NotTim
A small aircraft, on a heading of 225 degrees, is cruising at 150km/h. IT is encountering a wind blowing from a bearing of 315 degrees at 35km/h. Determine the aircraft's ground velocity.
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NotTim
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not physics
LordHades
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Vector addition?
NotTim
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application of vector addition, yah.
LordHades
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Draw it out.
Find it's components.
Add their x componants and y componants separately.
Put them back together in a triangle.
NotTim
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i somehow figured out how to get resultant, but not the angle...
NotTim
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im not suppose to use component method
LordHades
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Use tan(y/x)
LordHades
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Oh I see. THen I'm afraid I can be of no help to you.
NotTim
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ok...oh well. thx u
radar
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draw it out, and remember when directions are concerned, 0 degrees is North.
NotTim
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i did....but i didnt draw it here. i got the magnitude, but not the angle.
i think though that i might have read the answer wrong...im still double checking again. but thank you
NotTim
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yeah, i forgot it was bearings
radar
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O.K, good luck on this problem it is a good question
radar
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Oh yes, note that the wind is FROM a bearing of 315 degrees in other words the force is towards E 45 S.
radar
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Note also the angle between the aircraft and the wind vector is 90 degrees. as aircraft is flying on a bearing of W 45 S
NotTim
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hey, radar, i got 13.1 degrees (not considering bearings), as the angle between the resultant and the aircraft's cruising speed. Um. What should I do with that?
NotTim
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@Radar Hey....do you have time? I need you to please revisit this...
NotTim
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|dw:1333520833003:dw|
NotTim
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thats how i came to my angle. the correct answer is a BEARING of 211.9 degreees. my answer is not a bearing
lexi_rocks_lexi
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ur question is a hard 1. i tried 8 times and cant figure it out. sorry
NotTim
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ok. its alright.
bobobobobb
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Okay ive done this before. Let me dig up my notes.
NotTim
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hello?
NotTim
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computer went wonky.
bobobobobb
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Sorry. Were you given the answer?
NotTim
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yeah.its 154.0km/h at a bearing of 211.9 degrees.
bobobobobb
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Draw your version of the diagram. I wanna see if we are using the same one
NotTim
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i drew 2...
NotTim
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or do you want what the txtbook has>?
bobobobobb
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yeah that would be good
NotTim
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|dw:1333522792176:dw|
Hero
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I like how "not physics" is the first thing NotTim says.
NotTim
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you know, just in case.
bobobobobb
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So we are finding the resultant vector?
NotTim
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i already found it (somehow...I don't really know how). I'm looking for the angle now.
NotTim
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bump
experimentX
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find out the individual horizontal and vertical factors ... add them or substract them according to necessity ... and reconstruct them. this is how these kinds of problems are done
NotTim
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argghh. that's component method. im not suppose to use component method!
experimentX
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since they don't have same directions, i don't think you have other choice ...
NotTim
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what about application of vector additions? using sin adn cos law?
experimentX
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let's try it.
Dockworker
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\[\tan(\theta)=\frac{35}{150}, \theta=13.13\deg\]
subtract from 225
NotTim
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r^2=150^2+35^2-2(150)(35)cos90
Dockworker
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theta is angle between resultant and original flight
NotTim
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wait...so trig ratios work, even with non-right angle triangles?
experimentX
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|dw:1333520171086:dw|
Dockworker
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it is a right triangle because the plane is flying at 225 degrees, wind coming from 315 degrees or going at a bearing of 135
experimentX
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|dw:1333520250197:dw|
Dockworker
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225-135=90
NotTim
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its 315...
NotTim
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im sure thats not an error in my grammar
Dockworker
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blowing from a bearing of 315º is the same as blowing at a bearing of 135º
experimentX
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|dw:1333520364868:dw|
Dockworker
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if there was wind ahead of you at a bearing of 315 degrees and it passed by you, it'd be going at 135º
NotTim
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dock, i dont udnerstand. how would i derive that info?
experimentX
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i think that would be 180 - (315-225)
and apply cosine law ..
i think that might work
Dockworker
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if i tell you wind is coming from in front of you, which way is it going to push you?
Dockworker
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the same way its coming from?
NotTim
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no; I know the different of to and from (Radar explained that); how would I figure out:|dw:1333531452863:dw|
Dockworker
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you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector
NotTim
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yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?
Dockworker
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|dw:1333520788256:dw|
you could do as you did above.
Dockworker
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draw a line from 315 bearing through the origin
NotTim
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Yes, I understand that. I want to know how I can derive 135 from 315 when i get the equilibrant vector, like you did.
Dockworker
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its just intuitive, not really something you should have to prove
Dockworker
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just the wording of the problem
NotTim
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no, but how do i figure that out myself? to get 135 from the equilibrant of 315?
NotTim
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i dont know if im making sense here...
Dockworker
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they aren't giving you the true bearing of the wind's velocity
Dockworker
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they are giving it to you from a different reference
NotTim
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I understand that. Yes. Very much so. The difference between to and from.
But the angle, how did you derive your angle like that?
Dockworker
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its so far up now. what angle did i derive?
NotTim
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135
NotTim
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the wind blowing.
Dockworker
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its just a reference frame. from an observer's point of view, the wind is at a bearing of 315º. from the wind's perspective, the observer is at abearing of 135º
NotTim
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How did you calculate 135degrees?
Dockworker
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315-180
NotTim
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oohhhkay.
Dockworker
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the observer (plane) and wind, if looking at each other, form a 180º line of sight
Dockworker
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or a straight line
NotTim
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ok.
NotTim
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now that i figured that out, what's our next step?
Dockworker
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you know the magnitude of the plane's velocity and window's velocity. they form a right angle. the picture i drew above shows this. use arctan
NotTim
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we werent taught arctan (to add to my problems)
NotTim
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or, that was long before
Dockworker
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this will give you the angle between the resultant and the plane's direction
NotTim
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oh. trig ratios. tanangle=o/a right
Dockworker
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arctan(35/150) is approximately 13.13º
NotTim
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how do i know which one is my opposite and adjacent?
Dockworker
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|dw:1333521664877:dw|
NotTim
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oh bugger. I thought we were looking for another angle. this clarfies much,.
Dockworker
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as you can see, the magnitutde can also be calculated with pythagorean theorem
NotTim
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yes.
NotTim
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i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.
NotTim
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ok. we're near the end now. how do i figure out hte angle, as in bearings? we've got 13.13 degrees. wat next?
NotTim
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thx with sticking with me here folks, greatly appreciated.
Dockworker
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subtract that angle from the plane's bearing
Dockworker
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yw
NotTim
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thank you all. your assistance has been greatly appreciated. I wish you the best of luck in your studies, and that you are able to help others as greatly as you have done for me.
radar
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Had to leave, but I see that you got some excellent help and advice from Dockworker. I agree with the results. That angle of 13.13 degrees is not the bearing but it is the correct deviation from his original bearing. The bearing rose goes clockwise with 0 degrees north 180 degrees due south etc.