Question for Technician Class:(T3B06)

Started by Sunflower, October 09, 2012, 01:27:40 AM

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Sunflower

Reference: The No-Nonsense, Technician Class License Study Guide, by Dan Romanchik, KB6NU, PAGE 19.

Quoting part of fourth para:
"The formula for converting frequency to wavelength in meters is wavelength in meters equals 300 divided by frequency in megahertz. (T3B06)

Quoting fifth para:
"The approximate wavelength of radio waves is often used to identify the differnt frequency bands. (T3B07). For example, when we refer to the 2 meter band, we are referring to the amateur radio band that spans 144 MHz to 148 MHz. A radio wave with a frequency of 148 MHz, would have a wavelength of 2.02 meters. "

I do not understand this. Can someone please translate this into a math formula? or something to explain it better? I did not follow it at all.

....does this mean 300 divided by 144 MHz = meters? (2.08 meters)???
...or 300 divided by 148 =2.03 meters??
...Where does 2.02 come from? does it matter?

In the paragraphs directly above it, the approximate velocity of a radio wave as it travels through free space was noted as 300,000,000 meters per second.



Thank you,

gil

300/148=2.027 (2m band).

Let's try for 14.3MHz: 300/14.3=20.97 (20m band).

Sunflower

That was helpful.
I need to work on how to use my calculator more precisely.
Thank you.

Sunflower

Quote from: Sunflower on October 09, 2012, 01:48:43 AM
That was helpful.
I need to work on how to use my calculator more precisely.
Thank you.
Now I understand the " :-["
Very good. My calculatation was off. Goodness. I will be so happy when I get through this material. It is enjoyable, but it will be a whole lot more enjoyable to pass the test. I need to rethink which calculator I drag along.

Thank you for the help.

gil

You're welcome. I suggest practice tests on QRZ.com. They helped me a lot.

Gil.

Sunflower

#5
Quote from: gil on October 09, 2012, 01:36:33 AM
300/148=2.027 (2m band).

Let's try for 14.3MHz: 300/14.3=20.97 (20m band).
Not to be difficult, but why not ....call it 21 m band? or does it all get rounded up and down to the simpliest number?
Let's try for 14.3MHz: 300/14.3=20.97 (20m band).

--time for bed.

gil

Hello,

It's just rounded up for simplification. Even the 300,000,000 is rounded up. Doesn't matter...
The formula is just useful to calculate antenna length.
Say you need a half wave dipole for 7MHz: 300/7=42.85m.
In feet: 42.85*3.048=130.6ft.
Half wave: 130.6/2=65.3ft. of wire needed.

Usually, the formula used for a dipole though is 468/f
So, 468/7=66.85ft. You start with that and trim as needed.
Remember that formula for the exam.

Gil.

Paul

Converting frequency in Mhz to wave length in meters is only an approximation.  The commonly used 'band' numbers are not exact, and are a generalization.  Easier/shorter to say 2 meters than saying 2.0X meters of 14X.XXX Mhz.
The 'magic' number for finding lengths of antennas (234 or 468) are based on the 'general' numbers, 246 or 492, which are closer to lengths in 'free space'.  The other two, 234 or 468, factor in velocity factor for a small'ish conductor in air (typically used for wire antennas).  If you really want to know where all those 'numbers' come from I would suggest finding a 'Hand Book' and taking a look at the antenna section.
- Paul

White Tiger

It fits that Sunflower is struggling a bit with the loose factoring going on here - I hallen to know she teaches complicated and specific topics - like Algebra - so if you give her an equation...the math is supposed to work!

The good part (for me) is that after reading this thread, I no longer have to rely on memorization for this part of the exam!
If you're looking for me, you're probably looking in the wrong place.

Sunflower

Quote from: gil on October 09, 2012, 01:57:44 AM
You're welcome. I suggest practice tests on QRZ.com. They helped me a lot.

Gil.
Finally got on QRZ tonight. Wow. Went from 40 to 55 percent. Not very good.

I also spoke with the exam man for our area. Next test date is Nov 14th. I hope to be there. Much will depend on how the progress I make on QRZ.com.

gil

Keep doing them Tess, the same questions are going to come up over and over, and you'll remember them...

Gil.

White Tiger

#11
Gil's right Tess, take them OVER and OVER! That is how I passed Tech and General.

I am currently using this same strategy for the Extra exam (except there are 50 questions in the Extra class, question pool). When I first started taking the practice exams for Extra Class a few weeks ago, I was at 40%, Gil gave me the Extra class study book, and then I started taking the practice tests on www.hamtestonline.com (the non-study practice exams are free) - and I am now at 65%!

You can do it!
If you're looking for me, you're probably looking in the wrong place.

cockpitbob

Yes, keep grinding the QRZ.com practice tests over and over.  My son got his Tech and his General when he was 11 using this method.  At that age kids have a great memory.  No, he didn't understand the theory, but that's OK.  Learn the rules and procedures enough to stay out of trouble, and get on the air.  If ham radio is something you stick with, the theory will come with time.

White Tiger

It's actually surprising how much you retain - and then as you assemble your gear - how much you begin to associate between what you memorized and the practical application of putting it all together.

I can only imagine how much more you would learn actually operating for a couple of years - especially if you find someone that can help explain around some of the hindrances.

If I could teach the teacher - as I believe you are Tess - I'd say don't try to learn it - just memorize it.
If you're looking for me, you're probably looking in the wrong place.

buckeye43210

Here's a tip to help you solve problems of this type on the exams:

Draw a circle, then draw a horizontal line dividing the circle into an upper and lower half. Next draw a vertical line dividing the lower half of the circle into two parts. Write 300 in the upper half of the circle, then write 'f' in the lower left section of the circle followed by 'm' in the lower right half of the circle.

Now cover up the symbol you are trying to solve for. Suppose you are given the frequency in MHz and are asked to find wave length in meters. Cover up the m and you get m = 300 over f or 300 divided by f. Now suppose you are given the wavelength in meters and want to find the frequency in MHz. Cover up the f and you get f = 300 over m or 300 divided by m.

The same technique works for Ohm's Law, Power in Watts, etc. You can draw your memory aids on your scratch paper at the beginning of your exam.

An example using Ohm's law can be found here.