How big difference will it be between UV-5R 4W & newer UV-5R 8W (difference between 4W & 8W), in real life use 1 on 1 HT. In SAME conditions, how much different distance / range will we get? (simple example)

So I want to buy two HTs and change stock antennas with Nagoya NA-701, and major things that I want to use them for are:

  • Rare occasions use and for about 8-10 hours top max until it will be near charger;

  • will not use 8W all the time, just in case when other HT is out of reach with 4W (between two ski resorts, one car start the trip few minutes before other one..);

  • mountain - skiing, to get friends locations, to meet up;

  • while travelling by car, with friends (few km separated);

I read A LOT of topics 4W vs 8W, and most of people say it's too small difference, few are saying with hard evidence that it's a big difference (Link added).. here are some quotes (for helping other people that read this):

I hope that you will help me, and that this thread will help others while still there are not so much info & tests about "UV-5R 8W"

To simplify, here's the example situation:

Talk between two same HTs (4W to 4W or 8W to 8W), with line-of-sight (mountain to valley or mountain to another mountain) that means no earth curving, could we produce a situation when 4W to 4W is not enough and in same spots 8W to 8W will work?

And approximately what would be a difference in KM / Mile? or would it be more likely in Meters maybe?

Because I'm not interested in repeater just in one on one HT use, and use of 8W would only be when other HT is out of reach with 4W, also I would use 8W just to set meeting point... so in that kind of use, I don't need to worry about "Brain radiation" or whatever.

That is what I'm looking for here, in real life, will that work or not and what would be the difference in distance for the same spots if it works.

I'm new to radios / frequencies / dBs and all that comes to it, but thinking would it be good to spend little more $ for 4W more (for future needs) or it's a waste of money.

  • $\begingroup$ If you think of a signal meter on another type of receiver, they generally go up to 20. Every 3 is twice the power as the previous three. So From 4W to 8W is 3 dB. The biggest difference is in battery life, and the RF exposure to your brain. $\endgroup$
    – SDsolar
    Commented May 6, 2017 at 2:09
  • 2
    $\begingroup$ @SDsolar Please don't use comments to post short answers. $\endgroup$
    – Kevin Reid AG6YO
    Commented May 6, 2017 at 2:23
  • $\begingroup$ I didn't have time to post a very complete answer. But I had to throw in the RF exposure to the brain part. At UHF, it is not negligible to pump that much power into your cranium. Meanwhile, I see the answers are pretty good. So this post was worth @SixOne doing. Good info for the database. $\endgroup$
    – SDsolar
    Commented May 6, 2017 at 17:49

4 Answers 4


4W to 8W is a doubling of power, or a 3dB increase. That's 3dB you can add to your link budget. Provided of course that the limiting factor is the other station hearing you, not the other way around.

Let's say with 4W you can be heard up to 10 miles away. In idealized conditions (no terrain in the way, no interference, etc) 8W increases your range to 14.1 miles.

Why? According to the inverse square law, irradiance is inversely proportional to the square of the distance.

$$ \text{irradiance} \propto {1 \over \text{distance}^2} $$

That means multiplying the distance by $x$ will require an increase of $x^2$ in EIRP to maintain the same irradiance, which can be made by increasing transmitter power or antenna gain.

We must also consider the radio horizon. Even if there are no hills in the way, the curvature of the Earth will get in the way at some distance. The horizon (in miles) for a station at some height (in feet) is approximated by:

$$ \text{horizon} = 1.41 \sqrt{\text{height}} $$

For height in meters and horizon distance in kilometers, change the constant to 4.12.

This approximation takes into account the shape of a perfectly round Earth, and a "bonus" to account for atmospheric bending of the signal. It doesn't account for trees, buildings, or hills.

Let's say your HT is 4 feet high. Your horizon is $1.41\sqrt{4} = 2.82$ miles. And the repeater is on a 50 foot tower: $1.41\sqrt{50} = 9.97$ miles. The horizons add, so beyond about 12.8 miles you no longer have a line of sight, and adding more power won't do much to increase range.

