# EME path loss

The EME path loss relies on the Moon being an effective reflector. The transmitted power incident on the Moon is initially captured and then re-radiated. The Moon’s ability to reflect energy in this way is defined by its relative cross section area or echo area.

The scenario is like that used in billboard reflectors or passive reflectors in microwave radio. Until recently, large reflectors were installed on hilltops to reflect the microwave signal down into the next valley, thus avoiding costly relay stations.

In such a scenario, the path loss between transmitter and receiver comprises the loss in the first leg, the loss or gain of the reflection and the loss in the second or return leg.   I’ve described the scenario at the apogee in the figure below and described it further in the following text.

## The Moon as a reflector

The Earth to Moon path is modelled using the Free Space Path loss equation to give a one-way loss to the Moon of 193.7dBi at the apogee. Since the distance is less at the perigee, the loss is a little lower, but there’s only 1.2dB in it so I’ll proceed with the worst case. The return path is likewise 193.7dBi. That gives a total of 387.5dBi for go and return at the apogee.

I’ve assumed that there’s an isotropic radiator at the Moon end, doing the capture and re-radiation. But that’s overly pessimistic. In fact, the Moon is a better reflector that this – otherwise the huge path loss would render EME communications impossible.

## Billboard calculation

There are two parts to this reflection. First, I’ve modelled the Moon as a giant billboard of 3,475km diameter. The ‘gain’ of this billboard is the ratio of the power density from the passive repeater, to the power density which would exist at the same point if the passive repeater was replaced by an isotropic antenna.

I’ve calculated this gain using the effective area of the billboard in the equation Gain (dB) = 20log((πd)22) taken from microwave engineering. Variable d is the effective diameter and λ is the signal wavelength – 2 metres in my case.

It’s well established that the Moon reflects primarily from the area near the centre facing Earth. I can approximate this by assuming that the effective diameter is 0.7*3,475km, or 2,432km.   This gives me an effective gain of 143.3dB.   There is much discussion in the ham literature about how much of the Moon is an effective reflector. The 0.7 factor above comes from a paper given at an EME conference back in 2010.

Second, I need to consider that the Moon is not a perfect reflector, even in the area in which the bulk of the reflections occur. Past measurements have concluded a reflection coefficient of about 6.5%.  So, just 6.5% of the energy incident is reflected. There is some dispute in the literature suggesting a better refection coefficient (of up to 10%) at VHF frequencies, but a pessimistic 6.5% will be assumed here. This equates to a loss of 11.9dB.

## Round trip path loss

So, I’ve concluded that the total effective gain is therefore 143.3dB – 11.9dB or 131.4dB.   The EME path loss considering only the Free Space Path loss and the reflective capability of the Moon is therefore 387.5dBi + 131.4dBi or 256.1dBi. I’ve used this figure to calculate the path budget in an adjacent page.

As a final note, those interested in libration fading on EME circuits, I’ve described that here.