Great video thanks.. just one correction. 22:19 in a VOR receiver is not on the ground, but in the aircraft. Basic difference is that in an ADF the bearing is being measured at the aircraft receiver and in VOR bearing information is encoded in the signal transmitted from the ground itself. The aircraft receiver just demodulates that. Hence Variation is applied at the site. So a good way to remember would be that variation is applied wherever the bearing calculation is being done.
Thank you for the information. Sir can you please check 28:10 . The ADF is 190 so the aircraft should turn to RIGHT to get on track. please check and advise if I’m right or am I confused with the question. Thank you 😊
Great great great video. Really really helpful. Direly waiting for the VOR and to/from indicator video. Thank you so much for keeping your word!!! Thank you so very much!!!!
36 mins into the video in the type 5 intercept based example, to track Radial 100 inbound the aircraft has to track 280 (100+180). So the aircraft has to turn right initially from 240 Radial. I think you explained 30 degrees intercept to fly 280 Radial inbound. Can you please check and advise if I’m right or am I confused with the question. Thank you 😊
Oh yes, you will turn right and then intercept the radial 100. You are correct, I don’t know how I have done this. In the solution, I am actually tracking inbound R280!
Sir, "Track 100° inbound" is different from "Track inbound on radial 100°", I believe you have done the former one. But in this question we gotta find the Track inbound on radial 100°.
Intercept based: Inbound on radial 100 mean a/c will go towards the station while tracking on radial 100 as we do in holding paterns, coz radials are always away from the station or bearings from the station. I request you to kindly recheck or explain if I am wrong. Thanks
In question no. 1. If the wind is from the east.. why is it that the a/c si still going on to the right where it is supposed to move on to the left..unless the wind is pushing the tail.
The aircraft is pointed, or headed, at 10 degrees in order to correct for the wind. In other words, the pilot is crabbing the aircraft to the right so even though the wind is pushing him to the left his course or track remains at 360 degrees. If you don't crab the aircraft to the right, then his track won't be the required 360 degrees. In other words, the most important part of going from point A to point B is your track on over the ground. You're in the air trying to get there. So, in this case, you point the aircraft, as required, to correct for wind drift.
Bro In the 2nd numerical (of type 1: drawing based) I understood it by drawing but i am not getting the answer if i use the formula QDM= HDG(M) + R.B Since QDR is 200* then QDM will be 020* And by applying variation we get the magnetic heading that is 237* So by substituting, 020= 237 + R.B R.B= -217 🤷🏻♂️🤷🏻♂️🤷🏻♂️🤷🏻♂️🤷🏻♂️🤷🏻♂️
@@blueskiesandtailwinds6004 And in the last question (type:6 instrument based). a/c tracking away from an NDB on a track of 023*(T)….. in this 023* means- 23* or 230*???
Sir I believe, the question at 36 minutes has been solved for interception of inbound radial 280 instead of inbound 100. Also how can one decide that we are on radial 240 just by heading. Because we could be inbound stn with 60 hdg on any radial between 190 to 010. I think the question is incomplete.
Yes I’ve written about that correction in the description! But ya, in questions which only have such limited information, we can assume heading 060 inbound to be as on radial 240 inbound.
Yes I’ve written about that correction in the description! But ya, in questions which only have such limited information, we can assume heading 060 inbound to be as on radial 240 inbound.
PLEASE SOLVE THIS QUESTION At 1000 Z an aircraft is overhead NB PE enroute to NDB CN, Track 075(M), Heading 082(M) At 1029 Z NDB PE bears 176 Relative and NDB CN bears 353 Relative. The heading to steer at 1029 Z to reach NDB CN is:
This Question too plz. An aircraft heading 100° (M) has an ADF reading of 210° Relative. The alteration of heading required to intercept the 340° track inbound to the NDB at 60° is:
Great video thanks.. just one correction.
22:19 in a VOR receiver is not on the ground, but in the aircraft.
Basic difference is that in an ADF the bearing is being measured at the aircraft receiver and in VOR bearing information is encoded in the signal transmitted from the ground itself. The aircraft receiver just demodulates that. Hence Variation is applied at the site.
So a good way to remember would be that variation is applied wherever the bearing calculation is being done.
Thank you for the information. Sir can you please check 28:10 . The ADF is 190 so the aircraft should turn to RIGHT to get on track. please check and advise if I’m right or am I confused with the question. Thank you 😊
Great great great video. Really really helpful. Direly waiting for the VOR and to/from indicator video. Thank you so much for keeping your word!!! Thank you so very much!!!!
Thank you sooo much for the video.now it's very much clear to me.waiting for VOR.
2k soon...woo hoo 😍
The way you explain sir 🙌🏻 🙂
Greatly appreciated Sir!! Excellent explanation.
Watching this a day before my Nav exam.. really helpful...good job man.
Yo did you clear it ?
@ Bro 70/100 🤣🤣🤣 October OLODE ..
@@bigbang4425 damn !!!! Congratz aviator !
