# Wireless research

• Wireless sensitivity (10% PER)
```54 Mbps : ~ -74dBm
48 Mbps : ~ -77dBm
36 Mbps : ~ -83dBm
24 Mbps : ~ -86dBm
18 Mbps : ~ -90dBm
12 Mbps : ~ -91dBm
9 Mbps : ~ -93dBm
6 Mbps : ~ -94dBm
```

## Calculations

```Sensitivity (ex: Infinet)
-71dBm best case 26Mbps
-95dBm worst case 1 Mbps
Output power:
18 dBm (63mW)

in meters and megahertz
FSPL(dB) = 20log(d) + 20log(f) - 27.55

precalculated values of log()
15km = 83.52
30km = 89.54
40km = 92.04
50km = 93.97
60km = 95.56

5000 Mhz = 73.979
6000 Mhz = 75.56
2400 Mhz = 67.60
900 Mhz = 59.08

For example, for 60km, 5000 Mhz, two antenna 33dbm gain
141.989 db - 33 - 33 = 75.989 - 18dbm(power) = 57.989 db expected signal
Means link will work with maximum speed, with 13.02db signal margin.

If for example we put 23 dbm Infinet with 28db antenna:
141.989 - 28 - 28 - 18 = 62.989, we get 5db worse signal

This means it is better two infinet with 63mW and external antenna (rocketdish?) with 33dbm, than 200mW big infinet

Fresnel zone (we need it clear to get 100% of signal):

r = 17.31 * sqrt(N(d1*d2)/(f*d))
where r is the radius of the zone in meters, N is the zone to calculate,
d1 and d2 are distances from obstacle to the link end points in meters,
d is the total link distance in meters, and f is the frequency in MHz.
For 60km fresnel zone are 29.98 meters.

If fresnel zone blocked more than 40% signal can become unpredictable. This means we need clearance minimum 18 meters (29.98 * 0.6).

Earth curvature: distance to horizon [km] = 3.57*sqrt(h[meter]), distance to horizon from two towers: d[km] = 3.57*(sqrt(h[meter]) + sqrt(h[meter])

Means for 30+40 meter towers visual distance 42.129 km distance, and still we need fresnel zone to be clear.
Which means effective towers height lowered -18 meters, which means 12 and 22 meters effective height, which is only enough for 30km.

This is why all software giving very bad results.

FSPL = 92.66936920 + 73.979 - 27.55 = 139.098db