DISTANCES, FREQUENCES, POWERS MENTIONED ARE PURELY THEORETICAL AND SHOULD ONLY BE USED AS AN EXAMPLE TO AID STEERING
Hi everyone
Sorry for my bad english in first :)
I wanted to share with you how I manage to programme my flight in fpv so you don’t loose the signal when obstacles get in your way.
We all know that very high frequencies such as 2.4 ghz and 5.8 ghz just don’t go through any type of obstacle , all it takes is a mere leaf of paper and our signal is lost at certain distances.
We are however forced to pilot in a “visual line”, I mean we always have to be able to physically see our model aircraft. In other words, if we were to attach a invisible wire from the radio to our aircraft, it should never bend.
So, keeping in mind the question of obstacles, we tend to fly as high as possible, especially if our intention is to go as far as possible.
What can often happen is, we either guess or over estimate heights having less range and poor view.
In this paper I would like to explain how I calculate the precise minimum safe altitude at which I can fly without the worry of losing control due to obstacles.All we have to do is calculate the height of our first tall obstacle that we have in front of us and the distance between that obstacle and ourself.
In doing this we find two cathetuses of a triangle on which we can trace a hypothenus (cf. the invisible wire between the radio und ourselves that must not bend).
It follows that if ,for example ,at a distance of 2 cm from the first obstacle, another obstacle presents itself that is double the height of the first obstacle, the hypothenus is going to hit it! Therefore you need to do a survey of the area and locate the first tall object, after which the hypothanus will continue to travel at an altitude devoid of obstacles.
You will now get,,as you can see in the attached drawing, a first triangle rectangle (red caption).
Firstly, we have to calculate the angle of the triangle, from the given base and its height:
ʎ=arctan(c/b) which in my case is 11,30°
The smaller the angle, the lower you can fly; the further the distance between the obstacle and yourself ,the more acute the angle will be.
Our objective now is to know the maximum height (x) in relation to the angle(ʎ) that is formed and in relation to the maximum distance that our equipment (tx and VTX) allows us or in relation to the distance you want to keep within.
However, we must find the cathetus c’
If you are to be using a 5.8ghz 200mw kit and a radio frsky 2.4ghz on a aircraft model., the maximum safe distance that you want to reach and consequently set up as your warning range on the osd is 1000m(b’).
The trigonometrical calculation is c’=b’ tan ʎ that in my case is 199,81m. This is the distance that I will set up as the limit on my osd and it will be the mimimum safe distance that I will be able to fly. We hereby greatly reduce the risk of loss due to obstacles and I still advise use of circular polar antenna in order to avoid db cuts due to turns.
Note that this method does not however cover loss due to interference, hence the need to survey the area.
Personally, I found Google Earth a good way of measuring altitude, with the pro version you can scan a circular area with a radius in which you want to fly
I suggest you round up the measurement calculations for reassurance.
In the design I have also calculated the hypothenus that is always greater than the mazimum distance from the ground as you have probably noticed. Thus you will have noticed that there are 19 m extra due to inclination and altitude.
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