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Need to calculate length in metres

Posted on 2014-04-14
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Last Modified: 2014-04-22
I have a plan of a block of land as per the attachment. I need to calculate length X.
The Angles of the lines are degrees from the True North/South line.
Distances are in metres.
I don't need to see the working although it would be useful if I could check the calculations for accuracy.
The reason I need this is because the survey peg at the bottom left is missing and I need to triangulate to work out roughly where it should be. I don't want to pay AU$600+ for a surveyor as the precise location is not all that important.
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Question by:akb
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by:akb
ID: 39998714
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by:Dan Craciun
ID: 39998769
The bottom and top sides are almost parallel, so you can use Pitagora's theorem to calculate x:

x = sqrt( (34.56 - 24.48)^2 + 21.01^2 ) = 23.30m

It's not exact, but should be fairly close.

HTH,
Dan
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Expert Comment

by:d-glitch
ID: 39998788
You can convert polar coordinates (length, angle) to cartesian (x, y) coordinate.

You can find the coordinates of all the corners, and then the unknown side.

This would be exact.
Plan-for-ExEx.jpg
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by:Dan Craciun
ID: 39998797
Well, seems I was off by 11.05%. On the plus side, it only took 30 seconds :)
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by:d-glitch
ID: 39998805
You could make a better scale drawing to check the calculations with a big piece of graph paper and a protractor.

You could also use a computer drafting program to do the calculations.

Your figure does not seem to be to scale.  The whole drawing looks like it needs to be rotated clockwise by 45 degrees or so.
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by:awking00
ID: 39999412
Unfortunately, I don't have a means of graphically demonstrating the following calculations, so I'll try and express them in English. If we take the original drawing and rotate it clockwise 59 degrees and 27 minutes, it then becomes perpendicular (i.e. 90degrees) to true north. I now call this the bottom side. The 34.45 m side (now call it the right side) becomes 183 degrees and 47 minutes relative to true north or 93 degrees and 47 minutes relative to the bottom side. If we then extend the bottom line and also extend a line from the point where side x meets the now right side perpendicular (i.e. at a right angle) to the extended bottom line, we can determine the distance from the end of the original bottom line to the point we just created by applying the formula sin(of 3 degrees and 47 minutes = .0696) X 34.45 m to get a length of 2.28 m making our extended bottom line now 24.29 m. We can also compute the length of the new perpendicular line using the formula cos(3 degrees, 47 minutes) = .99840 34.45 to get a length of 34.4 m. Likewise, the line that is now on the left side (adding the 59 degrees and 27 minutes) becomes 179 degrees and 40 minutes relative to true north or virtually perpendicular to our extended line. So we now have, in essence a bottom line of 24.29 m with a perpendicular line of 24.48 m at one end and a perpendicular line of 34.4 m at the other. So, if we take away the 24.48 X 23.29 rectangle we are left with a right triangle at the top with one side of 23.29 m and one side of 9.92 m (34.4 - 24.48) and our hypotenuse x. Using the pythagorean theorem we determine x to be the square root of (23.29^2 + 9.92^2) or 25.3 m. Given the original drawing, I would have expected the length to be greater than the opposite side of 21.01 m, which it is.
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by:awking00
ID: 39999414
Sorry for the typos. Where I've shown the extended bottom line to be 24.29 m, it should be 23.29 m.
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by:Dan Craciun
ID: 39999426
@awking00: So basically you reached my result, using 1788 characters to explain it :)
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by:awking00
ID: 39999512
I guess if you consider 23.3 to be basically equivalent to 25.3, I suppose I did. :-)
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by:d-glitch
ID: 39999518
Rethinking ...  The Angles of the lines are degrees from the True North/South line.
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by:d-glitch
ID: 39999696
Your drawing is not to scale and your angles are not consistent.

If you are specifying from true North/South line, then is +5 deg East or West?
Is 0 deg North or South?

The sign of the angle depends on which end of the line you are using as the origin.

Converting your angles to standard polar form (East = 0 Deg), and using Excel to do the polar to Cartesian conversions, I get a figure that looks like yours.

I find the length of the unknown side to be 25.767
Plan-for-ExEx-2.pdf
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Accepted Solution

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d-glitch earned 250 total points
ID: 40000005
I found a 66' (1.1 deg) error.

