you've got a knack for this teaching thing. You made this so simple, even a student pilot like me can both understand it and know how to use it every time. Thank you from a fellow Aussie :)
Videos likes this makes me really proud for understanding english, all of your videos are by far the best explanations on internet. Best regards from a brazilian pilot.
Noticed there’s a triangulation during the landing path. The distance of the path the airplane going through is longer than the ground course that the airplane has, which means that you will arrive slower than calculated time. Same thing happens when you do the climbing.
My first time hearing about this rule in aviation. It is an interesting shorthand but remember folks, this rule is a consequence of small angle approximation in trigonometry so it will not work if correction angles are 10DEG or more.
Note that the actual track error in the last example is 3.63°, the correction angle is 9.09° (i.e. with 14° correction, you'd cross the direct line 3.14 nm earlier and miss the target by 0.57 nm in the y (north-south?) axis) and that the numbers obtained using the 1 in 60 rule without rounding would add up to 12.95° (i.e. missing the target by 0.62 nm on the x-axis and by 0.10 nm on the y-axis). Lesson learnt: if possible, use a calculator so that you don't have to round the numbers.
We have to be careful that this rule can only be a rough estimate and only works for a small off track variance. It won't work if it is so large. For example: if you are 20nm off track after travelling 20nm then the TE is 45 degrees. But using this 1 in 60 rule, it becomes 60 degrees and that is wrong. Times it by 2 for the total TE + CA if that is a halfway mark then it becomes 120 degree when it should have been only 90.
Its straight and clear explaination. Can I share something. What I do is carry a calculator on board. In this exqmple, if I found I am 3nm left of the lake and 30nm to go, I will use calculator enter: 1/tan 3/20, I will get the angle off track:8.5 degree, round to 9 degree, same theory for second half: 1/tan 3/30 is 5.7degree round to 6 degree. add them together: 9+6 is the angle I will be turning which is 15 degree
The equation for a situation where you have the drift and distance traveled and are looking for the distance off track is (Distance traveled X Drift) / 60 = distance off track
So basically we first need to find a ground reference point which is parallel to our flight plan right? And how do I estimate the off track distance from the ground reference point( In this case the lake) assuming my radio navigation aids are not functioning properly.
dear god may be simple but if you are traveling at say 60mph, a mile per minute, then by the time you've worked it out it will be wrong again. In any case I wouldn't be traveling 60 miles or what ever before checking my route. I would have more regular way markers. Also a lot of compasses wander too much to give a 2 degree accuracy
As a teacher looking for simple and fun math calculations, I love this! I have one question and I'm sure I'm way overthinking this but please help me understand so that I am not fumbling around with students. I'm looking at these drawings very literally as if it were all. happening within a single (geometric - no pun intended) plane in the air. In the drawings the plane ends up at a spot relative to the landmark forming a line that is exactly perpendicular to the intended original route. The drawings all make up right angles with the landmark. If the airplane discovers its error when it completes the distance it thought it would take to reach the landmark, wouldn't that occur earlier? In other words, how can the hypotenuse of the triangle be the same length as one of the other sides (i.e. 60 and 60)?
1/60 rule is an approximation, so it's there to do "somewhat quick math" to get close to the right answer. For a large horizontal distance and small vertical change it will work fairly well. We can pretend the elevation stays the same, so it's all on one flat plane (like a piece of paper). Adding the z (up/down) component honestly doesn't change much. You'd just have to descend (which a pilot would do toward the very end-- as in not gradually). You're on a flat path, until the very end just like in a real flight. As for making a right triangle with the starting position, the plane, and the intended landmark position, you can definitely do that--- When they recognize their error depends on what how they did. If they were doing it by distance traveled then yes they would not quite have reached the landmark, if they were doing it by looking out their window-- maybe they only noticed the error by saying hey, i should be flying over that lake not north of it. Then they can use the 1/60 rule to figure out how many degrees their current heading is off (assuming they started with 1 heading and kept it) and then use the 1/60 rule again to figure out how many degrees to get back on track. Again this "1/60 quick estimate rule" works for large horizontals with relatively small vertical deviations.
@@jilow That is true. The hypotenuse and base leg with small angles is a negligible distance difference. Wouldn't get you the precise answer on a math test but gets you close enough in flight.
