Hi folks,here is latest surveying problem,www.scsurveyjac.org/16.html. Here is my idea:Please refer to dotted lines from pt 8: I prolonged line 9-8 to create pt B on line 1-6 and 40' min on same line( perpendicular to line 1-6?) to create pt A. I would solve right triangle AB9 to determine 8 coordinate and proceed. Your feedback would be much appreciated,thanks much.
> Hi folks,here is latest surveying problem,www.scsurveyjac.org/16.html. Here is my idea:Please refer to dotted lines from pt 8: I prolonged line 9-8 to create pt B on line 1-6 and 40' min on same line( perpendicular to line 1-6?) to create pt A. I would solve right triangle AB9 to determine 8 coordinate and proceed. Your feedback would be much appreciated,thanks much.
The problem asks you to locate Pt.8 at a distance of 40.00 ft. or more from Line 1-6 and to place Pt.9 at a distance of 128 ft. from Pt.3. The missing piece of information is what a reasonable construction tolerance above 40.00 ft. would be. In practice, I'd think 0.10 ft. would be reasonable. In this case, the problem is more one of theoretical geometry than practicality, so you'd construct a line parallel with 1-6 and 40.00 ft. distant from it (at right angles) which would contain Pt. 8 and you'd construct a line parallel with that which would contain Pt.9. Find the intersection of that line with a circle of radius 128.00 ft. centered on Pt.3 and then calculate the building corners from that point.
In the real world, it is usually a poor idea to try to do zero-tolerance layouts with respect to boundaries and setback lines. Keep that in mind.
140-32=108'?
> 140-32=108'?
Uh, yes. :>
I don't like designers putting minimums on the plans...they should actually do their job and design the project.
At Parks the designers would put 5% maximum grade on the plans without actually checking to see if it is physically possible, like the 8' of proposed concrete walk with an 8' drop down to the existing ep.
[sarcasm][/sarcasm]> > Hi folks,here is latest surveying problem,www.scsurveyjac.org/16.html. Here is my idea:Please refer to dotted lines from pt 8: I prolonged line 9-8 to create pt B on line 1-6 and 40' min on same line( perpendicular to line 1-6?) to create pt A. I would solve right triangle AB9 to determine 8 coordinate and proceed. Your feedback would be much appreciated,thanks much.
>
> The problem asks you to locate Pt.8 at a distance of 40.00 ft. or more from Line 1-6 and to place Pt.9 at a distance of 128 ft. from Pt.3. The missing piece of information is what a reasonable construction tolerance above 40.00 ft. would be. In practice, I'd think 0.10 ft. would be reasonable. In this case, the problem is more one of theoretical geometry than practicality, so you'd construct a line parallel with 1-6 and 40.00 ft. distant from it (at right angles) which would contain Pt. 8 and you'd construct a line parallel with that which would contain Pt.9. Find the intersection of that line with a circle of radius 128.00 ft. centered on Pt.3 and then calculate the building corners from that point.
>
> In the real world, it is usually a poor idea to try to do zero-tolerance layouts with respect to boundaries and setback lines. Keep that in mind.
Kent:
Shouldn't your 128.00' distance be 108.00'? It seems that the controlling distance is the 32' Radial from Point 7 to Point 9. Another question comes to mind, What is Point 12? Appears to be something just stuck in there for confusion.
I see that the 128' was caught by someone prior to my getting this submittal written up and submitted. There were only two posts when I started this, the initial one and Kents reply.
Wouldn't that be an eight-foot high wall? Only way I know to drop eight feet with only eight feet of sidewalk.
There are an infinite number of answers, depending on how much you exceed the 40 ft minimum. I would expect for the purposes of the problem they want to actually use 40.000 ft, as that gives a unique answer that people might agree on, despite the impracticality of zero margin on setbacks.
So why didn't they say so? So many of the problems from that source are ambiguous.
I also note that the calculation of point 9 involves two lines intersecting at a very slight angle, so the answer in the real world would be subject to considerable magnification of measurement errors.
Kent's outline sounds good. In more detail:
-Find coordinates of points 2 and 4
-Find coordinates of point 3
-Note that 3-9 is 108 ft
-Calculate the right triangle to the left of point 8
-Now we know distance 13-9
-Construct a line parallel to 6-1 at the offset to make it pass through point 9
-find the point 108 ft from point 3 that falls on the offset line
-Calculate coordinates for points 8, 10, and 11
I think that these sorts of problems are most easily visualized as geometric constructions first. Once the construction is plain, the numerical solution follows.
They never actually looked at it.
I went as the result of a request to field review the site as part of a constructibility review. They should state where the building goes, not just give conditions.
Hi folks, and thanks so much for replies but I have a real dumb question; Why can't you just solve the right triangle to left of pt 8?
> Hi folks, and thanks so much for replies but I have a real dumb question; Why can't you just solve the right triangle to left of pt 8?
You can, but how are you going to calculate the distance 1-13?
Is point 7 the intersection of the arc and line 3-9 produced?
What is point 12? It appears to be on line 1-2 produced.
(I solved it without either point is why I am asking.)
Yes on 7.
Maybe 12 is the PI of the entire arc, just guessing, though.
> Is point 7 the intersection of the arc and line 3-9 produced?
>
> What is point 12? It appears to be on line 1-2 produced.
I think that Pt. 12 is intended to be the intersection of line 1-2 extended and a line parallel to 1-6 constructed from Pt.3, which would be one way to solve 14-3.
Uh, maybe I'm wrong on this, but, I get the distance from 1 to 8 to be precisely 47.00 feet which is 24.88 to the right and 39.87 down.
> > Is point 7 the intersection of the arc and line 3-9 produced?
> >
> > What is point 12? It appears to be on line 1-2 produced.
>
>
>
> I think that Pt. 12 is intended to be the intersection of line 1-2 extended and a line parallel to 1-6 constructed from Pt.3, which would be one way to solve 14-3.
Nope!
> > > Is point 7 the intersection of the arc and line 3-9 produced?
> > >
> > > What is point 12? It appears to be on line 1-2 produced.
> >
> >
> >
> > I think that Pt. 12 is intended to be the intersection of line 1-2 extended and a line parallel to 1-6 constructed from Pt.3, which would be one way to solve 14-3.
>
> Nope!
Yes, I typed Pt.3 when clearly Pt.9 was intended. An obvious scrivener's error. :>
You better record an affidavit right away!