Now I claim the length of the unknown side is  25.610
Plan-for-ExEx-3.pdf
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by:Dan Craciun
ID: 40000018
the precise location is not all that important
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by:awking00
ID: 40000104
d-glitch,
I agree with your method but I think the angle you show as 90.67 degrees should really be 89.33 degrees (119 degrees and 53 minutes minus 30 degrees and 33 minutes), which should make you end result be slightly smaller.
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by:akb
ID: 40000153
Wow! Thanks for the help. I thought my plan was reasonably to scale - maybe something got screwed up when I converted it from Corel Draw to JPG.
I had estimated it as close to 25m after drawing a rough scale drawing so I think awking00 may be close (or spot on). Seems this is confirmed by d-glitch.
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by:d-glitch
ID: 40000268
>> awking00

I agree that the angle between the 24.48 and the 21.01 meter sides should be 89.33 deg.
(I have the value for the complementary angle.)

But I did not use that value to calculate the result.
I have just added up the three known vectors.

One more time, in different order

Let the left most point be (0, 0)

Move 24.48 < +119.88  ==>  21.22 E  and  12.12 S

Move 21.01 < +30.55   ==>  10.68 E  and  18.09 N

We are at                  31.90 E  and   5.90 N

Move 34.56 < +304.33  ==>  28.54 W  and  19.49 N    The sign change here 
  or 34.56 <  -55.67                                is important.
=================================================
Now we are at               3.36 W  and  25.39 N

The length of this vector is the unknown side ==>  25.61m as before.

Open in new window

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Assisted Solution

by:awking00
awking00 earned 250 total points
ID: 40001429
d-glitch,
Actually, I was wrong. I realized about 3:00 this morning that adding the 59 degrees and 27 minutes to the left side meant the angle against true north would be 179 degrees and 40 minutes. Picturing that would mean that line would have a slight slant from upper left to lower right, meaning the angle created with my bottom line would be slightly greater than 90 degrees (90.67 as you stated). It also means that the bottom line would increase slightly by .285 m after plugging the new value into the equations for determining lengths of the sides of a right triangle knowing the length of the hypotenuse and two angles. The re-calculation of the length of x now shows a value of 25.58 m, roughtly equivalent to your determination of 25.61. I suspect the difference is due to the number of decimal places carried or any rounding that occurred. At any rate, I think it shows that both methods are valid. This was actually a fun exercise.

akb,
Hopefully we have given you enough information that you can feel relatively confident that side x is close to 25.6 m and should be adequate for your purposes. Thanks for the fun exercise.
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Expert Comment

by:d-glitch
ID: 40001978
I'm sure my method and awking00's answer are correct (so his method must be too).

If I keep my grubby hands off the data, and let the computer do all the processing,  I get
x = 25.5748 for the answer to the original question.

I have cleaned up my Excel Spreadsheet and I will attach it here.  It may be a useful toy for surveying and navigation calculations.  

It will calculate and plot up to ten linked (Dist, Bearing) segments, starting and finishing at the origin.  It also keeps track of the current distance from the origin.

Angles/Bearings must be in decimal degrees from due North:
  0.00 is due North   90.00 is due East   180.00 is due South   270.00 is due West

Excel will do the conversion from minutes to decimal if you enter the values as
                                          =DDD+MM/60

And if you find one your angles is backwards, try adding 180 (as I had to do for Segment 3
in the example).    =124+20/60+180
Surveying-and-Navigation-for-ExE.xls
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Expert Comment

by:awking00
ID: 40002144
Our final calculations were a grand total of .0029 meters different or about 1/10 of an inch which, I imagine, should be a lot smaller than the diameter of the peg akb needs to replace.
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Expert Comment

by:d-glitch
ID: 40002388
Even little tiny errors bother me, especially when I am the one making them.

If you look closely at the spreadsheet in my last post you will find not one but two data entry errors.
           =119+56/60      It should be 53
           =124+20/66      It should be 60

Two errors in six items is like a 50% error rate.
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Author Comment

by:akb
ID: 40009839
I have just finished recalculating it myself by using a triangle calculator at http://ostermiller.org/calc/triangle.html and I also came up with 25.61 metres. Thanks to d-glitch and awking00 for all your help and for the running commentary.
Hopefully I can do a little digging and maybe find the peg.
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Author Comment

by:akb
ID: 40009841
Whoops. I actually meant to split the points but somehow divided them by three. Is there some way I can fix that?
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by:d-glitch
ID: 40009874
You can hit the Request Attention Button and ask a moderator for help.
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Author Closing Comment

by:akb
ID: 40016243
Hopefully I have the point allocation right this time.
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