Nomography Main > Determinant 6
ρ=Σ+Ψ
Equation into Determinant
This method is from Construction Of Nomographic Charts by F.T. Mavis.
You can approach it like filling out a threedimensional crossword puzzle. Consider how the determinant expands:
A_{1} 
B_{1} 
C_{1} 
A_{2} 
B_{2} 
C_{2} 
A_{3} 
B_{3} 
C_{3} 

= 
+A_{1}B_{2}C_{3} 
+A_{3}B_{1}C_{2} 
+A_{2}B_{3}C_{1} 
A_{3}B_{2}C_{1} 
A_{1}B_{3}C_{2} 
A_{2}B_{1}C_{3} 

Lets put the parallel scale equation P + Q  R = 0 into the determinant. First we will
put P in the upper left corner and 1s down the diagonal. Then using the chart
above as a guide, we will fill out the equations on the right
( e.g., since P is in the A_{1} determinant box,
we will substitute P
in the equations wherever there is a A_{1} )


= 
+P * 1 * 1 
+? * ? * ? 
+? * ? * ? 
? * 1 * ? 
P * ? * ? 
? * ? * 1 

= P 
The 1s in the diagonal are to ensure that at least one term of the equation
has no unknowns and evaluates to P (it's the equation on the top).
When writing the expanded equation on the right, omit any terms that have question
marks in them.
Now, Q has to go in the middle row. We can place it in box
A_{2} or C_{2}. What happens if we
put Q in C_{2}?


= 
+P * 1 * 1 
+? * ? * Q 
+? * ? * ? 
? * 1 * ? 
P * ? * Q 
? * ? * 1 

Oh, no, there is a problem. Look at the line in red. This will make the expansion of
the determinant have a term P * Q. Remember that the idea is to make
the determinant expand to P + Q  R = 0. There is no P * Q in that equation
so putting Q into C_{2} is no good.
Therefore Q must go into A_{2}. 1s can be added to
the determinant to get a full Q value in two ways:


= 
+P * 1 * 1 
+? * ? * ? 
+Q * 1 * 1 
? * 1 * 1 
P * 1 * ? 
Q * ? * 1 

= P + Q 
or


= 
+P * 1 * 1 
+? * 1 * ? 
+Q * ? * ? 
? * 1 * ? 
P * ? * ? 
Q * 1 * 1 

= P + Q 
If we choose the first determinant, R has no choice but to
go into A_{3}. The rest of the determinant fills out like this:


= 
+P * 1 * 1 
+R * 0 * 0 (zeros out) 
+Q * 1 * 1 
R * 1 * 1 
P * 1 * 0 (zeros out) 
Q * 0 * 1 (zeros out) 

= P + Q  R = 0 
If we choose the second determinant, R still has no choice but to
go into A_{3} (if it goes into B_{3}
there is no way to make it expand properly).
There are two ways the rest of the determinant fills out:


= 
+P * 1 * 1 
+R * 1 * 0 (zeros out) 
+Q * 0 * 1 (zeros out) 
R * 1 * 1 
P * 0 * 0 (zeros out) 
Q * 1 * 1 

= P + Q  R = 0 
or


= 
+P * 1 * 1 
+R * 1 * 1 
+Q * 0 * 0 (zeros out) 
R * 1 * 0 (zeros out) 
P * 0 * 1 (zeros out) 
Q * 1 * 1 

= P + Q  R = 0 
Lets do an equation with multiplication: P * Q = R or
P * Q  R = 0.
P and Q must appear in the same diagonal:


= 
+P * Q * 1 
+? * ? * ? 
+? * ? * ? 
? * Q * ? 
P * ? * ? 
? * ? * 1 

= P * Q 
R may be placed in either A3 or B3 and completed with diagonal 1s.


= 
+P * Q * 1 
+ R * 1 * 1 
+? * ? * ? 
 R * Q * ? 
P * ? * 1 
? * 1 * 1 

= P * Q  R = 0 
or


= 
+P * Q * 1 
+? * ? * ? 
+1 * R * 1 
? * Q * 1 
P * R * ? 
1 * ? * 1 

= P * Q  R = 0 
Complete by placing zeros in the remaining spaces:


= 
+P * Q * 1 
+ R * 1 * 1 
+0 * 0 * 0 (zeros out) 
 R * Q * 0 (zeros out) 
P * 0 * 1 (zeros out) 
0 * 1 * 1 (zeros out) 

= P * Q  R = 0 
or


= 
+P * Q * 1 
+0 * 0 * 0 (zeros out) 
+1 * R * 1 
0 * Q * 1 (zeros out) 
P * R * 0 (zeros out) 
1 * 0 * 1 (zeros out) 

= P * Q  R = 0 
Let's try one more complicated. The equation is
V_{1}^{2}V_{2}^{2} / ( V_{1} + V_{2} )
= ( 32.2D_{1}V_{1} ) / 2
. First lets get it into a more convenient form.
V_{1}^{2}V_{2}^{2} / ( V_{1} + V_{2} )
= ( 32.2D_{1}V_{1} ) / 2
V_{1}^{2}V_{2}^{2} / ( V_{1} + V_{2} )
= 16.1D_{1}V_{1}
simplify
V_{1}^{2}V_{2}^{2}
= ( 16.1D_{1}V_{1} )( V_{1} + V_{2} )
multiply both sides by V_{1} + V_{2}
V_{1}^{2}V_{2}^{2}
= 16.1D_{1}V_{1}^{2} + 16.1D_{1}V_{1}V_{2}
simplify
V_{1}V_{2}^{2}
= 16.1D_{1}V_{1} + 16.1D_{1}V_{2}
divide both sides by V_{1}
V_{1}V_{2}^{2}
 16.1D_{1}V_{1} + 16.1D_{1}V_{2} = 0

V_{1} 
? 
? 
? 
V_{2}^{2} 
? 
? 
? 
1 

= 
+V_{1} * V_{2}^{2} * 1 
+? * ? * ? 
+? * ? * ? 
? * V_{2}^{2} * ? 
V_{1} * ? * ? 
? * ? * 1 

= V_{1}V_{2}^{2} 
The second term 16.1D_{1}V_{1} contains
both V_{1} and D_{1}. Since
V_{1} is in the first column, 16.1D_{1}
must go into the second column of its row (i.e., B_{3}).

V_{1} 
? 
? 
? 
V_{2}^{2} 
1 
? 
16.1D_{1} 
1 

= 
+V_{1} * V_{2}^{2} * 1 
+? * ? * ? 
+? * 16.1D_{1} * 1 
? * V_{2}^{2} * 1 
V_{1} * 16.1D_{1} * 1 
? * ? * 1 

= V_{1}V_{2}^{2}  16.1D_{1}V_{1} 
The third term 16.1D_{1}V_{2} contains both
V_{2} and D_{1}. There is only
one blank in the second row (A_{2}) and 16.1D_{1}
is already in the third row. So the solution is:

V_{1} 
0 
1 
V_{2} 
V_{2}^{2} 
1 
0 
16.1D_{1} 
1 

= 
+V_{1} * V_{2}^{2} * 1 
+0 * 0 * 1 (zeros out) 
+ V_{2} * 16.1D_{1} * 1 
0 * V_{2}^{2} * 1 (zeros out) 
V_{1} * 16.1D_{1} * 1 
 V_{2} * 0 * 1 (zeros out) 

= V_{1}V_{2}^{2}  16.1D_{1}V_{1}  16.1D_{1}V_{2} 