Vectors...

[Externet]

Been a long time since the university years, no books on hand, and with terminology in another language, this may be hard to explain/ask...

 

From a fixed point of view, say the tip of your tv antenna at the roof;

 

-The street hydrant is at a fixed direction; elevation and azimuth. Do not care about the distance.

 

-The sun is at another direction; elevation and azimuth.

 

How is the (bisectriz) bisecting vector elevation and azimuth calculated ? Simply sums divided by 2 ?

[timo]

Transform to cartesian, normalize the vectors, sum the entries for x-, y- and z-coordinate , convert back to angular coordinates.

 

You supposedly don't want to get an angle of 179° as a bisection for two vectors with 0° and 358°, respectively.

[Externet]

Thanks - Should I have said 'algebraic sums' ?

On your example of 0° and 358°

 

0°+ (-2°) = -2° ----> -2° ÷ 2 = -1° ----> ≠ 179°

[moth]

i'm just getting started with vectors but maybe this is what you're looking for?

[math]cos\theta[/math]=[math]\frac{\stackrel{\rightarrow}{A}\cdot\stackrel{\rightarrow}{B}}{\left|\stackrel{\rightarrow}{A}\right|\left|\stackrel{\rightarrow}{B}\right|}[/math]

as i said i'm a beginner but i don't think you need the actual length to the sun or hydrant if you use unit vectors.

[timo]

On your example of 0° and 358°

0°+ (-2°) = -2° ----> -2° ÷ 2 = -1° ----> ≠ 179°

Err, well .. that was more or less my point, wasn't it?

[Externet]

Yes. That was it.

 

...And how can I obtain the azimuth and elevation vectors of the sun from the observer (sidereal?) time, date and coordinates ? Do I have to use some tables or there is a formula ?

[moth]

Yes. That was it.

 

...And how can I obtain the azimuth and elevation vectors of the sun from the observer

 

it depends how accurate you want to be. you may be able to get away with using a compass for azimuth, and a protractor and a plumb bob for elevation (don't stare at the sun of course)

[Royston]

i'm just getting started with vectors but maybe this is what you're looking for?

[math]cos\theta[/math]=[math]\frac{\stackrel{\rightarrow}{A}\cdot\stackrel{\rightarrow}{B}}{\left|\stackrel{\rightarrow}{A}\right|\left|\stackrel{\rightarrow}{B}\right|}[/math]

as i said i'm a beginner but i don't think you need the actual length to the sun or hydrant if you use unit vectors.

 

That's for finding the angle between two vectors, the unit vector is defined as [math]\hat{\bf v}=\frac{1}{\bf|v|}\bf v[/math] it looks similar, but has a different use.

[moth]

That's for finding the angle between two vectors, the unit vector is defined as [math]\hat{\bf v}=\frac{1}{\bf|v|}\bf v[/math] it looks similar, but has a different use.

 

thanks, i'm new to vectors and spend as much time remembering algebra and trig as i do learning vectors.

i was trying to suggest in this case, where you just want the angle, you could measure the elevations and azimuth and forget about the distance to the sun or the hydrant by pretending the vectors are 1 unit long in the right direction.

does that change the angle somehow?

[Royston]

thanks, i'm new to vectors and spend as much time remembering algebra and trig as i do learning vectors.

 

Certainly remember the rules of algebra, trig identities not so much, unless you're that way inclined, but the latter is much harder to remember (personally speaking) and you will always have a book as reference.

 

i was trying to suggest in this case, where you just want the angle, you could measure the elevations and azimuth and forget about the distance to the sun or the hydrant by pretending the vectors are 1 unit long in the right direction.

does that change the angle somehow?

 

It doesn't change the angle, but with a unit vector you're not looking for an angle, the magnitude and direction has already been verified, that's what you plug into the equation to establish a unit vector...it's a reference for somebody reading the equation, that's all.

 

The problem with the equation you posted as a unit vector, is the term [math]cos\theta[/math], which means there's more information to grab out of the equation, we don't care about that, does that make sense ?

