Hey fine sir,
 I can see it all very complicated.
 Let's first understand the chart you have.

Log scale horizontal:
The flow numbers marked on the horizontal axis are 0.1, 1.0, 10.0, 100.0, 1000.0.
Note these each go up by a factor of 10, ie each gets another zero.
I can't measure the size of your paper. But lets say there are 1 inch between each of the  numbers.
Say
 flow 0.1 is at -1 inch
 flow 1.0 is at 0 inch
 flow 10.0 is at 1 inch
 flow 100.0 is at 2 inch
 flow 1000.0 is at 3 inch
Then the number of zeros in the flow is the number of inches along the horizontal.
Flow = 10^(# inches).

The interval from .1 to 1 is broken into 9 sub intervals.
These lines are at .1, .2, .3, ..., .9, 1.0
The interval from 1 to 10 is broken into 9 sub intervals.
These lines are at 1, 2, 3, ..., 9, 10

The lines get closer as you go right, but then jump bigger.
Consider the line labelled 10.0.
The line before it is 9,  ie 1 smaller.
The line after it is 20,  ie 10 bigger.

If your reading is between lines
  |       *   |
  3           4
Imagine breaking this in the ten pieces: 3.0, 3.1, 3.2, ... 3.9, 4.0
If this was linear, the ten pieces are evenly spaced. * might be 3.7.
But this is logarithmic, so the ten pieces are get smaller. * might be 3.5.

A normal graph goes from x cm on the horizontal and y cm on the vertical.
The equation of a line is x = m x + b,   where m is the slope and b is an additive constant.

The equation of a line in graph from x cm to Head log scale is  Head = 10^(m x + b).

But our graph goes from Flow log scale to Head log scale.
The equation of a line in graph from x cm to log scale is
Head = r Flow^c    for some exponent real number c and some multiplicative constant real number r.

Lets consider your left most blue line.
It crosses the horizontal axis one line before the flow number 1.0.
The bottom point of the line has
    Flow = 0.9  and Head = 0.1
The top point of the line has
    Flow = 50  and Head = 300

0.1 = r (0.9)^c  and 300 = r (50)^c
Take the Log on both sides  (say base e).
log(0.1) = c log(0.9) + log(r)   and  log(300) = c log(50) + log(r)
Subtract these equations to get rid of the log(r)
log(300)-log(0.1) = c log(50) - c log(0.9) = c [log(50)-log(0.9)]
Solve for c.
c = [log(300)-log(0.1)]/[log(50)-log(0.9)]
Drop this into google search gives  c = 1.9929. Likely they meant c = 2.
300 = r (50)^2
r = 300/(50^2) = 0.12
Check  r (0.9)^c =  0.12 (0.9)^2 = 0.0972  close enough to 0.1.

So the equation of your first line is
Head = r Flow^c = 0.12 Flow^2

Now lets consider the other blue lines.
They look parallel. Hence there c=2 is the same for all of them. We just need to find their r.
The bottom point of the line has
    Flow = F  for some f depending on the line
    and Head = 0.1
Head = r Flow^c
   0.1 = r F^2
   r = 0.1/F^2.

On the screen I measured the distance there was 5.63 cm between the 1.0 and the 10.0.
This gives me the equation Flow = 10^(# cm/5.63).
Then I measured the distance between the 1.0 and the bottom of each line.
That gave me the Flow at the bottom each line.
Lets name the ten lines If  0.5, 1.0, 1.5, 2.0, .... ,5
Then I did regression.
Then Flow = 1.161006*Turn + 0.338
  This is a fairly good approximation of all the lines except for the first and the last.
If I only pay attention to the first and the last line then this gives
  Flow = m turn + b
   0.9   = m  0.5  + b  and   5.8   = m  5.0  + b
  Subtract these equations to get rid of the b
   5.8-0.9 = m [5.0-0.5]
   m = [5.8-0.9]/[5.0-0.5] = 1.08. Surely they mean this to be 1.
   0.9   = 1.08* 0.5  + b
   b = 0.9  - 1.08* 0.5 = 0.36
Then Flow = 1.08*Turn + 0.36
Your call.

Recall
   r = 0.1/F^2 = 0.1/(1.08*Turn + 0.36)^2

Head = (0.1/(1.08*Turn + 0.36)^2) * Flow^2
          = 0.1 * (Flow/(1.08*Turn + 0.36)^2

Your table has values of Flow 0.9 to 400.
Let have 40 steps.
i = 40
Flow = q^40 = 400
q = 400^(1/40) = 1.06