Based On The Distance Of The Speed Of Light

YORK’S RF / LIGHT FORMULA (Based On The Distance Of The Speed Of Light) F=

528000 /λ 1863

or

γ=

528000 /F 1863

Simplified to 283.4138486 / X

HOW 283.4138486 / X WAS DERIVED AT INSTEAD OF THE MEANS 984 Step #1: Take the number of feet in one mile. 528000 ft. = 1 mile Step #2: Add two zeros on to 5280 as a scale modifier making it. 528000 Step #3: Take the distance light travels in one second. 186300 miles/sec Step #4: Subtract two zeros from 186300 as a scale modifier making it. 1863XX Step #5: Now use this formula:

FORMULA #1 #2 #3 528000 divided by 1863 divided by ( your frequency in question or your full wave length of the antenna ) One will give you the other. It's as simple as that. ONE, TWO, THREE!

EXAMPLES Example #1: What is the length of a full wave antenna for C/B (citizens band radio) that operates in the range of 27 MHz ? 528000 divided by 1863 divided by 27 = 10.49680921 ft. in length is the answer. Example #2: Now let’s work the same problem but this time we know the length of the antenna and are looking for what frequency range that full wave antenna will operate in at its optimum efficiency to check our first answer. We just simply substitute one in place of the other. 528000 divided by 1863 divided by 10.49680921 = 27. using ft. = 27 MHz as our answer.

HARMONICS One of the beauties of this formula other than the same exact formula to find the wavelength is also used to find the frequency if that is your unknown. On top of all this the formula is not only practical and simple but most important is based on the “distance of the speed of light”. One can also obtain the harmonics of a known frequency just by using the harmonic conversion guide shown below. To apply this sees the example shown on the prior page using 27 MHz for the CB Radio Antenna. We used feet which will give us MHz range for our CB radio frequency. If we use inches then we are talking about the GHz frequency range or if we use miles we are looking at the Hz frequency range. Just by changing scale we can find the harmonic of a frequency. The down side is some elements will have sub harmonics of a different element within that same window of operation range of target element under question. For now however we’ll just keep it simple and use the harmonic conversion guide shown below. This will give us the harmonics needed for the frequency under question. HARMONIC CONVERSION GUIDE Feet = MHz range Inches = GHz range Miles = Hz range To make the conversion we will have to take the know wavelength- (γ) used and convert that wavelength-(γ) to a different unit of measurement. Example once again using 27 MHz for our CB Radio frequency we get an answer in feet because “Feet” = MHz. The answer came out to be 10.49680921 ft. in length as our answer for the wavelength-(γ) of our antenna. If we change the feet units to units in inches then the answer will be the harmonic but in the GHz range scale. Feet for the CB Radio antenna will be our base wavelength-(γ) for 27 MHz. All other units will show the harmonic of that same base line wavelength-(γ). The symbol (γ) used for wavelength is called lambda-(γ).

FORMULA FOR FREQUENCY OR WAVELENGTH 528000 /1863 / ?( F )or ( L) = ANS

8.15 IS THE RF/LIGHT FORMULA BASED ON LIGHT OR JUST A MATHEMATICAL MANIPULATION?- (Fact) The difference between a mathematical manipulation and a mathematical formula is as clear as night and day. Example here is anybody can manipulate an arbitrary number like using 280 that when divided by another number will yield an answer within a proper mean range of probability. So if you divide 282 by 27 as in our RF/LIGHT formula you will get 10.44444, about the same length one would expect for the length of a 27MHz antenna however not exact. With our RF/LIGHT formula we get 10.44444444 feet. Again within the expected length for a 27 MHz antenna but this time it is exact. What is the difference other than the claim to be an exact answer? In electronics you will find an arbitrary manipulation used as a means value to obtain results. With our RF/LIGHT formula however we have figures not just arbitrary numbers that have a divertive base on light. Example our value 5280(00) is based on the number of feet in a mile. The (00) again is the unit scale adjuster. Our 1863-(00) value is based on the distance of the speed of light. Again the -(00) is a scale adjuster. The scale adjusters would be a manipulation but only one to set our values to proper scale and in agreement with each other. Both these values are based on factors of the distance of light and units light would have to travel for one mile. The arbitrary value 280 has no base derivative or factual bases for existence other than when a number is divided into it, it will yield a close facsimile of a correct answer for a set range of values. On the other hand we can show how our RF/ LIGHT formula is derived from actual values dealing with light while the 280 figure cannot. The hidden variable of this formula is that of temperature. A chart will need to be made to adjust for the temperature of the target.

Copywrite ©1989 Warren York Last revised: 10/24/08 Warren York: E-mail [email protected]