An aircraft departs from position A (04°10' S 178°22'W) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are? ua-cam.com/video/i2f-h0Dkh64/v-deo.html The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60° S 165° W) B (60° S 177° E), at the place of departure A, is: ua-cam.com/video/Fk3f0g-lXm8/v-deo.html Given: Waypoint 1. 60°S 030°W Waypoint 2. 60°S 020°W What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W? ua-cam.com/video/BcNEYDr45YM/v-deo.html What is the time required to travel along the parallel of latitude 60° N between meridians 010° E and 030° W at a groundspeed of 480 kt? ua-cam.com/video/vejFsDn_TmQ/v-deo.html Given: value for the flattening of the Earth is 1/298. Earth's semi-major axis, as measured at the equator, equals 6378.4 km. What is the semi-minor axis (km) of the earth at the axis of the poles? ua-cam.com/video/Z-gOj6s-gF4/v-deo.html Position A is located on the equator at longitude 130°00E. Position B is located 100 NM from A on a bearing of 225°(T). The coordinates of position B are: ua-cam.com/video/sssMAVAoCdU/v-deo.html The rhumb line track between position A (45°00'N, 010°00'W) and position B (48°30'N, 015°00'W) is approximately: ua-cam.com/video/2Ws7KIbrb4o/v-deo.html A Mercator chart has a scale at the equator = 1: 3 704 000. What is the scale at latitude 60° S? ua-cam.com/video/UiS8iB1qZN8/v-deo.html The distance measured between two points on a navigation map is 42 mm (millimetres). The scale of the chart is 1:1 600 000. The actual distance between these two point is approximately: ua-cam.com/video/t3rUqMjl_2w/v-deo.html The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N and 38°20' N. The constant of the cone for this chart is: ua-cam.com/video/Von47bYKCF4/v-deo.html
Nirvana Of Monkasar if you are using ifr flight you will have to have to use your flight instruments. So the answer is yes you can use it in ifr flights but just dont be confused that you are not acquiring a deviation angle off your course visually. Use your instruments to find out your heading correction angle
8 років тому+3
I still don't get one thing. How do I know how off track I am?
That's so much easier to understand! I was taught how to spin it up and all this technical garbage about angles and stuff, much simpler. Now I can apply that to what I already know!
Not quite, that's for 'correction angle'. You need to work out the 'track error' and 'correction angle', then add both together to obtain your heading correction.
Your videos are perfect I'm first officer iman Ghahremanzadeh Would you accept me for make very nice like as this video for you ? I'm ready to prepare video like this and better for free, I can send sample videos that I have made before for some aviation channels
In 3 minutes and 40 seconds, you taught me something I've been struggling with for months! Thank You!
you've got a knack for this teaching thing. You made this so simple, even a student pilot like me can both understand it and know how to use it every time.
Thank you from a fellow Aussie :)
Much appreciated!
The Best 1 in 60 rule explanation on the internet!
This means a lot to us. Thank you!
As a visual learner, I must admit your videos are an immense help!
Thanks, very helpful. The most straight forward, simple explanation out there! Regards from ATPL student in Greece.
Best video I’ve found on UA-cam on the 1 in 60 rule, thank you!
So far, best right to the point teaching videos available on UA-cam, thank you !
Thank you very much!
Thanks for the video! This has helped me to finally understand the 1 in 60 rule.
My god....This is golden! Best explanation of this rule ive come across on the interwebz. Thanks alot!
Very kind. Thank you.
This is super helpful for the commercial and fight instructor written. Thank you!
Glad it was helpful!
Videos likes this makes me really proud for understanding english, all of your videos are by far the best explanations on internet.
Best regards from a brazilian pilot.
Wow, thanks!
Thanks for the video!! I was like i need someone to explain it to me like i am 5 years old to understand it and you made it!!! Really helpful 👌
Simple lovely and clean explaining without making confused
Thank you so much 🙂
Noticed there’s a triangulation during the landing path. The distance of the path the airplane going through is longer than the ground course that the airplane has, which means that you will arrive slower than calculated time. Same thing happens when you do the climbing.
you have jsut made my life 10x easier with this video. thank you so much !
Glad it helped!
best explanation of the 1 in 60 i've ever watched /read! good job!
Thank you very much.
I love short explanations.
Just found this channel, watched 3 videos and subbed!! Great explanations and glad it's relative to Australia 😄👍
Welcome aboard! And thank you!
My first time hearing about this rule in aviation. It is an interesting shorthand but remember folks, this rule is a consequence of small angle approximation in trigonometry so it will not work if correction angles are 10DEG or more.
Before digging further deep in a subject, first address to visit to see what the hell it really is. thank you guys.
Note that the actual track error in the last example is 3.63°, the correction angle is 9.09° (i.e. with 14° correction, you'd cross the direct line 3.14 nm earlier and miss the target by 0.57 nm in the y (north-south?) axis) and that the numbers obtained using the 1 in 60 rule without rounding would add up to 12.95° (i.e. missing the target by 0.62 nm on the x-axis and by 0.10 nm on the y-axis). Lesson learnt: if possible, use a calculator so that you don't have to round the numbers.
it’s amazing how relevant a 9year old video is
That's so much work! Just use law of sines or law of cosines respectively to find the angle theta.
