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What Is The Range Of A Plane Radio?  
User currently offlineLenbrazil From Brazil, joined Apr 2006, 114 posts, RR: 0
Posted (8 years 6 months 2 weeks 4 days 7 hours ago) and read 8299 times:

What is the range of a plane radio broadcasting on a CTAF/Unicom frequency?

More specifically would it be possible for a transmission from a King Air at 1800 ft / 440 AGL be picked up at an airport (Sky Harbor) at 610 ft 45 – 50 miles away? It may have been snowing at the time.

Len

47 replies: All unread, showing first 25:
 
User currently offlineLuisca From , joined Dec 1969, posts, RR:
Reply 1, posted (8 years 6 months 2 weeks 4 days 7 hours ago) and read 8288 times:

It depends on several things, like the conditions of the atmosphere and the altitude of the plane, also the power of the transmitting radio is important.

Quoting Lenbrazil (Thread starter):
More specifically would it be possible for a transmission from a King Air at 1800 ft / 440 AGL be picked up at an airport (Sky Harbor) at 610 ft 45 -- 50 miles away? It may have been snowing at the time.

It is entirely possible.

A good rule of thumb is that the transmitter is line of sight with the receiver you can hear it. I have listened to MLB ATIS from 60 miles away.


User currently onlineMD11Engineer From Azerbaijan, joined Oct 2003, 14089 posts, RR: 62
Reply 2, posted (8 years 6 months 2 weeks 4 days 7 hours ago) and read 8286 times:

If it is VHF, then it is a direct line towards the electrical horizon (a little bit more than the optical horizon due to different difraction of EM waves at this frequency). So the range depends largely on altitude (up to 2-300 km).

HF (shortwave) transmissions can propogate due to two means:
Ground waves, were the signal follows the shape of earth with increasing loss until it peters out (disappears in the static background noise). This range is usually (depending on the antenna and transmission power) a few hundred kilometers.

Air waves: Depending on the frequency and time of day (solar activity) HF waves can be reflected at the ionosphere layers of the upper atmosphere. They can even (provided the angle of radiation at the transmitting antenna is low) bounce back and forth between the ionosphere and ground, so that HF waves can have a range from between 1000 km and global range.

Jan

[Edited 2006-04-17 19:57:50]

User currently offlineDTW757 From United States of America, joined Oct 2003, 1583 posts, RR: 4
Reply 3, posted (8 years 6 months 2 weeks 4 days 6 hours ago) and read 8255 times:

Oh I would think you should be able to hear an airplane using the CTAF 50 miles away with no problem at all. At my local airport KDUH we use a CTAF and I have heard calls being made from as far as 100nm while I was at pattern altitude.


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User currently offlineLenbrazil From Brazil, joined Apr 2006, 114 posts, RR: 0
Reply 4, posted (8 years 6 months 2 weeks 4 days 5 hours ago) and read 8181 times:

Most of you seem to be saying it's possible, how probable is it (definate, probable, possible)? Remember I'm asking about plane to airport not plane to plane. For those of you familiar with Northern Minn the plane in question would have been a few miles east of Eveleth the nearest airport with the same CTAF is Sky Harbor just SE of Duluth.

Len


User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 5, posted (8 years 6 months 2 weeks 4 days 5 hours ago) and read 8157 times:

Theoretical VHF reception range (nautical miles) is equal to 1.23 times the square root of the altitude (above the receiver). I presume you mean the aircraft was at 1800 feet above sea level ?

Then 1800 - 610 = 1190 feet above Sky Harbor

Reception range equals = 1.23 times (SQR 1190) = 42 nautical miles (48 statute miles).

JC

[Edited 2006-04-17 21:59:09]

[Edited 2006-04-17 21:59:55]

User currently offlineTimz From United States of America, joined Sep 1999, 6886 posts, RR: 7
Reply 6, posted (8 years 6 months 2 weeks 4 days 4 hours ago) and read 8146 times:

By "theoretical" you mean "assuming the earth is a sphere"-- right? In this case the aircraft was 440 AGL so we can't be sure the terrain between it and the airport would allow a line of sight. But presumably the airport antenna is not at ground level, and its height adds a few miles too.

