Skip to Content

APRS RF Bandwidth @ 1200 baud

Here is an explanation of APRS and the effects of channel traffic on it's performance. (Bandwidth)

In this example a mobile stations transmission rate is set to 30 sec.
(Not recommended in practice)

Packets are transmitted at a maximum speed of 1200 bits per second.

So assuming an APRS posit (position packet) is 50 bytes in size from a non compressed tracker or IGate (IGates decompress compressed packets before retransmission) the packet will consists of 400 bits of data plus (from memory) 296 bits for ax.25 protocol overheads.

Total number of packet bits 696

The time taken to transmit a packet is DataTime + TXDelay +TXTail In this case 0.58 + 0.35 +0.04 = 0.97 Sec ~ 1 Second per packet.

This packet is transmitted 3 times assuming the longest mobile path of WIDE1-1,WIDE2-2. So in effect this one packet is taking up 3 seconds of RF time or available bandwidth.

Therefore you could theoretically have a maximum of 20 (60sec / 3) mobile stations transmitting one posit every minute or 10 mobile stations transmitting one posit every 30 seconds before all available bandwidth would be used. ( No transmissions from home stations or objects.)

In practice and due to the nature of AX25 (as a CSMA protocol with hidden transmitters etc.) the best bandwidth you could expect for maximum efficiency (90% of txed data delivered from point A to point B) is 30% to 40% of the total bandwidth. So in reality you could have 4 mobile stations transmitting posits through 2 hops (digipeaters) every 30 seconds before you start creating packet collisions and loosing data on the network. Or you could have 24 stations transmitting through 2 hops every 3 minutes. This again does not include home stations or objects.

This is how we get the figure of 100 packets (of 80 bytes) per 10 min as an average for the VK network. (equals 33% of the bandwidth)

As you can see packet radio is not all that efficient especially with a large number of station or digis on a network. As APRS uses a the non-connected (UI) feature of ax.25 it is possible to get away with using a bit more of the bandwidth (approx. 50%) with reasonable efficiency (maybe 70%). This would be considered the Peek Traffic Load of the RF network.

If the load went to around 60% of the bandwidth you would find that only one in 5 or 6 of the txed packets would get through one hop. (In our mobile example, txing one posit every 30 sec in this network would mean that a posit would be rxed by the receiving station one digipeater away on average every 2.5 to 3 minutes. A receiving station two digipeaters away would rx one posit every 5 to 6 minutes)

Note that all these figures are approximations to give you an understanding of how a packet network could perform. So much of the practical side of the systems is dependent on local conditions.

A few more thoughts:

At a tx rate of 30 sec you will get an updated position every 0.83 Kms when traveling at 100 kph, at 3min its 5kms per update. What is the resolution and accuracy of the maps and equipment you are using? I Km per pixel, 500 mtrs per pixel or 1 mtr per pixel.

 

Copyrightpage |