 InssomniakThe GlitchPremium
join:2005-04-06
Cayuga, ON
kudos:1 | Calculating beam width at X Distance Anyone know how I can do that? I have 6 degree beam on a dish, And I need to know how wide it is at X distance. Is there an online calculator for that? Small idea to put 2 devices on a single dish if it fits in the beam, distance is 8.34 kilometers, its 270 meters away from existing link at that distance. --
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 John GaltForward, MarchPremium
join:2004-09-30
Happy Camp
kudos:5 | ~700 ft. perhaps...?? |
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Also keep in mind you are no longer covered under PtP rules, but instead PtMP rules for TX power in that case. At least in the U.S. you would be.
The easiest way to figure this out is to print out a map. Put a protractor on it and extend the degree lines. IMO, any way.. Not everything needs a program 
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public
join:2002-01-19
Santa Clara, CA
1 edit | reply to Inssomniak
said by Inssomniak:Anyone know how I can do that?
I have 6 degree beam on a dish, And I need to know how wide it is at X distance. Is there an online calculator for that?
Small idea to put 2 devices on a single dish if it fits in the beam, distance is 8.34 kilometers, its 270 meters away from existing link at that distance.
pi*6/180*8340 |
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LLigetfa
join:2006-05-15
Fort Frances, ON
kudos:1 | reply to Inssomniak
Should be basic math...
Circumference of a circle can be had using pi so 8340m radius = 52375m
Circumference has 360 degrees so 52375/360 = 145.486m per degree
Multiply times 6 for your degrees and there is your answer.
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Strange as it seems, no amount of learning can cure stupidity, and formal education positively fortifies it. -- Stephen Vizinczey |
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InssomniakThe GlitchPremium
join:2005-04-06
Cayuga, ON
kudos:1 | Haha thanks LLigetfa. I failed math.  . |
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WHT
join:2010-03-26
Rosston, TX
kudos:5
1 edit | reply to Inssomniak
Simple trig, where adjacent side is 8.34 km and angle is 6º/2. then x2 (for 6º) and result is 874 meters width.
/edited to add
said by LLigetfa:Multiply times 6 for your degrees and there is your answer.
6 x 145.486 = 872 meters
Hrmmm...872 meters is the length of an arc, which should be longer than the straight line distance. Check my math please. |
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InssomniakThe GlitchPremium
join:2005-04-06
Cayuga, ON
kudos:1 | Arc, straight line, close enough for me.. Lol. Thanks!
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OptionsDSL Wireless Internet
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said by WHT:Hrmmm...872 meters is the length of an arc, which should be longer than the straight line distance. Check my math please. Your math is good (well close enough). Your thinking the arc should be longer is what's flawed. The circle would be drawn with a radius equal to the length of the adjacent side, not the hypotenuse. Had me thinking for a minute too... |
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lutful... of ideasPremium
join:2005-06-16
Ottawa, ON Reviews:
 ·TekSavvy DSL
2 edits
1 recommendation | reply to Inssomniak
said by Inssomniak:Small idea to put 2 devices on a single dish if it fits in the beam, distance is 8.34 kilometers, its 270 meters away from existing link at that distance. The second device is approx (270/8340)*60 or 2 degrees away from first device.
You can keep aiming the center of the beam at the first device, or you can adjust aim to the middle of the two locations. The antenna's actual beam pattern will tell you the expected decibel value at 1 degree vs 2 degrees.
*** BTW I used to do this trick with 5Ghz backhauls - it works.
photo shows a grid: »Re: low on budget middle of nowhere
which is feeding two other grids: »Re: Shot through glass. |
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WHT
join:2010-03-26
Rosston, TX
kudos:5 | reply to Smokeshow
said by Smokeshow:Your thinking the arc should be longer is what's flawed. The circle would be drawn with a radius equal to the length of the adjacent side, not the hypotenuse. OK...Makes sense. Should have thought it out more, but with dogs whining for attention... You know. |
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