User currently offlineJspitfire From Canada, joined Feb 2005, 308 posts, RR: 2
Reply 7, posted (8 years 6 months 2 weeks 4 days 4 hours ago) and read 8138 times:

Quoting JetCaptain (Reply 5):
Theoretical VHF reception range (nautical miles) is equal to 1.23 times the square root of the altitude (above the receiver). I presume you mean the aircraft was at 1800 feet above sea level ?

Then 1800 - 610 = 1190 feet above Sky Harbor

Reception range equals = 1.23 times (SQR 1190) = 42 nautical miles (48 statute miles).

JC

I was just trying to post this exact reply, but the tread was moved here from the Civ Av forum, so you beat me to it. Just one small correction, however, it was actually recently changed to 1.25 instead of 1.23, not that it makes too much of a difference.

Jason


User currently offlineLenbrazil From Brazil, joined Apr 2006, 114 posts, RR: 0
Reply 8, posted (8 years 6 months 2 weeks 4 days 4 hours ago) and read 8133 times:

Quoting JetCaptain (Reply 5):
Theoretical VHF reception range (nautical miles) is equal to 1.23 times the square root of the altitude (above the receiver). I presume you mean the aircraft was at 1800 feet above sea level ?

Then 1800 - 610 = 1190 feet above Sky Harbor

Reception range equals = 1.23 times (SQR 1190) = 42 nautical miles (48 statute miles).

So 42 nautical miles is a best case senario? according to airnav.com it's 46.1 nautical miles so it would have been just out of range? They also say it's 37.1 from DLH to EVM the plane had been talking to Duluth ATC were they going out of range?


User currently offlineTimz From United States of America, joined Sep 1999, 6886 posts, RR: 7
Reply 9, posted (8 years 6 months 2 weeks 4 days 4 hours ago) and read 8128 times:

No, nobody knows whether there actually was a line of sight in this case, since nobody knows about the terrain between. Could be a bit better, could be a bit worse. And in any case you have to apply the formula twice, once for the aircraft and once for the ground antenna height.

User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 10, posted (8 years 6 months 2 weeks 4 days 1 hour ago) and read 8059 times:

>> So 42 nautical miles is a best case senario? <<

Not necessarily, but under normal conditions with typical equipment it should be close. How high was the receiving antenna ? A good exterior antenna 30 feet above the gound would add about 7 nautical miles to the reception range.

JC


User currently offlineLenbrazil From Brazil, joined Apr 2006, 114 posts, RR: 0
Reply 11, posted (8 years 6 months 2 weeks 4 days ago) and read 8036 times:

What is the range of a plane radio broadcasting on a CTAF/Unicom frequency?

More specifically would it be possible for a transmission from a King Air at 1800 ft / 440 AGL be picked up at an airport (Sky Harbor) at 610 ft 45 – 50 miles away? It may have been snowing at the time.
Sorry if I’m being a bit dense but wouldn’t increasing the height of the transmitter reduce the range? Let’s suppose it’s 30 feet off the ground that would place it at 640 ft. ASL so it would be 1800 – 640 = 1160 Sq. root = 34 x 1.25 = 42.5 (or x 1.23 = 41.8).

But let’s suppose the receiving antenna did add about 7 miles the maximum would be 49 miles but could be less.

If the plane was 44 – 46 miles from Sky Harbor people at that airport might have its transmissions but then again might not depending on terrain weather and other factors correct?

I have no idea what kind of antenna the have there, other than getting in touch with them how could I find out?

