Measuring a Distance with an Object in the Way
A Surveying Problem
Refer to the picture above. Here’s the situation: We are given a line and we are to determine a distance, X, from the Observer to the Survey Point. There is one problem though, there is a tree in the way, and the distance we are to measure goes directly through the tree.
Perhaps we can estimate the distance. Pick a point where the line of sight is not blocked. Let this new point be the Auxiliary Point and let C be the distance between the Observer and the Auxiliary Point.
We can also measure the distance between the Auxiliary Point and Survey Point, name this distance H. We can use the “right triangle” created to find the original distance X.
By the Pythagorean Theorem, we have:
X^2 + H^2 = C^2
X^2 = C^2 – H^2
X = √(C^2 – H^2) (only positive roots make geometric sense)
What if the calculator device does not have the square root function? Or if you are working by hand? In the 1938 book, The Principles and Practice of Surveying, Volume I (see the Source section), we can estimate the length of X without the need of taking a square root. Here is the derivation:
X^2 + H^2 = C^2
H^2 = C^2 – X^2
Observe that for any t, s: t^2 – s^2 = (t – s) × (t + s), and:
H^2 = (C – X) × (C + X)
If H is small, then C is close to X. Assume that H is small. As a result, C ≈ X. This derivation applies this approximation only to the term C + X:
H^2 ≈ (C – X) × (C + C)
Solve for X:
H^2 ≈ (C – X) × 2 × C
H^2 ÷ (2 × C) ≈ C – X
X ≈ C - H^2 ÷ (2 × C)
How good is the estimated formula?
Testing the Approximation
I used the HP 15C Collector’s Edition for testing the approximation formula. The code used to compare the approximate to the actual result:
Line #; Key Code; Key
Stack: y: hypotenuse (C), x: side length (H); (y > x)
01; 42, 21, 11; LBL A // approximate calculation
02; __, 44, _1; STO 1
03; __, __, 34; x<>y
04; __, 44, _2; STO 2
05; __, 45, _2; RCL 2
06; __, 45, _1; RCL 1
07; __, 43, 11; x^2
08; __, __, _2; 2
09; __, __, 10; ÷
10; __, 45, _2; RCL 2
11; __, __, 10; ÷
12; __, __, 30; -
13; __, 44, _3; STO 3
14; __, __, 31; R/S
15; __, 45, _2; RCL 2 // actual calculation
16; __, 43, 11; x^2
17; __, 45, _1; RCL 1
18; __, 43, 11; x^2
19; __, __, 30; -
20; __, __, 11; √
21; __, 44, _4; STO 4
22; __, __, 31; R/S
23; __, 45, _3; RCL 3 // absolute error
24; __, 45, _4; RCL 4
25; __, __, 30; -
26; __, 43, 16; ABS
27; __, 43, 32; RTN
(27 steps, 34 bytes)
Sample Data (rounded to 7 digits)
C |
H |
Approx. |
Actual |
Abs. Error |
5 |
1 |
4.9 |
4.8989795 |
0.0010205 |
6 |
1 |
5.9166667 |
5.9160798 |
0.0005869 |
7 |
1 |
6.2985174 |
6.9282032 |
0.0003682 |
8 |
1 |
7.9375 |
7.9372539 |
0.0002461 |
9 |
1 |
8.9444444 |
8.9442719 |
0.0001725 |
10 |
1 |
9.95 |
9.9498744 |
0.0001256 |
5 |
2 |
4.6 |
4.5825757 |
0.0174243 |
6 |
2 |
5.6666667 |
5.6556842 |
0.0098124 |
7 |
2 |
6.7142857 |
6.7082039 |
0.0060818 |
8 |
2 |
7.75 |
7.7459667 |
0.0040333 |
9 |
2 |
8.7777778 |
8.7749644 |
0.0028134 |
10 |
2 |
9.8 |
9.7979590 |
0.0020410 |
5 |
3 |
4.1 |
4 |
0.1 |
6 |
3 |
5.25 |
5.1961524 |
0.0538476 |
7 |
3 |
6.3571429 |
6.3245553 |
0.0325875 |
8 |
3 |
7.4375 |
7.4161985 |
0.0213015 |
9 |
3 |
8.5 |
8.4852814 |
0.0147186 |
10 |
3 |
9.55 |
9.5393920 |
0.0106080 |
20 |
19 |
10.975 |
6.2449980 |
4.7300020 |
20 |
1 |
19.975 |
19.9749844 |
0.0000156 |
50 |
49 |
25.99 |
9.9498744 |
16.0401256 |
50 |
1 |
49.99 |
49.9899990 |
0.0000010 |
From the data above, it seems that the greater the difference between C and H, in general, the better the approximation.
Source
Breed, Charles B. and George L. Hosmer. The Principles and Practice of Surveying: Volume I. Elementary Surveying. 7th Edition. John Wiley & sons, Inc. London, Chapman & Hall, Limited. 1938. pg. 14
Eddie
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