Sunday, January 25, 2015

Using Algebra to Derive New Formulas

Using Algebra to Derive New Formulas and Solve others

To follow this you are going to need to know some algebra.  If you need a refresher, I recommend  I am first going to show how to use algebra to find the temperature at which both a Fahrenheit and Celsius thermometer would be showing the same number.  That is, what temperature is the same in degrees Fahrenheit and Celsius.  You can think of this like using 2 different rulers to measure your height.  One in feet and one in meters.  Your height is the same no matter which you use, just the number will be different, i.e. 6 feet or 182.88 centimeters.

First we start with the conversion formulas.  To convert from degrees Celsius (C) to Fahrenheit (F), we multiply the temperature by 1.8 (also regularly stated as 9/5) then add 32.  So 20 degrees C = 68 degrees F.

Converting from F to C the formula is degrees F minus 32 then that result divided by 1.8:

To calculate the temperature where these are equal to each other, we set the 2 formulas equal to each other, use just one variable for temperature, e.g. “t” and solve for that variable:

The idea is to get the t all by itself and we do this by performing the inverse operations on various parts of the equation, remembering to do the same operation on BOTH sides!  First we’ll get rid of the 1.8 on the bottom of the right side by multiplying both sides by 1.8:

 The 1.8’s on the right side cancel leaving just the t - 32 and we multiply the 2 terms inside the parenthesis on the left side each by 1.8, giving us:

Now subtract t from both sides:

Doing the math this leaves us with:

Next we subtract 57.6 from both sides:

This gives us:

Next we divide both sides by 2.24 to get the t by itself (this is the same as multiplying by 1/2.24):

The result of this is:

You can check this by inserting minus 40 into the 2 equations above and you’ll see the conversions both come out to minus 40.

Now as they say, I showed you that so I can show you this, specifically, how to derive a new formula.

In the license books we’ve seen examples, usually poorly explained, where a new formula is derived by substituting values.  In the case of the power formula, P = E I, we are told to substitute the ohms law equivalents of E and I to get 2 new formulas.

The example I’m going to show you is from the ARRL Extra Class license manual which is only shown worked through superficially.  The assumption being that either the student is well versed in algebra, or will go to one of the many resources available online or elsewhere to bone up on it.

However, I have found that many who come to ham radio, especially those who come to it after a prolonged time away from high school or college, may have forgotten most, if not all that they’ve learned.  So I’d like to build on the previous example and work through the derivation of the formula for calculating resonant frequency of LC circuits.

The definition of LC circuit resonance is given as the frequency where both inductive reactance (XL) and capacitive reactance (XC) are equal.  Each reactance being 180 degrees out of phase with the other means that their reactances will cancel out leaving only a purely resistive load.  Let’s first look at the formulas for each, then we’ll set them equal to each other and solve for frequency to get the resonant frequency formula.

Setting them equal:

We want to find “f” so we’ll multiply both sides by f to remove it from the bottom of the right side.

Combining the f’s on the left and cancelling on the right leaves us with:

Next I’m going to get rid of the 2πL on the left by dividing both sides by 2πL (this is the same as multiplying by the reciprocal 1/L)

When we do the math we cancel the 2πL on the left side and multiply the top and bottom on the right, combining the like terms to get:

We now take the square root of both sides…
 This gives us:

…and that’s the formula for resonant frequency!  Not too bad, right?

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