Civil drafting school project help

[warpath37]

I was too far along to try the DWGUNITS command. i just re-interpolated the entire thing and it looks correct compared to other screenshots on this thread. i am having trouble finding out the "boundary" for the cut and fill of the driveway (plate 2). i understand the idea of filling in elevations lower than 180 and cutting anything above it, but i dont know how/where to put the boundary nor do i understand the angle of repose. i've been reading/re-reading my study unit and still cant figure it out. any help? btw i have searched thru this entire thread and couldnt (maybe i overlooked it) find the answer to my question. thanks

[ReMark]

Plot the existing grade.

 

Plot the proposed grade.

 

Where the proposed grade is above the existing grade that area will have to be filled. Where the proposed grade is below the existing grade that area would have to be cut. If the two grades match then no action is required.

 

Remember that your stationing will run across the bottom of your profile (i.e. - horizontally) while your elevations run up the side of your profile (from lower to higher).

[ReMark]

See the images contained in this thread ( posts #17 & #18 ).

 

http://www.cadtutor.net/forum/showthread.php?72818-Penn-Foster-Civil-Drafting-project-plate-3-and-4/page2

[SLW210]

I was too far along to try the DWGUNITS command.

 

You can never be "too far along to try the -DWGUNITS command".

[cfarr]

Folks,

I have read and reread all the threads about the spot elevations and how to calculate them. But I am stumped.

my example:

192.0-192.6=.6

130.00-192.6=62.6

62.6x50/.6=521

 

When I try to put the 521 point on the grid it goes past the benchmark.

 

Can you please assist me???

Thanks

Cfarr

[ReMark]

Figure out the difference in elevation between the lowest and highest spot elevation.

 

Divide this by the distance between the two spot elevations. This will give you the percentage of slope.

 

Calculate the difference in elevation between the new spot elevation you wish to plot and the lowest spot elevation.

 

Divide this by the percent slope previously calculated. This will result in the distance from the lowest elevation to your new spot elevation.

 

When you are dealing with elevations in the range of 130 to 192.6 a calculation that results in a number of 521 is completely out of range.

[cfarr]

I am at the contour elevation 130,140, and so forth.

 

I have been using the formula that was posted here.

 

I have been trying to do this for 2 days. I have tried both examples in the project booklet also.

 

Sorry, I am just stumped or I am not getting it.

 

Thanks Cfarr

[ReMark]

If you are between the 130 and 140 contours why would you think an answer of "521" would be correct and try to plot it?

 

Let's try this example.

 

You have two spot elevations. We'll call them the high spot and the low spot. The high spot = 54.0 and the low spot = 32.7. What is the difference in elevation between the two?

 

Subtract the low spot of 32.7 from the high spot of 54.0 which equals 21.3.

 

What is the distance between the high spot and the low spot? On our drawing we can either measure it using the Distance command or read it directly if the spot elevations are plotted on our grid which uses a square pattern of 50x50. Just for the sake of argument lets say the spot elevations fall on the grid and our distance then equals 50.

 

Divide the difference in elevation (21.3) by the distance (50) and what do you get? The answer is 0.426 which is the percent of grade.

 

With me so far? I hope so.

 

Now let's say we want to know, and plot, the 40 foot contour. 40 minus 32.7 (our low spot) = 7.3.

 

Divide 7.3 by the percent of grade of 0.426 and what do you get? The answer is 17.136 which we can round off to 17.14. This is the distance FROM the low spot you have to measure out.

 

Draw a circle with a radius of 17.14 using as your basepoint the location (on the grid) of your low spot as it's center. Where the circle crosses the grid between the low and high spot elevations is where your 40 foot contour will be located. Place some sort of line or block or whatever you want to use at this point (preferably something you can 'snap' to later on).

 

That's it basically in a nutshell.

 

Personally I like to always work from the lowest elevation towards the highest elevation although it can be done the opposite way. No matter which way you use the new elevation (40) should be located in the same spot between the two known elevations you started with.

 

Now, it's your turn. Go. Do. Now.

Edited April 22, 2013 by ReMark

[cfarr]

Thank you so very much Remark. I have been using this forum as a guide through all my studies at Penn Foster. This site and everybodies knowledge has been so helpful.

