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 three-dimensional crossword puzzle. Consider how the determinant expands:
| A1 |
B1 |
C1 |
| A2 |
B2 |
C2 |
| A3 |
B3 |
C3 |
|
= |
| +A1B2C3 |
| +A3B1C2 |
| +A2B3C1 |
| -A3B2C1 |
| -A1B3C2 |
| -A2B1C3 |
|
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 A1 determinant box,
we will substitute P
in the equations wherever there is a A1 )
|
|
|
= |
| +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
A2 or C2. What happens if we
put Q in C2?
|
|
|
= |
| +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 C2 is no good.
Therefore Q must go into A2. 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 A3. 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 A3 (if it goes into B3
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
V12V22 / ( V1 + V2 )
= ( 32.2D1V1 ) / 2
. First lets get it into a more convenient form.
V12V22 / ( V1 + V2 )
= ( 32.2D1V1 ) / 2
V12V22 / ( V1 + V2 )
= 16.1D1V1
simplify
V12V22
= ( 16.1D1V1 )( V1 + V2 )
multiply both sides by V1 + V2
V12V22
= 16.1D1V12 + 16.1D1V1V2
simplify
V1V22
= 16.1D1V1 + 16.1D1V2
divide both sides by V1
V1V22
- 16.1D1V1 + 16.1D1V2 = 0
|
|
|
= |
| +V1 * V22 * 1 |
| +? * ? * ? |
| +? * ? * ? |
| -? * V22 * ? |
| -V1 * ? * ? |
| -? * ? * 1 |
|
= V1V22 |
The second term -16.1D1V1 contains
both V1 and D1. Since
V1 is in the first column, 16.1D1
must go into the second column of its row (i.e., B3).
|
|
| V1 |
? |
? |
| ? |
V22 |
1 |
| ? |
16.1D1 |
1 |
|
= |
| +V1 * V22 * 1 |
| +? * ? * ? |
| +? * 16.1D1 * 1 |
| -? * V22 * 1 |
| -V1 * 16.1D1 * 1 |
| -? * ? * 1 |
|
= V1V22 - 16.1D1V1 |
The third term 16.1D1V2 contains both
V2 and D1. There is only
one blank in the second row (A2) and 16.1D1
is already in the third row. So the solution is:
|
|
| V1 |
0 |
1 |
| -V2 |
V22 |
1 |
| 0 |
16.1D1 |
1 |
|
= |
| +V1 * V22 * 1 |
| +0 * 0 * 1 (zeros out) |
| + -V2 * 16.1D1 * 1 |
| -0 * V22 * 1 (zeros out) |
| -V1 * 16.1D1 * 1 |
| - -V2 * 0 * 1 (zeros out) |
|
= V1V22 - 16.1D1V1 - 16.1D1V2 |