Len


User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 12, posted (8 years 6 months 2 weeks 3 days 20 hours ago) and read 7983 times:

>> Sorry if I’m being a bit dense but wouldn’t increasing the height of the transmitter reduce the range? <<

No it increases it. You need to consider the reception range at each location, then add them together.

range = (1.23 times (SQR 1190)) + (1.23 (SQR 30))

which is the same thing as saying 1.23 times (SQR 1220)

Terrain could be a factor, but in this case there doesn't appear to be any that would impact the line of sight between EVM (elevation 1378') and DYT (elevation 610').

So I'd say it "could" have been theoretically possible if they had a good antenna and radio at DYT, but the signal would probably have been weak, and the squelch would have had to be turned down or off on their radio.

So what exactly are you getting at anyway ? I'm guessing something to do with this ?

http://www.ntsb.gov/publictn/2003/AAR0303.htm

JC


User currently offlineMir From United States of America, joined Jan 2004, 21730 posts, RR: 55
Reply 13, posted (8 years 6 months 2 weeks 3 days 17 hours ago) and read 7947 times:

Sky Harbor is a little airport with a little FBO that caters to small GA planes. At bigger FBOs that serve the more corporate crowd, each FBO may have its own radio frequency that pilots can call in on to tell them that Mr. Executive will be arriving in thirty minutes, so have the Mercedes waiting and make sure the air conditioning is set to 72 degrees. You're not going to find that at DYT - there will be a little radio in the FBO, and people will listen to it if they're in earshot, but more often than not they'll be out working on something else on the airport - it's small enough that they can pause what they're doing and help out if they see someone taxiing in. According to Airnav, DYT only gets 38 operations per day on average, so whoever works at the FBO probably has a lot of free time on their hands. I'd put money on DYT having no fancy reciever, and I'd also put money on them not hearing the transmissions that you refer to.

-Mir



7 billion, one nation, imagination...it's a beautiful day
User currently onlinePihero From France, joined Jan 2005, 4609 posts, RR: 77
Reply 14, posted (8 years 6 months 2 weeks 3 days 15 hours ago) and read 7926 times:
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Come on, guys !
The initial formula was wrongly used : 1.23 x SQRT (A) gives a rough estimate of the line-of-sight distance [i.e]the distance to the apparent horizon[/b]. and that's the effective transmission/reception range of any given station antenna
Now, another station is bound to the same earth curvature limitations, governed by the same formula.
That means that you'll have to add-up the two ranges in order to find the maximum distance at which the two stations could talk to each other
In the initial problem, D max = 1.23 x SQRT (1800) + 1.23 x SQRT (610),
or 1.23 x [SQRT (1800) + SQRT (610) ]
so, 1.23 x [42 + 25 ]---->1.23 x 67 = 82.4Nm

Quoting JetCaptain (Reply 12):
range = (1.23 times (SQR 1190)) + (1.23 (SQR 30))

which is the same thing as saying 1.23 times (SQR 1220)

That must be a typo...Right ?



Contrail designer
User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 15, posted (8 years 6 months 2 weeks 3 days 11 hours ago) and read 7880 times:

Quote:
The initial formula was wrongly used

No it wasn't.


Quote:
In the initial problem, D max = 1.23 x SQRT (1800) + 1.23 x SQRT (610),
or 1.23 x [SQRT (1800) + SQRT (610) ]
so, 1.23 x [42 + 25 ]---->1.23 x 67 = 82.4Nm

This is wrong because 1800 feet is the altitude of the aircraft above SEA LEVEL, you need to use the altitude above GROUND LEVEL. And 610 feet is the elevation of the DYT airport, where above ground level would be 0.

In this case the reception range from the aircraft will vary depending on the direction because the ground is not level in this area. The ground rises as you go from south to north. The elevation of DYT is 610 feet while the elevation of EVM is 1378 feet. However it is a fairly constant rise and would not impact the field of view from an aircraft that is 422 feet above EVM (1800 feet ASL).

To get the reception range to the south in the direction of DYT you need to consider the relative altitude between the aircraft and ground in that direction. ie: 1800 - 610 = 1190 feet above GROUND LEVEL. The 42 nautical miles in the first example would be correct looking from the aircraft towards DYT to the south (if the receiving antenna is at ground level).

