Position fix Methods: RADAR range and bearing 

Welcome sir... It is a journey that's gonna be remembered all through my life. ❤️

@@prasobhsugadhan Long way to go....keep it up...wages of hard work is sweet😇

Thank you sir for the support ❤️

Welcome sir

I just had a video yesterday. Still not uploaded. I will post it in a few days sir. Will keep you posted.

Copied sir. Will do. Thank you for briefing. Regards ♥️

@@prasobhsugadhan you still uploadinh mate

Want to. But, exams and sailing is making it difficult. Will upload a lot more once I'm free. Regards ♥️

@@prasobhsugadhan do you take colleges while on board, if so what class are you grinding

Yes sir. The logic is correct. But, in this situation, our ship's Gyro error os known to be .5 and the corresponding correction has been already applied to the radar feed during sync. Hence, it won't matter in this case. If you have not applied the correct earlier at radar gyro heading sync, the correct need to be applied at chart. Hope you got a correct answer for your query. Thank you sir, for pointing it out.

@@prasobhsugadhan Thank you very much Prasobh sir.. Sir their is one more thing which I would like a suggestion.. Sir suppose we take gyro bearing of that becon but that bearing is relative to our ship's heading (not a true bearing as per chart) so to convert in to true bearing - Relative Brg + Ship's heading= True bearing, so can we directly plot this bearing on chart? & Second - Also for getting True Bearing of that object (eg. Becon) To convert by CDMVT formula the C amount should be our bearing (becon bearing ) taken by gyro / repeater & Deviation is as per our ship's heading.. is that right sir ? & when we get T that should be our plotting bearing.. So in this case what bearing should we plot on chart by R+S.h=T or by CDMVT .. Please correct me if i misunderstood in my learning..

Sir, if you know the Relative Brg (R) and the ship's head (course) and the Gyro Error (GE), Then, you can simply +/- the R to Course and apply the answer to GE to obtain the True brg. Suppose 270 Course, R is 45 Green and GE is 1° High, the answer can be worked out as, 270+45=315° (G) 315-1° =314°(T) When Gyro is being used, there is no more Variation and Deviation in use. The Gyro works on a different principle (inertia & precision) and hence is not affected by ships / earths magnetism (deviation & variation respectively). Only when the compass is used to take brg, we will be applying Var & Dev to get the True brg. Our ship's repeaters are normally Slave gyro, including the steering repeater, and hence, any brgs taken will be direct Gyro Brgs. Is this what you needed clarification sir?