Mr. McLogan, If I may offer some advice, I'm sure you already know this but applying it to bearing (navigation) problems might save you a good number of mental gymnastics while explaining this type of problem to your students: bearings are always the complements of angles in standard position, i.e. 𝛉 = 90° - 𝛃 . As such the Cartesian components, x = r cos 𝛉 and y = r sin 𝛉 must be re-written as: x = r cos(90° - 𝛃) and y = r sin(90° - 𝛃) . From the angle addition formulas we find, x = r sin 𝛃 and y = r cos 𝛃 . This means that for bearings the x and y components of a given vector are swapped: (y, x). You can accommodate this fact in a couple of different ways: 1) Re-orient your axes such that x represents the vertical direction and y, the horizontal, or 2) Don't use x and y, but use N and E instead (and remember that in a given ordered pair the (N) North component comes first, and the (E) East component comes second).
A simpler way is to break c vector into components where one component will be parallel to b vector and the other component is perpendicular to b vector. The components of c would be 8cos60 and 8sin60 where cos component would be perpendicular and sin would be opposing in this case(isn’t always that way you just need to draw out the triangle to know which component is which) you just subtract the opposing vector and now you have the magnitude and you just use trig on the perpendicular to get the angle of track error. In summary: Component of current pushing the boat to a side= 8cos60= 4 Component of current opposing/ assisting the boat= 8sin60=6.93 Magnitude of b+c vector= 27-6.93=20.07 mph Angle of drift of boat= tan inverse of ( magnitude of perpendicular current/ magnitude of b vector)= tan^-1(4/27)= 8.427 degrees to the left Therefore the new vector is 20.07 mph and at a bearing of 360-8.427degrees which is a bearing of 351.573degrees
You can. However it will only work for magnitudes and you need to have 2 of three magnitudes in a triangle. You can use components of vectors as well as Pythagoras theorem for magnitude but for angles you’ll need inverse trig or trig as explained by professor
This is literally the best explanation of bearings and vectors. Thank you!!!
Mr. McLogan,
If I may offer some advice, I'm sure you already know this but applying it to bearing (navigation) problems might save you a good number of mental gymnastics while explaining this type of problem to your students: bearings are always the complements of angles in standard position, i.e. 𝛉 = 90° - 𝛃 . As such the Cartesian components, x = r cos 𝛉 and y = r sin 𝛉 must be re-written as: x = r cos(90° - 𝛃) and y = r sin(90° - 𝛃) . From the angle addition formulas we find, x = r sin 𝛃 and y = r cos 𝛃 . This means that for bearings the x and y components of a given vector are swapped: (y, x). You can accommodate this fact in a couple of different ways: 1) Re-orient your axes such that x represents the vertical direction and y, the horizontal, or 2) Don't use x and y, but use N and E instead (and remember that in a given ordered pair the (N) North component comes first, and the (E) East component comes second).
A simpler way is to break c vector into components where one component will be parallel to b vector and the other component is perpendicular to b vector. The components of c would be 8cos60 and 8sin60 where cos component would be perpendicular and sin would be opposing in this case(isn’t always that way you just need to draw out the triangle to know which component is which) you just subtract the opposing vector and now you have the magnitude and you just use trig on the perpendicular to get the angle of track error.
In summary:
Component of current pushing the boat to a side= 8cos60= 4
Component of current opposing/ assisting the boat= 8sin60=6.93
Magnitude of b+c vector= 27-6.93=20.07 mph
Angle of drift of boat= tan inverse of ( magnitude of perpendicular current/ magnitude of b vector)= tan^-1(4/27)= 8.427 degrees to the left
Therefore the new vector is 20.07 mph and at a bearing of 360-8.427degrees which is a bearing of 351.573degrees
not sure how i would've survived math without your instructions! contantly so helpful
How do you get the formula for the direction of the resultant vector? (tangent theta ...)
Please refer this link. www.cuemath.com/direction-of-a-vector-formula/
Best teacher ever
Wow this video is really well explained thank you so much
Thank you very much great video
talented teacher!!
I don’t understand the part cosine , sine? How it is cos , sin,? Can you please explain?
Its just like finding sides of a triangle, cosine for adjacent and sine for opposite sides.
but cant you only use pythagorean theorem when its a right angle triangle?
You can. However it will only work for magnitudes and you need to have 2 of three magnitudes in a triangle. You can use components of vectors as well as Pythagoras theorem for magnitude but for angles you’ll need inverse trig or trig as explained by professor
why does mine shows 21.93 instead of 24mph?
Wait I didn't know J-mac became a teacher.
Woww just wow🎉
Thanks for this video btw