I was reading about GPR devices recently.

Considering the operation frequency of these devices, how they can go that deep (up to 50 feet)? Normally you would only expect a Hz to KHz range signal to go that deep into the ground.


2 Answers 2


There is a special adaptation of the Friis equation, called the Radar Equation, that describes the range of a radar system. Here is a basic version that calculates the maximum range of a radar system:

$$R_{max}=\sqrt[4] {\frac {P_tG^2\lambda^2\sigma}{P{_{r(min)}}*(4\pi)^3*L}} \tag 1$$

where $R_{max}$ is the maximum range in meters, $P_t$ is the transmit power in watts, $G$ is the linear gain the combined transmit and receive antenna, $\lambda$ is the wavelength in meters of the frequency, $\sigma$ is the effective radar target cross section in square meters, $P_{r(min)}$ is the minimum receiver power required in watts, and $L$ is a linear loss factor to consolidate all losses other than free space effects encountered in the radar path.

So if we take a 2 GHz radar (0.15 meters) with 10 watts of transmit power, a receiver with 0.25 $\mu$V of sensitivity (1.25e-15 watts for a 50 ohm system), an antenna with a linear gain of 31.6 (15 dBi), a 1 square meter metal target and no other path losses, we calculate a maximum effective distance of 3,085 meters (10,121 feet).

We would now need to account for the losses of earth (rock, soils, sands and water) to apply this to a ground penetrating radar. This can be factored into $L$ in equation 1. I do not have data at hand to suggest what the possible attenuation factors would be for various forms of earth so instead, it may be instructive to rework equation 1 to solve for the maximum attenuation that can be accommodated over a specified distance:

$$L={\frac {P_tG^2\lambda^2\sigma}{P{_{r(min)}}*(4\pi)^3*R_{max}^4}} \tag 2$$

If we set $R_{max}$ to 15.24 meters (50 feet) as questioned by the OP and leave all other parameters the same, we find that an additional path loss ($L$) of ~92 dB (~1.68e9 linear attenuation) could be accommodated before the radar would not be able to detect a signal from a 1 meter square metal target 15.24 meters below the surface.

So armed with these basic formulas, one can run various "what if" scenarios to get a sense of what is possible. Do keep in mind that this is a very simplified form of the radar equation that does not 'reflect' (pun intended) all of the factors involved.


I did find some very basic soil attenuation data at https://archive.epa.gov/esd/archive-geophysics/web/html/ground-penetrating_radar.html. Here is an extract of attenuation in the 40 to 1,500 MHz range:

enter image description here

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    $\begingroup$ This! Exactly this! Together with significant processing gain. If interested, I recommend looking into the topic of Synthetic Aperture Radar imaging; the rough idea is that you take many "snapshots" with an antenna with a very wide aperture (i.e. little directivity) that mostly overlap, and then mathematically combine these many overlapping low-resolution snapshots to one large high-resolution snapshot. The overlapping is done by the satellite doing the imaging moving around earth on its orbit. $\endgroup$ Jan 4, 2018 at 17:37
  • $\begingroup$ So I was correct, the lower the frequency (longer wavelength) the further it penetrates. Also I noticed it is a bit counter intuitive, but makes sense if you think about it, that if it reflects more (more water content, or harder,denser rock for example) it will penetrate less. $\endgroup$ Jan 5, 2018 at 0:50

Although I admit more ignorance than knowledge in this area, I remember reading many places (don't remember where) about how deep radio waves penetrate the earth with regards to transmitting and earth losses (and how to avoid the losses). I remember that even at HF the penetration can easily be be several feet, and if memory serves correctly, it gets deeper with lower frequencies. Although it may not be related (again I plead ignorance) it reminds me of the skin effect, which causes the signal to be closer to the surface of the conductor as the frequency rises (less "deep" in the conductor). It is therefore not unreasonable to believe that at these ELF and VLF frequencies the signals can penetrate the ground to a very deep level. Hope this helps.

  • $\begingroup$ Wow, I'm really impressed by the detailed responses I received. I have to confess I barely understand most of these formulas but I can understand the logic for these calculations. Recently I become interesred in metal detection functions and their limits in detecting objects. I did a lot of research about major metal detector companies and I saw some infamous companies claim their metal detector can go very deep by implementing larger antennas but then I thought if it was that simple why I haven't seen such products from well-known companies? Are they lazy making such metal detectors? Or.... $\endgroup$
    – Samk80
    Jan 5, 2018 at 6:44

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