Comps Challenge

> A surveyor acting for the authority has set out baseline A-B on the centreline of the 70m major road and observed the edges of the tank, where possible, through gaps in the dense hedge surrounding the site as shown in the diagram.

With the non-existence of observational redundancy, there is no "reliable" solution. If any one of those angles had error the solution changes slightly. It's hard to testify to the exact location of something with fuzzy edges.

To answer your question, I believe the tank is "outside" of the 50m limit, only due to "significant figures". It is probably even outside of "50.0m" limits. Proving it to a court is a different matter.

PS - I would have observed at least three separate and distinct points on the tank (rusty bolts, etc.) from each of the two instrument locations to provide some sort of precision.

It'd be better if he had shot both edges from B. That would help confirm the diameter. Then you could bisect the observations from A and those from B. The intersection of those bisected angles would reveal the location of the center of the tank. I don't see how you can solve it from this unless I'm missing something.

I agree.

You would have to come up with a diameter of the tank (I get 7.67m +-), cheating with CAD.

From that point it can be calculated by hand. But when in court the owner states that the diameter is wrong so you are shot down.

I like that one. I'll try to figure it out when I get a chance.

If this is a practical problem, I agree with Paden that some redundancy of measurement is absolutely required to have any confidence in the answer. Not as a matter of getting a precise answer mathematically, but as a matter of detecting sensitivity to measurement errors and detecting blunders.

The edges of the tank, if it can be assumed circular, can probably be sighted precisely enough but the geometry may multiply the observational error considerably.

As a computational puzzle, my tip is to assume the tank has a radius of one arbitrary unit and work the problem from there. At the end you find out how many units long the baseline is, and convert your assumed units to meters.

It's a computational puzzle not an exercise in practical surveying.

> It's a computational puzzle not an exercise in practical surveying.

Yeah...sometimes we get all caught up in proper survey techniques instead of enjoying a good puzzle. I'm surprised that someone didn't mention a reflectorless edm....;-)

It's a computational puzzle not an exercise in practical surveying.

respectfully disagree, when regulators and courts are mentioned to surveyors, it then becomes practical 😉

> Regulations state that "no part of the structure is permitted within 50m of a public road".

Local governments can all be different. From this regulation, it sounds like the tank must be 50m from the right of way, not the centerline. Not having checked the computations,it looks like this could make a difference.

I assume that to calculate this you need to extend the line from A to the right side of the tank to intersect the line from B, then calculate that triangle and the horizontal curve formed by the two tangents and PI. Haven't had time to try to work through it, am I on the right track?

> It's a computational puzzle not an exercise in practical surveying.
>
> respectfully disagree, when regulators and courts are mentioned to surveyors, it then becomes practical 😉

Stating in court that measurement accuracy suffers, as others have stated, without direct access may become the required path. When the court gets involved the lack of access may go away and the entire puzzle changes.

So after sketching it out... and computing the angles from azimuths
The radius of the tank (not diameter) is 7.67m.

Tan(A1-A2)*b*SinC=(tank radius, either triangle works.

Good fun with a memory test of the law of sines.

Off the top of my head, I'm seeing an algebraic equation with the radius of the tank is a common variable for all three legs. Each leg being the the adjacent side of the Tan function of opposite/adjacent. Beyond that, I got nuthin.

It's actually a simple brg-brg intersect...

After you intersect the transit lines turned from the road...I used the law of cosines on that part.

Curve A-B and B-C share the same radius. D/2 intersects provide the radius point.

disclaimer: In checking this I discovered a few centimeters of error. One or more of the angles has to bear some 'fluff'. The tank calcs 49.625 m at its closest proximity to the RW.

> It's a computational puzzle not an exercise in practical surveying.
>
> respectfully disagree, when regulators and courts are mentioned to surveyors, it then becomes practical 😉

Regulators and courts = practical....

hmmm....I'll have to remember that one.o.O

It's actually a simple brg-brg intersect...

How can azimuths have any "fluff"?

I didn't see the obvious bearing/bearing for the radius point though. That is still law of sines isn't it?

It's actually a simple brg-brg intersect...

Extend the outer sights to intersect then bisect this intersection angle and the two sights from A to meet at the tank centre.
After that it's just sine rule and trig.

Paden has the right answer but did it a different way.

Here's another.

A triangular shaped area on the corner of North and East Road is to be acquired by the road authority as shown in the diagram.
The circular arc AC starts at the tangent point A in North Road 88.44m west of the intersection and cuts the East Road boundary at C 27.52m from the corner.

Using geometry and trigonometry find:-
1. Radius, Arc Angle, Arc Length, Chord Bearing and Distance of the arc AC.
2. The area of land ABC cut off by the arc AC.

I worked that one.
Radius=153.5, Arc-length=93.02, Chord azimuth 123º24'25", Chord dist = 91.58, area 784.8 meters-squared.

If I didn't get any early answers wrong making all the rest wrong anyway.

I got the distance of the long-chord by the law of cosines (on the triangle) since the opposite angle wasn't 90, and a bearing by the law of sines. For the area, I got the area of the circle-segment, and subtracted the area of the triangle made by using the long-chord.