Piece of knowledge. Thanks man ❤️
36 mins into the video in the type 5 intercept based example, to track Radial 100 inbound the aircraft has to track 280 (100+180). So the aircraft has to turn right initially from 240 Radial. I think you explained 30 degrees intercept to fly 280 Radial inbound. Can you please check and advise if I’m right or am I confused with the question. Thank you 😊
Oh yes, you will turn right and then intercept the radial 100. You are correct, I don’t know how I have done this. In the solution, I am actually tracking inbound R280!
Sir, "Track 100° inbound" is different from "Track inbound on radial 100°", I believe you have done the former one. But in this question we gotta find the Track inbound on radial 100°.
@@avinashnatraj9552 yes. In the solution I have actually tracked inbound R280. I should’ve turned to the right.
Thank you for asking this question so it could be clarified.
@@blueskiesandtailwinds6004 thanks i tought i was going crazy...
Congrats for 2k ❤️
Very well explained. Good job and keep it up! 🔥
Brooooo waiting for your next video!!!!!!💯💯
Oh i saw these same solutions in my academy's notes as well
long video but very very helpful. thank you !
Good work brother
Waiting for next part
Sir Waiting for your RADIO navigation next video 🙏🙏
You're a saviour thank you so muchhhhhhhh
Please make more videos on general navigation .
Sir if moving to beacon winds from left track will be 010 and if wind from right track 350
Super❤❤
36:06
Wouldnt inbound on radial 100 be a heading of 280?
that picture looks like inbound radial 280...?
Yes. In the solution I have actually tracked inbound R280. I should’ve turned to the right.
Hey sir , want to ask u a question,
At which age should we start a ground classes for dgca exams? And from where we can start
Hi, can you please make a video on convergency?
Please make a video on propagation
Pleade Complete the radio nav series!!!!!!!!!!!!
Thank you!
Thank you so much
Sir I think in last que 33+180=213 so option D is fit for that , QDR is 33 .
You explain very well and calm thanks to you man 🫂
Intercept based:
Inbound on radial 100 mean a/c will go towards the station while tracking on radial 100 as we do in holding paterns, coz radials are always away from the station or bearings from the station.
I request you to kindly recheck or explain if I am wrong.
Thanks
Yes there’s a correction in that question. Read the description!
In question no. 1. If the wind is from the east.. why is it that the a/c si still going on to the right where it is supposed to move on to the left..unless the wind is pushing the tail.
Same here bro
Same question
The aircraft is pointed, or headed, at 10 degrees in order to correct for the wind. In other words, the pilot is crabbing the aircraft to the right so even though the wind is pushing him to the left his course or track remains at 360 degrees. If you don't crab the aircraft to the right, then his track won't be the required 360 degrees. In other words, the most important part of going from point A to point B is your track on over the ground. You're in the air trying to get there. So, in this case, you point the aircraft, as required, to correct for wind drift.
Bro
In the 2nd numerical (of type 1: drawing based) I understood it by drawing but i am not getting the answer if i use the formula
QDM= HDG(M) + R.B
Since QDR is 200* then QDM will be 020*
And by applying variation we get the magnetic heading that is 237*
So by substituting,
020= 237 + R.B
R.B= -217
🤷🏻♂️🤷🏻♂️🤷🏻♂️🤷🏻♂️🤷🏻♂️🤷🏻♂️
-217 when added to 360 will give you 143, which is the ADF reading (the RB)!
@@blueskiesandtailwinds6004 oh alright got it!!!!💯💯💯
@@blueskiesandtailwinds6004
And in the last question (type:6 instrument based).
a/c tracking away from an NDB on a track of 023*(T)….. in this 023* means- 23* or 230*???
@@dccreations6910 23 degrees
@@blueskiesandtailwinds6004 👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻
Anyhope for indigo Cadet program ?
Sir I believe, the question at 36 minutes has been solved for interception of inbound radial 280 instead of inbound 100. Also how can one decide that we are on radial 240 just by heading. Because we could be inbound stn with 60 hdg on any radial between 190 to 010. I think the question is incomplete.
Yes I’ve written about that correction in the description!
But ya, in questions which only have such limited information, we can assume heading 060 inbound to be as on radial 240 inbound.
Yes I’ve written about that correction in the description!
But ya, in questions which only have such limited information, we can assume heading 060 inbound to be as on radial 240 inbound.
I see. Thanks!@@blueskiesandtailwinds6004
PLEASE SOLVE THIS QUESTION
At 1000 Z an aircraft is overhead NB PE enroute to NDB CN, Track 075(M), Heading 082(M) At 1029 Z NDB PE bears 176 Relative and NDB CN bears 353 Relative. The heading to steer at 1029 Z to reach NDB CN is:
This Question too plz.
An aircraft heading 100° (M) has an ADF reading of 210° Relative. The alteration of heading required to intercept the 340° track inbound to the NDB at 60° is:
@@AyushGupta-dx3bu is the answer 280 deg
👏🏻👏🏻👏🏻
❤
In the first example you state that QDM is magnetic “track” to the station but isn’t QDM magnetic “bearing “ to the station
QDM is the magnetic heading and QDR is bearing.
QDM-> magnetic TO the station.
QDR-> magnetic FROM the station.
🔥👍
💯
❤️❤️
Taylor Anthony Allen Patricia Anderson Susan
this is quite hard to understand