How Plot Digitizer’s Calibration Works
Plot Digitizer allows the user to calibrate an image using any three non-collinear points. The calibration is
actually a transformation of the image data between two reference frames. Call the reference frame of the
scanned image the S' frame and the reference frame of the computer monitor the S frame (see Figure 1 below).
To make the transformation completely general, consider the possibility that the scanned image may be rotated
( θ ≠ 0 ) and that the axes of the image may be non-orthogonal ( φ ≠ 90o ) .
S' frame (scanned image)
•P
y'
S frame (computer monitor)
φ
G
r'
x'
θ
y
G
r
G
O
x
Figure 1: The reference frames.
G
Now consider some point P. The position of this point in the S' frame is given by the position vector r ' where
G
r ' = x 'iˆ '+ y ' ˆj' . Writing the unit vectors î ' and ĵ' in terms of the S frame we get:
ˆi ' = cos(θ) ˆi + sin ( θ ) ˆj
ˆj' = cos(φ + θ) ˆi + sin ( φ + θ ) ˆj
G
This allows us to rewrite r ' in terms of the S frame unit vectors. This gives us:
G
r ' = [ x 'cos(θ) + y 'cos(φ + θ) ] ˆi + [ x 'sin(θ) + y 'sin(φ + θ) ] ˆj
G
G G
G
Notice from Figure 1 that the position of the point P in the S frame is given by r = O + kr ' where O is a vector
pointing to the origin of the S' frame, and k is a scale factor to transform the physical units of the scanned
image to the dimensionless pixels of the computer monitor.
Finally, we can write the equations for transforming the ( x ', y ') image coordinates of the point P to the
( x, y ) coordinates of the computer monitor.
x = O x + kx 'cos(θ) + ky 'cos(φ + θ)
y = O y + kx 'sin(θ) + ky 'sin(φ + θ)
Note that we have 2 equations with 6 unknowns ( O x , O y , cos(θ), cos(φ + θ),sin(θ),sin(φ + θ) ) . To solve for the
6 unknowns, we therefore must calibrate the image by selecting 3 points. This will give us 6 equations and 6
unknowns.
Writing out all 6 equations we get:
x1 = O x + kx1 'cos(θ) + ky1 'cos(φ + θ)
y1 = O y + kx1 'sin(θ) + ky1 'sin(φ + θ)
x 2 = O x + kx 2 'cos(θ) + ky 2 'cos(φ + θ)
y 2 = O y + kx 2 'sin(θ) + ky 2 'sin(φ + θ)
x 3 = O x + kx 3 'cos(θ) + ky3 'cos(φ + θ)
y3 = O y + kx 3 'sin(θ) + ky3 'sin(φ + θ)
The points ( x1 ', y1 ') , ( x 2 ', y 2 ') , and ( x 3 ', y3 ' ) are the physical values (entered by the user during the
calibration procedure) of the three calibration points. The points ( x1 , y1 ) , ( x 2 , y 2 ) , and ( x 3 , y3 ) are the pixel
coordinates of the mouse as recorded by the computer when the user clicks the point during the calibration.