  • $\begingroup$ Godlike :) perfectly clear. $\endgroup$
    – SxOne
    Commented May 6, 2017 at 12:32

The exact answer is given by the Friis equation. If you double the power, your range will increase by 1.413 times whatever it was before doubling the power if you change nothing else. This assumes you do not violate line of sight rules and that there are no additional obstructions in the longer path.

You could get this same effect by doubling the gain of one of the antennas.

On the other hand, if you doubled the gain of both antennas and left the power as it was (4 watts), your distance is nearly doubled and you do not have the additional battery drain. The extra distance is because you are increasing the effective transmit power and the effective receive gain.

  • $\begingroup$ but then we can increase distance with both antenna changes, and with both 8W HTs right ? with your first paragraph in mind, if the distance with 4W is 5km, then with 8W it would be 7W (roughly of course)? $\endgroup$
    – SxOne
    Commented May 6, 2017 at 12:15
  • $\begingroup$ That would increase your range by 2.82 times (same qualifiers as above) from the base conditions. $\endgroup$
    – Glenn W9IQ
    Commented May 6, 2017 at 13:06
  • $\begingroup$ "line of sight" not "line of site" - but great answer! $\endgroup$ Commented Nov 18, 2017 at 14:49
  • $\begingroup$ The Friis equation only really applies in free space so is it really an "exact" solution? $\endgroup$
    – AG5CI
    Commented Nov 18, 2017 at 23:42

Of your quoted testimonials, the one I'd put the least credence in is the 10 miles at 4w and 16 miles at 8W, because I'm not sure what they mean by the "elevation difference" disclaimer; at these frequencies even a small difference in location, especially in elevation, can make much more difference than doubling wattage. The rest seem pretty much spot on.

4W vs 8W is slightly more than mosquito-fart difference, but not much. Most of the time it will just be a waste of battery. If the price difference means you can get the 4W plus a better antenna instead of the 8W, that's what I'd do. If you can afford both, get the 8W radio and a better antenna. Options are good to have even if you seldom use them, and those extra few dB might make the difference one day. You can always transmit at medium or low power when battery life matters.

  • $\begingroup$ here is liink with details of that story about "elevation difference" forums.qrz.com/index.php?threads/… $\endgroup$
    – SxOne
    Commented May 6, 2017 at 9:30
  • $\begingroup$ yeah, antenna for the price difference is a good thinking. can you simplify situation, with same conditions, how much different distance / range will we get 4W vs 8W? $\endgroup$
    – SxOne
    Commented May 6, 2017 at 9:32

The correct answer to your question is, I believe, that you should expect an increase of between 20-40% in range, depending on the height of the repeater(s) you are trying to hit as well as your local propagation environment (e.g. urban, suburban, free space, etc.).

I think it is a trickier question than it seems because it depends on how marginal you are with 4W to begin with. Your question can't really be answered without taking the sensitivity of the receivers of the repeaters (or other stations) you want to hit into account.

Receiver sensitivity is usually expressed in terms of some Minimum Discernible Signal (MDS) level for a given level of noise at the receiver. For FM receivers, this level is usually expressed in terms of some acceptable SINAD - SIgnal[-to-]Noise-And-Distortion ratio. A typical SINAD benchmark for a VHF FM receiver is 12 dB. One example of a typical VHF MDS can be found in this IC-R8500 Test Report from AB4OJ, which shows a 12 dB SINAD FM MDS of -121 dBm at 147 MHz (VHF). The same report shows an MDS of -114 dBm at 446 MHz (UHF). It turns out, though, that we don't need the absolute sensitivity to determine how much our range might increase.