[Bignose]

The problem with the equation you posted as a unit vector, is the term [math]cos\theta[/math], which means there's more information to grab out of the equation, we don't care about that, does that make sense ?

 

No, the equation he posted is fine. It is the formula to find the angle between given vectors. And, I can see how that would be useful to find the bisecting angle because if you knew that the angle was [math]\theta[/math] then obviously the bisecting vector makes the angle [math]\frac{1}{2}\theta[/math] with both of the original angles.

 

In fact, I think that in 2-D just that knowledge would be enough to solve for the bisecting vector:

 

Given vectors a and b:

 

and [math]\cos{\theta} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} [/math]

 

there is a unique bisecting vector, call it x from which we can make 2 equations:

 

[math]\cos{\frac{\theta}{2}} = \frac{\mathbf{a}\cdot\mathbf{x}}{|\mathbf{a}||\mathbf{x}|} [/math]

 

and

 

[math]\cos{\frac{\theta}{2}} = \frac{\mathbf{x}\cdot\mathbf{b}}{|\mathbf{x}||\mathbf{b}|} [/math]

 

If the vector is restricted to 2-D, there are only two unknowns in this case, the two vector components [math]x_1[/math] and [math]x_2[/math]. Which are going to be part of [math]\mathbf{a}\cdot\mathbf{x}[/math] and [math]\mathbf{x}\cdot\mathbf{b}[/math]. Two equations, two unknowns should be solvable.

 

The bisecting vector would, of course, only be known to a constant as was mentioned. I.e. if vector x bisects a and b, then so does 0.5x and 158.9x and so on.

[Royston]

No, the equation he posted is fine.

 

That wasn't what he was asking though, numerically the result will be the same, it doesn't matter, a unit vector gives orientation, as I'm sure you well know. If I use the term [math]cos\theta[/math] at the beginning of the equation, then there's still an angle I need to grab from that, which isn't what we're looking for. They're equivalent numerically, but so what.

 

It is the formula to find the angle between given vectors. And, I can see how that would be useful to find the bisecting angle because if you knew that the angle was [math]\theta[/math] then obviously the bisecting vector makes the angle [math]\frac{1}{2}\theta[/math] with both of the original angles.

 

We're not looking for an angle, just orinentation...perhaps I'm missing something ?

[moth]

i think it's my terminology that is flawed. i didn't mean a unit vector i meant a vector that is one unit long in the same orientation as the vectors to the sun and the fire hydrant.

O.P. "How is the (bisectriz) bisecting vector elevation and azimuth calculated ? Simply sums divided by 2 ?"

sounds like he's looking for the vector halfway between the sun and hydrant vectors?

[Bignose]

We're not looking for an angle, just orinentation...perhaps I'm missing something ?

 

If you calculate both components [math]x_1[/math] and [math]x_2[/math] of the vector, you have everything. The angle, the orientation (which maybe I am missing something, aren't these the two words pretty much synonymous?), whatever you want calculated. The angle comes from converting the components of the angle from Cartesian to cylindrical or spherical or whatever specific angle you want.

 

I am saying that in 2-D, you can find the bisecting angle, by calculating [math]\cos\theta[/math] and the using that [math]\theta[/math] to calculate [math]\cos\frac{1}{2}\theta[/math] for use in those two equations for the two unknown components of the bisecting vector.

[Royston]

If you calculate both components [math]x_1[/math] and [math]x_2[/math] of the vector, you have everything. The angle, the orientation (which maybe I am missing something, aren't these the two words pretty much synonymous?)

 

Sorry Bignose, I was on the verge of sleep when I wrote that post last night, (we're talking about two different things) what I meant by orientation is say, the direction of a field e.g in the opposite direction of [math]\hat{\bf r}[/math] so the unit vector provides that information, e.g

 

[math]{\bf F} = -G\frac{m_1m_2}{|\bf{r}|^2}\hat{\bf r}[/math]

 

I certainly don't disagree with what you were saying. I think I did a good job at confusing matters. I thought moth was stating that the formula he posted was a unit vector...but he wasn't, and then maximum confusion ensued. I'll errr, shut up now, before I make something basic even more confusing.

Edited February 16, 2009 by Snail