trial and error and I'm able to get close.  However, when I get the outside lines to match, the center line doesn't fit.  Either I'm off more than I think or the plat is incorrect.  


The only  reason I'm trying to get it nailed down is b/c the house needs to be 16' from each sideline.  Therefore every inch counts.  

Thanks for taking the time.  If the math help forum any any help, I may have to wait on the homeowner to get his lot surveyed.


(my best guess is pasted on to your file - attached)

Binary Data site_cory.spd43.8 KB
Bill Wimberley
Bill Wimberley's picture

Ok, I got an answer on how to locate the line. The image below is what I submitted on the math help forum. The solution is as follows:

start at  C

walk  N 17°47 a distance  =189.460

stop. you are at a point, call it  E

now draw a circle center  E radius  =AC=163.393

B must be on this circle

so  B is at the intersection of the circle with center  E and the circle with center  D


cbe's picture
I think I got it thanks to

I think I got it thanks to that solution.  When I drew the 2 circles, they intersected at 2 different points.  I drew the site plan from each of those points and one of them lined up perfectly.  Thanks for the help once again Bill.

Binary Data site_plan.spd35.91 KB
Bill Wimberley
Bill Wimberley's picture

Yep, interestingly there are 2 solutions for the location of the north-west boundary. But only one of them works with the center line.

ChrisStewart's picture
This is why I was trying to

This is why I was trying to show the solution to finding the arch cord when you know the arc length and radius.


d=delta   (I see now that I should have clarified that this is delta angle)


L=arc length 

ac=arc chord



d = (L x 360) / 2 x (Pi) x r     (oops -I see that I inserted the wrong formula) 

the correct formula is: d = (180/Pi) * (L/r)

57.288 * (118.33/403.98)

d=16.7825 degrees 


ac = 2 x r x sin(d/2)



       807.96 * .1459 



The formula for calculating the length of an arc is 

L = (d/360)*(2*Pi*r)

      (16.78/360) * (2*3.142*403.98)

       .0466 * 2538.6



So some of these numbers are rounded which will cause an discrepancy but if you use these three formulas in your spread sheet you will cover the basic needs of most situations we run into.

Unless I am missing something I think that without actually making that calculation in this case all you can do is make wild guesses.