> You better record an affidavit right away!
Either that or build a fence to show "acquiescence" later.
Point 12.
"Maybe 12 is the PI of the entire arc, just guessing, though." Dave Karoly Previously
His deduction appears to be the only logical response regarding the questionable Point 12.
The only way that I've been able to derive an answer was to plot a 32' offset from the perimeter curve, plot a 135' line and move it to a point in the arc of the 32' offset curve and get it close to the 40' minimum for the NW corner of the building. The 40' Min. is an arbitrary dimension to determine the building corner. The 32' radial is rhe controlling dimension for locating the NE corner of the building. There is a lack of any data to really use for calculation purposes.
I will scan my plot and post it a little later.
I could have tweaked this a little, but the 40.04' meets the 40' Minimum.
>
Except you can get an exact solution without CAD (exact in the sense of Pt. 8 being exactly 40.00 ft. distance from 1-6 and Pt.9 being exactly 32.00 ft. distant from Arc 2-4 with Center at Pt.3.
You can easily solve the distance that a line parallel to 1-6 would need to be to pass exactly through Pt.9. Likewise, you can solve the distance that Pt.3 is from 1-6. Knowing those, the rest of the solution falls into place.
The CAD equivalent would be to draw rectangle 8-9-10-11 with Pt.8 lying at any point on a line parallel to and 40.00 ft. distant from 1-6. Then draw a line parallel to 1-6 through Pt. 9 and find the point at which it intersects a curve concentric with Arc 2-4 and 32.00 ft. distant from it. Shift 8-9-10-11 so that Pt.9 coincides with that point of intersection. You'll note that Pt. 8 remains exactly 40.00 ft. distant from 1-6.
I agree with 47.00 ft. I didn't work it with trig formulas and calculator, but I think if I was motivated, I could. I used my LS software to fit an answer.
Did you find a shortcut way to get that value, or did you have to solve for most of the rest of the figure first?
Dumb Question: I'm not familiar with the coordinate over coordinate notation at point 1. Where is that convention defined, and which is N and which E?
Kent:
Yep. Using the 40' Minimum at the Point 1 & Point 8 corner area you have a parallelogram with all 4 sides being equal, which will give both a 40' offset from the North line and the West Line. Then Point 9 can be determined using a bearing distance intersection.
Except that point 8 is only 39.971 from line 1-2.
Dave:
How come? Rectangular/Polar using the angle 26°35'54" and a hypotenuse distance of 44.734' gives a perpendicular or h distance of 39.99968 footies.
But! Using the parallelogram dimensions to fix Points 13 & 8 and doing a bearing-distance intersection for Point 9 results on a line length 8-9 of 135.0868'.
Nothing in the problem says that point 8 is a given distance from the line 1-2. It just happens that from the radii specified over east of it, the distance works out close to 40 ft. I got 39.971 ft also.
If there is a setback requirement on that side, they forgot to tell us about it, but even so, 40 ft wouldn't be consistent with the specified distance 9-7.
The angle 26°35'54" is 13-8-(NW perpendicular point I call 14) and I don't find the same angle anywhere else I've looked. Angle 14-8-1 is not the same angle at 31 40 0 and 1-8-(North perpendicular) is 31 44 6.
What I get from my plotted drawing from point 8 to the line 1-2 is 39.985'.
From all indications the 40' Min. is a floater distance from the Westerly Line with the 32' Radial being the governing positon of Point 9 and Point 8 having a Min. offset from line 1-6 of 40'. Slightly more than 40' appears to be no biggie. Nothing really says that the Min. offset from line 1-2 needs to be a Min. of 40'.
You seem to be going about it backwards.
First, get coordinates for points 2,3,4,5,&6.
Hold the 40 foot distance as fixed, not variable.
The hypoteneuse is then 44.7344, which added to 135.00 gives a perpendicular distance from point 6 to a line through point 9 of 160.7126.
From the foot of the perpendicular (N 4957.4299, E 5119.5902)do a bearing-distance of North 26°21'47" East and a distance of 108 from point 3. That will give you the radial bearing to point 9. The rest is simple traverse.
Hi Chas, I believe your pt 1 Easting shows 5050.0649,should be 5050.0549. Possible sight error on my part,thanks.
Mike Burkes
626-833-1521
I cheated. Used AutoCad. Picked a spot on one offset and checked to see what the excess/shortage was at hitting the other offset. Several iterations later the excess/shortage was less than 0.01.
Way more math involved in doing it the old way than I care to do anymore. Forty years or more ago I would have enjoyed doing it the hard way just to prove to myself that I could do it. Now I have to eat a pint of prunes every day because I just don't give a cr@p.
5050.0649 Easting is what is on the original drawing that was posted by Mike, or rather posted by Kent. Which also is on the drawing I downloaded from the link in the original post.
Hi Chas 5050.0649 most certainly is the discrepancy was the original posting 5 yrs ago sorry about that.
Mike Burkes
Except that if we allow that distance to be more than 40, then we can all come up with answers that would work but are different. If we assume it is 40.000 then we can compare answers. Usually a test question needs to have a unique answer.
Hi Bill93,great question according to my surveying books usually North is listed first akin to Latitude and I would assume coord over coord notation is utilized to save space. Anyone else have info regarding that? If I find additional info I will post.
Hi folks, amidst all the problem's lack of info here is contact data if one wishes to notify them:
www.scsurveyjac.org and email;[email protected]. Phone nos are included.