My experience with lower VHF is that obstacles in fresnel zone and even some geographic formations in line-of-sight won’t stop the signal and you still get comms pushing 55km
Let's say an antenna transmitter 209 meters high :What would be the range of a signal...? Specially if the recipient is 30 meters higher than the ground at 400 km..away ?
If I want to use LoRaWAN for transmitting data from meters in buildings mostly located in the basement, I have totally other conditions regarding the fresnel zone.
sorry how i can answer this question please help me:Choose two tower locations (The minimum distance between the two location should not be less than 5km) Take GPS coordinates of each location (Longitude, Latitude, and Altitude) Go to Google Earth app and mark the locations as site A (location 1) and Site B (the other location) Choose frequency of your choice within the microwave band. Use the above informations to generate the Fresnel zones of the LoS link between the two locations.
Excellent videos, so what would be the advantage of Lora compared to other technologies apart from the battery life? in practice, the penetration of obstacles does not seem so
what if the end node is underground? like -5 floors from ground level. i know there will be packet loss and i will have to consider r instead of r+h but is there any alternative?
Hi Namratha, I have done several experiments: 1) Setup: Basement = -1, ground floor = 0, first floor = 1, second floor = 2 I had placed a gateway (INDOORS) on de second floor of a residential apartment and an end node in the basement, straight down in the same building. The gateway could pick up the transmitted signal. The end node used a PCB antenna. 2) I had placed a gateway (INDOORS) on de second floor in front of a window. I went into bicycle tunnel made of concrete (distance 500 m from the gateway) approx 4 meter below ground. The gateway could not pickup the signal. The end node used a PCB antenna. If I used a sleeve dipole antenna, it may work, but I have not tested this. I have given you two examples, hopefully it gives you a better idea what the possibilities are. Note: If you do tests: a) If possible attach an external antenna to your end node for example a sleeve dipole antenna. b) Make sure your gateway has a good antenna.
Your radius of the earth is way off. does this mean all the calculations are wrong in your video? en.wikipedia.org/wiki/Earth Physical characteristics Mean radius 6371.0 km (3958.8 mi)[6] Equatorial radius 6378.1 km (3963.2 mi)[7][8] Polar radius 6356.8 km (3949.9 mi)[9]
It is assuming because of the earth curvature. 6378km*(4/3)=8500 km. When you assume this rule, you can take two point as if this points see each other without any obstacles.
@@MrFloneil The rule, doesn't include the earth curvature. When you multiply the radius times 4/3 (k: curvature earth factor), you turn your problem in a flat earth problem again.
It is because radio waves diffract farther around the earth than light waves do due to their lower frequency. Because of this, the radio horizon is farther than the line of sight (light) horizon is. The Earth appears to have a larger radius to radio waves than to light waves. This is the 4/3 rule.
Thanks a lot sir..ur vedios are helping me a lot in my exams
My experience with lower VHF is that obstacles in fresnel zone and even some geographic formations in line-of-sight won’t stop the signal and you still get comms pushing 55km
Excellent explanation!!!
Really helpful - many thanks.
Flat Earth confirmed! 2:34 😂
Great videos! Thank you for these tutorials!
best presentation thanks a lot sir.
Most welcome
Excellent video!!!!
Let's say an antenna transmitter 209 meters high :What would be the range of a signal...? Specially if the recipient is 30 meters higher than the ground at 400 km..away ?
the earth radius given in that equation is different from what i found online.
If I want to use LoRaWAN for transmitting data from meters in buildings mostly located in the basement, I have totally other conditions regarding the fresnel zone.
sorry how i can answer this question please help me:Choose two tower locations (The minimum distance between the two location should not be less than 5km)
Take GPS coordinates of each location (Longitude, Latitude, and Altitude)
Go to Google Earth app and mark the locations as site A (location 1) and Site B (the other location)
Choose frequency of your choice within the microwave band.
Use the above informations to generate the Fresnel zones of the LoS link between the two locations.
great video.
Excellent videos, so what would be the advantage of Lora compared to other technologies apart from the battery life? in practice, the penetration of obstacles does not seem so
LoRa has advantages and disadvantages compared to other LPWAN technologies.
Search Google: LPWAN technologies comparison
I only focus on LoRa.
what if the end node is underground? like -5 floors from ground level. i know there will be packet loss and i will have to consider r instead of r+h but is there any alternative?
Hi Namratha,
I have done several experiments:
1) Setup: Basement = -1, ground floor = 0, first floor = 1, second floor = 2
I had placed a gateway (INDOORS) on de second floor of a residential apartment and an end node in the basement, straight down in the same building.
The gateway could pick up the transmitted signal. The end node used a PCB antenna.
2) I had placed a gateway (INDOORS) on de second floor in front of a window.
I went into bicycle tunnel made of concrete (distance 500 m from the gateway) approx 4 meter below ground.
The gateway could not pickup the signal. The end node used a PCB antenna.
If I used a sleeve dipole antenna, it may work, but I have not tested this.
I have given you two examples, hopefully it gives you a better idea what the possibilities are.
Note: If you do tests: a) If possible attach an external antenna to your end node for example a sleeve dipole antenna. b) Make sure your gateway has a good antenna.
👍
The S in Fresnel is silent. It is pronounced Fruhnel.
Go f*ck yourself. The "U" is not silent.
@@st105900 try decaf next time.
Btw “Fruhnel” is spelled _Fyurer_ XD
Your radius of the earth is way off. does this mean all the calculations are wrong in your video?
en.wikipedia.org/wiki/Earth
Physical characteristics
Mean radius 6371.0 km (3958.8 mi)[6]
Equatorial radius 6378.1 km (3963.2 mi)[7][8]
Polar radius 6356.8 km (3949.9 mi)[9]
Yes, my calculations are wrong! Instead of changing the video, i changed my presentation. See: www.mobilefish.com/download/lora/lora_part7.pdf
Earth radius is not equal 8504 km
It is assuming because of the earth curvature. 6378km*(4/3)=8500 km. When you assume this rule, you can take two point as if this points see each other without any obstacles.
@@MrFloneil The rule, doesn't include the earth curvature. When you multiply the radius times 4/3 (k: curvature earth factor), you turn your problem in a flat earth problem again.
It is because radio waves diffract farther around the earth than light waves do due to their lower frequency. Because of this, the radio horizon is farther than the line of sight (light) horizon is. The Earth appears to have a larger radius to radio waves than to light waves. This is the 4/3 rule.
Please research a little more thoroughly. There are several mistakes. The Antennas being at the focal points of the elipsoid, for instance.
Care to elaborate, please?