To make solving this set of equations a bit less cumbersome, I performed a change of variable:
a = k cos(θ)
b = k cos(φ + θ)
c = Ox
d = k sin(θ)
e = k sin(φ + θ)
f = Oy
This gives:
x1 = x1 'a + y1 'b + c
y1 = x1 'd + y1 'e + f
x 2 = x 2 'a + y 2 ' b + c
y 2 = x 2 'd + y 2 'e + f
x 3 = x 3 'a + y3 ' b + c
y3 = x 3 'd + y3 'e + f
Writing the set of equations as a matrix equation ( Ax = B ) in terms of the redefined variables gives:
0 0 ⎤ ⎡ a ⎤ ⎡ x1 ⎤
⎡ x1 ' y1 ' 1 0
⎢ 0
0 0 x1 ' y1 ' 1 ⎥⎥ ⎢ b ⎥ ⎢⎢ y1 ⎥⎥
⎢
⎢ ⎥
⎢ x 2 ' y2 ' 1 0
0 0⎥ ⎢ c ⎥ ⎢ x 2 ⎥
⎢
⎥•⎢ ⎥ = ⎢ ⎥
0
0
0
x
'
y
'
1
2
2
⎢
⎥ ⎢d ⎥ ⎢ y 2 ⎥
⎢ x 3 ' y3 ' 1 0
0 0⎥ ⎢ e ⎥ ⎢ x 3 ⎥
⎢
⎥ ⎢ ⎥ ⎢ ⎥
0 0 x 3 ' y3 ' 1 ⎥⎦ ⎣ f ⎦ ⎣⎢ y3 ⎦⎥
⎣⎢ 0
Using Maple® to solve the set of equations, I finally get:
y1 ' x 3 − y1 ' x 2 + y3 ' x 2 − y 2 ' x 3 + y 2 ' x1 − y3 ' x1
⎡
⎤
⎡a ⎤
⎢
⎥
⎢b⎥
− x 1 ' x 3 − x 3 ' x 2 + x 2 ' x 3 + x 3 ' x1 + x1 ' x 2 − x 2 ' x1
⎢
⎥
⎢ ⎥
⎢ c ⎥ 1 ⎢ y1 ' x 3 ' x 2 − x1 y 2 ' x 3 '− y3 ' x1 ' x 2 + y3 ' x 2 x1 + x 3 y 2 ' x1 '− y1 ' x 2 ' x 3 ⎥
⎥
⎢ ⎥= ⎢
y3 ' y 2 + y 2 ' y1 + y1 ' y3 − y3 ' y1 − y 2 ' y3 − y1 ' y 2
⎥
⎢d ⎥ D ⎢
⎢
⎥
⎢e⎥
− x 2 ' y1 + x 2 ' y3 + x 3 ' y1 − y3 x1 '− y 2 x 3 '+ y 2 x1 '
⎢
⎥
⎢ ⎥
⎢⎣ − x 2 ' y1 ' y3 + x 2 ' y3 ' y1 − x 3 ' y 2 ' y1 + y3 y 2 ' x1 '− y 2 y3 ' x1 '+ x 3 ' y1 ' y 2 ⎥⎦
⎣f ⎦
where D = − y 2 ' x 3 '+ y 2 ' x1 '− y3 ' x1 '− y1 ' x 2 '+ y 2 ' x 2 '− y1 ' x 3 ' .
The variables a − f are calculated at the end of the calibration procedure when the user clicks the “Calibrate”
button in the calibration pop-up window.
Finally, to get the physical values ( x ', y ') of a clicked point ( x, y ) , I solve the pair of equations
x = x 'a + y 'b + c
y = x 'd + y 'e + f
for ( x ', y ') to get:
a(y − f ) − d(x − c)
ea − db
x − by '− c
x'=
a
y' =
These equations calculate ( x ', y ') as the mouse is moved in the calibrated image and the results are displayed in
the status bar at the bottom of the screen. When the user clicks a point to digitize, the value ( x ', y ') is copied to
the data table.
Although it is not necessary to know the values of φ and θ for Plot Digitizer to work, this information is given
to the user as a tool to assess the quality of the scanned image and its calibration.
( a ) and φ = tan ( e b ) − θ . To avoid division by zero errors, I used
From the definitions of a − f , θ = tan −1 d
−1
⎛
⎞
⎛
⎞
and tan −1 (x) = cos −1 ⎜ 1
to finally get:
the trig. identities tan −1 (x) = sin −1 ⎜ x
2 ⎟
2 ⎟
1+ x ⎠
1+ x ⎠
⎝
⎝
⎛
⎞
⎛
⎞
θ = sin −1 ⎜ d 2
φ = cos −1 ⎜ b 2
−θ .
2 ⎟ and
2 ⎟
a +d ⎠
e +b ⎠
⎝
⎝