Quote:
which is the same thing as saying 1.23 times (SQR 1220)

Yes that was a typo.

Quote:
I'd put money on DYT having no fancy reciever, and I'd also put money on them not hearing the transmissions that you refer to.

I would agree that this is probably the more "practical" answer. In my experience little FBO's don't have the greatest radio setup and would have the squelch turned up to block out weak signals. But with good equipment and the squelch turned off I'd say it could be possible.

JC


User currently offlineKELPkid From United States of America, joined Nov 2005, 6415 posts, RR: 3
Reply 16, posted (8 years 6 months 2 weeks 3 days 9 hours ago) and read 7848 times:

Don't forget, too, that a King Air is most likely to have "big" airplane radios that run off the 400 Hz AC bus, and are more powerful (i.e. the radio has a higher output wattage) than your typical GA aircraft radio...

 stirthepot 



Celebrating the birth of KELPkidJR on August 5, 2009 :-)
User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 17, posted (8 years 6 months 2 weeks 3 days 9 hours ago) and read 7843 times:

Quote:
Don't forget, too, that a King Air is most likely to have "big" airplane radios that run off the 400 Hz AC bus, and are more powerful (i.e. the radio has a higher output wattage) than your typical GA aircraft radio...

Yeah but on the other hand in this particular case the antennas were probably iced up and the aircraft was out of control in a 90 degree bank with the antennas possibly pointed away from DYT.

JC

[Edited 2006-04-18 17:57:03]

User currently offlineSlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 68
Reply 18, posted (8 years 6 months 2 weeks 3 days 8 hours ago) and read 7823 times:

Quoting Jspitfire (Reply 7):
it was actually recently changed to 1.25 instead of 1.23,

The geometry has not changed. Probably someone started teaching 1.25 instead of 1.23 because anyone can do the 'one and a quarter' math in their head.

My nephew is a container ship captain and he says maritime use is 1.17 X the square root of your height above the sea. Perhaps that accounts for the bending of light by the increased water vapor in the air just above the sea, I don't know.

The 1.23 is probably not going to work for PHX though, as there are mountains in every direction within the distance he stated.



Happiness is not seeing another trite Ste. Maarten photo all week long.
User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 19, posted (8 years 6 months 2 weeks 3 days 8 hours ago) and read 7820 times:

Quote:
The 1.23 is probably not going to work for PHX though, as there are mountains in every direction within the distance he stated.

The last time I checked the earth hasn't changed shape, so the 1.23 factor hasn't changed. He's not talking about the Sky Harbor in Phoenix, he's talking about Sky Harbor, KDYT, in Minnesota (near Duluth).

JC


User currently offlineSlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 68
Reply 20, posted (8 years 6 months 2 weeks 3 days 6 hours ago) and read 7798 times:

Quoting JetCaptain (Reply 19):
The last time I checked the earth hasn't changed shape, so the 1.23 factor hasn't changed. He's not talking about the Sky Harbor in Phoenix, he's talking about Sky Harbor, KDYT, in Minnesota (near Duluth).

Well, my mistake. (that's why they call this forum "tech/oops"

I agree. 1.23 is the proper constant for the formula. I just meant that in mountainous terrain it might not be reasonable to expect the full service volume on radio reception.



Happiness is not seeing another trite Ste. Maarten photo all week long.
User currently onlinePihero From France, joined Jan 2005, 4609 posts, RR: 77
Reply 21, posted (8 years 6 months 2 weeks 3 days 5 hours ago) and read 7780 times:
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JetCaptain,
The formula used is just theoretical.
It takes into account the altitudes -MSL - of both the transmitter and the receiver.Obstacles do not come in it, and neither does the terrain slope. That's for local conditions only.
Earth curvature line-of-sight has MSL as level zero. Just imagine you're standing on the shore, watching the ships roll in and a friend of yours is in the near-by light tower.There is a yellow rubber dinghy on the horizon, coming towards you.Who is going to see it first :You or your friend ?
Now if that dinghy has a mast, how much farther will you be able to see it before you spot the dinghy underneath ?
That's what the maths say, and nothing else.
If I take your understanding of the formula further,an aircraft flying at Mt Everest altitude will never receive a transmitter held by an alpinist who's just made the top, right ? (because the "height" of the aircraft would be zero).
First you reinvent Mathematics, then you come out with a flat earth theory...
Well done, sir!



