Just a small but important clarification at 3:34 - You mean “numbers less than ONE but greater than zero” can be expressed using negative powers of ten. Correct? Great video BTW! Thank you!
Hello great video.. don’t hammer me but just wondering if the # 2 video has been produced ? I’ve looked but haven’t been able to find it. Thanks in advance.
Hi, are you referring to microwave transmission between towers? There are not any coax or fiber cables that would link from tower to tower. It would be a transmitter on one end and an antenna on the other with a distance of up to 30 miles or so. Please help to clarify your question for us to be able to provide you with the best response.
Perhaps I should have said pole to pole not tower. or two poles located on a hillside. There is a calculation to account for the sag i the cable between the two p points (I ionce had to work such a problem in a calculus class a few decades ago. But on a larger scale when your going between twenty or thirty poles you don't sit down ad integrate each sagment for each pair of poles (You could). Or is it an estimated value based on prior experience or rule of thumb? There should also be an adjustment for the type of cable and its tensile strength as well I would think.
Sag formula x = ½ pole spacing y= amount of sag in the line L = actual length of cable Arc Length = (Pi (x^2 + y^2) ArcSin[(2xy)/(x^2 + y^2)]) / (180y) If you were to take 100ft pole-to-pole spacing with 1.5ft of sag in the middle then the actual length of the cable needed would be 100.06ft meaning you would add roughly ¾” to the pole-to-pole distance in this scenario. Once you add in the fact that the cable would actually experience some elongation under its own weight, this compensation becomes even smaller. This is based on the Arc Length Calculation. Here are some additional calculations used in our industry. To Calculate the spacing between two (2) amplifiers, you need to know the following: Highest frequency of the system in MHz Type of cable The loss in dB per foot of the cable at the highest frequency Operating gain of the amplifier, at the highest frequency Example Calculation: Highest frequency is 870 MHz. .625 cable 1.95 dB loss per 100 feet 36 dB gain Amplifier Formula = Gain of amp/loss per 100 feet of cable x 100 Distance calculation: 36 / 1.95 = 18.367 X 100 feet = 1,836.73 feet between amplifiers. Answer = 1,836.73 feet between amplifiers
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Just a small but important clarification at 3:34 - You mean “numbers less than ONE but greater than zero” can be expressed using negative powers of ten. Correct? Great video BTW! Thank you!
Hello great video.. don’t hammer me but just wondering if the # 2 video has been produced ? I’ve looked but haven’t been able to find it. Thanks in advance.
Thanks for watching! We have not done part two yet. We have some great videos to come. Be sure to subscribe!
I learnt alot from this.good work
Glad it was helpful!
These are really good videos!
Thanks for watching!
Can we get a lesson on how to calculate the length of cable needed between two towers?
Hi, are you referring to microwave transmission between towers?
There are not any coax or fiber cables that would link from tower to tower. It would be a transmitter on one end and an antenna on the other with a distance of up to 30 miles or so.
Please help to clarify your question for us to be able to provide you with the best response.
Perhaps I should have said pole to pole not tower. or two poles located on a hillside. There is a calculation to account for the sag i the cable between the two p points (I ionce had to work such a problem in a calculus class a few decades ago. But on a larger scale when your going between twenty or thirty poles you don't sit down ad integrate each sagment for each pair of poles (You could). Or is it an estimated value based on prior experience or rule of thumb? There should also be an adjustment for the type of cable and its tensile strength as well I would think.
Sag formula
x = ½ pole spacing
y= amount of sag in the line
L = actual length of cable
Arc Length = (Pi (x^2 + y^2) ArcSin[(2xy)/(x^2 + y^2)]) / (180y)
If you were to take 100ft pole-to-pole spacing with 1.5ft of sag in the middle then the actual length of the cable needed would be 100.06ft meaning you would add roughly ¾” to the pole-to-pole distance in this scenario. Once you add in the fact that the cable would actually experience some elongation under its own weight, this compensation becomes even smaller. This is based on the Arc Length Calculation.
Here are some additional calculations used in our industry.
To Calculate the spacing between two (2) amplifiers, you need to know the following:
Highest frequency of the system in MHz
Type of cable
The loss in dB per foot of the cable at the highest frequency
Operating gain of the amplifier, at the highest frequency
Example Calculation:
Highest frequency is 870 MHz.
.625 cable
1.95 dB loss per 100 feet
36 dB gain Amplifier
Formula = Gain of amp/loss per 100 feet of cable x 100
Distance calculation: 36 / 1.95 = 18.367 X 100 feet = 1,836.73 feet between amplifiers.
Answer = 1,836.73 feet between amplifiers
Very nice designing
Better explaining logs than my school
Thank you for watching! Be sure to subscribe. We have some great stuff coming.
Awesome video dude ✌️😊
💯 🔥 💯 🔥 💯 🔥
Noise return path find and trouble shoot training
Training how to solve noise problem at return path
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1 kilo meter... = 1 million meters wait....
Still don’t get iiiiiiiittt😩
What do you require clarification with?