On a log scale, the equation that gives us the link range is, in general:

$$MDS = P_{T,dBm} + G_{T,dB} - L_{dB}(f,h_B,h_M,d_{max}) + G_{R,dB}$$

where $G_T$ is the antenna gain at the transmitter (e.g the UV-5), $G_R$ is the antenna gain at the receiver (i.e. repeater), $P_T$ is the transmitter power, and $L$ is the propagation loss factor. The latter depends, in general, on the frequency of operation, the antenna heights, and the distance between the two stations. The M subscript stands for our portable ("mobile") and the B subscript stands for our target repeater ("base").

If frequency and antenna heights are held constant, then, for a given MDS, we would have a low power case of:

$$MDS = P_{T,dBm} + G_{T,dB} - L_{dB}(f,h_B,h_M,d) + G_{R,dB}$$

and a high power case:

$$MDS = P_{T,dBm} + 3 dB + G_{T,dB} - L_{dB}(f,h_B,h_M,d') + G_{R,dB}$$

To calculate the increase in range, we subtract the first equation from the second to obtain:

$$ 3 dB - L_{dB}(f,h_B,h_M,d') + L_{dB}(f,h_B,h_M,d) = 0 $$

"Real World" propagation

Others have suggested that $L$ is proportional to $1/d^2$. This is certainly true for free space (viz. the Friis equation), but in general the dependency on distance can vary widely depending on the propagation environment. The Okumura-Hata model includes a comprehensive set of empirical equations based on field measurements under various conditions. According to the Okumura-Hata model:

$$L_{dB}(f,h_B,h_M,d) = K_0(f,h_B,h_m) + K_1(h_B)\log_{10}d$$

where, with $h_B$ in meters:

$$ K_1(h_B) = 44.9 - 6.55\log_{10}h_B $$

Since the $K_0$ term cancels, we need to solve:

$$ 3 - K_1(h_B)\log_{10}d' + K_1(h_B)\log_{10}d = 0 $$


$$ \log_{10}d' - \log_{10}d = \log_{10}\left(\frac{d'}{d}\right) = \frac{3}{ K_1(h_B)} $$

If we raise both sides to the power of 10 we get:

$$ \frac{d'}{d} = 10^{3/K_1(h_B)} $$

You can substitute your own local values, but assuming if the repeater tower is 20 m tall, then:

$$K_1(h_B) = 44.9 - 6.55*\log_{10}(20) = 36.4$$

so that:

$$ \frac{d'}{d} = 10^{3/36.4} = 1.21$$

Thus, in this particular scenario, doubling your power from 4 W to 8 W would result in your increasing your range - in the sense of reaching new repeaters - by about 20%.

Note that all of the above analysis presumes (1) that the Okumura-Hata model is accurate and (2) that the propagation characteristics are uniform in all directions. Neither of these, of course, will be strictly true, but at least the model provides a non-arbitrary means with some grounding in field truth of answering your question.

Comparison to free space ("line-of-sight")

If we are in free space, then, as others have suggested, the Friis equation applies and we have:

$$L_{dB} = 10\log_{10}4\pi + 20\log_{10}d$$


$$K_1(h_B) = K_1 = 20$$


$$ \frac{d'}{d} = 10^{3/20} = 1.41$$

which implies that our reach would increase by about 40%.

Thus, overall we can say that doubling your power should increase your range (again - meaning the range within which you are able to "hit" repeaters) by between 20% and 40%, depending on the extent to which you have "line-of-sight".

(I am assuming I didn't screw up the math along the way. If I did, hopefully someone will correct me).

  • $\begingroup$ +1 for an a great answer going back to first principles! (I haven't double-checked your math.) Thanks for the introduction to the Okumura–Hata model. As a suggestion, before posting a long answer you might go back and re-read the question. In this case you didn't explicitly address the OP's questions about antennas, although others did of course. Anyway, welcome belatedly to the group, and I'll look forward to more from you! $\endgroup$
    – rclocher3
    Commented Nov 18, 2017 at 17:27

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