Transposition and evaluation of formulae Transposition of formulae

Transposition and evaluation of formulae Transposition of formulae

We now consider the transposition of formulae and although this subject does not form a mandatory part of this outcome, it is considered a vital tool for all prospective technicians and engineers. You will see how some of the mathematical methods we have learnt up to now may be used to manipulate these formulae. These same techniques may also be used to simplify and solve linear, simultaneous and quadratic equations analytically, as you will see when we look at these equations later on.

As mentioned earlier formulae, provide engineers with a method of writing down some rather complex relationships and ideas in a very precise and elegant way. For example the formula, v = u + at tells us that the final velocity (v) of, say, a car is equal to its initial velocity (u) plus its acceleration (a) multiplied by the time (t) the car is accelerating. If the car is neither accelerating nor decelerating, then v = u because the acceleration a = 0 and 0 X t = 0, as you already know. I think you are already beginning to realize that to explain the meaning of one simple formula requires rather a lot of words! It is for this reason that formulae are used rather than just words, to convey engineering concepts. If you have met the transposition of formulae before, then this section will serve as good revision and consolidation and enable you to solve the equations you will meet later.

Before considering the techniques needed to manipulate or transpose formulae, we first need to define some important terminology. We will use our equation of motion v = u + at for this purpose.

Term, this is defined as any variable or combination of variables separated by a +, a - or an = sign. You have already met this definition in our study on the laws of arithmetic. Therefore, in our formula, according to the definition, there are three (3) terms, they are: v, u and at.

Variable, these are represented by literal numbers, which may be assigned various values. In our case, v, u, a and t are all variables. We say that v is a dependent variable because its value is determined by the values given to the independent variables, u, a and t.

Subject, the subject of a formula sits on its own on one side of the equals sign. Convention suggests that the subject is placed to the left of the equals sign. In our case, v is the subject of our formula. However, the position of the subject whether to the left or to the right of the equals sign makes no difference to the sense of a formula. So, v = u + at is identical to u + at = v, the subject is simply pivoted about the equals sign.

In the following examples we will transpose simple formulae, using the basic arithmetic operations of addition, subtraction, multiplication and division, to rearrange the subject of a formula.

(1) In this formula we are required to make b the subject; therefore, b needs to sit on its own on the left-hand side (LHS) of the equals sign. To achieve this we need to remove a term from the LHS. We ask the question, how is a attached to the LHS? It is in fact added, so to remove it to the right-hand side (RHS) of the equals sign we apply the inverse arithmetic operation, that is, we subtract it. To maintain the equality in the formula we need in effect to subtract it from both sides,                    that is, a - a + b = c - a, which of course gives

b = c – a

You will remember this operation as: whatever we do to the LHS of a formula or equation, we must do to the other or, when we take any term over the equals sign we change its sign.

(2) Applying the procedure we used in our first example to y - c = z, then                we subtract y from both sides to give y - y - c = z - y, which again gives -c = z - y. Now unfortunately in this case we are left with -c on the LHS and we require +c, or just c as we normally write it, when on its own. Remembering from your study of fundamentals that a minus multiplied by a minus gives a plus and that any number multiplied by one is itself then:

(-1)(-c) _ (-1)(z) - (y)(-1) or c = -z + y

and exchanges the letters on the RHS gives

c = y –z

Now all that we have done, in this rather long-winded procedure, is to multiply every term in the formula by (-1) or as you may remember it, we have changed the sign of every term in order to eliminate the negative sign from the subject of our formula.

(3) Now with the formula, x = yz we have just two terms and our sub_ject z is attached toy by multiplication. So, all we need to do is divide it out. In other words, apply the inverse arithmetic operation, then we get:

and reversing the formula about the equals sign gives x

(4) With the formula  , then b is attached to a by division, so we multiply it out to give:

This leaves