Contrail designer
User currently offlineLenbrazil From Brazil, joined Apr 2006, 114 posts, RR: 0
Reply 22, posted (8 years 6 months 2 weeks 3 days 4 hours ago) and read 7763 times:

Quoting JetCaptain (Reply 12):
So what exactly are you getting at anyway ? I'm guessing something to do with this ?

http://www.ntsb.gov/publictn/2003/AA...3.htm

Yes, this is regarding the Wellstone crash.

Conspiracy theorists esp. the author of a book called "American Assassination" make much of the lack of a distress call.

As a layman it seems to me and to pilots on this and other forums that an SOS call would have been unlikely, the pilots who had only flow together 4 times and had no CRM would have been unlikely to have made one trying to save the plane first and a drop from 440 ft. AGL after a stall at 76 KCAS wouldn't give them much time.

Later it occured to me that even IF one of the pilots had made a distress call it's unlikely anyone would have heard it. They were on the EVM CTAF frequency and the only person at EVM went outside to 'do some chores'. The closest airport that uses the same frequency is Sky Harbor so I wondered if it was possible that the call would have been picked up there. That from what I gather here is theoretically possible but unlikely.

Other factors to consider is that the plane would have been about 135 degrees behind the plane and it was heading towards the ground at (approx) 26 degrees. I said the plane was at 1800 ft. ASL but that was before the plane stalled, if the call presumably would have been made only after stall so the altitude probably would has been less.

I imagine line of sight could have been a problem. The plane was only about 440 feet AGL and 1300 feet above Sky Harbor and about 269170 feet (44.3 nautical miles) away. Unless the terrain suddenly dropped off near Eveleth chance that the LOS would have been blocked is quite high.

One more question (OK 2) doesn't the FAA or FCC space airports that use the same frequency so that one won't pick up broadcasts intended for the other*? If not for example wouldn't a plane clicking on the lights at one turn on the lights at another as well?

*If there is a rule regulating this please let me know.


User currently offlineJetCaptain From Canada, joined Dec 2000, 236 posts, RR: 1
Reply 23, posted (8 years 6 months 2 weeks 3 days 3 hours ago) and read 7756 times:

Pihero,

Quote:
Earth curvature line-of-sight has MSL as level zero. Just imagine you're standing on the shore, watching the ships roll in and a friend of yours is in the near-by light tower.

Your boat on the ocean logic doesn't work here because RADIO WAVES CAN NOT PASS THROUGH THE GROUND, airplanes are flying over the GROUND here, not the OCEAN, therefore you have to take this GROUND into account, and use the height above GROUND and not above sea level as a reference. Again 42 nautical miles from my first example is correct. Your answer that the reception range is 82 nautical miles is ridiculous.

Quote:
If I take your understanding of the formula further,an aircraft flying at Mt Everest altitude will never receive a transmitter held by an alpinist who's just made the top, right ? (because the "height" of the aircraft would be zero).

In the Everest example the maximum reception distance between an aircraft at say FL350 and an Alpinist at the top at 29,000 feet would be about 410 nautical miles. I don't know what you are talking about when you say I think it would be zero !?

JC


User currently onlinePihero From France, joined Jan 2005, 4609 posts, RR: 77
Reply 24, posted (8 years 6 months 2 weeks 3 days 3 hours ago) and read 7749 times:
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What do you make of this statement then ?