 

Thanks,

cfarr

[cfarr]

I have followed the directions given here step by step. My answer is still the same 521. what am I doing wrong??

 

Do I go left to right (90,100,110,120,130,140,150,160,170,180,190) or top to bottom (90,80,70,60,50,40,30,20,10)???

 

I have asked Penn Foster and still have not received an answer.

 

Thanks for all your help.

cfarr

[ReMark]

Could you have misplaced a decimal point and the distance be 5.21 feet?

 

What is the contour interval you are attempting to locate?

 

What are the two spot elevations it will fall between?

 

I assume the distance between the two spot elevations is 50 feet is that correct?

[cfarr]

ReMark,

I sent the problem in on my first thread.

yes you are correct in the 50 ft.

 

thanks

[ReMark]

ReMark,

I sent the problem in on my first thread.

yes you are correct in the 50 ft.

 

thanks

 

Humor me and post the answers to my questions in the order listed. Thank you.

 

I went looking for the first two spot elevations, mentioned in the post you refer to, of 192.0 and 192.6 and found them at the top of Plate 1. Trouble is there is no contour interval being asked for in that location that falls between those two points.

 

The contour interval for Plate 1 is 10 feet. On this particular drawing contours run from an elevation of 10 feet to an elevation of 200 (going from left to right on the drawing). There are no contours in the 500 foot range unless you are working on a drawing different than the one I am familiar with. You wouldn't be working on the civil "elective" project would you?

[cfarr]

okay

 

problem:

hi elev. 192.6 - low elev. 192.0 = .60

.60/50ft distance = .01

130.00 contour interval - low elev. 192.0 = -62.00

 

-62.00/ .01 = -6200.00

 

Does this help??

[ReMark]

Only to the point where you mention the 130 contour level then everything goes out the window.

 

Stop and think. Why would you attempt to locate a 130 contour if the two spot elevations you are working with at 192.6 and 192.0? The only contours between those two points are 192.1, 192.2, 192.3, 192.4 and 192.5 and none of those are on 10 foot intervals. 170, 180, 190, 200 would be an example of a range of contours having a 10 foot interval but they fall outside of your two spot elevations.

 

FYI - There is a 130 contour that runs through the drawing but it does not fall anywhere near your two spot elevations.

[ReMark]

See the white rectangle? That's where I found the two spot elevations you mention.

See the very thick, dashed, polyline? That's the 130 contour level. Note that it does NOT run between the two spot elevations. It's impossible.

 

Now I have to go for a walk. I'll check back later.

[Organic]

I am at the contour elevation 130,140, and so forth.

 

I have been using the formula that was posted here.

 

I have been trying to do this for 2 days. I have tried both examples in the project booklet also.

 

Sorry, I am just stumped or I am not getting it.

 

Thanks Cfarr

 

I don't have the pdf you are working from although it is simply linear interpolation.

 

Assume you are designing/constructing a new driveway. You know the existing elevations at either end of the driveway (top 100.25m, bottom 99.69m) which you have to connect into. You know the length of the driveway is to be 22m.

Calculate the grade (slope) of the driveway and the height of the finished surface a quarter of the way up the driveway.

 

100.25 - 99.9 = 0.56m total height difference

0.56m/22m * 100 = 2.55%

0.0255 * 5.5 (halfway) = 0.14m

=> FS height (a quarter of the way up the dwy from the bottom) = 100.04m

 

Obviously this example is trvial although you should be able to see the principle.

 

Of all the math I use this is the most frequent. On a side note the amount of math you have to take in an engineering degree is ridiculous almost. I understand why it was important to do so for the small percent who will use it, although personally I have very rarely used much of it.

[cfarr]

I would like to thank all the folks that helped me. I emailed p-f and still to this day have not received any help. If it was not for you folks helping I would still be stuck on the interpolating contour lines. I finally finished that part late last night.

 

thanks,

cfarr

[ReMark]

To add insult to injury you paid them. You got our help for free.

[PotGuy]

To add insult to injury you paid them. You got our help for free.

 

I think you should be on a payroll for P-F with the help and patience you give!