Finding Components for Displacement Vectors (Physics)

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TheJason

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Let;s say you were given 4 displacements with angles (directions) and magnitudes (measured in m)..

75 degrees southeast (A) 56 m
10 degrees northwest (B) 19 m
80 degrees northeast (C) 34 m
33 degrees southwest (D) 12 m

Well, you can't put those trig values in the component equations for each one, respectively. They have to be modified.

I know how to modify the northeast and southwest one

Northeast:

For C: 90 deg - 80 deg = 10 degrees

Southwest:

For D: 180 deg + 33 deg = 213 degrees

How would we calculate A and B?

Well, with C:

Cx=Ccos⁡θCC_{x} = C \cos \theta_{C}
Cx=(34)cos⁡(10)C_{x} = (34) \cos (10)
Cy=Csin⁡θCC_{y} = C \sin \theta_{C}
Cy=(34)sin⁡(10)C_{y} = (34) \sin (10)


Dx=Dcos⁡θDD_{x} = D \cos \theta_{D}
Dx=12cos⁡(213)D_{x} = 12 \cos (213)
Dy=Dsin⁡θDD_{y} = D \sin \theta_{D}
Dy=12sin⁡(213)D_{y} = 12 \sin (213)
 
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I'm a pilot and a sailor and I've never seen headings given like this before.

If I had to guess. I'd say you measure these from the x-axis in the appropriate direction.

So using standard nautical headings we'd have

75 degrees southeast would be a heading of 165 degrees
10 degrees northwest would be a heading of 280 degrees

then to find out the cartesian displacements, using nautical headings we have

[MATH](D_x,D_y) = R(\sin(hdg), ~\cos(hdg))[/MATH]
Of course you can choose to do it in standard coordinates as well where the positive x-axis is 0 degrees and the positive y-axis is 90 degrees.

In this case the 2 examples above are now

75 degrees southeast is -75 degrees = 285 degrees
10 degrees northwest is 170 degrees

Using standard coordinates the displacements are given by

[MATH](D_x,~D_y) = R(\cos(hdg),~\sin(hdg))[/MATH]
both coordinate systems are perfectly valid. Standards coordinates are used more often in math.
Nautical coordinates are used in navigation.

see if you can finish this up now.
 
Also, need values within the quadrant - as in the 1st go around. For instance, northwest is II quadrant with values between 90 and 180.
 
TheJason said:
But how did you calculate those values?
Click to expand...

east corresponds to 90 degrees. heading south from that would be moving in the clockwise direction i.e. you would add the 75 to 90 to get 165

west corresponds to 270 degrees. heading north from that would also be moving clockwise so you add the 10 degrees to 270 to get 280
 
Romsek said:
east corresponds to 90 degrees. heading south from that would be moving in the clockwise direction i.e. you would add the 75 to 90 to get 165

west corresponds to 270 degrees. heading north from that would also be moving clockwise so you add the 10 degrees to 270 to get 280
Click to expand...

I mean for northwest - you could also add from 90, or subtract from 180 - and for southeast you could subtract from 360 or add from 270. This would be the 1st spin around. How would you do that? I think you can do something along this line.
 
TheJason said:
I mean for northwest - you could also add from 90, or subtract from 180 - and for southeast you could subtract from 360 or add from 270. This would be the 1st spin around. How would you do that? I think you can do something along this line.
Click to expand...

Well as I said I've never seen headings given in this manner. They are ambiguous. Generally when someone says a rough direction and degrees they mean with respect to your current heading. No current heading is given in the problem.

All you really need to know is that 0 degrees is either North or East depending on which system you use. Going clockwise decreases the angle, going counterclockwise increases it. When your angle gets larger than 360 degrees, you'd subtract 360 degrees from it. When it gets less than 0 you would add 360 degrees to it.

0 degrees and 360 degrees are the same value.
 
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Romsek said:
Well as I said I've never seen headings given in this manner. They are ambiguous. Generally when someone says a rough direction and degrees they mean with respect to your current heading. No current heading is given in the problem.

All you really need to know is that 0 degrees is either North or East depending on which system you use. Going clockwise increases the angle, going counterclockwise decreases it. When your angle gets larger than 360 degrees, you'd subtract 360 degrees from it. When it gets less than 0 you would add 360 degrees to it.

0 degrees and 360 degrees are the same value.
Click to expand...

What was given is from a Physics textbook. University Physics - Young Friedman. The calculations for the northeast and southeast angles.
 
TheJason said:
10 degrees northwest (B) 19 m
80 degrees northeast (C) 34 m
33 degrees southwest (D) 12 m
Click to expand...
I agree with Romsek. These aren't given right.

You have to start somewhere so for the sake of argument I would say that the first one should be
10 degrees north of west
Similarly
80 degrees north of east
33 degrees south of west.

So the first one would be in the second quadrant making the angle from the positive x axis to be 180 - 10 = 170.

Try it out and see what you get.

-Dan
 
I guess what I'd do is pretty much what I said in post 1.

The cardinal direction, northeast, southwest, whatever, will tell you the quadrant the displacement angle is located in.

Then measure the degrees from the x-axis that's in the quadrant above.

Using standard coordinates this time.

75 degrees southeast puts us in quadrant IV , and we measure -75 degrees from the positive x-axis, i.e. 0 degrees
to get -75 degrees, and normalizing that by adding 360 degrees it becomes 285 degrees.

10 degrees northwest puts us in quadrant II. We measure -10 degrees from the negative x-axis, 180 degrees, to obtain 170 degrees.
 
The textbook problem. Not one I made up, has the following:

32 deg northeast

36 deg southwest

straight south

32 deg becomes (90 - 32) = 58 deg

36 deg becomes 180 + 36 = 216 deg

Straight south is simply 270 degrees

So, using this logic - if the problem had degrees in the southeast and northwest, how would they be changed to find the component?

Let's say they were 30 and 40, respectively.

TheJason said:
What was given is from a Physics textbook. University Physics - Young Friedman. The calculations for the northeast and southeast angles.
Click to expand...

Mistake here. Not really given as in numbers - just the logic behind what is done.
 
TheJason said:
The textbook problem. Not one I made up, has the following:

32 deg northeast

36 deg southwest

straight south

32 deg becomes (90 - 32) = 58 deg

36 deg becomes 180 + 36 = 216 deg

Straight south is simply 270 degrees

So, using this logic - if the problem had degrees in the southeast and northwest, how would they be changed to find the component?

Let's say they were 30 and 40, respectively.



Mistake here. Not really given as in numbers - just the logic behind what is done.
Click to expand...

Well if that's the way they want to do it. It's their method. I don't happen to agree with it.

I'd say that 32 deg northeast is just 32 deg. You move 32 deg counterclockwise from the x-axis

I agree with what they did for 36 deg southwest.

and yeah straight south is 270 deg.
 
Well, looking at it again: the example I gave from the book, I see that reference angles cannot be put into the vector calculations. In that case, they have to be de-referenced (so to speak). However, the angle in the 1st quadrant is de-referenced and I don't understand that one. Why is 32 taken away from 90 making 58 degrees?

@MarkFL ?
 
If we begin at north (90°) and then move 32° towards east (0°) then the angle we want for our computations becomes:

[MATH]\theta=(90-32)^{\circ}=58^{\circ}[/MATH]
 
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