Quoting JetCaptain (Reply 15):
To get the reception range to the south in the direction of DYT you need to consider the relative altitude between the aircraft and ground in that direction. ie: 1800 - 610 = 1190 feet above GROUND LEVEL. The 42 nautical miles in the first example would be correct looking from the aircraft towards DYT to the south (if the receiving antenna is at ground level).

You converted the airplane altitude - 1800 ft - into a height over the airfield - 610 ft-. and you come out with height = 1190 ft, which is correct, value upon which you base your computation,
Or am I seeing things ?

Therefore, following this kind of logic, an airplane at 29,000 ft is at 0 ft AGL, relative to Mt Everest ---> 1.23 x SQRT (0) = 0 Nm.

Quoting JetCaptain (Reply 23):
RADIO WAVES CAN NOT PASS THROUGH THE GROUND

And as I'm only human, neither does my vision.



Contrail designer
25 JetCaptain : This is correct, but only for the area behind the mountain below 29,000 feet, this area is blocked by the mountain therefore reception range is equal
26 JetCaptain : Lenbrazil, It sounds like you have a pretty good handle on the reception issue. I don't know what distances the FAA uses when pilot controlled lightin
27 Mir : Airports certainly can (and do) share CTAF frequencies. In fact, it's good policy not to click on the airport lights from too high up, as you'll turn
28 Timz : Assuming no refraction, the actual figure would be between 1.22 and 1.23, so presumably they decided to allow for some refraction. That's nautical mi
29 Timz : Ah-- now I notice he said Assuming no refraction, and assuming the WGS84 spheroid, the figure for statute miles is between 1.22 and 1.23, depending on
30 JetCaptain : Yes, except it is NAUTICAL miles. JC
31 SlamClick : Okay, here's how you really do it. 1. Find the distance, in feet, from the center of the earth to the radio transmitter site. Multiply this distance b
32 Timz : So now you know not to trust your source. You can do the Pythagoras for yourself and confirm that 1.22-1.23 is the range for 5280-foot miles. Assumin
33 Bhill : Lenbrazil..I dont' mean to digess from your original question regarding commonly used freqencies, but with all of the "smarts" being built into airpla
34 Post contains links JetCaptain : Hmmmmm. Every source I've ever seen has used ..... 1.23 for nautical miles 1.41 for statute miles 2.28 for kilometers As just one example, the websit
35 Timz : You can see that explanation needs work-- if transmitter height equals receiver height, reception range comes out zero. Like I said, it's 1.22 to 1.2
36 Post contains images Bond007 : I don't think so. Jimbo
37 SlamClick : Okay. What do you think?
38 Post contains images SlamClick : Here's what I mean: The circle (1) is the earth. The radio transmitter is on the surface at A Your airplane receiver is at Q The line PQ is your heigh
39 Timz : Good-- a diagram. Now all you need to add is another radial line, angled downward from the center of the earth ("downward" on the diagram, not on the
40 David L : This sounds remarkably like the way maximum radar range is calculated at sea. In that case you assume (not unreasonably) that the horizon is at sea l
41 SlamClick : Which I assume to be, on a smoothly rounded earth to be the same line-of-sight we use for radio reception. Man I hope so! Hey! Maybe if I'd been able
42 Timz : That does sound unlikely, depending on where you were exactly. There are lots of 7000-9000 ft mountains, closer to you-- even if Mt Hood weren't hidde
43 David L : But cut-and-paste in our day involved scissors and glue. I don't know about you but we weren't allowed scissors and back then we hadn't discovered wh
44 SlamClick : Yeah, it is really not precisely the height above the transmitter site but the absolute difference between airplane (receiver) altitude and the nomin
45 Post contains images David L : Calculator Syndrome! If you have to work out each decimal place yourself you tend to think about how many are relevant.
46 Timz : Supposing you were at 41000 ft at 42.5 N, 114.5 W, and assuming no refraction, you would see Steens Mtn 181 nm in azim 274 degrees Strawberry Mtn 214
47 SlamClick : Well if I can't see Three-Fingered Jack I'm not going up there!
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