Concept of meridian zenith distance ( MZD). MZD is taken from celestial body to observer ( opposite to bearing). Relation ship between MZD & Declination to obtain latitude.
To be simplified when sun is SOUTH of you , you have to subtract declination from Zenith distance, if sun is North of you you have to subtract Zenith distance from declination in order to get Latitude
at 1:20 in this video you say," in this case, the bearing will be either north or south." I'm a bit confused here. Which is it in this case.?It should be south, no? Also in this example at this point in the video, the zenith distance is on the equator, no? This is very confusing as the observer is not facing any cardinal point in this diagram so I can't make sense of this? Please explain. Thank you.
Bearing is the direction from observer to the object. @ 1.20 the object that is Sun lies south of observer. The statement that bearing will be north or south is general and has caused some confusion. Similarly , if you look around 3.10, sun is north of observer and in this case bearing is North. To make it simple, bearing is going to be either north or south when meridian passage occurs. One needs to figure it out which one will it be ? Hope this clears your doubt. Do let me know if you need further clarification. We can work to provide some more clarity as needed.
@@sailorstube3259 Thank you for your response, this part makes sense now, It's just your diagram is a little difficult to understand. I struggle to think 2 dimensional at times. What might be helpful is if you had the two diagrams side by side on the same image and referred over and back as you are explaining it. Thank you, Paul
Hi..,thank you for the video..really helpful could you please also solve a practical problem using this method..as one would solve on a ship. step by step .thank you.
Thank you sir. Can you please make videos about LONG BY CHRONO AND INTERCEPT METHODS. We will be very helpful . Please do Assist over the above request Captain.
HOW do we name the true altitude drom the bearing. the ones shown here are 000 digree and 180 digree, what if say, eg it is 66digree true or 112 digree true? how can we name it ? nort or south ??
+Akshaye M In case of latitude by meridian altitude a celestial body will be at the obsrevers zenith that is with maximum altitudes attained at the meridian passage. In such cases the bearing will always be either north or south only and hence position line is always east -west only that is at right angle to the bearing , a very reason to obtain pl through this technique. Trust this clarifies. If still you have doubts we can put some more basic vidoes on the topic or just te watch the present one. Cheers.
To be simplified when sun is SOUTH of you , you have to subtract declination from Zenith distance, if sun is North of you you have to subtract Zenith distance from declination in order to get Latitude
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Very precisely explained cleared all doubt please enlarge timing for thinking the concept insted of making pause by viewers #awesome sir thats alot
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This is a beautiful explanation!
at 1:20 in this video you say," in this case, the bearing will be either north or south." I'm a bit confused here. Which is it in this case.?It should be south, no? Also in this example at this point in the video, the zenith distance is on the equator, no? This is very confusing as the observer is not facing any cardinal point in this diagram so I can't make sense of this? Please explain. Thank you.
Bearing is the direction from observer to the object. @ 1.20 the object that is Sun lies south of observer. The statement that bearing will be north or south is general and has caused some confusion. Similarly , if you look around 3.10, sun is north of observer and in this case bearing is North. To make it simple, bearing is going to be either north or south when meridian passage occurs. One needs to figure it out which one will it be ? Hope this clears your doubt. Do let me know if you need further clarification. We can work to provide some more clarity as needed.
@@sailorstube3259 Thank you for your response, this part makes sense now, It's just your diagram is a little difficult to understand. I struggle to think 2 dimensional at times. What might be helpful is if you had the two diagrams side by side on the same image and referred over and back as you are explaining it. Thank you, Paul
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Capt. Rizvi sir your work is amazing, your video cleared all my doubts
pls... make a video on lights and shapes
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Sir Thanks for the video . Really Helpful . Just wanted to ask when is part 2 of this video would come? .It will be really beneficial .
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Hi..,thank you for the video..really helpful could you please also solve a practical problem using this method..as one would solve on a ship. step by step .thank you.
For second mate/DNS/Bsc NS prepration join kundan patna #kundanpatna
Thank you sir. Can you please make videos about LONG BY CHRONO AND INTERCEPT METHODS. We will be very helpful . Please do Assist over the above request Captain.
For second mate/DNS/Bsc NS prepration join kundan patna #kundanpatna
HOW do we name the true altitude drom the bearing. the ones shown here are 000 digree and 180 digree, what if say, eg it is 66digree true or 112 digree true? how can we name it ? nort or south ??
+Akshaye M In case of latitude by meridian altitude a celestial body will be at the obsrevers zenith that is with maximum altitudes attained at the meridian passage. In such cases the bearing will always be either north or south only and hence position line is always east -west only that is at right angle to the bearing , a very reason to obtain pl through this technique. Trust this clarifies. If still you have doubts we can put some more basic vidoes on the topic or just te watch the present one. Cheers.
thank u verymuch
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hi can you please give the other ways of calculation in celestial navitgation just like latitude by meridian altitude =)
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Like 312 don👌
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solution please..
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