# Civil project Penn Foster Plate 1

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Hello,

I would like some help with starting this project. I finished the title block and border. I read halfway through the instructions for plate 1.

The bullet point 1: " Therefore, the origin (or point 0,0) for all surveys is at that intersection, and any X and Y coordinates tell the distance in feet from the intersection. If, for example, a benchmark is at point 11675,39532, the benchmark is 2.21 miles east of the intersection and 7.49 miles north of the intersection" makes me feel like I need to set the upper and lower limits to 0,0. Also how does a benchmark with points at 11675,39532, be 2.21 miles east of the intersection and 7.49 miles north of the intersection? Where do they base these calculations off? It was easier to understand the booklet info and I thought i would find this project easier to understand than the confusing structural project I just finished. I got a 100% for the online test which I knew all answers to without much reference since I read it well and understood it. Why are they making the project a new study topic and not based on the information we just learned from the book is beyond my understanding.

Also bullet point 2: "A benchmark (BM 312) exists at the northwest corner of the survey area. BM 312 has an elevation of 88.9 feet, and its coordinates are 521662,44895." So is this what I have to do? : I set the limits to 0,0 in upper and lower corners. Then I measure 88.9 ft from x axis and draw a line for the BM312. What are these coordinates for if the elevations helped me draw the line above the x axis?

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Okay, I drew the Benchmark. I am trying to form the array now. Closing for the night.

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A benchmark is a place that any surveyor can locate. We use a benchmark as the origin for a project's coordinates, the same way AutoCAD uses (0,0,0) as an origin; everything is located in relation to that point. In the US, the Geodetic Survey has a system of benchmarks that are often used to locate project benchmarks (or to be the benchmark itself).

Your project is measuring distances in feet, so 11,675 feet is 2.21 miles (11675/5280). In this case the "intersection" (whatever that is) is your origin, so all your coordinates are measured from that point. If you know about Universal Coordinate Systems (UCS) in AutoCAD, you can redefine the origin to match your project benchmark, your geodetic benchmark, or anything else that gives you the results you need. Typically, each state has its own "origin," so every location in the state can be expressed in State Plane Coordinates; this is handy if (for instance) you have data from several different surveys and want to combine them.

The point BM312 is a project benchmark. The surveyor knows where BM312 is in relation to the first benchmark. You can then adjust the points from the survey to express them in relation to the first benchmark, the project benchmark, or any other origin.

Finally, "elevation" refers to the Z coordinate. A surveyor also describes the elevation of points in relation to the benchmark.

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Posted (edited)

The benchmark should be drawn at the coordinates given in the instructions (521662,44895).  Your drawing limits should be set at 521500,44000 for the lower left hand corner and 523000, 45100 for the upper right hand corner.  The grid starts at the benchmark and goes left to right, then top to bottom.  The student is then asked to plot the survey points on the grid.  The survey points will be indicated by a block constructed of two lines and shaped like the letter "X".  Next to each survey point the student will indicate the elevation (i.e. - the height of the land at that particular point).  After doing that for all the points given the student then will interpolate the contours between points based upon the survey elevations.  The contour "interval" will be 10 feet.  Contours will start at an elevation of 10 feet on the left side of the drawing and progress to the right ending in a contour of 200 (in 10 foot intervals...10, 20, 30........all the way to 200).  BTW...your North arrow should go in the upper left hand corner of the drawing and not at the bottom of the drawing.  The scale bar you are asked to draw should go towards the bottom right hand corner of the drawing.  Your also asked to create a "tree" block much like you did in the Oleson Village project.

Edited by ReMark

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Thank you

This clears a lot of things up for me to start understanding the beginning of this project. I just completed the Grid and now interpolating contours. When I try to interpolate how doI use the points command to get the countour point to go to the exact point? It just pots a point in the vicinity for now and not on the exact location. Why are the instructions asking the student to use the point command? I tried using osnap in different ways, and it still doesnt help to allign the interpolated contour point to its exact location, sinceupon zooming close , i can still see it not aligned well. What am i doing wrong?

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You cannot accurately snap to a POINT if you have not enabled NODE under OSNAPS.

Personally, I wouldn't use the POINT command.  I would first do the required math.  Next I would insert my "X" block at the intersection of two grid lines then I would move the block the required distance which can be input via your keyboard (make sure Orthomode is enabled).  That will be the location of your contour interval.  When you have all the spot elevations plotted you can draw a polyline from point-to-point followed by using the PEDIT command to FIT a curve to the polyline.

Edited by ReMark

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I drew the polylines. The next part ofdrawing cedar park drive is confusing. Is it an arc with a radius and a centre point? If so, when i try drawing an arc , the centre point positions the arc very far outside the grid. I used the center point coordinates from the following bullet point in pdf (Also after i enter the center point, it asks me where the starting point of arc is, i dont know what to enter there, and it has not asked me to enter radius yet. So I am stuck at this point.

"It’s at the bottom of a hillside. The centerline radius of Cedar Park Drive is 7115.37′, and the center point is at located at 528257.05,47668.89. The distance from the centerline to the curb on each side is 25′, and the distance from the centerline to the right-of-way on each side is 40′."

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OK..I found the instructions you quoted above.  Yes, the centerpoint of Cedar Park Drive (a curve with a very large radius) is far outside the limits of the drawing (up and to the right) which is not all that unusual.  Cedar Park Drive will cut across the left side of your drawing from top to bottom.  Locate the coordinates given in your instructions and use them as the centerpoint for a CIRCLE with a radius of 7115.37.  Where the circle crosses the top and bottom grid lines use the TRIM command to remove the part of the circle that is outside of the grid lines then OFFSET the portion of the circle that remains a distance of 40 units to both sides of the centerline.  That will define the right-of-way.  Done.

This is what it looks like before the Circle trimmed.

Edited by ReMark

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On 8/5/2020 at 6:48 AM, ReMark said:

OK..I found the instructions you quoted above.  Yes, the centerpoint of Cedar Park Drive (a curve with a very large radius) is far outside the limits of the drawing (up and to the right) which is not all that unusual.  Cedar Park Drive will cut across the left side of your drawing from top to bottom.  Locate the coordinates given in your instructions and use them as the centerpoint for a CIRCLE with a radius of 7115.37.  Where the circle crosses the top and bottom grid lines use the TRIM command to remove the part of the circle that is outside of the grid lines then OFFSET the portion of the circle that remains a distance of 40 units to both sides of the centerline.  That will define the right-of-way.  Done.

Thanks a lot. That helped. I am now where I am drawing the line from the start of the cul desac- in the counterclockwise . I am asked me to mark delta. I calculated it using the formula:

L = (3.14 x R x delta) / (180)

144.32" = (3.14 x 35 x delta) / 180

delta = (144.32 x 180)/(3.14 x 35) = 236.37

Is this correct?

For reference I am on this step in pdf:

"4. From where you left off, draw a 144.32′ segment in a counterclockwise direction around the cul-desac. Show the delta angle, the length, and the radius of this curve. You would usually show this information on the inside of the curve, but, in this case, show it on the outside."

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Delta = (180L)/(3.1415R)

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Is L= 144.32 and R =35?

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Is L= 144.32' and R =35'?

Yes, it is.

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