The irony 😅
i've been forgotten this formula since my cpl exam :D, thanks for the video.
Happy to help!
Thank you so much for making it this simple. Thank you, thank you
You are so welcome!
Wow brilliantly well explained
Glad you liked it
Very helpful. Thank you!
We have to be careful that this rule can only be a rough estimate and only works for a small off track variance. It won't work if it is so large. For example: if you are 20nm off track after travelling 20nm then the TE is 45 degrees. But using this 1 in 60 rule, it becomes 60 degrees and that is wrong. Times it by 2 for the total TE + CA if that is a halfway mark then it becomes 120 degree when it should have been only 90.
If you are 20 NM off course after flying 20 NM, then you probably shouldn't have gotten licensed haha. But yes that trig is correct.
Helpful stuff. Very clearly illustrated.
very good job done , am really benefitted by your video , thank you
Glad to hear that. Thank you!
perfectly taught!!! thanks
You're very welcome!
Its straight and clear explaination. Can I share something. What I do is carry a calculator on board. In this exqmple, if I found I am 3nm left of the lake and 30nm to go, I will use calculator enter: 1/tan 3/20, I will get the angle off track:8.5 degree, round to 9 degree, same theory for second half: 1/tan 3/30 is 5.7degree round to 6 degree. add them together: 9+6 is the angle I will be turning which is 15 degree
The equation for a situation where you have the drift and distance traveled and are looking for the distance off track is
(Distance traveled X Drift) / 60 = distance off track
So basically we first need to find a ground reference point which is parallel to our flight plan right?
And how do I estimate the off track distance from the ground reference point( In this case the lake) assuming my radio navigation aids are not functioning properly.
I have the same question !
Fantastic video. Thank you
Oh my godnesss!!!! Best video ever!!! I finally understand
Very helpful!
So glad!
Best explanation. Bravo.
Thank you so much .
You're most welcome
Wonderful and easy to understand ...Well done guys..thumbs up!
VOILA ! Merci ! 😁😁😁
*every kid always said we wouldn't need trig in the real world, but they were right for them, wrong for me, because I went on to become pilot.*
Excellent video. Helped lot to understand this!
Thank you!
Is there a similar rule for kilometres?
great help! thank you
Glad it helped!
How do you know your distance to the lake?
Good question! You look, you reference your chart, and you estimate the distance.
dear god may be simple but if you are traveling at say 60mph, a mile per minute, then by the time you've worked it out it will be wrong again. In any case I wouldn't be traveling 60 miles or what ever before checking my route. I would have more regular way markers. Also a lot of compasses wander too much to give a 2 degree accuracy
Really great explanation 😍
Glad it was helpful!
Best explanation I've seen. Meeting in 30mins for my BFR review... 😬
Thank you. All the best!
Superb.. Couldn't have explained it better ! In 3 and Half minutes.. u've mastered the one in sixty rule !
Ranjit Jackson Thank you Ranjit.
Is the 60 to 1 rule in any faa publications im having trouble finding it? Or is it just a rule of thumb.
THANK YOU !!!!
Thank you for taking the time to comment.
noob here, how do I estimate my distance from the lake?
Good question. Pilots can estimate the distance off track visually and with reference to a chart (map).
@@flightclubonline makes sense, thank you! I just started flying properly on flight Sims and learning radio vor navigation
Thank you!
As a teacher looking for simple and fun math calculations, I love this! I have one question and I'm sure I'm way overthinking this but please help me understand so that I am not fumbling around with students. I'm looking at these drawings very literally as if it were all. happening within a single (geometric - no pun intended) plane in the air. In the drawings the plane ends up at a spot relative to the landmark forming a line that is exactly perpendicular to the intended original route. The drawings all make up right angles with the landmark. If the airplane discovers its error when it completes the distance it thought it would take to reach the landmark, wouldn't that occur earlier? In other words, how can the hypotenuse of the triangle be the same length as one of the other sides (i.e. 60 and 60)?
1/60 rule is an approximation, so it's there to do "somewhat quick math" to get close to the right answer. For a large horizontal distance and small vertical change it will work fairly well.
We can pretend the elevation stays the same, so it's all on one flat plane (like a piece of paper). Adding the z (up/down) component honestly doesn't change much. You'd just have to descend (which a pilot would do toward the very end-- as in not gradually). You're on a flat path, until the very end just like in a real flight.
As for making a right triangle with the starting position, the plane, and the intended landmark position, you can definitely do that--- When they recognize their error depends on what how they did. If they were doing it by distance traveled then yes they would not quite have reached the landmark, if they were doing it by looking out their window-- maybe they only noticed the error by saying hey, i should be flying over that lake not north of it. Then they can use the 1/60 rule to figure out how many degrees their current heading is off (assuming they started with 1 heading and kept it) and then use the 1/60 rule again to figure out how many degrees to get back on track. Again this "1/60 quick estimate rule" works for large horizontals with relatively small vertical deviations.
