This segment of calculating distances by hand will work with perspective and projected geometry. In particular this segment will discuss obtaining accurate distance calculations from the use of photographs. For these calculations to work, a few control points are needed. These control points will be a known distance apart from each other. As a working example, let’s say we want to know how high off the ground a point is off the ground. For this example, we will use a flag pole with our inquiry being how high off the ground is the bottom of the flag.
For this example, our control points will be the pole’s access port. We will assume this has been field measured (RED) and that this port is 36” above ground level and 9” in height. We will need to calculate the remaining distance between the top of the access port and the bottom of the flag. A
In order to achieve the calculation needed to provide us with the missing dimension, we will gather some data from the picture, specifically the same distances measured in pixels (BLUE) where 36″ = 55px, 9″ =15px and X = 240px.
The Cross Ratio Formula
AC · BD = A’C’ · B’D’
BC · AD B’C’ · A’D’
We will utilize a practical concept in projective geometry known as cross-ratios to accurately provide the unknown distance in inches (X) utilizing the below equation.
The terms (AC, BD, etc) refer to the photo distances
between points in question (pA, pB, etc) [Blue]
The terms (A’C’, B’D’, etc) refer to the real distances
between points in question (pA, pB, etc) [Red]
Photo (pixels) Real (inches)
70 · 255 = 45 · ( 9 + X)
15 · 310 9 · (45 + X)
17,850 = 45 · ( 9 + X)
4,650 9 · (45 + X)
3.8387 = 45 · ( 9 + X)
9 · (45 + X)
This simplifies to;
X = 1149.677145
10.451619
Or X = 109.999991” (X = 9’-2”)
For this example, the bottom of the flag would be 12’-11” above
ground level.
Below is the expanded mathematics.
55px 15px 240px Referencing the photo above;
dA dB dC pA = ground level
|--------+----+-----------------------| pB = bottom of port
pA pB pC pD pC = top of port
dA’ dB’ dC’ pD = bottom of flag
36in 9in X in
AC · BD = A'C' · B'D'
BC · AD B’C’ · A’D’
AC = 70px [ dA+dB ] A’C’ = 45in
BD = 255px [ dB+dC ] B’D’ = X + 9in
BC = 15px [ dB ] B’C’ = 9in
AD = 310px [dA+dB+dC] A’D’ = X + 45in
Using the information diagram above; the long hand math follows below.
First, we will calculate the cross ratio for the photo or known quantities. We only need a ratio so we do not need to worry over the unit of measure (pixels, inches, etc.)
(55 + 15)· (15 + 240) => 70 · 255 => 17,850 => 3.83870967741
15 · (55 +15 + 240) 15 · 310 4,650
Next, we will analyze the real-world quantities.
(36 + 9) · (9 + X) => 45 · (9 + X)
9 · (36 + 9 + X) 9 · (45 + X)
Since the cross ratios of both units of measure remain the same, we can now say;
45 · (9 + X) = 3.83870967741
9 · (45 + X)
A) Firstly, we will simplify the problem by removing the fraction.
We will multiply both sides by the denominator. [ 9 * (45 + X) ]
Side A:
45 · (9 + X) · (9 · (45 + X))
9 · (45 + X)
= 45 · (9 + X)
Side B:
3.838709 · (9 · (45 + X))
= 34.548381 · (45 + X)
We now have the equation; 45 · (9 + X) = 34.548381 · (45 + X)
B) Now we can expand both sides of the equation.
Side A:
45 · (9 + X)
= ((45 · 9) + (45 · X))
= 405 + 45X
Side B:
34.548381 · (45 + X)
= 34.548381 · ((34.548381 · 45) + (34.548381 · X))
= 1554.677145 + 34.548381X
We now have the equation; 405 + 45X = 1554.677145 + 34.548381X
C) We will now remove the known term from the left side of the equation.
Subtracting this quantity from both sides of the equation. [ 405 ]
Side A:
405 + 45X – 405
= 45X
Side B:
1554.677145 + 34.548381X – 405
= 1149.677145 + 34.548381X
We now have the equation; 45X = 1149.677145 + 34.548381X
D) We will now remove the multiplier to X from the right side of the equation.
Subtract the X quantity from both sides of the equation. [ 34.548381X ]
Side A:
45X - 34.548381X
= 10.451619X
Side B:
1149.677145 + 34.548381X - 34.548381X
= 1149.677145
We now have the equation; 10.451619X = 1149.677145
E) We can now simplify the equation by dividing each quantity by the multiplier of X.
We are left with our solution in the form of; X = Missing Quantity.
Divide each side by the X multiplier. [ 10.451619 ]
Side A:
10.451619X = X
10.451619
Side B:
1149.677145 = 109.999910
10.451619
We now have the equation; X = 109.999910”
F) We now have the solution for dC’ ( X ) for real-world distances.
This amount is in the original unit of measure for the real world distances.
Round this number to the nearest whole number for ease. [ 110″]
110in/12in = 9.166659ft => 9ft + 0.166659ft
12in · 0.166659 = 1.999in
We now have the simplified answer; dC’ & X = 9ft 2in
G) We can now say that the total real-world distance;
dA’ + dB’ + dC’ => 36in + 9in + 110in = 155in
155in / 12in
= 12.916666ft
= 12ft + 0.916666ft
=> 12in · 0.916666 = 11in
H) The full answer to our inquiry regarding the height of point D (
12ft 11in above ground