Step 1: Specify the maximum through-view or open area distance Dmax
For our B3 baby monitor the Dmax=3000 (ft)
Step 2: Calculate the total attenuation due to obstacles between the camera and the monitor;
|
attenuation factor for each obstacle |
attenuation factor(short for AF) |
|
A thin wood-brick wall |
8 |
|
A concrete wall |
16 |
|
A plaster walls |
4 |
|
A glass doors and windows |
2 |
|
|
|
|
Total attenuation factor TAF=∏AF(n) |
Multiple all the ATs. If there are two or more the same obstacles just multiple several times. |
Notes：
The attenuation of wood will be less than brick and concrete, the attenuation of metal will be the biggest one.
Step 3: Evaluate the wireless communication distance
The actual communication distance is equal to Dmax divided by the total attenuation factor.
D=Dmax / TAF
For example:
(1) There are 2 plaster walls and a glass door from the living room to the baby’s bedroom, so the distance will be:
D=3000/4/4/2=93.75(ft)
It is enough for most family use cases.
(2) There are 2 Plaster walls and a Concrete wall from the owner bedroom to the baby’s bedroom, so the distance will be:
D=3000/4/4/16=11 (ft)
So it is not enough for this use case.
(3) There are a plaster wall and a concrete wall from the yard to the baby’s bedroom, so the distance will be:
D=3000/4/16=44 (ft)
So it is some near for this use case.
(4) While if there are a plaster wall and a the glass window from the yard to the baby’s bedroom, so the distance will be:
D=3000/4/2=375 (ft)
So it is good for this use case.
If the glass window is embedded in the brick wall so the distance will be between the distance for glass and the distance for brick wall.