The Wind Triangle - Navigation
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- เผยแพร่เมื่อ 22 ส.ค. 2024
- This video explains the two main methods by which the wind triangle can be constructed and how to interpret it.
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It's always appreciated to get more instruction on things unfamiliar and things familiar. Math is math, however, and I must therefore say:
A 5kt crosswind (or current) doesn't imply a 5° degree drift. Not even close. Where x is the angle of degrees of drift, tan(x) is the ratio of the drift (in nm) from the target divided by the distance to the target.
But hey, that's complicated, so let me simplify it for you with examples:
You're in a B737 going 500kt. A 5kt crosswind will affect your course by 1%. 1% of 360° is 3.6°. Next example is you're in a Robinson R22 helicopter going 60Kts. A 5kt cross wind will affect your course by (5/60) 8% which of 360° is 29°. So drift is entirely related not just to the crosswind component, but to the speed of travel (groundspeed, waterspeed, airspeed, whatever you're measuring crosswing in reference to.)
The same holds true to the end. If you're going to be 5kt off from your target in 60 minutes then we can safely assume you're going 60kt to 0° (heading 360). But is that a safe assumption?
When you start off describing wind triangles to learners, it helps to explain that like a chess-knight moves two squares forward and one to the side, if you increase the forward speed but don't change the sideways speed, it no longer is the same chess-knight move.
The critical factors are only distance, rate, and crosswind rate and direction. Everything else is a distraction. Why do I say distraction? What if your next VFR reporting point is right at the end of that distance, but the airport is another 5nm past it. If you've corrected for the reporting point you've undercorrected for the final destination.
Anyway I'm not trying to be harsh. Good on you for A)Education, and B)Taking the time, and C)Putting this video explanation out here for us to enjoy. The math is a bit too simple so if there was no GPS and no VORTAC and no NDB and no visual references and we really were just dead reckoning it... this would be off.
So much more better than using the e6b at your exam
This channel is amazing, I use the English channel and the Spanish channel. It has helped me a lot to understand part of my PPL and without a doubt it is a great tool to help me understand the things easy.
5:33 - 8:10 I don't quite understand how we found 9 degrees, what is the formula?
@@1Mr.Legend1yeah idk how he found 100mm and 9° wtf
Never in my life did I expect that I would voluntarily watch a video on trigonometry… yet here we are haha
I agree with Summer, amazing channel. I am preparing for the written and besides the normal software you get this is great. Keep up the good work. Will join channel. Thank you.
Very Helpful, Thank you!! Good video and easy explained. Ok, still i have to learn, but i think i understand ir
Excelent video. Thanks. Simple and very helpful.
5:33 I don't quite understand how we found 9 degrees, what is the formula?
Very well explained, thank you.
5:33 I don't quite understand how we found 9 degrees, what is the formula?
By measuring the triangle using a plotter you can obtain the DA and Track.
so Isn't there another easy mathematical formula for this?@@RamiLaw11
Thank you for providing us these videos.
5:33 - 8:10 I don't quite understand how we found 9 degrees, what is the formula?
In this example it's 90°-81°= 9°
Thanks!
What is the 'next video' called, please?
With the river and 5 KTS current example, the boat would be shy of the far shore after sixty minutes. And between the 000 and 005 radials should be arcs, not lines. This is radial geometry.
Super helpful!
The first example excludes current and wind. Since wind also impacts surface watercraft, it deserves a mention as well.
Superb Thanks.
Tommorow an exam. Starting eary
🏆 wonderful 🎉
Thanks a lot , so helpful !!
Thanks a bunch
What about, how to determine the magnetic heading?
Fantastic. Give us some trigonometry videos :)
You have to consider also that wind changes always after inversion layer( increases and veers) so you should add + 30 degrees and wind x2 otherwise 9 degrees correction will be not enough. Same Situation for WCA in holdings when you planning a holding above the layer.
Cool !!!
best
Math is a little off, I think.
calculating an wca to achieve 90 heading...
WCA = arcsin (windspeed / truairspeed * sin (windangle - truecourse)) = -9.059
Resultant ground speed = sqrt (windspeed**2 + trueairspeed**2 - 2*windspeed*trueairspeed*cos(truecourse-windangle+wca)) = 98.628
Bruh why ain't my ground school teach me dis?
I can only suggest to ask your air school this question
I think your triangles are confusing. Start with wind, put wind arrow at destination drawn to scale, mark with 3 headed arrow pointing at destination, wind comes from that direction. Next draw track line to destination, any length, mark double arrow. Now measure out speed to scale and join end of wind line, scribe till it intersect track line. Track line is now ground speed to scale.
All easier to draw than say.
Automatic thumbs down for the AI voice.
I’m not sure why I’m still confused 🫤
Using a confusing method.