I am wanting to setup a simplex path that is 7KM from base to base in relatively flat ground.

How can I estimate the path loss, and what do the numbers mean when I find them?

I am wanting to do it at the 2 meter band, and there are houses and buildings in between

My latitude is 39.76256409 , longitude is -104.86949757, and my friends latitude is 39.73935726 and longitude is -104.95158389

We will probably use cheap easy-to-build 7db yagis on the roof (3 stories or so) if that will work

  • $\begingroup$ Tell us a bit more about band you plan to use, antenna height, buildings in the area... If it's just flat ground, then two-ray model could be useful. $\endgroup$
    – AndrejaKo
    Mar 9, 2014 at 14:32
  • $\begingroup$ ok, I added a few things $\endgroup$
    – Skyler 440
    Mar 9, 2014 at 14:47
  • $\begingroup$ OK, so urban area, low buildings, no large, high building in between. I'll see what I can dig up. $\endgroup$
    – AndrejaKo
    Mar 9, 2014 at 15:04
  • $\begingroup$ You can calculate the best case path loss with the Friis transmission equation. Then it's just a matter of estimating the additional attenuation due to other stuff in the way, which is highly variable. $\endgroup$ Mar 9, 2014 at 20:43
  • $\begingroup$ Related: What is a link budget, and how do I make one? $\endgroup$
    – user
    Mar 10, 2014 at 13:27

1 Answer 1


How can I estimate the path loss

The answer seems to be to pick one from a catalog of available path-loss models and apply the formula.

Here, I'd try using the Hata model for urban areas. The model was originally made for portable cellular systems, so it uses a bit different terminology.

The formula is not so complicated:

$$L_{50} =69.55 +26.16 \cdot \log_{10}f -13.82 \cdot \log_{10}h_b -C_h +(44.9-6.55 \cdot \log_{10}h_b) \log_{10}d$$

and for 2 m band the correction factor is

$$ C_h = 8.29(\log_{10}(1.54 \:h_m))^2 - 1.1 $$


  • $f$ is the frequency in MHz
  • $h_{b}$ is the height of the base station antenna in meters
  • $h_{m}$ is the height of the mobile station antenna in meters
  • $d$ is the distance between the base antenna and mobile receiver antenna in kilometers

what do the numbers mean when I find them?

Basic formula for power at the receiver is:

$$ P_{rx} = P_{tx} - P_L $$


  • $P_{rx}$ is the power received, and
  • $P_{tx}$ is the transmit power, and
  • $P_{L}$ is all losses (path loss, feedline loss, etc).

The Hata model I've shown here will provide path-loss estimate for 50% of locations on the diameter of the circle with center in the "base station" and radius $d$. I personally, as a rough guide, would just use that number as $P_L$.

Now let's try to calculate the number. I'll take the frequency to be 145.5 MHz, which is a bit below the optimal range for the use of this model. The given height is 3 stories plus roof, so I'll round that to 10 meters.

First, the correction factor:

$$ \begin{align} C_h &= 8.29\left(\log_{10} \left(1.54 \cdot \frac{10 \ \mathrm{m}}{1 \ \mathrm{m}}\right)\right)^2 - 1.1 \\ &= 10.6 \ \mathrm{dB} \end{align} $$

Next, the main formula:

$$ L_{50} =69.55 +26.16 \cdot \log_{10}\left(\frac{145.5 \ \mathrm{MHz}}{1 \ \mathrm{MHz}}\right) -13.82 \cdot \log_{10}\left(\frac{10 \ \mathrm{m}}{1 \ \mathrm{m}}\right) -10.6 +\left(44.9-6.55 \cdot \log_{10}\left(\frac{10 \ \mathrm{m}}{1 \ \mathrm{m}}\right)\right) \log_{10}\left(\frac{7 \ \mathrm{km}}{1 \ \mathrm{km}}\right) $$ $$L_{50}=69.55+56.58-13.82-10.6+38.35 \cdot 0.85 $$ $$L_{50}=134.3 \ \mathrm{dB}$$

So what does this number give us and how is it helpful?

The antennas are given as 7 dB Yagi, so I'll take that to be 7 dBi and I'll assume that you have 10 m of say RG-8 coaxial cable (a calculator gives me 0.5 dB loss for that) and that you're transmitting at 35 W.

The transmitter EIRP would be then:

$$P_{tx} =10 \cdot \log_{10}\left(\frac{35\mathrm W}{1\mathrm W}\right) +7-0.5 =21.94\:\mathrm{dBW} =51.94\:\mathrm{dBm}$$

Our estimate of path loss is 134.3 dB and we have a same Yagi at the receiver plus same coax losses. This gives us received power of: $$P_{rx}=51.94-134.3+7-0.5=-75.86 \ \mathrm{dBm}$$ which should be above S9 at the receiver, if my calculations are correct and the S-meter is properly calibrated.

I hope that my description of the process will let you do your own more exact calculations and help you a bit.

  • $\begingroup$ @Phil Frost I actually picked 7 by looking at neighboring building and the result was bad... At some locations, it's 7 meters for the antenna height and at some other locations, it's 7 km as the distance between the Tx and Rx. Also at the same time, completely removed units because they could complicate the logarithms. That wasn't the best idea, so I'll fix that now. $\endgroup$
    – AndrejaKo
    Mar 9, 2014 at 21:53
  • $\begingroup$ I think the EIRP calculation is wrong; I get $10\cdot \log_{10}(35\:\mathrm W / 1\:\mathrm{mW}) + 2.5 - 0.5 \approx 47.44 \:\mathrm{dBm}$. $\endgroup$ Mar 10, 2014 at 3:33
  • 1
    $\begingroup$ Also, 2.15dBi is the gain of a dipole in free space. Over Earth the gain is more. The question suggests 7dB antennas (though, 7dBi, 7dBd...it doesn't specify). You could go with that (though that's not better, and maybe worse, than a favorably positioned dipole). $\endgroup$ Mar 10, 2014 at 3:40
  • $\begingroup$ @Phil Frost Thanks for data about the dipole. For dB, I missed the 10 times factor :). $\endgroup$
    – AndrejaKo
    Mar 10, 2014 at 8:10

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