@@jilow That is true. The hypotenuse and base leg with small angles is a negligible distance difference. Wouldn't get you the precise answer on a math test but gets you close enough in flight.
An aircraft departs from position A (04°10' S 178°22'W) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are?
ua-cam.com/video/i2f-h0Dkh64/v-deo.html
The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60° S 165° W) B (60° S 177° E), at the place of departure A, is:
ua-cam.com/video/Fk3f0g-lXm8/v-deo.html
Given: Waypoint 1. 60°S 030°W Waypoint 2. 60°S 020°W What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W?
ua-cam.com/video/BcNEYDr45YM/v-deo.html
What is the time required to travel along the parallel of latitude 60° N between meridians 010° E and 030° W at a groundspeed of 480 kt?
ua-cam.com/video/vejFsDn_TmQ/v-deo.html
Given: value for the flattening of the Earth is 1/298. Earth's semi-major axis, as measured at the equator, equals 6378.4 km. What is the semi-minor axis (km) of the earth at the axis of the poles?
ua-cam.com/video/Z-gOj6s-gF4/v-deo.html
Position A is located on the equator at longitude 130°00E. Position B is located 100 NM from A on a bearing of 225°(T). The coordinates of position B are:
ua-cam.com/video/sssMAVAoCdU/v-deo.html
The rhumb line track between position A (45°00'N, 010°00'W) and position B (48°30'N, 015°00'W) is approximately:
ua-cam.com/video/2Ws7KIbrb4o/v-deo.html
A Mercator chart has a scale at the equator = 1: 3 704 000. What is the scale at latitude 60° S?
ua-cam.com/video/UiS8iB1qZN8/v-deo.html
The distance measured between two points on a navigation map is 42 mm (millimetres). The scale of the chart is 1:1 600 000. The actual distance between these two point is approximately:
ua-cam.com/video/t3rUqMjl_2w/v-deo.html
The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N and 38°20' N. The constant of the cone for this chart is:
ua-cam.com/video/Von47bYKCF4/v-deo.html
can it apply to ifr flights? i mean how will you know your error off track?
Nirvana Of Monkasar if you are using ifr flight you will have to have to use your flight instruments. So the answer is yes you can use it in ifr flights but just dont be confused that you are not acquiring a deviation angle off your course visually. Use your instruments to find out your heading correction angle
I still don't get one thing. How do I know how off track I am?
André Ranulfo you are right i think it is used for vfr flight rules
Use the VFR charts. it will give you an approximation.
important question but this channel didn't explain this
@@aviator6547i need the formulas, not too clear the calculations without formula
THANKS!!
Welcome!
thank u so much
You are most welcome! Thank you.
you are a genius my friend
Haha, not quite... but thank you!
Nice and simple explanation. Good job :)
Thank you!
Really great! Thank you so much :-)
That's so much easier to understand! I was taught how to spin it up and all this technical garbage about angles and stuff, much simpler. Now I can apply that to what I already know!
thank you very much for the formula
• TKE (track error) = Distance off X 60 / Distance gone = 9
• CA correction error = Distance off X 60 / Distance to go = 6
but why dont we just simply find the tangent of angle?
It is the tangent of the angle, in radians. The ratio between degrees and radians is 57, using the value of 60 is close enough.
This has been killing me for 2 months, should have watched this earlier lol i
this is basic trigonometry
great explanation
Thank you Stephen.
So its the same as saying
Track error Angle = (Dist. off track x 60)/Dist to go
sometimes when it says use the 1 in 60 rule it doesn't work for me
Not quite, that's for 'correction angle'. You need to work out the 'track error' and 'correction angle', then add both together to obtain your heading correction.
+flight-club so can i just use it twice?
Thank you GPS people for not making me go through this math.
This math could be handy if your GPS fails ;)
Nice work!!
Thank you.
Still I'm Confused Would you please Help Me
good animation!
netflasher Thank you.
Forget all that old-school stuff we are in 2024 so how about I use all the satellites that are in space and foreflight GPS 😊
Or, just use you vor or gps
I love you ❤
your drawing of angles is incorrect!
sorry I accidentally hit the dislike button but I always enjoy your videos
Bro you can press the like button again 😂
Your videos are perfect
I'm first officer iman Ghahremanzadeh
Would you accept me for make very nice like as this video for you ? I'm ready to prepare video like this and better for free, I can send sample videos that I have made before for some aviation channels
böyle video olmaz kardşim lütfen düzeltin
Furkan Eken totally agree with you 👍🏻👍🏻👍🏻 not sufficient enough
Simple and best explanation. Thank you
Glad you liked it. Thank you.